Introductory General, Organic, and Biochemistry Applications to HIV Treatment and Prevention

Citation preview

Introductory General, Organic, and Biochemistry - Applications to HIV Treatment and Prevention -

Prof. Lawrence Mailloux CHM1033 Miami Dade College

Paying for this Book – Help STOP HIV

This book is a labor of love. It is unacceptable that in one the richest cities, in the wealthiest country in the world, that a treatable and preventable disease like HIV is still a major public health concern. Because education is part of the solution to this problem, I am providing this textbook to you free of charge. HIV advocacy and prevention have been an important cause to me for many years. Starting in 2010 when I moved to Miami I began participating in the SMART Ride, an HIV charity bike ride based in south Florida. The SMART Ride (Southern Most AIDS/HIV Ride), is a two day 165 mile bicycle ride from Miami to Key West. Since its inception in 2002, the SMART Ride has raised over $10 million dollars for HIV/AIDS related charities in South Florida. The SMART Ride is a very efficient charity, raising about one million dollars for charity each year. All labor is volunteer, and all overhead costs are covered by either corporate donations, or rider registration fees; as a result, 100% of all donations go to charity and are tax deductible. Every year each of the over 500 riders must raise a minimum of $1250 in donations. If you wish to show your appreciation for getting a free textbook, while at the same time helping stop HIV in South Florida, then please make an anonymous donation to the SMART Ride in my name.

To learn more about the SMART Ride visit: https://thesmartride.org/about/what-is-it/

To make a donation visit this page https://thesmartride.org/donate/

This book is copyrighted by Prof Larry Mailloux Educators and students are free to copy and distribute this book for noncommercial educational purposes so long as attribution is given.

This book is dedicated to my partner Steven and to my father Mark Mailloux who passed away during the writing of this book.

Table of Contents Part I – Survey of General Chemistry Chapter 1 – Introduction and Scientific Method Chapter 2 – Measurement and Conversions Chapter 3 – Matter and Energy Chapter 4 – Nuclear Chemistry Chapter 5 – From Atoms to Compounds Chapter 6 – Grams, Moles, and Chemical Reactions Chapter 7 – Acids, Bases, and Solutions

Part II – Survey of Organic Chemistry Chapter 8 – Drawing Structures of Covalent Compounds Chapter 9 – Drawing Structures of Organic Compounds Chapter 10 – Alkanes, Alkenes, and Alkynes Chapter 11 – Reactions of Alkanes, Alkenes, and Alkynes Chapter 12 – Reactions of Alcohols Chapter 13 – Reactions of Carboxylic Acids

Part III – Biochemistry and Application to HIV Chapter 14 – Carbohydrates Chapter 15 – Lipids Chapter 16 – Proteins and Enzymes Chapter 17 – DNA, RNA, and the Genetic Code Chapter 18 – DNA, RNA, and Antiviral Medications

Copyright Prof. Lawrence Mailloux, 2021

Chapter 1 Introduction and the Scientific Method History and Development of the Scientific Method As recently as a couple hundred years ago, if you were to suggest that lifeforms too small to be seen were causing us to get sick and die you probably would have been laughed at. For most of human history superstition held sway over medicine. The mechanisms that cause diseases were not understood; germs and viruses were unknown, and spontaneous generation of life was believed to be possible. All of this began to change however in 1668 when an Italian physician named Francesco Redi; applied the scientific method to advance our understanding of disease. The scientific method is a systematic series of steps consisting of observation, measurement, experiments, testing, refinement, and publication. While scientists perform the scientific method frequently, nonscientists apply the scientific method frequently as well. Doctors use a form of the scientific method when diagnosing a patient, and you, as a medical professional, will play a key part in that process. A flow chart of the scientific method is shown on the next page.

1-1

Copyright Prof. Lawrence Mailloux, 2021

Observation

Ask Questions

Background research

Construct Hypothesis

Test w/ Experiments

Try Again

Analyze Results

Correct Hypothesis

Incorrect Hypothesis Publish Results

All science starts with observations. In Redi’s time it had been observed that maggots would live on garbage, however how they got there was not correctly understood at the time. Although the maggots were of course observed by people using their eyes, “observations” need not be limited to vision; observations can also be made using our other senses as well. In the 1600’s belief in spontaneous generation, that is the formation of life from non-living matter, was common place. Luckily for humanity, Redi was skeptical of this; he hypothesized that spontaneous generation was incorrect. A hypothesis is a tentative explanation of the natural world; similar to an educated guess. All scientific hypothesis must be falsifiable, that is an experiment, in principle, would be able to disprove it. To test his hypothesis, Redi performed an experiment. Redi’s experiment, while simple by today’s standards, was revolutionary at the time. All good science experiments must hold all conditions constant except for one. Redi’s experiment is a great example. Into three separate jars, Redi placed meatloaf and an egg. One jar was left open, the second was sealed shut, and the third was covered with gauze that would allow air in, but nothing else. The jars were then allowed to sit for several days. The open jar was what 1-2

Copyright Prof. Lawrence Mailloux, 2021

is known as a control. In an experiment, a control is performed to eliminate the impact of other variables. For example, in a clinical research study to test a new medication, half the participants are given the active drug under development, while the other half are given a sugar pill as a placebo. In the case of Redi’s meatloaf experiment, the open jar was to control for the effect of free access to the air. After sitting for several days, maggots were observed on the meatloaf and egg only in the open jar. Inside the closed jar there were no maggots, and while maggots were observed on top of the gauze on the third jar, they were not present inside the jar with the meatloaf and egg. This meant that air itself was not responsible for the appearance of maggots. These results were an important step in the abolition of the belief in spontaneous generation, and in the development of our modern understanding of disease. Around the same time as Redi, a Dutch scientist named Antonie van Leeuwenhoek was also making major contributions to our understanding of disease. Leeuwenhoek was one of the first scientists to study the world through a microscope, and in the course of his studies, he discovered bacteria. Leeuwenhoek’s work paved the way to the development of our current germ theory of medicine, although the role bacteria played in disease would not be understood for nearly another 200 years when another scientist build upon Leeuwenhoek’s work.

1-3

Copyright Prof. Lawrence Mailloux, 2021

About two centuries later, during the early 1860’s, work by a scientist named Louis Pasteur would revolutionize the world of medicine. By the time of Pasteur, the existence of bacteria and other microorganisms (except for viruses) was well known; however, the fact that they caused disease was not known at the time. Pasteur hypothesized that microbes were responsible for causing disease and performed a series of famous pasteurization experiments to test his hypothesis. The figure below, obtained from Wikipedia, illustrates Pasteur’s experiment.

Pasteur filled three flasks with a nutrient rich broth and pasteurized (sterilized) each one by applying heat; the milk you buy at the store is pasteurized. During the heating, each of the three flasks was protected from the environment by a long narrow glass neck as can be seen in the above diagram. This long neck would allow air to travel in and out, but not allow for any particles in the air to make their way into the flask. After the heating was complete one flask was simply allowed to sit undisturbed with the long neck left intact. This flask acted as the control. 1-4

Copyright Prof. Lawrence Mailloux, 2021

On the second flask the neck was broken off after heating, this would allow microbes on particles in the air to enter the broth. Lastly, in the third flask, the neck was left intact; however, the flask was tilted so that the tube was full of liquid. This would allow any microbes from the environment to enter via the tube of liquid. Even after siting for some time, the control flask with the neck intact did not have any bacteria growing in it. Because the long neck allowed for the passage of air into the flask, the lack of bacterial growth meant that something other than the air itself was responsible for the growth of bacteria. This was an important realization, as during the 1800’s it was commonly believed that “bad air,” and not microbes, made people sick. In the remaining two flasks, bacteria grew readily demonstrating that bacteria could enter the system via particles either in the air, or through the solution itself, but not simply via pure air. These results, taken with the combined work of others, lead to the development of what is known as the germ theory of disease; which is the currently accepted scientific theory that many diseases are caused by microorganisms. In science when a hypothesis is confirmed by repeated experiments, by multiple independent researchers, it can be promoted to what is known as a scientific theory. In science, a theory is a well-established explanation of the natural world. Note that this is unlike our everyday use of the word “theory,” which is more like a hunch or guess. That is our everyday use of the word “theory” is more like what a scientist would call a hypothesis. Just like hypothesis, all scientific theories must be falsifiable. Just because a scientific theory has been well established does not mean that it does not undergo refinement. Most scientific theories are modified overtime as more and more information is gathered; however, it is quite rare for a scientific theory to be completely overthrown. The germ theory of disease is no different. For example, despite his best efforts, Pasteur was unable to find any microorganisms responsible for rabies. To his credit, Pasteur postulated the existence of microorganisms too small to be seen using a regular microscope, however it would not be until the early 1890’s that a Russian biologist named Dmitri Ivanovsky solved the mystery.

1-5

Copyright Prof. Lawrence Mailloux, 2021

Dmitri was experimenting with tobacco plants that were infected with a disease that is now known as the tobacco mosaic virus. Because viruses are up to 100 times smaller than bacteria they cannot be seen with normal light microscopes, especially those available in the 1800’s. Dmitri took samples of infected plant, crushed them up, and extracted the ground up material to obtain a solution. Dmitri observed that these plant extract solutions could be used to infect other healthy plants. This meant that the infections agent was present in the extract solution. Next Dmitri passed the extracted solution using a recently invented filter with very small holes that were too small for bacteria to pass through. After passing the extract through the filter, Dmitri could be certain that any bacteria that might have been present in the extract initially had been removed. However, despite filtering out all of the bacteria, the solution still was capable of infecting other plants. Therefore, Dmitri concluded that some form of life, far smaller than a bacterium, was still present in the solution after filtering, and it was this form of life that was making the tobacco plants sick. We now refer to this form of life as a virus, and it is to viruses that we will devote our attention in this course. Fast forward one hundred years later; it was 1981 and large numbers of injectable drug users and gay men were showing up in hospitals in major cities with Pneumocystis carinii (a very rare form of pneumonia referred to as PCP) as well as Kaposi’s sarcoma (a very rare form of cancer referred to as KS). Both PCP and KS are known as opportunistic infections. An opportunistic infections is an infection that under normal circumstances is very uncommon in healthy people, but under the correct circumstances can become a problem. For example, healthy dry skin is a very effective barrier against many diseases, whereas soft moist skin is not. This means that after taking a shower the skins resistance to infection is temporarily diminished while the feet are moist and soft, creating a perfect opportunity for infection to take root. This is why walking around barefoot in a locker-room puts one at high risk of athlete’s foot and/or plantar warts, compared to walking around elsewhere without shoes. The conditions of a locker room give infections an opportunity to take hold when they otherwise probably would not be able to, hence it is smart to wear sandals in a public lockeroom.

1-6

Copyright Prof. Lawrence Mailloux, 2021

Because PCP and KS are opportunistic infections that are only seen in people with greatly weakened immune systems, doctors hypothesized that some kind of pathogen was attacking the immune systems of these patients. Furthermore, because the illness did not respond to antibiotics, doctors hypothesized that a virus may be the culprit. Two years later in 1983, independent research teams from the United States and France both determined that a previously unknown virus was responsible for the illnesses being observed at the time. This virus of course was the human immunodeficiency virus, better known as HIV. If left untreated HIV will continually weaken the immune system of its host. Although it takes longer for some people than others, eventually (for 90% of people the average time is 10-12 years)1 an HIV patient’s immune system becomes too weak to prevent illness. At this point, the patient is said to have acquired immunodeficiency syndrome, or AIDS. If not brought under control quickly, an AIDS patient will develop infections that their weakened immune system is unable to fight off. Death will follow shortly thereafter. Until the development of the first antiHIV medication in 1986, HIV was a death sentence. Since then advancements in testing, treatment, and prevention now allows HIV positive persons to live healthy normal lives. We will be learning more about the science of HIV and the medications used to treat it in this course. The proceeding narrative was to illustrate the major steps of the scientific method and to introduce the concepts of hypothesis and theory as used in science. Both hypothesis and theories explain why something happens; with a hypothesis being a tentative explanation, and a theory being a well-established explanation of observed natural phenomena. In addition to hypothesis and theories, there are also scientific laws. In science, a law is a summary of observations that have been repeated time and again, with no explanation given as to why. For example, the law of conservation of energy states that “in any process energy is neither created nor destroyed, but rather changes form.” This law tells us for example that if a car moving down the road has 100 units of kinetic energy, 100 units of energy must be dissipated as heat by the breaks to stop the car. The law of conservation of energy tells us that the total energy at the beginning of a process must be equal to the total energy at the end of a process, but it does not say why. Lastly it is worth noting that a law in NOT a “super-theory.” With sufficient confirmation a hypothesis can be promoted to a theory, however a theory can not be promoted to a law. Hypothesis and theories explain why things happen, while laws do not explain; laws summarize. 1.

Fan, H. Y., Conner, R. F., Villarreal, L. P. AIDS Science and Society 7 th ed. Jones and Bartlett Learning, 2014

1-7

Copyright Prof. Lawrence Mailloux, 2021

End of Chapter Problems 1. Write definition for each of the following a. Hypothesis b. Theory c. Law 2. What is an opportunistic infection? 3. What is the main difference between a theory and a hypothesis? 4. What is the main difference between a law and a theory of hypothesis? 5. With sufficient support from the work of multiple researches a scientific __________ can be promoted to a(n) _____________, however a ___________ can not be promoted to a __________. 6. Observations include more than just what we see. We use all of our senses to make observations everyday. Which sense is being used for each of the following observations? a. A patient’s urine is dark yellow b. A patient’s breath smells like almonds (a sign of cyanide poisoning) c. A patient with a pneumonia hears fluid in their lungs. d. A patient’s skin feels warm and flushed. e. A patient reports that their breath tastes like nail polish remover. (sign of ketosis) 7. Why were viruses not discovered until well after bacteria? 8. All good scientific theories and hypothesis must be ____________. 9. Everyday usage of the word “theory,” is very different that the scientific use of the word theory. The everyday use of the word theory is more like a(n) _____________ in science. 10. Cryptococcus is a very common fungus that grows in soil. For people with healthy immune systems, Cryptococcus is not a threat. However, when ones immune system is compromised, such as with an AIDS patients, or people taking medications to prevent organ rejection, this fungus can cause serious illnesses like Meningitis. Such an illness is known as a(n) ____________ infection. 11. What does HIV stand for? 12. What does AIDS stand for? 13. If left untreated HIV will usually develop into AIDS within about _____ to _____ years. 1-8

Copyright Prof. Lawrence Mailloux, 2021

1-9

Copyright Prof. Lawrence Mailloux, 2021

Answers to End of Chapter Problems 1. Write definition for each of the following. a. Hypothesis – tentative explanation of natural events. b. Theory – well established explanation of natural events. c. Law – summary of repeated observations in science. 2. What is an opportunistic infection? An infection that takes advantage to special circumstances to take hold when it otherwise would not be able to. This is common in people with weakened immune systems including cancer patients. Persons taking medications to prevent organ rejection and persons with immune systems weakened by untreated HIV. 3. What is the main difference between a theory and a hypothesis? A hypothesis is a tentative explanation while a theory is a well-established explanation of natural events. 4. What is the main difference between a law and a theory or hypothesis? Both hypothesis and theories explain why something happens, while laws do not explain why. Laws simply state what will happen based upon repeated observations of similar events. 5. With sufficient support from the work of multiple researches a scientific __hypothesis__can be promoted to a(n) _theory_, however a theory can not be promoted to a _law_. 6. Observations include more than just what we see. We use all of our senses to make observations everyday. Which sense is being used for each of the following observations? a. A patient’s urine is dark yellow - sight b. A patient’s breath smells like almonds (a sign of cyanide poisoning) - smell c. A patient with a pneumonia hears fluid in their lungs. - hearing d. A patient’s skin feels warm and flushed. - touch e. A patient reports that their breath tastes like nail polish remover. (a sign of ketosis) - smell

1-10

Copyright Prof. Lawrence Mailloux, 2021

7. Bacteria are small, but viruses are extremely small. We will see just how small in the next chapter on measurement. Because they are so much smaller than bacteria, the technology to see viruses was not developed until much later. 8. All good scientific theories and hypothesis must be __falsifiable_. 9. The everyday usage of the word “theory,” is very different that the scientific use of the word theory. The everyday use of the word theory is more like a(n) hypothesis_in science. 10. Cryptococcus is a very common fungus that grows in soil. For people with healthy immune systems, Cryptococcus is not a threat. However, when ones immune system is compromised, such as with an AIDS patients, or people taking medications to prevent organ rejection, this fungus can cause serious illnesses like Meningitis. Such an illness is known as a(n) _opportunistic infection. 11. HIV stands for human immunodeficiency virus 12. AIDS stands for acquired immunodeficiency syndrome 13. For the majority of people if left untreated HIV will develop into AIDS within about __10__ to __12_ years.

1-11

Copyright Prof. Lawrence Mailloux, 2021

Chapter 2 Measurement and Conversions Scientific Notation Oftentimes scientists work with very large numbers, like 37,200,000,000,000 (37.2 trillion, the number of cells in the human body) or 0.00000472 inches (the size of a virus particle). Because all of the zeros make these numbers cumbersome to write, scientists use a special notation known as scientific notation. When using this notation, scientists refer to the number part as the coefficient, and the small number on top of the times ten part as the exponent as shown below: 6.02 x 1023 Coefficient

Exponent

In this notation the non-zero numbers are written and the zeros are condensed as the number ten raised to a power equal to the number of places the decimal point needs to be moved to put the decimal between the first and second non-zero number. For small numbers less than one the ten is raised to a negative power, and for large numbers the ten is raised to a positive power. To clarify here are a few examples. Look carefully at how the sign of the exponent and the direction the decimal moves are related. 0.000123 → 1.23 x 10-4

123,000→ 1.23 x 105

Decimal moved five places to the left

Decimal between 1st and 2nd number

- Large number -positive exponent

Decimal moved four places to the right

2-1

Decimal between 1st and 2nd number

-Small number -negative exponent

Copyright Prof. Lawrence Mailloux, 2021

Problem 1 – Viruses are very small and 37.2 trillion is a lot of cells. Rewrite these two quantities from the opening paragraph in scientific notation. a) Number of cells in human body = 37,200,000,000,000 = ________________ b) Diameter of a HIV virus particle - 0.00000472 in = ______________ Problem 2 – Write the following numbers in scientific notation a) 87,800,000 b) 1,460,000,000 c) 1300 d) 0.000005703 e) 0.0000000201 Problem 3 –Write the following numbers in regular notation a) 5.44 x 10-3 b) 3.05 x 108 c) 1.89 x 10-6 d) 2.22 x 109 e) 4.29 x 10-8

Measurement, Conversions, and the Factor Label Method (FLM) Unit Conversions with the Factor Label Method The scientific method starts with observations; many of which take the form of numerical measurements. For example, using a speedometer we can measure how fast a car is driving. All measurements require the use of units. In the case of a speedometer the units we would use in the United States are miles per hour; or in any other country in the world, kilometers (km) per hour. Because units are so important in science, several systems of units have been developed over the years. Unfortunately, there is no one single agreed upon set of units, therefore we must learn how to use multiple systems of units and how to convert measurements from one system to another.

2-2

Copyright Prof. Lawrence Mailloux, 2021

Scientists, engineers, and nurses use relationships between units (known as conversion factors) to convert measurements from one unit of measure to another. To do this, scientists use a method known as the Factor Label Method, or FLM for short. Some books may refer to this as dimensional analysis or the line mole method, these are just more names for the same method. Whatever you want to call it, we will be using this method throughout the entire semester. You MUST learn this method because we will be using it to calculate dosages of medications.

Let us consider an example. Below is a picture (obtained from Wikipedia) of Cream Puff, the oldest cat ever to live.

Amazingly Cream Puff lived to be 38 years old! Let’s find out how many seconds that is. If you happen to know how many seconds are in a year you could simply multiply that number by 38 to get the answer. However, the number of seconds in a year is not a number that most people know off the top of their head. Instead we will use conversion factors that we already know to build a bridge between our known (38 years) and our unknown (number of seconds). In this case we will use the following relationships:

1 yr = 365 days

1 day = 24 hr

1 hr = 60 min

1 min = 60 sec

By putting these conversions together, we can use relationships we already know to solve this problem. To begin our calculation we set up a fraction with our known (38 years) on top and a one (which is usually not written) on the bottom. 38 𝑦𝑒𝑎𝑟𝑠 ( ) 1

2-3

Copyright Prof. Lawrence Mailloux, 2021

Next we use our conversion factors such that the unit on top is canceled by the same unit on the bottom, the one is understood to be there and is not written. 38 𝑦𝑒𝑎𝑟𝑠 365 𝑑𝑎𝑦𝑠 ( )( ) = 13,870 𝑑𝑎𝑦𝑠 1 𝑦𝑒𝑎𝑟

If we were to stop at this point we would obtain the cats age in days: 38 x 365 = 13870 days. However, since we are interested in seconds, we continue to use conversion factors, canceling units along the way, until the desired unit is all that is left. 38 𝑦𝑒𝑎𝑟𝑠 365 𝑑𝑎𝑦𝑠 24 ℎ𝑜𝑢𝑟𝑠 60 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 60 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 ( )( )( )( )( ) = 1,198,368,000 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 1 𝑦𝑒𝑎𝑟 1 𝑑𝑎𝑦 1 ℎ𝑜𝑢𝑟 1 𝑚𝑖𝑛𝑢𝑡𝑒

While the age of the cat example is simple enough that many students can perform the calculations without using the factor label method, it is imperative that you learn this technique. The calculations later in the course will not be so simple and they will not involve units that you are familiar with.

Standard Units (a.k.a. English or Imperial Units) The standard unit system is the system of units used in the United States. Despite the fact that the United States is essentially the only country that still uses these units, they are skill known as standard units. The table below summarizes the commonly encountered standard units used in the United States. Units of Volume

Units of Mass

Units of Length

Gallon (gal)

Pound (lbs)

Mile (mi)

Quart (qt)

Ounce (oz) (a.k.a. ounce mass)

Foot (ft)

Pint (pt)

Inch (in)

Cup

Yard (yd)

Ounce (oz) (a.k.a. fluid ounce) Teaspoon (tsp) Tablespoon (tbsp) 2-4

Copyright Prof. Lawrence Mailloux, 2021

The relationships between these units are as follows

Volume Conversions 1 gal = 4 qts

Mass Conversions 1 pound = 16 oz

Length Conversions 1 mi = 5,280 ft

1 qt = 2 pts

1 ft = 12 in

1 pt = 2 cups

1 yd = 3 ft

1 cup = 8 oz 1 tbsp = 3 tsp 1 tsp = 5 mL

The process of converting between these units is exactly the same method we used to calculate Cream Puff’s age. For an example, prof Mailloux rides his bike to work seven miles each way. How many inches is that? 7.0 𝑚𝑖

(

5280 𝑓𝑡

)(

1 𝑚𝑖

12 𝑖𝑛

) ( 1 𝑓𝑡 ) = 443,520 𝑖𝑛

That is a lot of inches, which is reasonable given how small an inch is and how far seven miles is in comparison. It is important that you check that your answer is reasonable. This is easy to do using units you are familiar with like feet and inches; however, doing so with units you are less familiar with will take some getting used to. For a second example let’s calculate how many ounces are in a 12.0 gallon tank of gas. 12.0 𝑔𝑎𝑙

(

4 𝑞𝑡𝑠

2 𝑝𝑡𝑠

2 𝑐𝑢𝑝𝑠

) (1 𝑔𝑎𝑙 ) ( 1 𝑞𝑡 ) (

1 𝑝𝑡

8 𝑜𝑧

) (1 𝑐𝑢𝑝) = 1,540 𝑜𝑧

2-5

Copyright Prof. Lawrence Mailloux, 2021

Problem 4 – Perform the following conversions: a) 0.557 quarts to ounces b) 75046 in to yards c) 1995 miles to inches d) 4.1415 gallons to cups e) 13.58 oz to pints

The Metric System Feet and miles may be good enough for everyday usage, but for scientific work a more systematic and uniform system is necessary. Through international agreement in 1790 a more systematic set of units was created giving rise to what is known as the metric system. These units are listed below:

Unit

Symbol

Used to Measure

Meter

m

Distance

Gram

g

Mass

Liter

L

Volume

To give you an idea of the size of these units, Prof Mailloux is about two meters tall, a candy bar is about 50 grams, and everyone is familiar with a two-liter bottle of soda. Because Prof. Mailloux is just over six feet tall, one meter must be about three feet long. Therefore, to measure very small or very large things with meters is not always practical. To adjust for the size appropriate for a given measurement, metric units are modified by prefixes that are based upon powers of ten. There are many of these prefixes, the most common ones that we will make use of in this class are given in the following table. You will not be given this information on exams, so put it in your brain. Memorize it!

2-6

Copyright Prof. Lawrence Mailloux, 2021

Prefix Symbol

Meaning

Example of Usage

kilo

k

1,000 (or 103) x base unit

265 km, dist. from Miami to Key West

deci

d

1/10th (or 10-1) x base unit

1 dm ≈ width of a fist

centi

c

1/100th (or 10-2) x base unit

1 cm ≈ width of pinkie finger

milli

m

1/1,000th (or 10-3) x base unit

1 mm ≈ thickness of few sheets of paper

micro

μ or mc

1/1,000,000th (or 10-6) x base unit

1-10 μm ≈ size of a bacteria cell

nano

n

1/1,000,000,000th (or 10-9) x base unit

120 nm ≈ with of an HIV virus particle

Converting between measurements in the metric system is easy once you get used to it. You always convert from the first prefix back to the base unit, and then convert to the second prefix: 1st prefix → base unit → 2nd prefix

For an example let us convert 183 cm to mm 183.2 𝑐𝑚

(

1𝑚

1000 𝑚𝑚

) (100 𝑐𝑚) (

1𝑚

) = 1,832 𝑚𝑚

By relating everything back to the base unit we can minimize the amount of material that we need to memorize. If you know that there are 100 cm in a meter and that 1000 mm equals one meter, it is not necessary to know that there are 10 mm in a cm, the FLM takes care of it for you. Let’s do another example. Professor Mailloux is 193 cm tall. How many kilometers is this? 193 𝑐𝑚

(

1𝑚

1 𝑘𝑚

) (100 𝑐𝑚) (1000 𝑚) = 0.00193 𝑘𝑚

Note that 0.00193 is a very small number. This is because a km is a fairly big unit (1 mi = 1.609 km) and is far too big a unit to be appropriate for measuring something the size of a person. You should always keep the size of the units you are working with in mind so you know if an answer is reasonable or not. In the case of measuring a person’s height in km, a small number is to be expected. 2-7

Copyright Prof. Lawrence Mailloux, 2021

Problem 5 – Perform the following conversions a) 43.5 km to dm b) 0.005643 cm to mcm c) 45,400 µL to mL d) 2250 dm to cm e) 475 ns to µs f) 1.14 km to cm g) 1777 cL to kL h) 0.00545 ds to ks

Metric to Standard Conversions It would be nice if the whole world used only one system of units, unfortunately there is simply too much institutional inertia to allow the United States to easily switch to the metric system. As a result you will need to learn to convert between metric and standard units. To do this you will need to memorize three conversions factors to convert between metric and standard; one for mass, one for volume, and one for distance. Below are the conversions that we will make use of in this class:

Mass 1 kg = 2.2 pounds

Volume

Distance

0.946 L = 1 quart

2.54 cm = 1 inch

453.59 g = 1 pound For an example, lets convert 3.1 miles to km (this is the length of a “5 K race”) 3.10 𝑚𝑖 5280 𝑓𝑡 12 𝑖𝑛 2.54 𝑐𝑚 1𝑚 1 𝑘𝑚 ( )( )( )( )( )( ) = 4.99 𝑘𝑚 1 𝑚𝑖 1 𝑓𝑡 1 𝑖𝑛 100 𝑐𝑚 1000 𝑚

Note that the last three steps are simply a metric-metric conversion like the ones you just finished in the prior section. Let’s consider a couple more examples:

2-8

Copyright Prof. Lawrence Mailloux, 2021

Convert 150 pounds to kg 150 𝑙𝑏𝑠 453.59 𝑔 1 𝑘𝑔 ( )( )( ) = 68 𝑘𝑔 1 𝑝𝑜𝑢𝑛𝑑 1000 𝑔

Convert 18.5 gallons to mL 18.5 𝑔𝑎𝑙 4 𝑞𝑡𝑠 0.946 𝐿 1000 𝑚𝐿 ( )( )( )( ) = 70004 𝑚𝐿 = 7.00𝑥105 𝑚𝐿 1 𝑔𝑎𝑙 1 𝑞𝑡 1𝐿

Problem 6 – many things nurses work with are very small. To give you an idea of just how small things can get, convert the following measurements into inches and express your answer in scientific notation. a) A human cell is about 100 μm in diameter. b) A bacteria cell is about 10 μm across. c) A virus is about 0.1 μm long. d) A single water molecule (H2O) is about 2.82 nm wide. e) A single oxygen atom is about 0.12 nm in diameter. Problem 7 – Perform the following conversions. a) A 6’4” patient is ________ m tall b) A 176 pound patient weights ________ kg c) A person has about 4.7 L of blood in their body, which is approximately _______ gal. d) A 9 pound 8 oz baby weights ________ kg. Problem 8 – Perform the following conversions. a) 4,005 yards to dm b) 12.5 gal to mL c) 0.00434 kL to quarts d) 75.4 ft to dm e) 1.25 miles to km

2-9

Copyright Prof. Lawrence Mailloux, 2021

Fractional Conversions Sometimes it is necessary to convert two units in a single conversion calculation, for example we may wish to convert miles per gallon to km per liter. Fractional conversions are no more difficult than regular conversions. For an example Prof Mailloux’s car gets 32.0 miles per gallon, what would this be in km per liter? To do this conversion we will need to convert miles to km and gallons to liters. You may convert the miles first or the gallons first, the choice is up to you. The calculations below demonstrate how to perform this conversion both ways, to help you see the two different parts of the conversion, I have colored the text for you:

Miles first 32.0 𝑚𝑖 5280 𝑓𝑡 12 𝑖𝑛 2.54 𝑐𝑚 1𝑚 1 𝑘𝑚 1 𝑔𝑎𝑙 1 𝑞𝑡 ( )( )( )( )( )( )( )( ) = 13.6 𝑘𝑚/𝐿 1 𝑔𝑎𝑙 1 𝑚𝑖 1 𝑓𝑡 1 𝑖𝑛 100 𝑐𝑚 1000 𝑚 4 𝑞𝑡𝑠 0.946 𝐿

Gallons first 32.0 𝑚𝑖 1 𝑔𝑎𝑙 1 𝑞𝑡 5280 𝑓𝑡 12 𝑖𝑛 2.54 𝑐𝑚 1𝑚 1 𝑘𝑚 ( )( )( )( )( )( )( )( ) = 13.6 𝑘𝑚/𝐿 1 𝑔𝑎𝑙 4 𝑞𝑡𝑠 0.946 𝐿 1 𝑚𝑖 1 𝑓𝑡 1 𝑖𝑛 100 𝑐𝑚 1000 𝑚

Problem 9 – Perform the following conversions a) 155 gal/hr to cups/s b) 62 mi/hr to ft/s c) 0.005456 pounds/day to g/wk d) 7526 kg/L to cg/qt e) 33.4 dm/yr to km/hour

2-10

Copyright Prof. Lawrence Mailloux, 2021

Dosage Calculations Dosage errors cost lives! In this class you are learning the factor label method, a powerful tool for performing many calculations, including dosages. You may learn other methods for doing these calculations in future courses, but you must learn this method because it’s easy for another person to follow as it has a standardized format and it works for many different types of problems.

Oftentimes when using the FLM to perform dosage calculations, the conversion factors needed are given to you in the problem. Consider the following example:

Dosage Example 1 - If the dosage of a medication is 15 mg/kg, and the medication comes in 500 mg tablets, how many tablets would need to be given to a 150 pound patient?

First we find the conversion factors that are hidden in the problem

15 mg drug = 1 kg patient

500 mg drug = 1 tablet

The problem asks for “how many tablets,” which is a one-unit answer, tablets. To obtain a one-unit answer we start with a one-unit starting point. Oftentimes the starting point is either an ordered dosage, or a patient’s body weight. For this problem we start with 150 pounds. Then we convert the patients weight from pounds to kg. Next we use our two conversion factors as shown below and round our answer to the nearest tablet or half tablet as appropriate: 150 𝑙𝑏𝑠 𝑝𝑎𝑡𝑖𝑒𝑛𝑡

(

1 𝑘𝑔 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 15 𝑚𝑔 𝑑𝑟𝑢𝑔 1 𝑡𝑎𝑏𝑙𝑒𝑡 )( )( ) 2.2 𝑝𝑜𝑢𝑛𝑑 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 1 𝑘𝑔 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 500 𝑚𝑔 𝑑𝑟𝑢𝑔

)(

2-11

= 2.04545 = 2 𝑡𝑎𝑏𝑙𝑒𝑡𝑠

Copyright Prof. Lawrence Mailloux, 2021

Here is another, slightly more complex example.

Dosage Example 2 - Emtricitabine is a nucleoside reverse transcriptase inhibitor (a type of anti HIV medication we will learn about in Chapter 18). For children it is supplied as a liquid suspension with a strength of 10mg of medication per mL of liquid. The recommended dosage is 6 mg of medication per kg of body weight. How many tablespoons of medication should a 25 kg patient be given?

Once again we find the conversion factors hidden in the problem:

6 mg medication = 1 kg patient

10 mg medication = 1 mL of liquid

In addition to these conversion factors, we will also need some conversion factors that we memorized from earlier in the chapter:

3 tsp = 1 tbsp.

1 tsp = 5 mL

Starting once again with the patient’s body weight we calculate the needed dosage: 25 𝑘𝑔 𝑝𝑎𝑡𝑖𝑒𝑛𝑡

(

6 𝑚𝑔 𝐸𝑚𝑡𝑟𝑖𝑐𝑖𝑡𝑎𝑏𝑖𝑛𝑒 1 𝑚𝐿 𝑙𝑖𝑞𝑢𝑖𝑑 1 𝑡𝑠𝑝 1 𝑡𝑏𝑠𝑝. ) (10 𝑚𝑔 𝐸𝑚𝑡𝑟𝑖𝑐𝑖𝑡𝑎𝑏𝑖𝑛𝑒) ( 5 𝑚𝐿 ) ( 3 𝑡𝑠𝑝 ) 1 𝑘𝑔 𝑝𝑎𝑡𝑖𝑒𝑛𝑡

)(

= 1 𝑡𝑏𝑠𝑝

Here are some dosage problems for you to try.

Problem 10 - The order for a medication is 1200 mg. If the medication is supplied as a liquid, with a concentration of 40 mg/mL, how many tablespoons of this medication must be given?

Problem 11 - A child accidently consumes a bottle of 50 acetaminophen (name brand Tylenol®) tablets. Note this is very dangerous as acetaminophen overdose can easily be fatal. If each tablet contains 500 mg of acetaminophen, how many grams of acetaminophen did the child consume?

2-12

Copyright Prof. Lawrence Mailloux, 2021

Problem 12 - A patient is prescribed 750 mg of a medication per day, to be given in three separate equal doses, every eight hours. If the medication is supplied as a syrup with a strength of 12.5 mg/mL answer the following (for best practice do a full separate factor label method calculation for each one). Round answer to the nearest tsp or half tsp as appropriate. a) How many tsp should be given per each dose? b) How many tsp should be given per day? c.) How many tsp should be given per week?

Problem 13 - A European patient states that they weight 97 kg. Given this information, how many pills should this patient be given each day if the dosage of the medication is 15mg/lbs/day. This medication comes as 200 mg pills and is to be given four times each day in equally sized dosages. Round your answer to the nearest pill or half pill as appropriate.

Problem 14 - A patient was originally taking 30 mL of a medication each dose with a strength of 10 mg/tsp. The clinic runs out of liquid form of the medication so the script is changed to tablets. If each tablet contains 20 mg of the medication, how many tablets should this patient get each dose?

Problem 15 - An 8.5 kg dog is prescribed amitriptyline. The dosage for this medication is 2.0 mg of amitriptyline per pound. If the medication is supplied in 25 mg pills, how many pills should the dog be given? Round your answer to the nearest pill or half-pill as applicable.

Problem 16 - A 330 pound man is prescribed amitriptyline. The dosage for this medication is 0.5 mg of amitriptyline per kg body mass. If the medication is supplied in 25 mg tablets, how many tablets should he be given? Round your answer to the nearest tablet or half-tablet as applicable.

Problem 17 -The script is for amitriptyline, an antianxiety medication. If the dosage for this medication is 1.0 mg of amitriptyline per pound of body weight, how many 50 mg tablets should be given to a 68 kg patient? Round your answer to the nearest tablet or half-tablet as applicable.

2-13

Copyright Prof. Lawrence Mailloux, 2021

Problem 18 - Cats are sometimes prescribed tricyclic antidepressants to help stop them from urinating outside their litterbox. If the dosage for one such medication is 2.0 mg of medication per pound, how many 12.5 mg tablets should be given to an 8.5 kg cat?. Round your answer to the nearest tablet or half-tablet as applicable.

Problem 19 - A 22 pound pediatric patient is prescribed ampicillin, an antibiotic. The dosage for this medication is 100 mg of ampicillin per kg body mass. If the medication is supplied as a suspension with a strength of 250 mg ampicillin per 5 ml of syrup, how many tsp should be given? Round your answer to the nearest tsp as applicable.

Temperature Scales and Conversions There are three temperature scales in common usage; Fahrenheit, Celsius, and Kelvin. Measuring temperature is one of the most fundamental measurements that scientists can make. When one measures the temperature of something, in Kelvin, they are actually measuring the kinetic energy (energy of movement) of the particles that make up the substance whose temperature is being measured. For example, imagine a cup of water. The hotter a cup of water is, the faster the water molecules are moving. The faster the molecules move the harder and more frequently they will collide with a thermometer, and the more the thermometer will rise. Conversely, the colder the water is, the slower the molecules move. If we were to continue to cool our cup of water more and more it would eventually reach a point (long after it had frozen solid) where the water molecules stop moving entirely. This temperature, at which all molecular motion stops, is known as absolute zero. Absolute zero is the coldest temperature that is possible and is the basis of the Kelvin scale; absolute zero is zero Kelvin. The Kelvin scale is the official scientific temperature scale as it has its zero point based upon zero energy. In addition to the Kelvin scale, we will also be making use of the other two commonly used temperature scales the Celsius and Fahrenheit. For convenience the Celsius scale has the freezing and boiling points of water set at zero and 100 degrees. The Celsius scale is the most commonly used temperature scale the world over. The Fahrenheit scale is a very old scale that is only used in the United States where it is commonly used in daily life. It is not however used by scientists very often because the numbers on the

2-14

Copyright Prof. Lawrence Mailloux, 2021

scale (32 and 212) are not nice round numbers. The figure below summarizes the important values on all three temperature scales.

Water Boils

212 -

100 -

373.15 -

Water Freezes

32 -

0-

273.15 -

Absolute Zero

-459.67 -

-273.15 -

0-

°F

°C

K

To convert between one scale and another we will make use of the following formulas

F = 1.8C + 32

K = C + 273.15

Problem 20 - Human body temperature is 98.6 ℉. Convert this temperature into degree Celsius and Kelvin. Problem 21 – A warm comfortable room is about 25 degrees Celsius. Convert this in to degrees Fahrenheit and Kelvin. Problem 22 – convert 45.5 °F to °C and K Problem 23 – Convert -65.7 °C to °F and K Problem 24 – Convert 355 K to °F and °C

2-15

Copyright Prof. Lawrence Mailloux, 2021

Reading Laboratory Equipment Whenever one reads an instrument you must record all digits that are certain, plus one last digit that must be estimated. To estimate the last digit in any measurement you must always estimate between the smallest marks on any measuring device. As an example, consider the two photographs of Prof. Mailloux’s speedometer shown below:

In the case of the speedometer shown above, the smallest markings are the white numbers marked every ten miles per hour. Therefore, since the speedometer is marked to every ten miles per hour (tens place) when you estimate between the marks you are estimating to the nearest mile per hour (ones place). In the case of the speedometer on the left, the needle is between 30 and 35, lying closer to 30 than 35. Therefore, a reasonable reading of this speedometer would be 31 or 32 miles per hour. In this case the thirty is certain but the one or two is estimated. This means that two different people reading the same speedometer may arrive at a slightly different reading because the ones place must be estimated. The same is true with the speedometer on the right, here reasonable readings would be 53 or 54, with the five (tens place) being certain and the three or four (ones place) being estimated. In both cases this speedometer measures speed to the nearest mile per hour.

2-16

Copyright Prof. Lawrence Mailloux, 2021

In a lab you will not be using speedometers, however you will be obtaining measurements using equipment like beakers, graduated cylinders, and rulers which are read the same was as a speedometer. When reading volumes of liquids, it is important to read the volume at eye level. Failure to do so results in what is known as parallax error. Also because water is attracted to glass, when it is placed into a small narrow container (like a graduated cylinder) the water forms a curve called a meniscus; always read from the bottom of the meniscus as shown below.

Below are some cartoon examples of laboratory equipment with different markings. When reading the scales be careful to notice the direction of the markings. Laboratory equipment can be designed to either measure the amount of liquid they are holding, these are marked TC (to contain), or the amount of liquid they have dispensed, these are marked TD (to deliver). As a result, the scales of some equipment are reversed. Compare the direction of the markings on the four cartoon examples shown below. Notice that the markings on the third one runs in the opposite direction than the others, be careful and watch out for this as you read lab equipment. Regardless of the direction of the markings, you must estimate between the smallest marks as illustrated below: If your equipment is marked every 10 mL – estimate to nearest mL. -

Lies between 40 mL and 50 mL, lying closer to 50

-

Reasonable readings 46 mL to 48 mL

If your equipment is marked every 1 mL – estimate to the nearest 0.1 mL

-

Lies between 36 mL and 37 mL, lying near the middle

-

Reasonable readings 36.4 mL to 36.6 mL

2-17

Copyright Prof. Lawrence Mailloux, 2021

If your equipment is marked every 0.1 mL – estimate to the nearest 0.01 mL

-

Lies between 20.3 mL and 20.4 mL, lying closer to 20.4.

-

Reasonable readings 20.37 mL to 20.39 mL

-

Be careful, notice this one is marked in the other direction

If your measurement lands right on the mark – put a zero to indicate the level of precision with which the measurement was made. This lets the reader know that the zero was actually measured and not a result of rounding.

.

-

Lies right on the 6.6 mark

-

Correct reading is 6.60 mL

-

Simply putting 6.6 is wrong! The zero at the end allows the reader to know that this measurement was accurately made to the nearest 0.01 mL

Reading rulers, thermometers, and other laboratory equipment is done in a similar way except there is no meniscus to worry about. For example, to use the ruler below (not to scale) to read the length of the arrow shown below, look closely and carefully to estimate between the smallest marks. As shown by the dotted line, the arrow falls between 4.1 and 4.2 cm, lying closer to 4.1 cm than it is to 4.2 cm. Therefore, 4.11 or 4.12 cm would be a reasonable measurement.

cm

2-18

Copyright Prof. Lawrence Mailloux, 2021

Problem 25 - For each of the following pieces of laboratory equipment give a reasonable reading on the line below. Your answer may vary from the answer listed in the answer key by +/- one decimal at the last place. For example if the listed answer is 24.3 mL a reasonable reading is anything between 24.2 mL and 24.4 mL, if the answer listed is 15.43 mL, than 15.42 mL to 15.44 mL would be acceptable answers as well.

a) ____________

b) ___________

2-19

c) ____________

d) ____________

Copyright Prof. Lawrence Mailloux, 2021

Problem 26 -Now consider the following photographs of actual laboratory equipment shown below. Give a reasonable reading for each piece of equipment on the lines given. Reading real lab equipment can be tricky – a trick used by chemists is to hold a sheet of paper behind the equipment when reading to increase contrast. As with the prior cartoon examples, your reading may vary from the answer by +/- one.

a) ______________

b) ______________

c) _____________

d) _______________

e) _______________

f) ______________

2-20

Copyright Prof. Lawrence Mailloux, 2021

Density and Specific Gravity Density Is the ratio of mass to volume in a substance; it describes how compactly the matter in a substance is arranged. As a formula this is expressed as

𝑑=

𝑚 𝑣

where d is the density (in units of mass/volume, often g/mL or g/cm3), m is the mass (often in grams), and v is the volume (often in mL or cm3).

Grams, mL, and cubic centimeters are only the most commonly used units, any units of mass or volume may be used however the units on the left side of the equal sign must be the same as the units on the right side of the equal sign. In other words, if the density is given in pounds/gallon, then mass must be expressed in pounds and volume in gallons. If not all of your units match, you will need to convert them into the same units prior to putting them into the density formula.

Less dense materials, like oil, float on top of more dense materials, like water. Densities of substances are something that scientists look up, however there is one material for which all scientists know its density, water. Water has a density of 1.0 g/mL and as a consequence the following very important relationships are true 1 g water = 1 mL water = 1cm3 water = 1 cc water (cc = cm3)

2-21

Copyright Prof. Lawrence Mailloux, 2021

In this course you will use density in a variety of ways: 1. As a conversion factor between mass and volume using the factor label method – for example Lead has a density of 11.3 g/cm3. What would be the mass (in pounds) of one gallon of lead? 1.0 𝑔𝑎𝑙 4 𝑞𝑡𝑠 0.946 𝐿 1000 𝑚𝐿 1 𝑐𝑚3 11.3 𝑔 1 𝑝𝑜𝑢𝑛𝑑 ( )( )( )( )( )( ) = 94 𝑝𝑜𝑢𝑛𝑑𝑠 )( 1 𝑔𝑎𝑙 1 𝑞𝑡 1𝐿 1 𝑚𝐿 1 𝑐𝑚3 453.59 𝑔

2. As a formula with: a. Values simply given – these are often short story problems - for example A piece of copper has a mass of 75.4 grams and a density of 8.96 g/cm3. What would be the volume of this piece of copper? 𝑚

𝑑 = (𝑣)

𝑔

75.4 𝑔 ) 𝑣

8.96 𝑐𝑚3 = (

𝑣 = 8.42 𝑐𝑚3

Get out your calculator and solve this problem yourself. If you get 0.119 as your answer you got the reciprocal of the answer, in other words you solved for 1/v. If you get 634.9, you have also made an algebraic error by multiplying 75.4 by 8.96 when you should have divided 75.4 by 8.96.

b. Volume obtained by water displacement. This is done for objects with irregular shape for which there is no simple formula for their volume. Instead the object is dropped into water and the volume of the object is equal to the change in volume of the water from its initial level (vi) to its final level (vf). Thus our formula becomes:

𝑑=

2-22

𝑚 𝑚 = 𝑣 𝑣𝑓 − 𝑣𝑖

Copyright Prof. Lawrence Mailloux, 2021

These are also usually story problems. For example, a key (irregular shape, no formula for its volume) is placed into a graduated cylinder causing the water level to rise from 36.61 mL to 42.82 mL. What is the mass of the key? The density of copper is 8.96 g/cm3

𝑑=

𝑚 𝑚 = 𝑣 𝑣𝑓 − 𝑣𝑖

8.96

𝑔 𝑚 𝑚 = = 𝑐𝑚3 42.82 𝑚𝐿 − 36.61 𝑚𝐿 6.21 𝑚𝐿 𝑚 = 55.6 𝑔

c. With volume obtained via geometry formulas. Although we could consider all sorts of shapes (spheres, cylinders, etc) in this class we will limit ourselves to box shaped solid with the formula: 𝑣𝑏𝑜𝑥 = 𝐿 𝑥 𝑤 𝑥 ℎ

Where: h is the height, L is the length, and w is the width

These problems also tend to take the form of story problems. For example, consider a box shaped piece of lead metal (density of lead 11.34 g/cm3). If the lead box has a mass of 1343 g, a length of 2.322 cm and a height of 4.122 cm, what is the width of the box?

This problem is most easily solved in two steps. First we use the mass and density to calculate the volume of the lead box: 𝑑=

𝑚 𝑣

11.34

𝑔 1343 𝑔 = 𝑐𝑚3 𝑣

𝑣 = 118.43 𝑐𝑚3

Then using the volume we just calculated, we then solve for the width of the box. Notice that the only unit left is cm, which is our desired unit of length. 𝑣𝑏𝑜𝑥 = 𝐿 𝑥 𝑤 𝑥 ℎ 118.43 𝑐𝑚3 = 2.322 𝑐𝑚 𝑥 𝑤 𝑥 4.122 cm 𝑤=

118.43 𝑐𝑚3 = 12.37 𝑐𝑚 2.322 𝑐𝑚 (4.122 𝑐𝑚) 2-23

Copyright Prof. Lawrence Mailloux, 2021

Problem 27 – Solve the following mixed density problems. a) Gasoline has a density of 0.789 g/mL. Given this fact, how many pounds would a 12.5gallon tank of gas weigh? b) Given that ocean water has a density of 1.05 g/mL, how many kg would 4.5 quarts of ocean water weigh? c) A metal cube has a mass of 725.5 grams and a density of 5.56 g/cm3. Given this information, what is the length of the metal cube? d) A piece of wood has dimensions of 25.4 cm x 6.42 cm x 8.11 cm. If the wood has a mass of 1.211 kg, what is the density of the wood in g/cm3? Remember wood floats in water. Is your answer reasonable? e) A toy dinosaur has a density of 1.24 g/cm3. When this toy is placed into a graduated cylinder of water, the level of the water rose from 4.81 mL to 5.66 mL. What is the mass of the toy dinosaur? f) A piece of metal has a mass of 37.41 grams and a density of 3.21 g/cm3. What is the volume, in mL, of this piece of metal?

Specific Gravity The specific gravity of a substance is a way of reporting the density of a substance by comparing its density to that of water. To calculate specific gravity all one does is divide the density of the substance under consideration by the density of water. Because water has a density of 1.0 g/mL the units will cancel, however the numbers do not change; specific gravity therefore is simply density reported without units.

For an example of a medical application, many HIV medications require routine urine tests to check for kidney damage as some of the medications can be hard on some peoples kidneys. Because urine is basically just water, it is common to report its specific gravity as opposed to its density. Let’s assume a sample of urine has a density of 1.02 g/m. How would we “calculate” its specific gravity? Healthy urine has a specific gravity between 1.002 and 1.030. If the specific

2-24

Copyright Prof. Lawrence Mailloux, 2021

gravity is outside this range it could be an indication of a more serious health problem. What is the specific gravity of our 1.02 g/mL urine?

𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐺𝑟𝑎𝑣𝑖𝑡𝑦 =

𝑑𝑒𝑛𝑠𝑖𝑡𝑦𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 1.02 𝑔/𝑚𝐿 = = = 1.02 𝑑𝑒𝑛𝑠𝑖𝑡𝑦𝑤𝑎𝑡𝑒𝑟 1.0 𝑔/𝑚𝐿 1.0 𝑔/𝑚𝐿

Notice that the specific gravity is exactly the same as the density without the units. A specific gravity of 1.02 simply means that a sample of urine is 2% more dense than pure water. If a problem gives you the specific gravity, then you know the density as it is numerically the same. Problem 28 – Given the following densities, give the specific gravity of the following substances a) lead = 11.34 g/cm3 → Specific gravity = ___________________ b) Alcohol = 0.789 g/mL → Specific gravity = ___________________ c) Copper = 8.96 g/cm3 → Specific gravity = ___________________ d) Oil = 0.821 g/mL → Specific gravity = ___________________ Problem 29 – Given the following specific gravities, give the densities of the following substances. For liquids give the density in g/mL and for solids give the density in units of g/cm3. a) Specific gravity of mercury 13.6 = → density = ____________________ b) Specific gravity of aluminum 2.7 = → density = ____________________ c) Specific gravity of ocean water = 1.03 → density = ____________________ d) Specific gravity of gasoline 0.77 = → density = ____________________ Problem 30 – Solve the following specific gravity questions. Remember specific gravity is just density without the units, so being given the specific gravity is effectively the same as being given the density. a) Alcohol has a specific gravity of 0.789. Given this fact, how many grams would 44.5 mL of alcohol weigh? b) A 2.34 cm x 12.1 cm x 32.4 cm piece of Styrofoam has a mass of 104.4 grams. What is the specific gravity of the Styrofoam? c) When a piece of plastic with a specific gravity of 1.12 is placed into a graduated cylinder the water level rises from 12.51 mL to 16.72 mL. What is the mass (in g) of the plastic? 2-25

Copyright Prof. Lawrence Mailloux, 2021

Answers to In Chapter Problems 1. a. 3.72 x 1013 cells b. 4.72 x 10-6 in 2. a. 8.78 x 107 b. 1.46 x 109 c. 1.3 x 103 d. 5.703 x 10-6 e. 2.01 x 10-8 3. a. 0.00544 b. 305,000,000 c. 0.00000189 d. 2,220,000,000 e. 0.0000000429 4. a. 17.8 oz b. 2084.6 yds c. 126,400,000 in or 1.264 x 108 in d. 66.264 cups e. 0.8488 pints 5. a. 435,000 dm b. 56.43 mcm c. 45.4 mL d. 22,500 cm e. 0.475 μs f. 114,000 cm g. 0.01777 kL h. 5.45 x 10-7 kg 2-26

Copyright Prof. Lawrence Mailloux, 2021

6. a. 0.00394 in b. 0.000394 in c. 0.00000394 in = 3.94 x 10-6 in d. 0.000000111 in = 1.11 x 10-7 in e. 0.0000000047 in = 4.7 x 10-9 in 7. a. 1.93 m (note: 6’4” = 6.333 ft, not 6.4 ft at there are 12 inches in one foot) b. 80.0 kg c. 1.24 gal d. 4.31 kg (note: 9 lbs 8 oz = 9.5 lbs, not 9.8 lbs as there are 16 oz in a pound) 8. a. 36,622 dm b. 47,300 mL c. 4.59 quarts d. 230. dm e. 2.01 km 9. a. 0.689 cups/s b. 91 ft/s c. 17.3 g/wk d. 712,000,000 cg/qt = 7.12 x 108 cg/qt e. 0.000000381 km/hr = 3.81 x 10-7 km/hr 10. 2 tablespoons of medicine 11. 25 grams of acetaminophen 12. a. 4 tsp/dose b. 12 tsp/day c. 84 tsp/wk 13. 16 pills 14. 3 tablets 2-27

Copyright Prof. Lawrence Mailloux, 2021

15. 1.5 pills of amitriptyline 16. 3 pills of amitriptyline 17. 3 tablets 18. 3 tablets 19. 4 tsp 20. 37.0°C, 310.2 K 21. 77 °F, 298 K 22. 7.5 °C, 280.7 K 23. -86.3 °F, 207.5 K 24. 81.9 °C, 179.3 °F 25. Last digit may vary by ± 1 a. 2.65 mL b. 21.5 mL c. 44.9 mL d. 24.0 mL 26. Last digit may vary by ± 1 a. 82 mL b. 10.48 mL (Note direction of scale, NOT 11.52 mL) c. 24.2 mL d. 23 mL e. 14.67 mL (Note direction of scale, NOT 15.33 mL) f. 7.15 mL 27. a. 82.3 lbs b. 4.5 kg c. 5.07 cm (solve for v = 130.48 cm3, then take cube root by raising to 1/3 power) d. The density of wood is 0.916 g/cm3, which is reasonable as wood floats and water has a density of 1.0 g/cm3. e. 1.05 grams f. 11.7 mL (of 11.7 cm3)

2-28

Copyright Prof. Lawrence Mailloux, 2021

28. The specific gravities are the same as the density without the units a. 11.34 b. 0.789 c. 8.96 d. 0.821 29. The densities are the same as the specific gravity with units of g/mL or g/cm3 added as appropriate. a. 13.6 g/mL b. 2.7 g/cm3 c. 1.03 g/mL d. 0.77 g/mL 30. a. 35.1 grams b. 0.114 (which is reasonable given that Styrofoam floats on water, which has a specific gravity of 1.0) c. 4.72 grams

2-29

Copyright Prof. Lawrence Mailloux, 2021

Chapter 3 Energy and Matter Everything around you is made up of either matter or energy. In this chapter we will learn about the units used to describe energy and how those units are used on the nutrition labels on the foods that we buy at the store. Then we will turn out attention to matter and the terminology that scientists use to classify and describe it.

Energy Units of Energy There are two things in the universe, matter and energy. Matter is anything that has mass and takes up space. If something is not matter, it’s energy. Energy is defined as the ability to do work. Energy comes in many forms including heat, light, radio waves, ultraviolet radiation, and more. Broadly speaking there are two types of energy, potential and kinetic. Potential energy is energy that exists due to position (like a ball on top of a hill) or condition (like a compressed spring). Another form of potential energy is the energy stored in chemical bonds, such as in gasoline or food. Kinetic energy is energy due to motion, such as a moving car. Furthermore, recall from Chapter 2 that when we measure the temperature of a substance we are really measuring the kinetic energy of the particles that make up that substance. Over the past couple hundred years’ scientists have observed that energy is neither created nor destroyed; it can only change forms. This is known as the Law of Conservation of Energy. For an example of how energy can change forms consider the following narrative. Energy from the sun is caused by nuclear reactions inside the sun. This energy is released as sun light which travels through space and reaches the Earth. Upon reaching the Earth, some of this energy is absorbed by plants through photosynthesis which converts the sunlight energy into potential energy which is stored in the chemical bonds in sugars made by the plants. A pig then eats the plants and stores some of this energy as fat. We eat the pig (fat) when we eat bacon. If after eating the bacon you were to go for a run that energy would be converted into kinetic energy. While it may be difficult to believe, the total amount of energy does not change throughout this entire series of events, it only changes form. 3-1

Copyright Prof. Lawrence Mailloux, 2021

Energy can be measured with a variety of units. In everyday life we often measure energy in Calories (Cal); while in science energy is often measured in another unit called Joules (J, named in honor of James Prescott Joule, an English scientist and mathematician). If you look at the previous sentence you will notice that the word Calorie was capitalized, that was not a typo. It turns out that a calorie is actually not a very large unit of energy. Thus when you read the Calorie content listed on the nutrition label of foods it turns out that what you are reading is not actually calories, but rather kilocalories. However, because Americans are not used to the metric system, you will not see kcal written on a nutrition label, instead a capital C is used. Below are the definitions and conversions between the various units of energy: 1 food Calorie (Cal, capitalized) = 1000 cal (lower case) = 1 kcal. Additionally 1 cal = 4.184 Joules or 1 Cal = 1 kcal = 4.184 kJ. Before applying the units to reading actual nutrition labels, familiarize yourself with these units by doing the conversion practice problems below. Problem 1 – One serving of chicken (about the size of a deck of cards) contains 335 Calories. Notice how much smaller this is than a normal American “serving” of meat. Convert this into cal, kcal, Joules, and kJ. Problem 2 - When Prof Mailloux rides his bike to work everyday he burns about 420 Calories. Express this in kJ, J, cal, and kcal.

Energy, Food, and Nutrition Whenever one is dealing with food, nutrition, or exercise, you are talking about Calories with a capital C. There are three substances in food that contribute Calories. These are known as macronutrients. “Macro” means large, thus these are substances in our food that we need in large quantities. These macronutrients are fat, protein, and carbohydrates (carbs). This is in contrast to micronutrients that we need in relatively small quantities, such as sodium, potassium, iron, vitamins, etc. While we need both micro and macronutrients to live, only macronutrients contribute Calories. The number of Calories contributed by each of these 3-2

Copyright Prof. Lawrence Mailloux, 2021

macronutrients is given below. Using these values one can estimate the total calories contained in a given amount of food. Note that due to rounding, and differences between one batch of food and another, your answer is only an estimate and may vary from the total listed on the package by a fair amount. The amount of Calories contained in fats, carbohydrates, and proteins are as listed below: 1 gram fat = 9 Cal 1 gram protein = 4 Cal 1 gram carbs = 4 Cal Note: fiber is a type of carbohydrate needed in our diets to promote regularity, however because our bodies cannot breakdown fiber it does not contribute any significant amount of calories. To take this into account you must subtract the grams of fiber from the total grams of carbohydrates to obtain what is known as the “net carbs,” the carbs that contribute calories, before multiplying by 4 Cal/g. We will now use the calorie content of fats, carbs, and protein to estimate the total in some common commercial food products. For an example, consider the following nutrition label from a Snickers® bar shown below:

12 g fat x 9 Cal/g = 108 Cal 4 g protein x 4 Cal/g = 16 Cal 32 g total carbs – 1 g fiber = 31 g net carbs x 4 Cal/g = 124 Cal Estimated total Calories = 108 + 16 + 124 = 248 Cal Which is in close agreement with the 250 Calories listed on the label. 3-3

Copyright Prof. Lawrence Mailloux, 2021

Problem 3 -Below are some nutrition labels from a variety of products with the totals blocked out. For each product estimate the total number of Calories in the product by using the information contained in the nutrition label. Do not forget to subtract the fiber. See the answer key at the end of the chapter to see how close your estimated values are to those listed on the product label.

3-4

Copyright Prof. Lawrence Mailloux, 2021

Matter States of matter Scientists frequently describe matter according to its state, that is whether it’s a solid, liquid, or gas. Solids are the lowest energy state of matter in which the particles (atoms, molecules, or ions) are packed closely together and are not free to move around. They have a definite shape and volume and are incompressible. The next higher energy state for a material is the liquid state in which the particles are still closely packed, but are able to move around relative to each other. Because the particles are close together, liquids are also incompressible, however because the particles are able to move next to each other, liquids flow and take the shape of their container. Liquids have a definite volume but an indefinite shape. The highest energy state of matter is the gas state. In a gas the particles are widely separated and are able to move freely, and independently, of one another. Gases take both the shape and volume of their container and are easily compressible. The three figures compare the structures these three states of matter. The little “tails” on the circles are intended to show the movement of the particles.

Solid

Liquid State

Gas State

Matter can convert between solid, liquid, and gaseous states by absorbing or releasing energy. Each of these changes has its own name, most of which you already know. When a solid absorbs heat is can turn into a liquid, we call this process melting. If a liquid absorbs heat it can turn into a gas. If this occurs quickly we call it boiling, if it occurs slowly we call it evaporation. Going in the opposite direction if a gas releases heat it can turn back into a liquid, a process known as condensation; and if a liquid loses heat it can turn back into a solid by freezing. The phase changes just described are the ones which we have all observed in our daily lives because water and other common substances frequently undergo these transformations. There are

3-5

Copyright Prof. Lawrence Mailloux, 2021

two additional changes that can occur with which most people are not familiar, sublimation and deposition. There is nothing chemically special about these two phase changes, however there are not very many substances that we encounter in our daily lives that undergo these changes and as a result they seem special. Some substances, such as solid carbon dioxide (dry ice) and solid iodine do not melt if heated. Instead they go directly from solid to gas (skipping liquid entirely) in a process known as sublimation. The opposite of sublimation is Deposition; it is when a substance goes directly from gas to solid. Examples of deposition include frost forming on grass on a cold morning, snowflakes forming in a cloud, and soot from a burning candle depositing on a nearby surface. Problem 4 – fill in the boxes in the following diagram using the vocabulary discussed above.

Solid Freeze

3-6

Copyright Prof. Lawrence Mailloux, 2021

Classification of Matter In addition to their state of matter, chemists classify substances according to the number of components they contain. Chemists call a substance with only one component in a definite fixed composition a pure substance. Water is an example of a pure substance. It contains only water molecules, (H2O) with two hydrogen atoms and one oxygen atom. On the other hand, if there are multiple components physically combined together then the composition can vary and we have what is known as a mixture. Vodka is an example of a mixture of alcohol and water, which can come in a variety of proofs or concentrations.

Pure substances can be further broken down into elements and compounds. An element is the simplest building block of matter. There are a little over 90 naturally occurring elements (iron, copper, carbon, hydrogen, oxygen, etc.) in existence and they cannot be broken down into simpler, smaller parts by ordinary physical or chemical means. Additionally one element cannot be turned into another. As the saying goes, “you can’t turn lead into gold.” All of the elements are arranged on the periodic table. Each element has either a one or two letter abbreviation. For elements with two letter abbreviations, the first letter is always capitalized and the second letter is always lowercase. For elements with single letter abbreviations, the letter is always capitalized. In this class you do NOT need to memorize the entire periodic table; you will always be given a periodic table on test days when necessary. The periodic table you will be given will only have the symbols of the elements, not their names. This means you will have to memorize the names and abbreviations for the most common elements. Some of these symbols are familiar to us, like hydrogen (H) and oxygen (O); while some are less commonly known. Many of the strange seeming abbreviations, such as Au, Pb, and Sn are based on ancient Latin names. For example, gold (Au) comes from aurum (similar to oro in Spanish), lead (Pb) comes from plumbum (similar to plomo is Spanish), and tin (Sn) comes from stannum (similar to estaño in Spanish); thus the abbreviations used today. You are expected to know the names and symbols for the elements that are shaded in on the following periodic table. Google “element symbols and names” to look up their names and start making flash cards today!

3-7

Copyright Prof. Lawrence Mailloux, 2021

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

IA

IIA

IIIB

IVB

VB

VIB

VIIB

----

VIIIB

----

IB

IIB

IIIA

IVA

VA

VIA

VIIA

VIIIA

H

He

Know names/symbols of shaded elements

Li

Be

Na

Mg

K

Ca

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Rb

Sr

Y

Zr

Nb

Mo

Tc

Ru

Rh

Pd

Cs

Ba

*

La

Hf

Ta

W

Re

Os

Ir

Fr

Ra

**

Ac

Rf

Db

Sg

Bh

Hs

Mt

* **

B

C

N

O

F

Ne

Al

Si

P

S

Cl

Ar

Zn

Ga

Ge

As

Se

Br

Kr

Ag

Cd

In

Sn

Sb

Te

I

Xe

Pt

Au

Hg

Tl

Pb

Bi

Po

At

Rn

Ds

Rg

Cn

Nh

Fl

Mc

Lv

Ts

Og

Ce

Pr

Nd

Pm

Sm

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

Lu

Th

Pa

U

Np

Pu

Am

Cm

Bk

Cf

Es

Fm

Md

No

Lr

A compound is made up of two or more elements chemically bonded together. Compounds can be broken down into simpler substances. Water is an example of a compound, made up of two atoms of hydrogen and one atom of oxygen. Because water is a compound it can be broken down into simpler substances. Below is a picture showing water being broken down in to hydrogen and oxygen using electricity supplied by a battery. The bubbles seen in the picture are the hydrogen and oxygen gas produced by the decomposition of the water molecules.

3-8

Copyright Prof. Lawrence Mailloux, 2021

Mixtures can also be broken down into subcategories depending upon their uniformity. If a mixture is not uniform throughout it is said to be heterogeneous, while mixtures that are uniform throughout are known as homogenous. Homogenous mixtures are also known as solutions and a homogenous mixture of two or more metals is often called an alloy. Clear ocean water, vodka, homogenized milk, and air are all examples of homogenous mixtures/solutions, while murky swamp water and oil/vinegar salad dressing are all heterogeneous mixtures. Examples of alloys would be brass, steel, and pewter. The following flow chart summarizes this classification system for matter.

3-9

Copyright Prof. Lawrence Mailloux, 2021

Carbon Asprin Stainless Steel Mercury Olive Oil Diesel Radon Aluminum Pepto-Bismol® Iron Pure H2O Coffee (without grounds) Clean ocean water Biscayne Bay water Tequila (no worm) Tequila (with worm) Sand stirred into water NaCl Coca-Cola Coffee (with grounds) Orange Juice (with pulp) Orange juice (without pulp) Tap water 14 carat gold 24 carat gold Calcium chloride (CaCl2) Air Iron (III) oxide (Fe2O3) Gasoline AZT (first HIV medication)

3-10

Heterogeneous

Homogenous

Mixture

Compound

Element

Pure Substance

Problem 5 –Practice classifying substances by filling in the table below.

Copyright Prof. Lawrence Mailloux, 2021

The Bohr Model of the Atom If one were to take a sample of an element and cut it into smaller and smaller pieces you would eventually arrive at a point where you could not cut anymore. You would have arrived at the atom. An atom is the smallest piece of an element that still has properties of that element. Although an oversimplification, scientists often use what is known as the Bohr model to visualize the atom. Most of you are familiar with the Bohr model; it is the “solar system” model of the atom that we were all taught in middle school and high school. Although this model is not 100% correct, it is good enough to explain many of the phenomena that we will observe in this class. In the Bohr model the small central portion of the atom is called the nucleus and it makes up over 99.9% of the atom’s mass, but nearly none of the volume. Inside the nucleus are two kinds of particles, protons and neutrons. Protons (p+) have a charge of plus one and neutrons (n0) are electrically neutral, meaning they have no charge. Protons and neutrons have about the same mass and together these two particles make up nearly all the mass of the atom. Surrounding the tiny dense nucleus is a cloud of small, lightweight, negatively charged particles named electrons (e-) making up what is known as the electron cloud. The electron cloud has very low density, meaning that while the electron cloud makes up the vast majority of the volume of the atom, it contains almost none of the mass. This means that atoms are mostly empty space. The Bohr model treats the atom like a miniature solar system with the positively charged protons and uncharged neutrons in the nucleus at its center acting like the sun, and the very small lightweight negatively charged electrons orbiting in circular orbits inside the electron cloud like planets. Atoms are very small, as are the particles that make them up. Even the smallest drop of water contains trillions and trillions of water molecules. To give you an idea how small these particles are, protons and neutrons each weigh 1.67 x 10-27 kg, while electrons weigh 9.11 x 10-31 kg. For comparison, a can of soup weighs about one kilogram. Because these numbers are not easy to work with scientists invented a unit known as the atomic mass unit (amu) to describe the relative masses of these particles. Setting the mass of the proton and neutron equal to one amu leaves the 3-11

Copyright Prof. Lawrence Mailloux, 2021

electron with a mass of only (9.11 x 10-31/1.67x10-27) = 0.000546 amu = 1/1833th of an amu. In other words it would take the combined mass of 1833 electrons to equal the mass of just one proton or neutron, as a result scientists usually ignore the mass of the electron and say they weight essentially zero amu. Problem 6 - You must know the basic structure of the atom, including the location, relative masses, and charges of the three fundamental particles. Fill in the following tables regarding the three fundamental subatomic particles. Particle

Symbol

Charge

Mass (amu)

Location

Makes Up Mass

1/1833 amu +1

Nucleus

Initially, all atoms start off electrically neutral; that is, the number of protons is equal to the number of electrons. Additionally, each of the different “orbits” around the atom is, in fact, an energy level which increases in energy as the levels get farther from the nucleus. According to the Bohr model, electrons can only exist in the energy levels, not in-between them. Also, each energy level can only hold a certain number of electrons before it is full. According to the Bohr model, each energy level can hold the following number of electrons: First energy level = 2 electrons Second energy level = 8 electrons Third energy level = 18 electrons Fourth energy level = 32 electrons The figure below shows the Bohr Model for an atom (not to scale) with eleven electrons (sodium). Notice that there are two electrons in the first energy level, eight in the second, and one in the third for a total of eleven electrons. The protons and neutrons are in the nucleus and will be discussed momentarily.

3-12

Copyright Prof. Lawrence Mailloux, 2021

e-

e-

Nucleus

e-

- 11 protons – light circles

eee e

-

e

-

ee-

e-

- 12 neutrons – dark circles

-

Electron Cloud – 11 electrons - 1st energy level – 2e - 2nd energy level – 8e - 3rd energy level – 1e

Subatomic Particles - Protons, Neutrons, and Electrons The number of protons in the nucleus of an atom is known as the atomic number. The atomic number is very important. It determines what element we have and it is also how the periodic table is arranged. Hydrogen (H) is the first element with an atomic number of one; thus it has one proton in its nucleus. Helium (He) is the second element, it has two protons, lithium (Li) has three protons, and so on. Furthermore, since elements all start out electrically neutral, the atomic number is also equal to the number of electrons that element possesses. Thus, hydrogen has one electron, helium has two electrons, and lithium has three electrons. The next important number is the mass number. The mass number is the sum of the protons and neutrons in the nucleus of an atom. It is called the mass number because, as mentioned previously, protons and neutrons make up over 99.9% of the atom’s mass. Some atoms of the same element (i.e. same atomic number) will have a different number of neutrons. These different versions of an element are known as isotopes; atoms with the same number of protons but different numbers of neutrons. Different isotopes of an element are virtually chemically identical. The only differences between isotopes that we will concern ourselves with is their difference in mass and later in Chapter 4 we will see that some isotopes of elements are radioactive, which have numerous medical applications.

3-13

Copyright Prof. Lawrence Mailloux, 2021

When writing chemical symbols, chemists often include the atomic number and mass number on the symbol to convey additional information. By convention, the mass number is written on the upper left, while the atomic number is written on the bottom left:

mass # atomic#

X

For example, carbon (C) is element number six, thus its atomic number is six, meaning it has six protons. There are two different isotopes of carbon, carbon-12 and carbon-14. The symbols for these two different isotopes of carbon are shown below. 12 6𝐶

14 6𝐶

Because the mass number is equal to the number of protons plus neutrons, we can determine the number of neutrons in an isotope by subtracting the atomic number from the mass number. Thus carbon-12 has 12 – 6 = 6 neutrons, while carbon-14 has 14 – 6 = 8 neutrons. Also, if we remember that elements start off electrically neutral (protons = electrons) we can also know that each carbon atom must have six electrons as well.

3-14

Copyright Prof. Lawrence Mailloux, 2021

For another example let’s reconsider the Bohr model diagram shown previously. What element is this? What would be the symbol for the isotope represented by this diagram.

e-

e-

Nucleus

e-

- 11 protons (sodium) – light circles

ee-

e

-

e

e-

- 12 neutrons – dark circles

-

eee-

Electron Cloud – 11 electrons - 1st energy level – 2e - 2nd energy level – 8e - 3rd energy level – 1e

To determine what element this is we count the number of protons in the nucleus. In this case there are 11 protons in the nucleus, and looking at a periodic table we can see that sodium is element number 11. So we will put an 11 in the bottom left of our symbol, 11Na. To find the mass number we count the neutrons in the nucleus and add them to our number of protons. In this case there are 12 neutrons giving a mass number of 11p+ + 12n0 = 23. Therefore, the symbol for this isotope is 23 11𝑁𝑎 .

3-15

Copyright Prof. Lawrence Mailloux, 2021

Problem 7 – Fill in the table below concerning the following isotopes. The first row has been done for you as an example. Atomic #

Mass #

# Protons

# Electrons

# Neutrons

Isotope Name

11

23

11

11

12

sodium-23

Symbol 23 63

Na

Cu

25

12

92

146 Copper-65 Uranium-235

16

32 112

48

30

35

19

20

6

14 83

1

126

1

1

1 1

2

3-16

Copyright Prof. Lawrence Mailloux, 2021

Average Atomic Mass Because atoms are so small, chemists work with trillions and trillions of atoms at a time. Therefore, instead of speaking of an individual atom’s mass, scientists generally talk about an element’s average atomic mass, which is the weighted average mass of the various isotopes of an element. The average atomic mass is the non-integer number found on your periodic table. A weighted average is an average where the importance (weight or abundance) is taken into account. Grades are often calculated this way. Consider a hypothetical class with three exams worth 20% each and a final exam worth 40%. In this class a student named Evan earned the following grades: Test One = 75%

Test Two = 84%

Test Three = 88%

Final exam = 85%

To calculate their final grade we simply multiply each test score by its weighting as a decimal and then take the sum. Using Evan’s grades as an example: Class grade = (test-1 score) x (0.20) + (test-2 score) x (0.20) + (test-3 score) x (0.20) + final score x (0.40) Class grade = 75 (0.20) + 84 (0.20) + 88 (0.20) + 85 (0.40) = 83.4% For a chemistry related example consider potassium. Potassium has two naturally occurring isotopes, potassium-39 with a mass of 38.96 amu and potassium-41 with a mass of 40.96 amu. Of these two isotopes, potassium-39 is far more common making up 93.26% of all naturally occurring potassium; the remaining 6.74% is potassium-41. Calculating the average atomic mass of potassium is no more difficult that calculating Evans class grade. Average atomic mass = (mass isotope 1) x (abundance isotope 1) + (mass isotope 2) x (abundance isotope 2) Average atomic mass = (38.96 amu) x (0.9326) + (40.96 amu) x (0.0674) Average atomic mass = 39.1 amu Looking at a periodic table we see that potassium has an average atomic mass of 39.098 amu which agrees with our calculation within rounding error. Note that it was NOT necessary for me to give you the abundances of all the isotopes of potassium because all the percentages must add

3-17

Copyright Prof. Lawrence Mailloux, 2021

to 100%. Therefore had the abundance of potassium-40 not been given, all you would have needed to do was subtract the abundance of the other isotopes (postassium-39, 93.26%) from 100% to arrive at the abundance of potassium-40 because 100% - 93.26% = 6.74%. It is important that you understand that mass numbers of individual isotopes are NOT given on a periodic table. Because atoms are so extremely small, even a very small sample of an element will have trillions and trillions of atoms. As a result, scientists always work with averages when dealing with elements, so each box on the periodic table represents the average of all the isotopes of that element. Notice with the example of potassium given above, the average atomic mass of potassium listed on the periodic table (39.098 amu) is much closer to the mass of potassium-39 (38.96 amu, 93.26%) than it is to the mass of the less common isotope potassium-40 (40.96 amu, 6.74%). The same is true of grades in a college course. Because the final exam in most college classes is worth more than any other graded items, a student’s final class grade will be closest to their final exam score than it will be to their scores on less important items. With this concept in mind answer the following two problems. Problem 8 – Given that bromine has only two isotopes, bromine-79 and bromine-81. Looking at a periodic table what can you say about the relative abundances of these two isotopes of bromine? a) Bromine-79 is much more abundant b) Bromine-81 is much more abundant c) Bromine-79 and bromine-81 are of approximately equal abundance d) Bromine-79 has an abundance of nearly 100% e) Bromine-81 has an abundance of nearly 100%.

3-18

Copyright Prof. Lawrence Mailloux, 2021

Problem 9 – When introducing element symbols on page 3-14 carbon-12 and carbon-14 were used as examples. Assuming that these are the only two isotopes of carbon that exist, what can you say about the relative abundance of these two isotopes? a) Carbon-12 is much more abundant b) Carbon-14 is much more abundant c) Carbon-12 and carbon-14 are of approximately equal abundance d) Carbon-12 has an abundance of nearly 100%. e) Carbon-14 has an abundance of nearly 100%. Now it’s your turn to practice performing these calculations. You may need a periodic table to look up the average atomic mass of the elements to solve these questions. Also keep in mind that all percentages need not be given as they must always add to 100%. Problem 10– Chlorine has two major isotopes; chlorine-35 and chlorine-37. Of these two isotopes chlorine-35 is more common, making up 75.8% of naturally occurring chlorine. Given that chlorine-35 has a mass of 34.97 amu and chlorine-37 has a mass of 36.97 amu, calculate the average atomic mass of chlorine. How close does your answer come to the value given on a periodic table? Problem 11 – Boron has two common isotopes, boron-10 and boron-11. Given that boron-10 has a precise mass of 10.01 amu and an abundance of 19.9%, what is the precise mass of boron-11? Sometimes teachers use make-believe elements when writing average atomic mass questions to prevent you from being able to look up the average atomic mass of a given element on a periodic table. The following is an example of such a problem. Problem 12 – A fictitious element X has three isotopes with the following masses and abundances. Given this information, what is the average atomic mass of element X? Isotope

Abundance

Mass

X-204

17.3%

204.36 amu

X-206

60.1%

206.41 amu

X-207

??? %

207.31 amu

3-19

Copyright Prof. Lawrence Mailloux, 2021

The Periodic Table The modern periodic table is arranged according to atomic number. As we proceed from left to right across the table, each element has one more proton, and thus one more electron, than the element before it. Horizontal rows on the periodic table are known as periods, and are numbered one through seven. Vertical columns are known as either groups or families, and can be numbered in two different ways. The newer way is to simply number the columns one through 18. The older system numbers the periodic table using roman numbers and the letters A and B. Because the A/B system is historical there are a few weird things about it. For example, notice that IB starts in column 11 (as opposed to column three, where it would seem to belong) and that columns 8-10 together constitute VIIIB. You may use either system you wish, as both systems are still in common usage. Like groups of friends, or relatives in a family, elements in the same column have similar chemical properties. We will learn why this is the case later in the course. A typical periodic table is shown below. The two rows of elements at the very bottom are known as the innertransition metals. They are shown here to make the periodic table fit on paper better; technically they belong inside sixth and sevenths rows of the table as indicated by the * and **. 1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

IA

IIA

IIIB

IVB

VB

VIB

VIIB

----

VIIIB

----

IB

IIB

IIIA

IVA

VA

VIA

VIIA

VIIIA

H

He

Li

Be

B

C

N

O

F

Ne

Na

Mg

Al

Si

P

S

Cl

Ar

K

Ca

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Kr

Rb

Sr

Y

Zr

Nb

Mo

Tc

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Xe

Cs

Ba

*

La

Hf

Ta

W

Re

Os

Ir

Pt

Au

Hg

Tl

Pb

Bi

Po

At

Rn

Fr

Ra

**

Ac

Rf

Db

Sg

Bh

Hs

Mt

Ds

Rg

Cn

Nh

Fl

Mc

Lv

Ts

Og

* **

Ce

Pr

Nd

Pm

Sm

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

Lu

Th

Pa

U

Np

Pu

Am

Cm

Bk

Cf

Es

Fm

Md

No

Lr

3-20

Copyright Prof. Lawrence Mailloux, 2021

Over time the various parts of the periodic table have been broken up into different parts based on their properties. The simplest way that the periodic table is divided up is into metals, nonmetals, and semimetals (aka. metalloids). The dividing line between metals and non-metals is marked by the bold “staircase” on the periodic table. To the left of the “staircase” are the metals. Metals are elements that are: shiny and good conductors of heat and electricity. In addition to luster and conductivity, metals are also malleable, meaning they can be beaten into thin sheets, and ductile, meaning they can be drawn into wires. Elements to the right of the staircase are known as non-metals. Non-metals are elements that are poor conductors of heat and electricity (insulators), are brittle, and dull. Along the staircase are a few elements with properties inbetween those of metals and non-metals. These elements are called semimetals or metalloids. Because there is no magic change upon reaching the semimetal part of the periodic table the classification of a few elements may vary from book to book. For example, some textbooks/websites list boron (B) as a semimetal, while other list it as a non-metal. In this book I have chosen to list it as a semimetal. Do not concern yourself with these details as your test questions will be written to avoid such ambiguity. In addition to metals vs non-metals, the periodic table is further broken down as indicated below. Be familiar with the location and information regarding these regions as given below. H B

Ge e

Halogens

Alkali Metals

Alkaline Earth Metals

Si

Other nonmetals As Sb

Te

Nobel (Inert) Gases

Semi-metals in grey

Transition Metals Other/poor Metals

At

New Elements Properties Unk.

Lanthanides – together with actinides make up inner-transition metals Actinides - together with lanthanides make up inner-transition metals

3-21

Copyright Prof. Lawrence Mailloux, 2021

Taken together the alkali metals, alkaline earth metals, poor metals, semi-metals, and non-metals make up what are know as the main group or representative elements. Their columns are numbered IA – VIIIA using the old numbering system so scientists sometime refer to them collectively as “Group A Elements.” The properties of the various regions of these elements are listed below. •

Alkali Metals o Very soft, can be cut with a knife o Very reactive - react violently with water ▪

On your phone look up “sodium in water” on YouTube to see what I mean

o Reactivity increases down the column •

Alkaline Earth Metals o Harder than alkali metals o Less reactive that alkali metals – although still reactive o Be and Mg do not react with water, however Ca, Sr, and Ba all react with water

•

Halogens o Very reactive and poisonous o Exist as diatomic elements (F2, Cl2, Br2, I2)

•

Noble (Inert) Gases o Chemically very unreactive o Full outer shell of electrons o Very rarely form chemical compounds (for purposes of this class they never react)

•

Other (Poor) Metals o Because these metals are close to the semi-metal border they are relatively poor conductors of heat and electricity compared to most other metals.

3-22

Copyright Prof. Lawrence Mailloux, 2021

There is no special name for the elements other than the main group elements. The properties of the remaining groups of elements are as follows: •

Transition Metals o Properties of these elements are not as easy to predict as the main group elements o The “octet rule” (discussed shortly) does not apply here.

•

Lanthanides and Actinides o Together make up inner-transition metals o Written at bottom of periodic table to save space o Other than names of a few elements (U, Pu, and Am) you don’t need to know anything about these elements in our class

•

New Elements o All elements past uranium (U, atomic number 92) are ▪

Radioactive

▪

Unstable

▪

Synthetically prepared with nuclear reactors and/or particle smashers using advanced physics

o As the atomic number increases past uranium the elements become increasingly unstable and radioactive. o These last six elements (Nb, Fl, Mc, Lv, Ts, and Og) are extremely unstable and decay in fractions of a second after being made. In addition, only a few atoms of these elements have ever even been made. o It is predicted that these elements would be chemically similar to the elements above them however weird physics may come into play at this extreme end of the periodic table. o Other than knowing they are synthetically prepared and unstable you do not need to worry about these elements.

3-23

Copyright Prof. Lawrence Mailloux, 2021

Miscellaneous Final Things About Elements The following seven elements are never found alone in nature: Bromine, Iodine, Nitrogen, Chlorine, Hydrogen, Oxygen, Fluorine Instead they are always found in groups of two. These elements are known as the diatomic elements and are always encountered in pairs. A mnemonic commonly used by chemists to help remember these seven elements is the name of a fictitious person Mr. BrINClHOF which stands for: Br2 I2 N2 Cl2 H2 O2 F2 On the next page you will find a blank periodic table. Color in this periodic table and use it to help summarize all the proceeding material regarding the periodic table that you are required to know. Make this table neat and organized, we will be referring back to it, and adding more information, as the course progresses.

3-24

Copyright Prof. Lawrence Mailloux, 2021

3-25

Copyright Prof. Lawrence Mailloux, 2021

Electron Configurations The energy levels that the electrons in an atom reside in is known as an atoms electron configuration. Recall that according to the Bohr model, each energy level can only hold a certain maximum number of electrons. The first can hold two electrons, the second can hold eight electrons, the third 18 and the fourth can hold 32. Because the Bohr model is only partly correct, the order in which the electrons enter these energy levels is a bit strange. Therefore, we will keep things simple by only looking at the electron configuration of the first 20 elements, hydrogen through calcium. To go beyond calcium would result in entering the transition metal region of the periodic table where things become more complex than is needed for this course. Each row of the periodic table represents an energy level. Electrons in the outermost energy level are known as valance electrons, while those in the inner energy levels are known as core electrons. Because the valance electrons are on the outside of the electron cloud, they are the ones responsible for how an element behaves chemically. The core electrons are buried deep within the electron cloud and do not have much impact on an elements properties. Thus we will be focusing most of our attention on the valance electrons. The figure on the next page shows Bohr model diagrams for elements hydrogen through calcium. Because we are not concerning ourselves with the transition metal part of the periodic table, it (as well as the inner-transition metals) has been omitted to save space. Study this image carefully, you are expected to be able to give the electron configuration for any of the first twenty elements (H through Ca). You should notice a relationship between the column of the periodic table an element resides in and the number of valance electrons (shown in bold) that element has.

3-26

3-27

Cl (2-8-7)

S (2-8-6) P (2-8-5)

Si (2-8-4)

Al (2-8-3)

Mg (2-8-2)

Ca (2-8-8-2)

Na (2-8-1)

K (2-8-8-1)

Ar (2-8-8)

Ne (2-8)

He (2)

Going beyond calcium we reach the transition metal part of the periodic table and things become too complex to illustrate with simple Bohr diagrams, however the same general patterns continue. All elements below K have one valance electron, elements below Ca have two valance electrons, elements below Al have three valance electrons, elements below Si have four valance electrons, elements below P have five, elements below S have six, elements below Cl have seven, and elements below Ar have eight valance electrons.

F (2-7) O (2-6)

N (2-5)

C (2-4)

B (2-3)

Be (2-2)

Li (2-1)

H (1)

Electron Configurations of the First Twenty Elements

Copyright Prof. Lawrence Mailloux, 2021

Copyright Prof. Lawrence Mailloux, 2021

Notice how the number of valance electrons an atom has depends upon the column of the periodic table in which it resides. All elements in column IA (Li, Na, K…) have one valance electron, elements in column IIA (Be, Mg, Ca) have two valance electrons, and so on all the way across to column VIIIA, the noble gases. The noble gases all have their valance energy level filled with eight electrons. Having a full valance energy level with eight electrons makes an atom extra stable. Chemists refer to the importance and special stability of having eight valence electrons as the octet rule. This is the reason that the noble gases do not undergo chemical reactions. They already have all the electrons they need and as a result have no desire to gain or lose any of their valance electrons and therefore they are chemically unreactive.

3-28

Copyright Prof. Lawrence Mailloux, 2021

Formation of Ions It turns out that the other elements are envious of the noble gases, as they want to have an octet of electrons too. Under the right circumstances, atoms can gain or lose electrons to fill their outermost energy level with eight electrons. The gaining or losing of electrons results in the formation of charged species known as ions. When it comes to forming ions, metals and nonmetals behave differently: •

Non-metals gain electrons to fill their valance energy level with eight electrons. This gives non-metal ions the electron configuration of the next noble gas. Because electrons are being gained the resulting ions have a negative charge and are called anions. When non-metals form anions, they change their suffix to “ide.” For example chlorine (Cl) forms chloride (Cl-) and sulfur (S) forms sulfide (S2-)

•

Metals lose their valance electrons to leave behind a full outer shell with eight electrons. This gives metal ions the same electron configuration as the prior noble gas. Because electrons are being lost, the resulting ions have a positive charge and are called cations. Metal cations do not have special names. For example, Na+ and Mg2+ are simply called sodium and magnesium ions.

Let us consider some examples. Fluorine has the electron configuration 2-7. This means that fluorine must gain one electron to have an octet of electrons. In so doing fluorine forms the F(fluoride) ion with the electron configuration 2-8 (same as Ne). Oxygen has the electron configuration 2-6 and thus needs to gain two electrons to have an octet. In so doing oxygen forms the O2- (oxide) ion, also with the same electron configuration as Ne. Nitrogen has the electron configuration 2-5 and therefore must gain three electrons giving the N3- (nitride) ion. This information is summarized in the following table.

3-29

Copyright Prof. Lawrence Mailloux, 2021

Fluorine (F)

Electron Configuration of Element 2-7

F- (fluoride)

Electron configuration of anion 2-8

Gains 1 e-

Ne

Oxygen (O)

2-6

Gains 2 e-

O2- (Oxide)

2-8

Ne

Nitrogen (N)

2-5

Gains 3 e-

N3- (Nitride)

2-8

Ne

Chlorine (Cl)

2-8-7

Gains 1 e-

Cl- (Chloride)

2-8-8

Ar

Sulfur (S)

2-8-6

Gains 2 e-

S2- (sulfide)

2-8-8

Ar

Phosphorus (P)

2-8-5

Gains 3 e-

P3- (phosphide)

2-8-8

Ar

Element

Action

Ion formed

Same electron configuration as

Metals lose their valance electrons to leave behind the underlying full energy level. Lithium has the electron configuration 2-1. Lithium loses its one valance electron to form the Li+ ion, which has the same electron configuration as He. Likewise, sodium that has the electron configuration 2-8-1, loses its one valance electron to form the Na+ ion giving it the same electron configuration as Ne (2-8). Elements in the alkaline earth column form +2 ions. Beryllium has the electron configuration 22, so it loses its two valance electrons to achieve the same electron configuration as He and in so doing forms the Be2+ ion. Magnesium has the electron configuration 2-8-2, it loses its two valance electrons, forming the Mg2+ ion which has the same electron configuration (2-8) as Ne. A summary of this information is in the following table.

Lithium (Li)

Electron Configuration of Element 2-1

Li+ (Lithium ion)

Electron configuration of cation 2

Loses 1 e-

He

Sodium (Na)

2-8-1

Loses 1 e-

Na+ (Sodium ion)

2-8

Ne

Potassium (K)

2-8-8-1

Loses 1 e-

K+ (Potassium ion)

2-8-8

Ar

Beryllium (Be)

2-2

Loses 2 e-

Be2+ (Beryllium ion)

2

He

Magnesium (Mg)

2-8-2

Loses 2 e-

Mg2+ (Magnesium ion)

2-8

Ne

Calcium (Ca)

2-8-8-2

Loses 2 e-

Ca2+ (Calcium ion)

2-8-8

Ar

Aluminum (Al)

2-8-3

Loses 3 e-

Al3+ (Aluminum ion)

2-8

Ne

Element

Action

Ion formed

3-30

Same electron configuration as

Copyright Prof. Lawrence Mailloux, 2021

To summarize, using the octet rule we are able to predict the charge ion that many elements form simply based upon their location on the periodic table. The partial periodic table below summarizes this information for the elements who’s charge can be predicted using the octet rule. IA 1 valence e-

IIA 2 valence e-

IIIA 3 valence e-

IVA 4 valence e-

VA 5 valence e-

VIA 6 valence e-

VIIA 7 valence e-

VIIIA 8 valence eHe

H+ Li+

Be2+

Na+

Mg2+

K+

Ca2+

Rb+

Sr2+

Cs+

Ba2+

Octet Rules does not apply to transition metals

no ions

Al3+

N3-

O2-

F-

Ne

nitride

oxide

fluoride

no ions

P3-

S2-

Cl-

Ar

phosphide

sulfide

chloride

no ions

As3-

Se2-

Br-

Kr

arsenide

selenide

bromide

no ions

Te2-

I-

Xe

telluride

iodide

no ions

Rn no ions

3-31

Copyright Prof. Lawrence Mailloux, 2021

Problem 13 – Fill in the table below. Unlike in problem seven, which dealt with neutral atoms, the number of protons and electrons are NOT the same in this problem. The first row has been done for you as an example Atomic #

Mass #

# Protons

# Electrons

# Neutrons

Ion Name

11

23

11

10

12

Sodium ion

10

12

12

16

2

18

Se2-

81

Oxide

20 20

35

Calcium ion

36 P3-

31 35

17

18

14

Nitride 9

10

10 I-

131 52

Na+

4 8

19

23

Sulfide

44 3

Symbol

128

54 33

36

42 88

Sr2+

27

37

86

36

3-32

Al3+

Copyright Prof. Lawrence Mailloux, 2021

Answers to In Chapter Problems 1. 335 Cal = 335 kcal = 355,000 cal = 1,402 kJ = 1,402,000 J 2. 420 Cal = 420 kcal = 420,000 cal = 1,757 kJ = 1,757,000 J 3. a. Beans 1 g fat x 9 Cal/g = 9 Cal 6 g protein x 4 Cal/g = 24 Cal 29 g total carbs – 5 g fiber = 24 g net carbs x 4 Cal/g = 96 Cal Estimated total Calories = 9 + 24 + 96 = 129 Cal b. SPAM® 4.5 g fat x 9 Cal/g = 40.5 Cal 9 g protein x 4 Cal/g = 36 Cal 1 g total carbs – 0 g fiber = 1 g net carbs x 4 Cal/g = 4 cal Estimated total Calories = 40.5 + 36 + 4 = 80.5 Cal c. Peanut M&M’s® 13 g fat x 9 Cal/g = 117 Cal 5 g protein x 4 Cal/g = 20 Cal 30 g total carbs – 2 g fiber = 28 g net carbs x 4 Cal/g = 112 Cal Estimated total Calories = 117 + 20 + 112 = 249 Cal

Spam and M&M wrappers on next page

3-33

Copyright Prof. Lawrence Mailloux, 2021

4.

Gas

Melt Solid

Liquid Freeze

3-34

Copyright Prof. Lawrence Mailloux, 2021

x

Asprin

x

x

Olive Oil

x

x

Diesel

x

x

Compound

x

x

Stainless Steel Mercury

x

x

Radon

x

x

Aluminum

x

x

Pepto-Bismol®

x

Iron

x

Pure H2O

x

Heterogeneous

x

Homogenous

Element

Carbon

Mixture

Pure Substance

5.

x

x x

Coffee (without grounds)

x

x

Clean ocean water

x

x

Biscayne Bay water

x

Tequila (no worm)

x

Tequila (with worm)

x

x

Sand stirred into water

x

x

NaCl

x

x x

x

Coca-Cola

x

Coffee (with grounds)

x

x

Orange Juice (with pulp)

x

x

Orange juice (without pulp)

x

x

Tap water

x

x

14 carat gold

x

x

x

x

x

x

24 carat gold

x

Calcium chloride (CaCl2)

x

x x

Air Iron (III) oxide (Fe2O3)

x

x

Gasoline AZT (first HIV medication)

x

x

x

3-35

Copyright Prof. Lawrence Mailloux, 2021

6.

Particle

Symbol

Charge

Mass (amu)

Location

Makes Up

neutron

n0

0

1 amu

nucleus

Mass

electron

e-

-1

1/1833 amu

e- cloud

volume

proton

p+

+1

1 amu

Nucleus

mass

7.

Atomic #

Mass #

# Protons

# Electrons

# Neutrons

Isotope Name

11

23

11

11

12

Sodium-23

23

29

63

29

29

34

Copper-63

63

12

25

12

12

13

Magensium-25

25Mg

92

238

92

92

146

Uranium-238

238U

29

65

29

29

36

Copper-65

65Cu

92

235

92

92

143

Uranium-235

235U

16

32

16

16

16

Sulfur-32

32S

48

112

48

48

64

Cadmium-112

112Cd

30

65

30

30

35

Zinc-65

65Zn

19

39

19

19

20

Potassium-39

39K

6

14

6

6

8

Carbon-14

14C

83

209

83

83

126

Bismuth-209

209Bi

1

1

1

1

0

Hydrogen-1

1H

1

2

1

1

1

Hydrogen-2

2H

1

3

1

1

2

Hydrogen-3

3H

3-36

Symbol Na

Cu

Copyright Prof. Lawrence Mailloux, 2021

8. Answer C. The average atomic mass of bromine listed on a periodic table is 79.9 amu. This means that bromine-79 and bromine-81 must have about the same abundances. 9. Answer D. The average atomic mass of carbon listed on a periodic table is 12.01 amu. Because the average is so much closer to 12 than 14, carbon-12 must make up nearly 100% of naturally occurring carbon. 10. Avg = mass-1 x abundance-1 + mass-2 x abundance-2 Avg = 34.94 x 0.758 + 36.97 x 0.242 Avg = 35.43 amu Periodic table lists the average atomic mass of chlorine at 35.45 amu. The slight difference between our calculated value, and that listed on the periodic table is due to rounding error. 11. First you need to look up the average atomic mass of boron (10.811 amu) Avg = mass-1 x abundance-1 + mass-2 x abundance-2 10.811 = 10.01 x 0.199 + mass-2 x 0.801 Mass-2 = 11.01 amu 12. Avg = mass-1 x abundance-1 + mass-2 x abundance-2 + mass-3 x abundance-3 Avg = 204.36 x 0.173 + 206.41 x 0.601 + 207.31 x 0.226 Avg = 206.26 amu

3-37

Copyright Prof. Lawrence Mailloux, 2021

13.

Atomic #

Mass #

# Protons

# Electrons

# Neutrons

Ion Name

11

23

11

10

12

Sodium ion

12

24

12

10

12

Magnesium ion

24Mg2+

16

32

16

18

16

Sulfide

32 2-

34

78

34

36

44

Selenide

78

3

7

3

2

4

Lithium ion

7

8

16

8

10

8

Oxide

16

19

39

19

18

20

Potassium ion

39

20

40

20

18

20

Calcium ion

35

81

35

36

46

Bromide

15

31

15

18

16

Phosphide

17

35

17

18

18

Chloride

35

7

14

7

10

7

Nitride

14

9

19

9

10

10

Fluoride

19 -

53

131

53

54

78

Iodide

131 -

52

128

52

54

76

Telluride

128

33

75

33

36

42

Arsenide

75

38

88

38

36

50

Strontium ion

88

13

27

13

10

14

Aluminum ion

27

37

86

37

36

49

Rubidium ion

86

3-38

Symbol 23

Na+

S

Se2-

Li+ O2K+

40

Ca2+

81

Br-

P3Cl-

N3F

I

Te2-

As3Sr2+

Al3+ Rb+

Copyright Prof. Lawrence Mailloux, 2021

Chapter 4 Nuclear Chemistry Introduction to Radiation What do bananas, cat litter, and your household smoked detector have in common? Answer, they are all radioactive! Radiation is one of the most highly studied phenomena in the history of science. Since the second world war scientists working for governments, private industry, and universities the world over have studied radiation for military, medical, and basic scientific research purposes. Despite all the billions of dollars spent, and countless hours dedicated to understanding radiation, it is still extremely poorly understood, and feared, by the vast majority of the general population. It is easy to understand why this is the case. Being invisible, odorless, tasteless, and potentially deadly radiation appears to most people to be a mysterious force to be feared. This is unfortunate, because accidents involving radiation are extremely rare, but when they do happen they can be very scary and captivating. This fear, combined with the rapid communication made possible by the internet, results in a very large amount of misinformation and unnecessary fear. People tend to fear very rare and shocking accidents (like plane crashes) much more than everyday accidents such as fatalities from drunk driving. Regardless of the fact that one is much more likely to die in a car accident than a plane crash, many people fear flying, despite the fact that it is a very safe means of travel. The same is true with radiation. The world was captivated by the Fukushima nuclear disaster and many responded with excessive fear. Despite the fact that coal mining is very dangerous, and that the air pollution released from coal power plants causes many more deaths per year than nuclear power1, many people overreacted and wanted to immediately stop nuclear power altogether. A more educated population would take a more rational approach and work to understand what caused the disaster in the first place as well as determine how such mistakes can be prevented in the future.

1. Daigle, D. Air Pollution is Shortening Lives Worldwide. Science News, September 2018, pp 32.

4-1

Copyright Prof. Lawrence Mailloux, 2021

The purpose of this chapter is by no means intended to make you an expert on radiation. I as an organic chemistry by training have a very limited background it in by myself, however I do know enough to know that radiation is a very important tool in our modern lives and that when used correctly it is very safe. Without radiation most of the amazing medical imaging technologies that we have today would not be possible. Today we routinely detect and treat many cancers early, which only a few decades ago would have been detected only after it was too far advanced to be treated. People associate the word radiation with cancer; and while it is true that radiation can cause cancer, the reality is that radiation saves many more lives from cancer than it causes. Radiation is a complex topic and there is a lot of misinformation out there, so it is important to carefully consider one’s sources and think critically when reading about hot button topics like radiation on the internet.

4-2

Copyright Prof. Lawrence Mailloux, 2021

Radioactive Decay Unstable Isotopes The Bohr model of the atom tells us that all atoms consist of a large electron cloud surrounding a small nucleus consisting of protons and neutrons. The number of protons (atomic number) determines the element, and the varying numbers of neutrons gives rise to the various isotopes of each element. Of course the nucleus can’t just contain some random number of protons and neutrons. Just like how the electrons in the electron cloud are arranged into energy levels and sublevels (giving the periodic table its shape), the protons and neutrons in the nucleus are also arranged into levels and sublevels. The details behind this are way too complex for this course, all you need to know is that this means that only some combinations of protons and neutrons give stable nuclei, others are unstable and are radioactive. Radioactive isotopes release this energy of instability in a variety of ways through a process known as radioactive decay. The following image, obtained from Wikipedia, is a chart of all the different isotopes of each element. There are about 90 naturally occurring elements, each with multiple isotopes. If you look carefully, you will see there are hundreds of little boxes. Each little box represents an

50 0

Number of Protons

100

isotope, An enlargement of the section circled in red is on the next page.

Number of Neutrons 0

90 4-3

180

Copyright Prof. Lawrence Mailloux, 2021

Below is an enlargement of the circled portion of the image on the prior page. So that the elements read from left to right (as opposed to bottom to top) I have placed the protons on the x-

Number of Neutrons

axis in my enlargement. 12

22

Ne

11

21

Ne

20

Ne

10

18

O

19

F

9

17

O

18

F

16

O

8

14

15

N

7

13

14

N

13

N

C C

6

10

5 4

7

Li

3

6

Li

2

3

H

4

He

1

2

H

3

He

0

1

H 1

2

3

Be

11

B

12

9

Be

10

B

11

7

Be

4

5

C C

6

7

8

9

10

Number of Protons

The first column represents all the isotopes of hydrogen, the second helium, the third lithium, and so on. Recall that on the periodic table each box represents the average of all the isotopes of each element. This means that on a periodic table, each box represents the average off all the isotopes listed in a given column on the figure above.

4-4

Copyright Prof. Lawrence Mailloux, 2021

The colors in the first image describe that particular isotopes stability. Only the narrow black strip down the center are stable isotopes, all the others are radioactive, with the color indicating what kind of radioactive decay that isotope undergoes. In the sections that follow we will learn a little bit about three of the most common types of radioactive decay: alpha particles, beta particles, and gamma rays. There are many other kinds of decay, however we will keep things as simple as possible by only looking at these three. Also, while there are ways for scientists to predict what kind of radioactive decay an isotope will undergo, you will NOT be expected to do this. There are many different radioactive isotopes that are used in medicine. They are often administered prior to various imaging procedures to act as dyes or tracers. By following the movement of the radioactive isotope inside a patients body the doctor can learn more about what’s going on inside. There are also many other applications for radioactive isotopes besides medicine. Carbon-14 is used to do carbon dating, a technique that allows scientists to find out how old things like mummies, shipwrecks, and other artifacts are. Americium-241 is used in smoke detectors. This is just a few of the many applications out there.

4-5

Copyright Prof. Lawrence Mailloux, 2021

Alpha Particle Decay The first kind of radioactive decay we will consider is alpha particle emission. An alpha particle consists of two protons and two neutrons bonded together. Because there are no electrons present, the particle has a charge of plus two. With an atomic number of two, an alpha particle is actually just a helium nucleus. Scientists use either 42𝐻𝑒2+ or the α (lowercase Greek alpha) symbol to represent an alpha particle when writing nuclear chemistry equations; the equations used to describe the transformations that occur when an isotope undergoes radioactive decay. Sometimes an isotope’s nucleus is unstable because it has too many protons and neutrons. When this is the case, the nucleus of the radioactive isotope ejects two protons and two neutrons together as an alpha particle. The loss of the high energy alpha particle releases energy causing the product (commonly referred to as the daughter particle) of the reaction to be more stable (of lower energy) than the starting isotope. Unlike the ordinary chemical equations covered in Chapter 6, and used throughout the rest of this book, nuclear chemistry equations can involve a change in the number of protons in the isotopes nucleus. This mean that the atomic number may change during nuclear reactions, meaning the identity of the elements themselves can change. For an example recall at the start of the chapter that I told you your smoke detector was radioactive, I wasn’t lying; next time you remove the cover of your smoke detector to change the battery look carefully and you may notice a warning label similar to the following one found in prof Mailloux’s smoke detector.

4-6

Copyright Prof. Lawrence Mailloux, 2021

Because of their larger size, mass, and charge, alpha particles have poor penetrating power. In fact, a sheet of paper is all that is needed to stop an alpha particle. Smoke detectors take advantage of an alpha particles poor penetrating power to do their job. When smoke from a fire enters the detector, it blocks the alpha particles emitted by the radioactive americium-241 from reaching the detector, which triggers the alarm. Below is the balanced equation for the decay reaction taking place in a smoke detector: 241 95𝐴𝑚

→

237 93𝑁𝑝

+ 42𝐻𝑒2+

Or if using the α symbol the same equation can be written as 241 95𝐴𝑚

→

237 93𝑁𝑝

+ 𝛼

Where it is the readers responsibility to know that α and 42𝐻𝑒2+ mean the same thing. Unless instructed otherwise you may use either notation you prefer. You will see both notations throughout this book Because of the Law of Conservation of Mass, nuclear equation must be balanced. This means that the sum of the superscripts and subscripts on the left of the arrow equal must equal the sum of the superscripts and subscripts on the right of the arrow. In the reaction above note that 241 = 237 + 4, and 95 = 93 + 2 (we ignore charges when balancing nuclear equations). Notice that 4-7

Copyright Prof. Lawrence Mailloux, 2021

because of the loss of two protons from the nucleus due to the release of the alpha particle, the atomic number in our above equation dropped by two, changing the element from americium to neptunium (Np). We have moved two places to the left on the periodic table. Let’s consider another example by filling in the reactions shown below. Given that uranium-235 also decays by emitting alpha particles fill in the blank in the following equation 235 92𝑈

→ ______+ ∝

To makes things easier, let us rewrite this equation using the 42𝐻𝑒2+ symbol instead giving: 235 92𝑈

→ ______ + 42𝐻𝑒2+

To balance this equation, we must determine what number would be needed to balance the mass numbers on top, and the atomic numbers on the bottom. The atomic number of the product must be 90 as 92 = 90 + 2 which balances the subscripts. Looking at a periodic table we see that element 90 is Th (thorium). Inserting this into our equations gives 235 92𝑈

→

? 90𝑇ℎ

+ 42𝐻𝑒2+

To balance the mass number, we must put in 231, as 231 + 4 = 235. Adding this to our equation gives our final balanced equation. 235 92𝑈

→

231 90𝑇ℎ

+ 42𝐻𝑒2+

4-8

Copyright Prof. Lawrence Mailloux, 2021

Let’s consider another example, where the problem is slightly different. Fill in the blank in the equation below ______ →

227

𝐴𝑐 + ∝

To simplify things I will again rewrite the equation using 42𝐻𝑒2+ instead of the α symbol to make the numbers more obvious. Also notice the atomic number for Ac was not given. Looking at a periodic table we find that Ac has an atomic number of 89. Filling in the atomic number and switching 42𝐻𝑒2+ for the alpha symbol gives ______ →

227 89𝐴𝑐

+ 42𝐻𝑒2+

The mass number of our reactant isotope must be 331 because 227 + 4 = 331 and our atomic number must be 91 as 89 + 2 = 91. Looking at a periodic table we see that element 91 is Pa. Filling this information in gives our final balanced equation: 331 91𝑃𝑎

→

227 89𝐴𝑐

+ 42𝐻𝑒2+

Problem 1 – Balance the following nuclear equations which occur by alpha decay. You may need to look up some atomic numbers using a periodic table. a) _______ → b)

144 60𝑁𝑑

c)

210

+ 𝛼

→ ______ + 42𝐻𝑒2+

𝑃𝑜 → ______ + ∝

d) _______ → e)

171 76𝑂𝑠

238 94𝑃𝑢

→

208

𝐵𝑖 + 42𝐻𝑒2+

234 92𝑈

+ ______

4-9

Copyright Prof. Lawrence Mailloux, 2021

Beta Particle Decay The second type of radioactive decay we will consider is beta decay. Beta decay involves the release of beta particles which are high energy electrons. Scientists use either the beta symbol 𝛽, or −10𝑒, to represent beta particles. Because the mass of an electron is so small compared to that of a proton or neutron (see Chapter 3) the beta particle is assigned a mass number of zero. Additionally, for purposes of balancing nuclear equations, beta particles are assigned an “atomic number” subscript of minus one (-1). The electrons emitted during beta decay are NOT from the electron cloud. Because nuclear decay is driven by an unstable nucleus, the electrons emitted are actually produced from within the unstable nucleus. Inside the nucleus of an isotope that undergoes beta decay, a neutron ( 10𝑛) spontaneously decays into a proton ( 11𝑝) and an electron ( −10𝑒). It is this electron that is ejected as the beta particle. The equation for this process is given below, notice that neutrons are given an atomic number of zero for purposes of writing these equations 1 0𝑛

→

1 1𝑝

+

0 −1𝑒

Before going further, we should take note of a few things regarding the above equation. First notice that it is balanced. The mass numbers balance (1 = 1 + 0) and the atomic numbers balance (0 = 1 + -1). Secondly, because protons and neutrons have both weigh one amu. (see Chapter 3) the mass number does not change during a beta decay reaction. Lastly, by having a neutron turn into a proton and an electron the atomic number increases by one during a beta decay reaction. This means we will move one element to the right on the periodic table. Let’s consider an example. Carbon-14 is a naturally occurring rare isotope of carbon. Scientists make use of this isotope when performing carbon dating to determine the age of artifacts which we will learn about more later in this chapter. For now, let us consider the beta decay of carbon14 by filling in the following equation. 14 6𝐶

→ _____ + 𝛽

First I will replace the beta symbol with 14 6𝐶

→ _____ +

0 −1𝑒

to make the numbers more obvious.

0 −1𝑒

4-10

Copyright Prof. Lawrence Mailloux, 2021

The mass number will be unchanged because 14 = 0 + 14. The atomic number on the other hand will increase to seven because 6 = 7 + (-1). Filling in this information gives our balanced equation. 14 6𝐶

→

14 7𝑁

0 −1𝑒

+

Looking at a periodic table we can see that nitrogen is indeed one box to the right of carbon. It’s important to remember that adding a negative number is the same as subtraction. A common error that students make is forgetting that the one is negative giving the result shown below. 14 6𝐶

→

14 5𝐵

+

0 −1𝑒

(Common wrong answer!)

Problem 2 – Balance the following nuclear equations which occur by beta decay. You may need to look up some atomic numbers using a periodic table. a) _______ → b)

3 1𝐻

c)

131

+ 𝛽

→ ______ +

0 −1𝑒

𝐼 → ______ + 𝛽

d) _______ → e)

32 16𝑆

63 28𝑁𝑖

→

36 18𝐴𝑟 63 29𝐶𝑢

+

0 −1𝑒

+ ______

4-11

Copyright Prof. Lawrence Mailloux, 2021

Gamma Emission Unlike alpha and beta decay which involve the release of particles, gamma emission involves the release of gamma rays which are simply very high energy forms of electromagnetic radiation. Electromagnetic radiation is simply energy. There are many different forms of electromagnetic radiation, some of which you have already heard of: radio waves, microwaves, infrared (heat), visible light (all colors of the rainbow), ultraviolet light, x-rays, and gamma rays. The only difference between these forms of electromagnetic radiation is the amount of energy they carry. Visible light (which is made up of all the colors of the rainbow) is only a tiny sliver of the entire spectrum. The only thing that is special about visible light is that our eyes have evolved to see it, there are animals that can see slightly outside of this range, for example bees can see slightly into the UV part of the spectrum. The image below, obtained from NASA’s website2, arranges these different forms of radiation in order of decreasing energy.

Starting on the right side of the diagram we have radio waves. These waves come from radio stations, are very low in energy, and pass through our bodies all day without us feeling anything. Infrared light is felt as heat, and visible light is what we use to see. All of these are harmless. At the higher energy side of the spectrum ultraviolet light (UV) is capable of damaging DNA (we will learn more about DNA in Chapter 17) and as a result excessive UV exposure can cause sun burns and skin cancer. Sunscreen works to protect one’s skin by blocking these rays. X-rays are even higher in energy, however the amount of x-rays one is exposed to during a medical x-ray is 2. National Aeronautics and Space Administration. Imagine the Universe. https://imagine.gsfc.nasa.gov/science/toolbox/emspectrum1.html (accessed April 12, 2019)

4-12

Copyright Prof. Lawrence Mailloux, 2021

very low and the benefits of obtaining such an x-ray greatly outweigh any very small risk from having the x-ray taken. Lastly at the far left we have gamma rays, which are the highest energy part of the spectrum. Recall from Chapter 3 that there are two things in the universe, matter and energy. Because gamma rays are only energy (that is they are not matter) they do not impact the balancing of nuclear equations. Scientists use the Greek letter gamma ( 00𝛾) to represent gamma rays in nuclear equations. Notice that both the mass and atomic numbers on the gamma symbol are zeros, thus they will not impacting the balancing nuclear equations. A small number of isotopes emit only gamma rays. This happens when the protons and neutron in the nucleus are arranged in an unstable way, giving rise to an unstable nucleus. To represent this scientists, use the letter m, to indicate an unstable/high energy form of an isotope. The “m” stands for “meta,” referring to the isotope being only “metastable.” When the protons and neutrons in the nucleus rearrange themselves into a more stable arrangement, the excess energy is released as a gamma ray, and the isotope left behind is more stable. Since only energy was released, the atomic and mass numbers are unaffected, the isotope is simply more stable so the letter “m” is removed. An example of this is an unstable form of the element technetium, which has several medical applications and is used as a tracer in many types of imaging studies. 99 m 43𝑇𝑐

→

99 43𝑇𝑐

+ 00𝛾

While the reaction of Tcm listed above is an important example, most of the time gamma rays occur at the same time as alpha or beta decay. However, since they don’t have any mass they don’t have an impact on balancing. The reaction below illustrates an important example of a gamma emitter, iodine-131 which can be used in the treatment of thyroid cancer. In this case the gamma ray is released along with a beta particle. 131 53𝐼

→

131 54𝑋𝑒

+

0 −1𝑒

+ 00𝛾

Another example is the decay of radium-226, in this case the gamma ray is emitted along with an alpha particle. 226 88𝑅𝑎

→

222 86𝑅𝑛

+ 42𝐻𝑒2+ + 00𝛾 4-13

Copyright Prof. Lawrence Mailloux, 2021

Problem 3 – Balance the following nuclear equations which involve the release of gamma rays. You may need to look up some atomic numbers using a periodic table. a)

11 5𝐵 *

b)

40

→ ______ + 00𝛾

𝐾 → ______ +

c) ______ → d)

214 84𝑃𝑜

→

e) ______ →

235 92𝑈 210 82𝑃𝑏 60 28𝑁𝑖

0 −1𝑒

+ 00𝛾

+ 42𝐻𝑒 + ______ + ______ + 00𝛾

+ 00𝛾

Radiation Safety and Shielding This book is NOT intended to make one an expert on radiation safety, however there are simple ways that the risks from radiation can be kept to a minimum. Three of those ways are: (1) shielding, (2) minimizing exposure, (3) keeping as large a distance from the source of the radiation as possible. As was mentioned earlier in the chapter, because alpha particles are relatively big and have a large charge of +2 they have very poor penetrating power. All that is needed to stop (shield one from) an alpha particle is a sheet of paper, or even a few inches of air. You skin will also stop alpha particles; however, if ingested alpha particles can be very hazardous. A famous example of this occurred in late 2006 when a former Russian KBG agent named Alexander Litvinenko was poisoned (likely by the Russian government) with polonium-210. Litvinenko died on November 22, 2006. His assassin was never found. Below are two photograph of Litvinenko, obtained from Wikipedia, one before his poisoning, and one shortly before his death in a London hospital.

4-14

Copyright Prof. Lawrence Mailloux, 2021

Beta particles are much less massive and less charged than alpha particles, therefore more shielding is needed to stop beta particles. Beta particles can travel many feet through the air, as opposed to the few inches an alpha particle can travel. A few inches of wood (like a door), or heavy clothing will stop beta particles. Gamma rays are very similar to x-rays. Because gamma rays are pure energy (no mass) they have great penetrating power and can travel a long distance (hundreds of feet) through the air. This is why when getting an x-ray taken a thick lead blanket is used to cover the parts of the patient’s body not needing to be x-rayed to minimize exposure for the patient. Additionally, the x-ray technician usually leaves the room, or walks behind a wall, before taking the x-ray to minimize their own exposure. Problem 4 – Fill in the following table regarding the concepts covered in the past few paragraphs. Type of Radiation

Charge

Symbol(s)

Shielding

Distance Traveled through air

____ or β 0

hundreds of feet 4 2+ 2𝐻𝑒

or ____

4-15

Copyright Prof. Lawrence Mailloux, 2021

An easy way to minimize radiation exposure is to keep as great a distance as possible between yourself and the radioactive substance. The blue letter “S” below represents a source of radiation. The radiation from this material will spread out in a sphere in all directions as shown in this image obtained from Wikipedia.

This means that the farther the radiation travels, the larger the area it will be spread over and the less intense it will be. Remember from geometry class that the surface area of a sphere is A = 4 π r2. This means that if we double the distance (radius) from a radioactive material, the surface area of the sphere with be 22 = 4 times larger and the radiation will be 1/4th as intense as it will be spread over four times the area. Likewise, if we were to triple our distance from the radioactive material the sphere over which the radiation is spread would be 32 = 9 times as large, so the radiation would be only 1/9th as intense. In more technical terms, the intensity decreases with the square of the distance from the source. Because the intensity decreases with the square of the distance, even a small increase in distance can result in a significant decrease in radiation intensity.

4-16

Copyright Prof. Lawrence Mailloux, 2021

Problem 5 – Fill in the following table regarding the relationship between distance and radiation intensity for a sample initially one meter away from you. Distance from Source

1m

Increase in area over which radiation is spread

12 = 1

Relative intensity of radiation

1

2m

4m 32 = 9

1/2

52 = 25 1/16

Units Used to Measure Radiation Measuring radiation is important for ensuring it’s safest and most effective use in medicine. When measuring radiation, we can do so in multiple ways. We can measure the activity, which is a measure of how many decay reactions occur in a given amount of time. There are two units used to measure radiation activity, the Curie, and the Becquerel. The Curie is named in honor of Marie Curie who was one of the greatest scientists off all time. She was the first woman to win a Nobel Prize, and also the only one two win two Nobel Prizes in two different fields, chemistry and physics. Element number 96 (Curium, symbol Cm) is named in her honor. She is entombed in the Pantheon in Paris and her life was the topic of the movie Marie Curie. A Curie (Ci) is defined as the number of disintegration that happen in one second for one gram of the element radium, which Curie discovered. In one second a one gram sample of radium will undergo 3.7 x 1010 disintegrations, so 1 Ci = 3.7 x 1010 disintegrations/sec. While the Curie is a commonly used unit for radioactive decay, the number 3.7 x 1010 is not easy to work with so the official metric scientific unit is the Becquerel (Bq), 1 Bq = 1 disintegration/sec.

4-17

Copyright Prof. Lawrence Mailloux, 2021

The picture below is a close up of the radiation warning on Professor Mailloux’s smoke detector. It lists the activity of the americium-241 inside as being 1.0 μCi, which is the same as 37 kBq as the following calculation shows.

1.0 𝜇𝐶𝑖 1 𝐶𝑖 3.7 𝑥 1010 𝑑𝑖𝑠𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛𝑠/𝑠𝑒𝑐 1 𝐵𝑞 1 𝑘𝐵𝑞 ( )( 6 )( )( )( ) = 37 𝑘𝐵𝑞 10 𝜇𝐶𝑖 1 𝐶𝑖 1 𝑑𝑖𝑠𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛/𝑠𝑒𝑐 1000 𝐵𝑞

In addition to measuring activity we can also measure how much radiation is absorbed, often by some biological tissue. The rad (radiation absorbed dose) is the unit used to measure the amount of radiation absorbed by one gram of the material under consideration. The related metric unit is the gray (Gy) 1 Gy = 100 rads. Once we know how much radiation has been absorbed the next logical question is how much damage would that cause? The amount of damage that occurs depends on the type of radiation (alpha, beta, gamma, x-ray…) as well as the amount of radiation that was absorbed and what kind of tissue was exposed. The amount of absorbed radiation, in rads, is multiplied by a factor (whose value depends on the type of radiation) as the formula below shows. This gives the rem as shown below. Rem = rad x factor The rem (roentgen equivalent man) is a unit used to measure the biological damage from radiation exposure. The related metric unit for biological damage is the sievert (Sv), 1 Sv = 100 rems.

4-18

Copyright Prof. Lawrence Mailloux, 2021

Problem 6 - fill in following table to summarize the units given in the proceeding paragraphs.

To Measure

Non-metric Unit

Metric Unit

1 Ci = 3.7 x 1010 disintegrations/sec Absorbed radiation

1 Gy = 100 rads rem = rad x factor

You will not be asked to do any calculations with these units. However, you will be expected to know what these various units are used for and which are metric units and which are not. Do the problem below for practice.

4-19

Copyright Prof. Lawrence Mailloux, 2021

Problem 7 – fill in the blanks in the following narrative. Marie ______ was an amazing double Nobel Prize winning scientist who discovered the element radium. The unit for radiation activity, which is the _________, is named in her honor and is equal to the number of disintegrations/second for a one-gram sample of radium. The symbol for this unit Ci. One Ci is equal to _________ disintegrations per seconds. The metric unit for activity is ___________ (Bq). One Bq = ________ disintegrations per second. The amount of absorbed radiation is measured using _______ which stands for radiation absorbed dose. The metric unit for absorbed radiation is the ______ (Gy), which is equal to _____ rads. After the amount of absorbed radiation has been measured, the biological impact can be measured by multiplying the number of _____ by a factor that depends on the kind of radiation that was absorbed, doing so gives the _____ which stands for roentgen equivalent man. The related metric unit of biological impact is the ______ (Sv) which is equal to 100 rems.

Half-Lives and Radioactive Decay The rate at which a given isotope decays radioactively is an intrinsic property of that isotope and cannot be speed up or slowed down. Some isotopes decay very rapidly in fractions of a second, while others can take millions, or even billions, of years to decay. To describe the rate at which a material decays scientists talk about the half-life of an isotope. The half-life of an isotope is the time needed for half of a sample of that isotope to decay radioactively. All isotopes have their own half-lives. You will either be given the half-life of an isotope or asked to solve for it, you do NOT need to memorize half-lives.

4-20

Copyright Prof. Lawrence Mailloux, 2021

To see how this works, let us consider an example mentioned earlier, iodine-131. Iodine-131 has a half-life of eight days. Therefore, if one were to start with a sample of 240 grams of iodine-131 after eight days (one half-life) only 120 grams would remain and after eight more days (16 days total, two half-lives) only 60 grams would remain, and so on. This is easily summarized using an arrow diagram. As the following diagram below shows, after a total of 32 days (four half-lives) only 15 grams of iodine-131 remain. 8 𝑑𝑎𝑦𝑠

240 𝑔 131 51𝐼 →

8 𝑑𝑎𝑦𝑠

120 𝑔 131 51𝐼 →

60 𝑔

131 51𝐼

8 𝑑𝑎𝑦𝑠

→

8 𝑑𝑎𝑦𝑠

30 𝑔 131 51𝐼 →

15 𝑔 131 51𝐼

For an example of how you could be asked about this consider the following story problem example. A sample of 500 grams of phosphorus-32 is allowed to sit for 42 days, after which time it is observed that only 62.5 grams of phorphorus-32 remains. What is the half-life of phosphorus-32? Start by writing an arrow diagram and filling in what you know starting with 500 and dividing by two until your reach 62.5. ?

?

500 𝑔 32𝑃 → 250 𝑔 32𝑃 → 125 𝑔

32

?

𝑃 → 62.5 𝑔 32𝑃

Look at your diagram. How many half-lives were required to go from 500 g to 62.5 grams? In this case three half-lives were needed. Given that the total time was 42 days, each half-life must be equal to 42/3 = 14 days. Note that these problems can be worded in many different ways. In the prior example you were asked to determine the half-life. Now consider this alternative example. A sample of 500 grams of phosphorus-32 is allowed to sit for 56 days. Given that the halflife of phosphorus-32 is 14 days, how much phosphorus-32 will be left after 56 days? Because the half-life is 14 days and 56 days have passed this means that 56/14 = 4 half-lives will have taken place over those 56 days. Start your arrow diagram with 500 grams and cut the amount in half four times until you reach your answer. In this case, only 31.25 grams of phosphorus-32 will remain. 14 𝑑𝑎𝑦𝑠

500 𝑔 32𝑃 →

14 𝑑𝑎𝑦𝑠

250 𝑔 32𝑃 →

125 𝑔

32

4-21

14 𝑑𝑎𝑦𝑠

𝑃 →

14 𝑑𝑎𝑦𝑠

62.5 𝑔 32𝑃 →

31.25 𝑔 32𝑃

Copyright Prof. Lawrence Mailloux, 2021

There are additional ways these questions can be worded as well. You could be given the total time, half-life, and final amount and asked to work backwards. Using phosphorus-32 as an example again, such a question might be worded like this. After sitting on a shelf for 56 days only 31.25 grams of phosphorus-32 remains. Given that the half-life of phosphorus-32 is 14 days, how much phosphorus-32 was there at the beginning? To solve this problem we would use the same diagram, but work backwards from 31.25 grams to 500 grams. 14 𝑑𝑎𝑦𝑠

? ? ? 𝑔 32𝑃 →

14 𝑑𝑎𝑦𝑠

250 𝑔 32𝑃 →

125 𝑔

32

14 𝑑𝑎𝑦𝑠

𝑃 →

14 𝑑𝑎𝑦𝑠

62.5 𝑔 32𝑃 →

31.25 𝑔 32𝑃

Problem 8 – Thallium-201 (201Tl) is used for heart stress tests. It has a half-life of 73 days. If a 500 gram sample of thallium-201 is allowed to sit for 219 days, how many grams will remain? Problem 9 – Technetium-99m (99Tcm) is one of the most commonly used isotopes in medicine and is used in a wide variety of scans. If only 112.5 grams of an originally 1800 grams sample of this isotope remains after 24 hours, what is the half-life of this technetium-99m? Problem 10 – Cobalt-60 (60Co) is used as a source of gamma rays for nuclear medicine. Given that the half-life of cobalt-60 is 5.25 years, how many grams of cobalt-60 would need to be purchased in order to have 10 grams remaining after 21 years? Problem 11 – Gallium-67 (67Ga) is a gamma-emitting isotope that is used in nuclear medicine to detect certain types of tumors. If only 12 grams of a 96 grams sample of gallium-67 remains after 9.9 years, what is the half-life of this isotope? Problem 12 – Fluorine-18 (18F) is a radioactive isotope that is used in PET (positron emission tomography) scans. Its half-life is 110 minutes. How many grams of an 80 gram sample of fluorine-18 would remain after 440 minutes?

4-22

Copyright Prof. Lawrence Mailloux, 2021

Beyond Radioactive Decay, Applications of Half-lives to Pharmaceuticals The concept of half-life is not limited to radioactive substances. The rate at which any substance breaks down can be described using half-lives. One important application of this is pharmaceuticals. The rate at which one’s body processes a medication can be described in terms of half-lives too. If a medication has a half-life of 12 hours this means that after 12 hours, half the medication will have been processed by the body and only half will remain. After 24 hours, only a quarter will remain; and after 36 hours only one eighth will remain and so on until all the drug is gone. (Note that half-lives of medications are not exact, like those for radioactive decay, and can vary due to complex reasons beyond the scope of this class, therefore some simplifications have been made to write this section and the half-lives stated here are only approximate.) Medications with longer half-lives stay in the body longer than those with shorter half-lives. There are several important consequences of this, one is dosage. A medication with a very short half-life may need to be taken two or three times a day, whereas a medication with a longer halflife may only need to be taken once a day. In addition to dosing, half-life can also have an impact on stopping medications too. Some medications can cause dependence, meaning that when a person stops taking a particular medication they feel unpleasant withdrawal symptoms. If a person changes their behavior regarding the use of a substance, despite the harm it causes them or those around them, we use the term addiction. It is possible to have dependence without addiction, although the two often go together. For example, antidepressant medications (Zoloft ®, Paxil ®, Effexor ®, Prozac ®, etc.) can cause withdrawal if stopped abruptly, however people stopping these medications don’t seek out these substances, or steal to obtain them, the same way an alcoholic might seek out alcohol, or an opiate addict might steal money to purchase OxyContin®. While covering substance abuse in detail is far beyond the scope of this class, it is important that you have a basic understanding of how half-lives and these concepts relate, as substance abuse has significant public health implications.

4-23

Copyright Prof. Lawrence Mailloux, 2021

As mentioned at the start of this book, there are three ways that people catch HIV: birth, blood, and sex. When people use drugs intravenously (like heroin) they often share needles between users. This can easily result in the spread of HIV and other blood borne diseases. Just like sex, addiction is a powerful driving force that can greatly influence a person’s behavior in ways that are self-destructive and illogical to an outside observer. Because these drives are so powerful, telling people not to have sex, or use drugs, is simply not an effective means of changing behavior. Shaming people is not effective either. Substance abuse is a complex social problem, with no simple solution. However, improving awareness about substance abuse amongst health care workers in an important step towards solving this problem. When a person stops taking any drug (legal or illegal) the level of that drug in their body begins to drop. The shorter a drugs half-life is, the more quickly it will leave the person’s body. This gives the body less time to adjust to not having the drug in their system and withdrawal will set in more quickly. In general, this means that drugs with shorter half-lives tend to produce quicker and stronger withdrawal symptoms than drugs with longer half-lives which leave a person’s system more gradually. This of course has serious implications for dependence and addiction, which in turn has implications for disease transmission.

4-24

Copyright Prof. Lawrence Mailloux, 2021

Problem 13 – Heroin is an opiate drug that has a very short half-life, of less than an hour. Methadone is also an opiate; however, it has a much longer half-life of about 24 hours. People who are addicted to heroin are sometimes proscribed methadone as an alternative opiate to prevent withdrawal and allow them function better in their daily lives. Explain using half-lives which of these two drugs (heroin or methadone) would cause withdrawal more quickly? What implications could this have for addiction treatment? Problem 14 – The half-lives for three common antidepressant medications are given below. Rank these medications in order increasing likelihood of causing withdrawal upon discontinuation.3 Citalopram (Lexapro®) = 33 hours Fluoxetine (Prozac®) = 5 days Paroxetine (Paxil®) 21 hours. Problem 15 – what is the difference between addiction and dependence? Can dependence occur without addiction?

3. Marken, P A.; Munro, J. S. Selecting a Selective Serotonin Reuptake Inhibitor: Clinically Important Distingusihing Features. J. Clin Psychiatry, 2 (6). 2000, 205-209.

4-25

Copyright Prof. Lawrence Mailloux, 2021

Answers to In Chapter Questions 1. a.

175

Pt

b.

140

Ce

c.

206

Pb

d.

212

At

e.

4 2+ 2𝐻𝑒

or α

2. a.

32

b.

3

c.

131

d.

36

e.

0 −1𝑒

a.

11

b.

40

c.

239

d.

4 2+ 2𝐻𝑒

e.

60

P

He Xe

Cl or β

3. B

Ca Pu or α

Nim

4. Type of Radiation

Charge

Symbol(s)

Shielding

Distance Traveled through air

beta

-1

____ or β

Heavy clothing

several feet

gamma

0

𝟎 𝟎𝜸

Thick lead

hundreds of feet

alpha

+2

sheet of paper

about an inch

4 2+ 2𝐻𝑒

or ____

4-26

Copyright Prof. Lawrence Mailloux, 2021

5. Distance from Source

1m

2m

3m

4m

5m

Increase in area over which radiation is spread

12 = 1

4

32 = 9

16

52 = 25

Relative intensity of radiation

1

1/2

1/9

1/16

1/25

6.

To Measure

Non-metric Unit

Metric Unit

1 Ci = 3.7 x 1010 disintegrations/sec

Becquerel (Bq)

activity

1 Bq = 1 disintegration/sec

Absorbed radiation

1 rad

1 Gy = 100 rads

biological damage

rem = rad x factor

sievert (Sv) 1 Sv = 100 rem

7. Marie _Curie_ was an amazing double Nobel Prize winning scientist who discovered the element radium. The unit for radiation activity, which is the __Curie___, is named in her honor and is equal to the number of disintegrations/second for a one-gram sample of radium. The symbol for this unit Ci. One Ci is equal to _3.7 x 1010_ disintegrations per seconds. The metric unit for activity is _becquerel_ (Bq). One Bq = _one_ disintegrations per second. The amount of absorbed radiation is measured using __rad__ which stands for radiation absorbed dose. The metric unit for absorbed radiation is the _Gray__ (Gy), which is equal to _100_ rads. After the amount of absorbed radiation has been measured, the biological impact can be measured by multiplying the number of _rads_ by a factor that depends on the kind of radiation that was absorbed, doing so gives the _rem_ which stands for roentgen equivalent man. The related metric unit of biological impact is the _sievert_ (Sv) which is equal to 100 rems. 4-27

Copyright Prof. Lawrence Mailloux, 2021

8. 62.5 grams 9. 6 hours 10. 160 grams 11. 3.3 years 12. 5.0 grams 13. Because methadone has a much longer half-life it will not cause as severe of withdrawl. It also will take longer since the last dose for withdrawal to start. This makes it much easier to someone taking methadone to function in their daily lives than if there were to take a quicker acting opiate like heroin. 14. Fluoxetine is the lease likely to cause withdrawal, followed by Citalopram, and then paroxetine which is most likely to cause withdrawal. The longer the half-life the less severe the withdrawal generally is. 15. Addiction is when a person changes their behavior because of craving for a substance despite the harm caused to their personal/professional lives and/or health. Dependence is when a person experiences unpleasant withdrawal upon stopping a drug, but does not experience craving for the substance as occurs with addiction. Dependence is mainly about the physical effects, while addiction is about the psychological effects of stopping a drug. It is possible to have dependence without addiction, however the two often occur together.

4-28

Copyright Prof. Lawrence Mailloux, 2021

Chapter 5 From Atoms to Compounds Covalent Vs. Ionic Compounds Recall that atoms want to have a full outer/valance energy level with an octet of eight electrons. There are two ways that an atom can do this; they can either share electrons, or transfer electrons. We will consider the sharing of electrons first.

Covalent Compounds When two or more non-metals form a compound they share electrons to form what are known as molecular or covalent compounds. At room temperature, these compounds can be solids, (like sugar, C6H12O6), liquids (H2O), or gases (carbon dioxide, CO2). Covalent compounds exist as distinct individual molecules. This means that if we were to look at a sample of a covalent compound under an “ultra-powerful” microscope, we would see separate/distinct molecules floating around everywhere as the images below illustrate.

O=C=O O=C=O O=C=O

O=C=O

O=C=O O=C=O

5-1

Copyright Prof. Lawrence Mailloux, 2021

I put “ultra-powerful,” in quotes because atoms are far too small to be seen with an ordinary microscope like the ones found in a biology class. Atoms are extremely tiny. To see atoms, a microscope about 1,000,000 times more powerful than a regular light microscope is needed. We will learn more about how to draw structures of these types of compounds later in the course. For now, we will simply learn how to name them. Naming of covalent compounds makes use of prefixes (listed below) to tell the reader how many of each atom are present in a given compound. Number 1

Prefix mono

Number 6

Prefix hexa

2

di (not bi)

7

hepta

3

tri

8

octa

4

tetra (not quad)

9

nona

5

penta

10

deca

To name a molecular compound follow the following formula 1st prefix (unless mono) + name 1st element + 2nd prefix + first part of 2nd element + “ide” Using this formula would give the following names N2S4 = dinitrogen tetrasulfide S2O3 = disulfur trioxide To shorten names, chemists do NOT include the prefix mono if there is only one of the first element. In addition, when adding a prefix would give double vowels like “oo” or “ao” the extra vowel is dropped. The following examples illustrate these common errors. CO = carbon monoxide, not monocarbon monoxide, or carbon monooxide CO2 = carbon dioxide, not monocarbon dioxide N2O4 = dinitrogen tetroxide, not dinitrogen tetraoxide 5-2

Copyright Prof. Lawrence Mailloux, 2021

Do the following practice problems before moving onto naming ionic compounds. Problem 1 – Fill in the following table of names and formulas of covalent compounds. Name

Formula

Nitrogen dioxide P2O10 Cl2O7 Phosphorus trichloride ICl3 Phosphorus pentabromide Sulfur hexafluoride SO

This marks the end of the section on naming of covalent compound. The rest of this chapter is dedicated to ionic compounds. In naming ionic compounds, prefixes like di, tri, tetra, etc. are NOT used. If you use a prefix to do any of the naming problems after this point in this chapter you are making one of the most common errors students make. Prefixes are for naming covalent compounds only. Do NOT use prefixes to name ionic compounds.

5-3

Copyright Prof. Lawrence Mailloux, 2021

Ionic Compounds When metals and a non-metals form compounds, they do so by transferring (gaining and losing) electrons. Recall that metals lose electrons to form positively charged cations, while non-metals gain electrons to form negatively charged anions. Because opposite charges attract, cations and anions will be attracted to each other and form what are known as ionic compounds. Ionic compounds are compounds made up of oppositely charged ions that are held together by ionic bonds; bonds that result from the attraction between oppositely charged ions. Ionic compounds are nearly always crystalline solids, and are often referred to as “salts.” In chemistry the word “salt,” can refer to any ionic compound. “Table salt,” refers specifically to sodium chloride, one of the most well-known ionic compounds. In the case of table salt, the sodium atom loses an electron to form the sodium ion (Na+), while the chlorine atom gains that same electron to form the chloride ion (Cl-); the attraction between these two oppositely charged ions forms the ionic bond. This process can be shown by writing a set of reactions where the symbol e- is used to represent the transferred electron: Na → Na+ + eCl + e- → ClIn looking at the reaction above you will notice that the number of electrons lost by sodium (one) is equal to the number of electrons gained by chlorine. This makes sense given that salt is a neutral compound. Chemists say that the charges must balance. With sodium chloride, this is easy because everything has a charge of one, but there are many other ionic compounds that can form where the charges may be different. For example, based on its location on the periodic table, magnesium forms a plus two ion (Mg2+). Therefore, if magnesium were to react with chlorine, we would have to write the formula as MgCl2 because two minus one chloride ions would be needed to cancel out the plus two charge from the magnesium ion. Unlike covalent compounds that exist as distinct molecules, ionic compounds exist in a repeating 3D array of billions and billions of cations and anions arranged into what is known as a crystal lattice. Because there are no distinct “molecules” for ionic compounds, scientists do not use the terms “molecule” or “molecular” when discussing ionic compounds, but rather use the term formula unit to refer to the formulas of ionic compound. Below is an image (obtained from 5-4

Copyright Prof. Lawrence Mailloux, 2021

Wikipedia) of the crystal lattice of sodium chloride, NaCl. If it were possible to look at a sample of NaCl under a ultra-powerful microscope this is what you would see; a seemingly endless array of zillions of cations and anions stacked neatly, and repeating over and over again. The purple spheres represent Na+ cation and the green spheres represent Cl- anions.

Notice that there is no one distinct NaCl “molecule.” Instead, the formula NaCl simply tells the reader that the ratio of Na+ cations to Cl- anions is 1:1, as a result there are equal numbers of green and purple spheres in the crystal lattice diagram above. For compounds where the ions are not 1:1 the crystal lattice looks a little different. Below is the crystal lattice of MgCl2. In this compound the ratio of Mg2+ cations to Cl- anions is 1:2 and thus we see one green (Mg2+) for every two purple spheres (Cl-). For simplicity, I have drawn the crystal lattice as 2D, however the actual lattice is three dimensional.

5-5

Copyright Prof. Lawrence Mailloux, 2021

Unlike with covalent compounds where prefixes are used to indicate the number of atoms of each element in a compound, in ionic compounds the charges of the ions determine the ratio of the elements in a compound. Because chemists know what charge ions the different elements form, they are able to use those charges to figure out the ratio of elements in an ionic compound and as a result, they do NOT use prefixes to name ionic compounds.

Summary of the Ion Charges of the Elements In the previous chapter, we learned how to predict the charge ion that many elements form based upon their location on the periodic table by applying the octet rule. For your convenience I have included that table again below. IA 1 valence e-

IIA 2 valence e-

IIIA 3 valence e-

IVA 4 valence e-

VA 5 valence e-

VIA 6 valence e-

VIIA 7 valence e-

VIIIA 8 valence eHe

H+ Li+

Be2+

Na+

Mg2+

K+

Ca2+

Rb+

Sr2+

Cs+

Ba2+

Octet Rules does not apply to transition metals

no ions

Al3+

N3-

O2-

F-

Ne

nitride

oxide

fluoride

no ions

P3-

S2-

Cl-

Ar

phosphide

sulfide

chloride

no ions

As3-

Se2-

Br-

Kr

arsenide

selenide

bromide

no ions

Te2-

I-

Xe

telluride

iodide

no ions

Rn no ions

All of the metal elements in the table above only form one charge of ion. In addition to these elements, silver, zinc, and cadmium only form the following ions: Ag+, Zn2+, Cd2+. Every other metal on the periodic table can form multiple charges. Chemists memorize those charges for important elements, and look up the others when needed. In this class, you will only be required to memorize the charges that the elements copper (Cu), iron (Fe), tin (Sn), and lead (Pb) form as summarized in the following table:

5-6

Copyright Prof. Lawrence Mailloux, 2021

Element Ions Formed Copper (Cu) Cu+

Iron (Fe)

Tin (Sn)

Lead (Pb)

Ion Names Copper (I)

Cu+2

Copper (II)

Fe2+

Iron (II)

Fe3+

Iron (III)

Sn2+

Tin (II)

Sn4+

Tin (IV)

Pb2+

Lead (II)

Pb4+

Lead (IV)

To summarize, you are expected to know the ions listed in the periodic table below:

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

IA

IIA

IIIB

IVB

VB

VIB

VIIB

----

VIIIB

----

IB

IIB

IIIA

IVA

VA

VIA

VIIA

VIIIA

N -3

O -2

F -1

P -3

S -2

Cl -1

As -3

Se -2

Br -1

Te -2

I -1

H +1 Li +1

Be +2

Na +1

Mg +2

K +1

Ca +2

Rb +1

Sr +2

Cs +1

Ba +2

MEMORIZE THESE IONS Al +3 Fe +2/3

Cu +1/2

Zn +2

Ag +1

Cd +2

Sn +2/4

Pb +2/4

* **

5-7

Copyright Prof. Lawrence Mailloux, 2021

Using Ion Charges to Write Formulas of Ionic Compounds Ionic compounds are neutral compounds. This fact requires that the formulas of ionic compounds are written such that the total positive charge from the cations is equal to the total negative charge from the anions. With compounds like sodium chloride (NaCl) or magnesium chloride (MgCl2) this is easy. Sodium has a charge of plus one and chloride has a charge of minus one. Therefore, one Na+ cation is needed for every one Cl- anions giving the formula Na1Cl1; which is always written as NaCl, as ones in formulas are understood, they are never written. With magnesium two chloride ions are needed to cancel out the positive two charge from the Mg2+ ion. The paper cutouts below, that were used in lecture, provide a visual representation of this process.

When the ion charges are not simple number like one or two, it can be useful to use the “crissycrossy” trick to help write correct formulas of ionic compounds. We will use aluminum and oxygen as an example. Aluminum forms a +3 ion (Al3+) and oxygen forms a -2 ion (O2-). Our goal is to write a formula that gives these ions in a ratio such that the charge cancels out. To do this we criss-cross the numbers (just the numbers, not the charges) as illustrated below.

Al3+

O2-

Al2O3

5-8

Copyright Prof. Lawrence Mailloux, 2021

What this formula says is that the combined positive charge to two Al3+ ions (+6) is canceled by the combined negative charge (-6) of three O2- anions. Because a 2:3 ratio can not be simplified the formula is written as Al2O3. For a slightly different example let’s consider the formula that results when magnesium and oxygen form a compound. We begin by writing our ions with their charges and then criss-cross the numbers as before Mg2+ O2-

Mg2O2

giving the formula Mg2O2. Because a 2:2 ratio can be simplified this formulas is written as MgO.

Mg2O2 → MgO

This last simplification step is important. Writing the formula as Mg2O2 is wrong. Chemists always simplify formulas of ionic compounds when possible.

5-9

Copyright Prof. Lawrence Mailloux, 2021

Problem 2- write formulas for the ionic compounds forms by the ions in the table below. F-

Cl-

O2-

S2-

Li+ Na+ K+ Mg2+ Ca2+ Zn2+ Cd2+ Fe2+ Fe3+ Sn2+ Sn4+ Cu+ Cu2+ Ag+

5-10

N3-

P3-

Copyright Prof. Lawrence Mailloux, 2021

Naming Ionic Compounds How ionic compounds are named depends upon the metal ion involved. If the metal can only form one charge of ion, then the charge is NOT given in the name. In an ionic compound, the charges of the ions determine their ratios in the formula of a compound. As a result prefixes are NOT used when naming ionic compounds because they are not necessary. To name ionic compounds we apply the following formula: Name of cation + 1st part of anion name + “ide” Look at the following examples and see how they fit the above formula NaCl – Sodium chloride KCl – Potassium chloride CaCl2 – Calcium chloride

(NOT calcium dichloride - no prefixes in ionic naming)

Na2S – sodium sulfide

(NOT disodium monosulfide – no prefixes in ionic naming)

Al2O3 – aluminum oxide

(NOT dialuminum trioxide – no prefixes in ionic naming)

ZnBr2 – Zinc bromide

(NOT zinc dibromide – no prefixes in ionic naming)

Ag2O – silver oxide

(NOT disilver oxide – no prefixes in ionic naming)

CdS – Cadmium sulfide If the metal in question can form more than one charge, it is necessary to specify the charge in the name. To keep things as simple as possible, we will only consider ions of Cu, Fe, Sn, and Pb as listed in previously. To name such compounds we apply the following formula: Name of cation + (charge in Roman Numerals) + 1st part of anion name + ide Below are some examples for iron: FeCl2 – Iron (II) chloride

(NOT iron dichloride – no prefixes in ionic naming)

FeCl3 – Iron (III) chloride

(NOT iron trichloride – no prefixes in ionic naming) 5-11

Copyright Prof. Lawrence Mailloux, 2021

Common Naming Errors for Ionic Compounds If both ions have a charge other than one, the formulas can become more complicated. When this happens students making some common naming errors. For example, when iron reacts with oxygen (O2-) two different compounds can be formed depending on if its Fe2+ or Fe3+ that combines with the oxygen:

FeO – Iron (II) oxide

Fe2O3 – Iron (III) oxide

(NOT Iron (I) oxide)

(NOT Iron (II) oxide)

Notice that there is only one iron atom in the formula of iron (II) oxide (FeO) and that there are two iron atoms in the formula of iron (III) oxide (Fe2O3). It is the charge on iron, not the number of iron atoms, that belongs inside the ( )’s. Another example of compounds that students often name incorrectly are the oxides of copper:

Cu2O – Copper (I) oxide

CuO – Copper (II) oxide

(NOT copper (II) oxide)

(NOT copper (I) oxide)

5-12

Copyright Prof. Lawrence Mailloux, 2021

Problem 3- Fill in the following table of names and formulas for ionic compounds. Name

Formula

Potassium bromide Na3N Ca3P2 Magnesium sulfide PbO Pb3As4 Fe2S3 Barium selenide Aluminum iodide AgF Copper (I) oxide Copper (II) oxide Li2S PbS PbO2 ZnCl2 Tin (II) phosphide Tin (IV) phosphide CdO

5-13

Copyright Prof. Lawrence Mailloux, 2021

Ionic Compounds Containing Polyatomic Ions Sometimes several atoms bond together to make one big ion known as a polyatomic ion. These multi-atom ions are so common that they are given their own names. Chemists memorize the names and formulas of dozens of these ions; luckily you will only have to memorize the nine common ones given below. Formula

Ion Name

+

Ammonium

NH4 OH-

Hydroxide

CN-

Cyanide

NO3-

Nitrate

HCO3-

Bicarbonate

C2H3O2-

Acetate

CO32-

Carbonate

SO42-

Sulfate

PO43-

Phosphate

Writing formulas for compounds containing these ions is no more difficult than writing formulas of any other ionic compound. We still criss-cross the numbers. If more than one of a polyatomic ion is needed we put the polyatomic ion in parentheses. As usual, the formula is simplified whenever possible.

5-14

Copyright Prof. Lawrence Mailloux, 2021

For an example consider the formula that results when aluminum reacts with nitrate (NO3-). Because nitrate has a minus one charge, three nitrates will be needed. Al3+

NO3-

Al(NO3)3 In the formula above the nitrate ion is enclosed in ( ) to indicate that the subscript three applies to the entire ion. When possible formulas are simplified. For example consider the compound of Al3+ and PO43Al3+

PO43-

Al3(PO4)3

which must be simplified to give AlPO4 as the final formula. Al3(PO4)3 → AlPO4

Notice that no ( ) were used in the formula AlPO4. Parentheses are only used if there are multiple polyatomic ions. To use parentheses when there is only one of a polyatomic ion, such as Al(PO4) is wrong.

5-15

Copyright Prof. Lawrence Mailloux, 2021

Problem 4- Write formulas for ionic compounds containing the following polyatomic ions. OH-

NO3-

SO42-

Li+ Na+ NH4+ Mg2+ Zn2+ Cd2+ Fe2+ Fe3+ Pb2+ Pb4+ Cu+ Cu2+ Ag+

5-16

CO32-

PO43-

Copyright Prof. Lawrence Mailloux, 2021

Naming compounds containing polyatomic ions is no different that naming any other ionic compound. For elements that can only form one charge, the name is simply the name of the cation followed by the name of the polyatomic ions. NaHCO3 – sodium bicarbonate (a.k.a. baking soda) KOH – potassium hydroxide Ammonium (NH4+) is the only polyatomic ion with a positive charge. This sometimes confuses students. Below are a couple examples containing the ammonium ion. In these compounds, the ammonium cation is playing the role that the metal ion usually plays. NH4Cl – ammonium chloride NH4NO3 – ammonium nitrate If multiple polyatomic ions are needed to balance the charges, then the polyatomic ion is enclosed in ( ). (NH4)2SO4 – ammonium sulfate Ca(OH)2 – calcium hydroxide Zn(C2H3O2)2 – zinc acetate If the metal in question can form multiple charges, the charge must be included in the name. Fe(NO3)2 – Iron (II) nitrate Fe(NO3)3 – Iron (III) nitrate SnSO4 – Tin (II) sulfate Sn(SO4)2 – Tin (IV) sulfate 5-17

Copyright Prof. Lawrence Mailloux, 2021

Problem 5- Fill in the following table regarding ionic compounds containing polyatomic ions. Name

Formula

Potassium bicarbonate (NH4)3PO4 Ca(NO3)2 Magnesium sulfate Pb(C2H3O2)2 Pb(OH)4 Barium carbonate Aluminum cyanide AgCN Copper (I) carbonate Copper (II) sulfate NH4NO3 Pb(C2H3O3)4 Pb(CN)4 ZnCO3 Tin (II) phosphate Tin (IV) bicarbonate CdSO4

5-18

Copyright Prof. Lawrence Mailloux, 2021

Answers to In Chapter Questions 1. Name

Formula

Nitrogen dioxide

NO2

Diphosphorus decoxide

P2O10

Dichlorine heptoxide

Cl2O7

Phosphorus trichloride

PCl3

Iodine trichloride

ICl3

Phosphorus pentabromide

PBr5

Sulfur hexafluoride

SF6

Sulfur monoxide

SO

5-19

Copyright Prof. Lawrence Mailloux, 2021

2. F-

Cl-

O2-

S2-

N3-

P3-

Li+

LiF

LiCl

Li2O

Li2S

Li3N

Li3P

Na+

NaF

NaCl

Na2O

Na2S

Na3N

Na3P

K+

KF

KCl

K2O

K2S

K3N

K3P

Mg2+

MgF2

MgCl2

MgO

MgS

Mg3N2

Mg3P2

Ca2+

CaF2

CaCl2

CaO

CaS

Ca3N2

Ca3P2

Zn2+

ZnF2

ZnCl2

ZnO

ZnS

Zn3N2

Zn3P2

Cd2+

CdF2

CdCl2

CdO

CdS

Cd3N2

Cd3P2

Fe2+

FeF2

FeCl2

FeO

FeS

Fe3N2

Fe3P2

Fe3+

FeF3

FeCl3

Fe2O3

Fe2S3

FeN

FeP

Sn2+

SnF2

SnCl2

SnO

SnS

Sn3N2

Sn3P2

Sn4+

SnF4

SnCl4

SnO2

SnS2

Sn3N4

Sn3P4

Cu+

CuF

CuCl

Cu2O

Cu2S

Cu3N

Cu3P

Cu2+

CuF2

CuCl2

CuO

CuS

Cu3N2

Cu3P2

Ag+

AgF

AgCl

Ag2O

Ag2S

Ag3N

Ag3P

5-20

Copyright Prof. Lawrence Mailloux, 2021

3. Name

Formula

Potassium bromide

KBr

Sodium nitride

Na3N

Calcium phosphide

Ca3P2

Magnesium sulfide

MgS

Lead (II) oxide

PbO

Lead (IV) arsenide

Pb3As4

Iron (III) sulfide

Fe2S3

Barium selenide

BaSe

Aluminum iodide

AlI3

Silver fluoride

AgF

Copper (I) oxide

Cu2O

Copper (II) oxide

CuO

Lithium sulfide

Li2S

Lead (II) sulfide

PbS

Lead (IV) oxide

PbO2

Zinc chloride

ZnCl2

Tin (II) phosphide

Sn3P2

Tin (IV) phosphide

Sn3P4

Cadmium oxide

CdO

5-21

Copyright Prof. Lawrence Mailloux, 2021

4. OH-

NO3-

SO42-

CO32-

PO43-

Li+

LiOH

LiNO3

Li2SO4

Li2CO3

Li3PO4

Na+

NaOH

NaNO3

Na2SO4

Na2CO3

Na3PO4

NH4+

NH4OH

NH4NO3

(NH4)2SO4

(NH4)2CO3

(NH4)3PO4

Mg2+

Mg(OH)2

Mg(NO3)2

MgSO4

MgCO3

Mg3(PO4)2

Zn2+

Zn(OH)2

Zn(NO3)2

ZnSO4

ZnCO3

Zn3(PO4)2

Cd2+

Cd(OH)2

Cd(NO3)2

CdSO4

CdCO3

Cd3(PO4)2

Fe2+

Fe(OH)2

Fe(NO3)2

FeSO4

FeCO3

Fe3(PO4)2

Fe3+

Fe(OH)3

Fe(NO3)3

Fe2(SO4)3

Fe2(CO3)3

FePO4

Pb2+

Pb(OH)2

Pb(NO3)2

PbSO4

PbCO3

Pb3(PO4)2

Pb4+

Pb(OH)4

Pb(NO3)4

Pb(SO4)2

Pb(CO3)2

Pb3(PO4)4

Cu+

CuOH

CuNO3

Cu2SO4

Cu2CO3

Cu3PO4

Cu2+

Cu(OH)2

Cu(NO3)2

CuSO4

CuCO3

Cu3(PO4)2

Ag+

AgOH

AgNO3

Ag2SO4

Ag2CO3

Ag3PO4

5-22

Copyright Prof. Lawrence Mailloux, 2021

5. Name Potassium bicarbonate

Formula KHCO3

Ammonium phosphate

(NH4)3PO4

Calcium nitrate

Ca(NO3)2

Magnesium sulfate

MgSO4

Lead (II) acetate

Pb(C2H3O2)2

Lead (IV) hydroxide

Pb(OH)4

Barium carbonate

BaCO3

Aluminum cyanide

Al(CN)3

Silver cyanide

AgCN

Copper (I) carbonate

Cu2CO3

Copper (II) sulfate

CuSO4

Ammonium nitrate

NH4NO3

Lead (IV) acetate

Pb(C2H3O3)4

Lead (IV) cyanide

Pb(CN)4

Zinc carbonate

ZnCO3

Tin (II) phosphate

Sn3(PO4)2

Tin (IV) bicarbonate

Sn(HCO3)4

Cadmium sulfate

CdSO4

5-23

Copyright Prof. Lawrence Mailloux, 2021

Chapter 6 Grams, Moles, and Chemical Reactions Chemical Reactions – Balancing and Terminology In the previous chapter we learned what happens when two or more atoms come together to form a compound. Now we will learn about what happens when one compound reacts with another. When you leave a piece of iron out in the elements it will react with oxygen in the air and form iron (III) oxide, aka rust. We would write a reaction for this as follows: Fe + O2 → Fe2O3 This reaction however is not correct because it violates the law of conservation of matter. It is impossible to start with one atom of iron and end with two. Likewise, one can’t turn two oxygen atoms into three. To make this reaction correct we must balance it. To balance the irons, we add a two on the left before the iron: 2 Fe + O2 → Fe2O3 Now we must balance the oxygens. There are two oxygens on the left and three on the right. The numbers two and three meet at six. Putting in a three and a two gives 2 Fe + 3 O2 → 2 Fe2O3 We are almost done. However, adding the two on the right unbalanced the irons. Now there are four irons on the right hand side, rebalancing the iron gives the final balanced equation. 4 Fe + 3 O2 → 2 Fe2O3 This is the equation for the rusting of iron. Balanced equations are to chemists what recipes are to a chef, they tell a chemist the ratio of ingredients needed for a given chemical reaction. In this case, our reaction tells us that four parts iron reacts with three parts oxygen to produce two parts iron (III) oxide. Chemists refer to the chemicals on the left side of the arrow as reactants, reagents, or starting materials, and chemicals on the right of the arrow are products. In the

6-1

Copyright Prof. Lawrence Mailloux, 2021

above reaction the reactants are iron and oxygen (Fe and O2) while the product is iron (III) oxide (Fe2O3). Balancing chemical reactions requires practice. There is no single list of steps to balance every reaction you will encounter. Therefore, what follows are three hints to make balancing easier, however it is only with practice that you will get the hang of it. Some kinds of reactions have clear patterns to them that make balancing easier. Some are just a pain. Below are some hints for balancing a variety of equations and some examples of each. At the end are some practice problems for you to do. The first kind of reaction we will balance is a combustion reaction; we start with these reactions because they are amongst the easiest to balance. Combustion means burning. Chemicals made of only carbon and hydrogen are called hydrocarbons; propane, butane, and gasoline are all example of hydrocarbons. Whenever a hydrocarbon burns (reacts with oxygen) the products will be carbon dioxide and water. Hint One - For reactions like those shown below (combustion reactions) C3H8 + O2 → CO2 + H2O C4H10 + O2 → CO2 + H2O a. Balance carbon first b. Balance hydrogen second c. Balance oxygen last d. Using half numbers if needed to temporarily balance the oxygens. e. Double everything to turn halves into integers

There are three carbons on the left and one on the right, so I put a three before the CO2.

C3H8 +

O2 →

3

CO2 +

H2O

Then to balance the hydrogens I put a four before the water.

C3H8 +

O2 →

3

CO2 +

6-2

4

H2O

Copyright Prof. Lawrence Mailloux, 2021

Now there are a total of ten oxygens on the right of the arrow, and two on the left. Take a moment to carefully count the oxygens on the product side of the arrow to confirm to yourself there are ten oxygens total; six in the three carbon dioxides and four in the four water molecules. Adding a five before the oxygen balances our reaction.

C3H8 +

5

O2 →

3

CO2 +

4

H2O

Because the total number of oxygens on the right was an even number (ten) it is not necessary to use any halves to balance the oxygens. The reaction is balanced.

For an example where we do need to use halves to balance the oxygens. Consider this equation: O2 →

C4H10 +

CO2 +

H2O

First we balance the carbons by putting in a four before the carbon dioxide O2 →

C4H10 +

4 CO2 +

H2O

Then we balance the hydrogens by putting in a five in front of the water O2 →

C4H10 +

4 CO2 +

5

H2O

This brings the total number of oxygens on the right to 13. Because this is an odd number of oxygens we must use a half number to temporarily balance the oxygens.

C4H10 +

6.5

O2 →

4

CO2 +

5 H2O

Next we double everything (including the implied one before the C4H10) to leave only integers and the correctly balanced equation

2

C4H10 +

13

O2 →

8

CO2 + 10 H2O 6-3

Copyright Prof. Lawrence Mailloux, 2021

Unfortunately most reactions don’t balance according to a set of steps like combustion reaction do. For most other reactions it’s simply a matter of practice; although the hints below may make things easier for you.

Hint Two - Start with the cations/metals first. For example when balancing 4 Fe + 3 O2 → 2 Fe2O3, on the first page of this chapter, I balanced the irons first.

Hint Three - Balance polyatomic ions as a unit if they remain intact.

For an example with intact polyatomic ions consider the following reaction: (NH4)3PO4 +

MgSO4 →

(NH4)2SO4 +

Mg3(PO4)2

We first balance the cation, which in this reaction is magnesium. (NH4)3PO4 +

3 MgSO4 →

(NH4)2SO4 +

Mg3(PO4)2

Next we start balancing the polyatomic ions as a unit. As a result of the three we just added there are now three sulfates on the left, so to balance this we must put a three on the right: (NH4)3PO4 +

3 MgSO4 →

3

(NH4)2SO4 +

Mg3(PO4)2

Now there are a total of six ammoniums on the right and three on the left, putting in a two gives our final balanced equation. 2

(NH4)3PO4 +

3 MgSO4 →

3

6-4

(NH4)2SO4 +

Mg3(PO4)2

Copyright Prof. Lawrence Mailloux, 2021

If a polyatomic ion does break apart, you must balance it atom by atom. For example, in the reaction below phosphite (PO33-) falls apart. We must balance it as separate phosphors and oxygen atoms. The manganese atoms are already balanced so we start with the phosphorus atoms. There are two phosphorus atoms on the left and one on the right so we put a two before phosphorus dioxide. Mn(PO3)2 →

MnO +

O2 +

2

PO2

Next we must balance the oxygens. There are six on the left and a total of seven on the right. To get around this we will temporarily balance the oxygens with a ½ just as we did with the first (combustion) example. Putting a ½ before the oxygen on the right reduces the total oxygens on the right to six. Mn(PO3)2 →

MnO +

1/2

O2 +

2

PO2

Lastly to removes the halves we double everything to give our final balanced equation. The resulting one before the O2 is understood and is not written.

2

Mn(PO3)2 →

2 MnO +

O2 +

6-5

4

PO2

Copyright Prof. Lawrence Mailloux, 2021

Problem 1 – Balance the following reactions. H2SO4 →

Fe +

Fe2(SO4)3 +

H2

NH3 +

O2 →

NO +

H2O

SnO2 +

H2 →

Sn +

H2O

KClO3 →

KCl +

C3H6N3O9 →

N2 +

CO2 +

H2CO3 →

KNO3 +

H3PO4 →

KOH +

P4 +

O2

O2 →

Al +

SeCl6 +

O2 → H2SO4 →

H3PO4 +

Ba(OH)2 →

(NH4)3PO4 +

HNO3

K3PO4 +

H2O

Al2O3 +

CaO +

KClO2 +

P4O10 +

K2CO3 +

P4O6

Fe2O3 →

Ca(OH)2 →

H2O

H2O →

Fe

H2O

SeO2 +

Cl2

K2SO4 +

HClO2

Ba3(PO4)2 +

H2O

H3PO4

Pb(NO3)4 → 6-6

Pb3(PO4)4 +

NH4NO3

Copyright Prof. Lawrence Mailloux, 2021

Problem 2 – Even more balancing fun

C6H12 +

O2 →

B2Br6 +

HNO3 →

Pb(NO3)2 →

BF3 +

CO2 +

B(NO3)3 +

PbO + Li2SO3 →

NO2

B2(SO3)3 +

O2 →

CO2 +

FeS2 +

O2 →

Fe2O3 +

Fe2(SO4)3 + _____ KOH → O2 →

HBr

O2 +

C5H12 +

S8 +

H2O

LiF

H2O

SO2

K2SO4 +

Fe(OH)3

SO3

_____ H2SO4 + ____ HI → _____ H2S + _____ I2 + _____H2O _____ N2 + _____O2 → _____N2O _____ Fe2O3 + _____ H2 → _____ Fe + ______ H2O _____ Na + _____H2O → ______ NaOH + ______H2 _____ C2H6 + _____ O2 → _____CO2 + ______ H2O

_____ Fe + _____ N2 → _____Fe3N2 _____ Na + _____ Cl2 → ______ NaCl 6-7

Copyright Prof. Lawrence Mailloux, 2021

Thermochemical Equations – Reactions that Gain or Lose Heat Some chemical reactions involve the gain or loss of energy (heat). Whether heat is gained or lost during a chemical reaction depends upon the relative stability of the reactants vs the products. If the products of a chemical reaction are lower in energy than the reactants, the chemical reaction will give off heat and the test tube will feel hot. Reactions that release heat are called exothermal reactions. A disposable heating pack is an example of an exothermic reaction. The picture below shows a commercial heating pack designed for lower back pain.

Inside a disposable heating pack is powdered iron, salt, and water. Because the iron is powdered (and covered with moisture and salt) it will rust very fast. When iron rusts, heat is released. Normally we cannot feel this because rusting is usually such a slow reaction. However because the iron inside the heating pack is a present as a fine, salty, damp powder, it will rust very quickly, allowing us to feel the heat. Below is the reaction that takes place inside a hot pack. Because the reaction releases energy, heat is shown as a product. 4 Fe + 3 O2 → Fe2O3 + Heat

6-8

Copyright Prof. Lawrence Mailloux, 2021

We can represent this using a diagram like the one shown below. Energy is plotted on the y-axis and time is plotted on the x-axis. Because heat is released, the products are of lower energy (more stable) than the reactants. The difference in energy between the reactants and products is equal to the amount of heat released by the reaction. The activation energy (Ea) represents the energy needed to get the reaction started. We will learn more about activation energy (and how it

Increasing Energy →

relates to catalysts/enzymes) in Chapter 16.

Ea

Reactants Fe, O2

Heat Released

Product(s) Fe2O3

Reaction Progress →

While less common, there are chemical reaction in which heat is taken in rather than released. These reactions are called endothermic reactions. In an endothermic reaction, the chemicals take in heat from their surroundings; this means that the test tube feels cold. Disposable cold packs are an example of an endothermic reaction. The disposable cold packs you will work with in a hospital contain solid ammonium nitrate and a small bag of water. The photograph below shows a disposable cold pack,

6-9

Copyright Prof. Lawrence Mailloux, 2021

When you squeeze the pouch to activate the cold pack, you are simply breaking the bag of water inside. When the water inside dissolves the solid ammonium nitrate, the pack gets cold. Below is the reaction for a disposable cold pack. Because energy is taken in, heat is a reactant in this reaction. Heat + NH4NO3(s) + H2O(l) → NH4NO3(aq) Below is the energy diagram for a cold pack. Compare this diagram to the diagram of the

Increasing Energy →

exothermic reaction of the hot pack on the prior page.

Ea Reactants NH4NO3(s), H2O

Products NH4NO3(aq) Heat Absorbed

Reaction Progress →

6-10

Copyright Prof. Lawrence Mailloux, 2021

Grams, Moles and Stoichiometry – Using Balanced Equations A balanced chemical equation is just like a recipe. It gives the ratios in which different chemicals react. Let’s consider our rusting equation once again 4 Fe + 3 O2 → 2 Fe2O3 This equation tells you that four atoms of iron combine with three molecules of O2 to form two formula units of Fe2O3. Recall however that atoms are super small. Even a tiny spec of iron contains trillions and trillions of iron atoms; counting them four at a time would not be very practical. We need to use a bigger unit. Just as bakers find it easier to count by the dozen, chemists find it easier to count by the mole. Just like a dozen of anything is 12 of that thing, one mole of anything is 6.02 x 1023 of that thing. The numbers we added to balanced the chemical reactions earlier in this chapter are in moles. The reaction above says that four moles of iron reacts with three moles of oxygen to make two moles of iron (III) oxide. Chemists call mathematical calculations performed using balanced equations stoichiometry. Stoichiometry is what allows us to answer questions like the following: •

How much product will a chemical reaction produce

•

How much of one chemical would be needed to react with another

When performing stoichiometry calculations we will take the following route. 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝐴

𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝐴 →

𝑏𝑎𝑙.𝑒𝑞.

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐴 →

𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝐵

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐵 →

𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝐵

Time to consider some examples. For starters, let’s look up and calculate the molar masses of all the chemicals involved in this equation as we will be needing these numbers. Looking up molar masses from the periodic table, we find Fe = 55.8 g/mol

O2 = 32 g/mol

Fe2O3 = 55.8 x 2 + 16 x 3 = 159.6 g/mol

6-11

Copyright Prof. Lawrence Mailloux, 2021

Example 1 – Moles A → Moles B How many moles of Fe2O3 can be made by reacting of 12.0 moles of Fe with excess oxygen? 12.0 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒 (

2 𝑚𝑜𝑙𝑒 𝐹𝑒2 𝑂3 4 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒

) = 6 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒2 𝑂3 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑

Example 2 – Moles A → Moles B How many moles of oxygen would be needed to react with 12.0 moles of Fe? 3 𝑚𝑜𝑙𝑒 𝑂

2 12.0 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒 (4 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒 ) = 9 𝑚𝑜𝑙𝑒𝑠 𝑂2 𝑛𝑒𝑒𝑑𝑒𝑑

Example 3 – Moles A → Moles B How many grams of Fe2O3 can be made by the reaction of 12.0 moles of Fe with excess oxygen? 12.0 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒 (

2 𝑚𝑜𝑙𝑒 𝐹𝑒2 𝑂3 4 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒

159.6 𝑔 𝐹𝑒 𝑂

) ( 1 𝑚𝑜𝑙𝑒 𝐹𝑒2𝑂3) = 958 𝑔 𝐹𝑒2 𝑂3 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 2 3

Example 4 – Moles A → Grams B How many grams of O2 are needed to react with 12.0 moles of Fe? (moles of A → grams of B) 3 𝑚𝑜𝑙𝑒 𝑂

32 𝑔 𝑂

2 12.0 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒 (4 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒 ) (1 𝑚𝑜𝑙𝑒 𝑂2 ) = 288 𝑔 𝑂2 𝑛𝑒𝑒𝑑𝑒𝑑 2

Example 5 – Grams A → Grams B How many grams of Fe2O3 could be produced by the reaction of 25.5 grams of Fe with excess oxygen? (grams of A → grams of B) 1 𝑚𝑜𝑙𝑒 𝐹𝑒

2 𝑚𝑜𝑙 𝐹𝑒2 𝑂3

25.5 𝑔 𝐹𝑒 ( 55.8 𝑔 𝐹𝑒 ) (

4 𝑚𝑜𝑙𝑒 𝐹𝑒

159.6 𝑔 𝐹𝑒 𝑂

) ( 1 𝑚𝑜𝑙 𝐹𝑒 2𝑂 3 ) = 36.5 𝑔 𝐹𝑒2 𝑂3 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 2 3

Example 6 - Grams A → Grams B How many grams of oxygen would be needed to react with 25.5 grams of Fe? (grams of A → grams of B) 1 𝑚𝑜𝑙𝑒 𝐹𝑒 3 𝑚𝑜𝑙 𝑂2 32 𝑔 𝑂2 25.5 𝑔 𝐹𝑒 ( )( )( ) = 11.0 𝑔 𝐹𝑒2 𝑂3 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 55.8 𝑔 𝐹𝑒 4 𝑚𝑜𝑙𝑒 𝐹𝑒 1 𝑚𝑜𝑙 2

6-12

Copyright Prof. Lawrence Mailloux, 2021

Problem 3 – Consider the balanced equation below then consider the following problems C3H8 + 5 O2 → 3 CO2 + 4 H2O a. Calculate the molar mass of each of the chemicals involved in the above reaction C3H8 = _______ g/mol O2 = _______ g/mol CO2 = _______ g/mol H2O = _______ g/mol b. How many moles of CO2 can be made from the burning of 2.2 moles of C3H8 with excess O2? c. How many moles of C3H8 would need to be reacted to produce 40 moles of water? d. How many grams of CO2 could be made by the combustion of 2.25 grams of C3H8? e. How many grams of O2 would be needed to react with 14.4 grams of C3H8 Problem 4 – Consider the balanced equation below then consider the following problems 2 NaI + Pb(NO3)2 → 2 NaNO3 + PbI2 a. Calculate the molar mass of each of the chemicals involved in the above reaction NaI = _______ g/mol Pb(NO3)2 = _______ g/mol NaNO3 = _______ g/mol PbI2 = _______ g/mol b. How many moles of PbI2 can be made from reacting of 4.5 moles of NaI with excess lead (II) nitrate? c. How many moles of lead (II) nitrate would be needed to react with 786 grams of NaI? d. How many moles of PbI2 could be made by the reaction of 885.6 grams of Pb(NO3)2 with excess NaI? e. How many grams of PbI2 could be made by reacting 35.6 grams of Pb(NO3)2 with excess NaI?

6-13

Copyright Prof. Lawrence Mailloux, 2021

Problem 5 – Consider the balanced equation below then consider the following problems (NH4)2CO3 + BaCl2 → BaCO3 + 2 NH4Cl a. Calculate the molar mass of each of the chemicals involved in the above reaction (NH4)2CO3 = _______ g/mol BaCl2 = _______ g/mol BaCO3 = _______ g/mol NH4Cl = _______ g/mol b. How many moles of BaCO3 can be made from reacting 7.5 moles of BaCl2 with excess ammonium carbonate? c. How many moles of BaCl2 would needed to be reacted to produce 40. moles of NH4Cl? d. How many grams of BaCO3 could be made by the reaction of 2.25 grams of BaCl2 with excess ammonium carbonate?

6-14

Copyright Prof. Lawrence Mailloux, 2021

Answers to In Chapter Problems 1. H2SO4 →

2

Fe +

3

4

NH3 +

5

O2 →

SnO2 +

2

H2 →

2

KClO3 →

2

C3H6N3O9 →

2

KNO3 +

3

KOH +

P4 +

2

3

2

H3PO4 +

P4O10 +

4

Sn +

3

3

N2 +

O2 →

6

CO2 +

H2O

K2CO3 + 2

HNO3

K3PO4 + 3

H2O

2

SeO2 +

H2SO4 →

H2O →

Fe

4

3

Cl2

K2SO4 +

Ba(OH)2 →

3

6

H2O (already balanced ☺, no 1’s needed)

O2 →

(NH4)3PO4 +

H2O

Al2O3 +

CaO +

6

2

H2O

P4O6

Fe2O3 →

3

6

H2

O2

H3PO4 →

SeCl6 +

2

NO +

KCl +

Ca(OH)2 →

KClO2 +

4

H2CO3 →

Al +

2

Fe2(SO4)3 + 3

2

Ba3(PO4)2 +

HClO2

6

H2O

H3PO4

Pb(NO3)4 → 6-15

Pb3(PO4)4 +

12

NH4NO3

Copyright Prof. Lawrence Mailloux, 2021

2.

C6H12 +

9

O2 →

B2Br6 +

6

HNO3 →

6

CO2 +

2

6

B(NO3)3 +

2

Pb(NO3)2 →

2

BF3 +

_

C5H12 +

8

O2 →

5

CO2 +

4

FeS2 +

11

O2 →

2

Fe2O3 +

2 PbO +

O2 +

Li2SO3 →

3

12

O2 →

8

4

3

6

6

HBr

NO2

B2(SO3)3 +

Fe2(SO4)3 + __6__ KOH →

S8 +

H2O

6

LiF

H2O

8

K2SO4 +

SO2

2

Fe(OH)3

SO3

_____ H2SO4 + __8_ HI → _____ H2S + __4__ I2 + __4__H2O __2__ N2 + _____O2 → __2__N2O _____ Fe2O3 + __3__ H2 → __2__ Fe + __3__ H2O ___2__ Na + __2__H2O → __2__ NaOH + _____H2 ___2__ C2H6 + __7__ O2 → __4___CO2 + ___6__ H2O

___3__ Fe + _____ N2 → _____Fe3N2 __2__ Na + _____ Cl2 → __2__ NaCl 6-16

Copyright Prof. Lawrence Mailloux, 2021

3. a. C3H8 = 44.0 g/mol, O2 = 32.0 g/mol, CO2 = 44.0 g/mol, H2O = 18.0 g/mol b. 6.6 mol CO2 c. 10. mol C3H8 d. 6.75 g CO2 e. 52.4 g O2 4. a. NaI = 149.9 g/mol, Pb(NO3)2 = 331.2 g/mol, NaNO3 = 85.0 g/mol, PbI2 = 461.0 g/mol b. 2.25 mol PbI2 c. 2.62 mol Pb(NO3)2 d. 2.67 mol PbI2 e. 49.6 g PbI2 5. a. (NH4)2CO3 = 96.0 g/mol, BaCl2 = 208.3 g/mol, BaCO3 = 197.3 g/mol, NH4Cl = 53.5 g/mol b. 7.5 mol BaCO3 c. 20. mol BaCl2 d. 2.13 g BaCO3

6-17

Copyright Prof. Lawrence Mailloux, 2021

Chapter 7 Acids, Bases, and Solutions Solutions, Colloids, and Suspensions It is very uncommon for a chemical reaction to occur when two solids are simply mixed together. Instead, most reactions occur with the chemicals dissolved in a solvent to give what is known as a solution. Solutions are homogenous mixtures. A solution has two parts. The part that is in the majority is known as the solvent, it is the chemical that does the dissolving. The part in the minority is as known as the solute, it is the chemical that is being dissolved. For example, in the ocean, water is the solvent, salt is the solute, and the ocean water is the solution. The solute need not be a solid, any state of matter (solid, liquid, or gas) can be used to make a solution. Vodka for instance is a solution made of alcohol (the solute) dissolved in water (the solvent). Most of mixtures we will be taking about in this chapter are true solutions. A true solution is a homogenous mixture in which the solute particles are small and fully dissolved on the molecular level. True solutions are always clear, although not necessarily colorless. The solute in a solution will not settle to the bottom of its container no matter how long it sits. Vodka and ocean water are true solutions. If the dissolved solute is of a medium size the mixture is known as a colloid. Because the solute particles are of moderate size they will scatter light; this causes the Tyndall effect which can be seen in the picture to the right where the laser pointer beam is made visible due to scattering. Milk and fog are both examples of colloids. Although large enough to scatter light waves, colloid particles will not settle on their own. Blood is an example of a colloid. Lastly, if the solute particles are large then the resulting mixture is heterogeneous and is known as a suspension. Here the particles are too large to be dissolved; given enough time they will

7-1

Copyright Prof. Lawrence Mailloux, 2021

sink to the bottom of the container. Pepto-Bismol ® is an example of a suspension; you must shake it before use. Another example of a suspension is milk of magnesia.

Qualitative and Quantitative Measure of Solution Strength Solutions are often described by their concentration, which is the amount of solute dissolved in a given amount of solution. A solutions concentration can be given qualitatively (without numbers) or quantitatively (with numbers). Solutions with only a small amount of dissolved solute are said to be dilute while solutions with a large amount of solute in them are said to be concentrated. The maximum amount of solute that can be dissolved in a given amount of solvent, at a given temperature, is known as a solutes solubility. For example, the solubility of sodium chloride (at 25°C) in water is 35.7 grams of NaCl per 100 mL of water. Any solution that has this amount of NaCl dissolved in it is said to be saturated. No amount of additional string will dissolve anymore NaCl. If one were to try to dissolve more NaCl, the extra NaCl would simply sink to the bottom of the container. If one were to dissolve less than the maximum amount of solute (for example five grams of NaCl dissolved in 100 mL of water) the resulting solution would be unsaturated. If one were to add more solute to an unsaturated solution and stir, the added solute would dissolve.

7-2

Copyright Prof. Lawrence Mailloux, 2021

Unsaturated and saturated solutions cover the vast majority of situations. Sometimes under the correct conditions it is possible to dissolve more than the usual maximum amount of solute. The resulting solutions are unstable and are said to be supersaturated. A fun application of a supersaturated solution is making rock candy. To make rock candy a large amount of sugar is dissolved in hot water and then allowed to cool creating a supersaturated sugar solution. Food coloring is often added for color. A string, or stick, is placed into the supersaturated solution and allowed to sit. Over the next few days, crystals of sugar will grow on the stick. Once the crystals are done growing the candy is ready to eat. Another example is an opened can of soda. The soda is supersaturated with CO2 when it is bottled under pressure. When you open the soda you release the pressure and the CO2 bubbles out of solution.

Sometimes it is necessary to use numbers when describing a solutions concentration. In this class, we will consider two quantitative units of concentration, percent and molarity. When working with percent concentration nurses, doctors, and scientists differentiate percent by mass and percent by volume depending on if the substances under consideration are solids or liquids. When a solid solute is dissolved in a liquid solvent, the concentration is usually given in units of mass/volume percent (m/v). Mass/volume percent refers to the number of grams of solute dissolved in 100 mL of solution. 𝑚 𝑣

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

% = (𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) 𝑥100

For example, if 15 grams of glucose, were dissolved in enough water so that the total volume is 300. mL the resulting solutions concentration would be 𝑚 𝑣

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

15 𝑔

% = (𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) 𝑥100 = (300 𝑚𝐿) 𝑥 100 = 5% (𝑚⁄𝑣)

We can use percent concentration as a conversion factor in calculations. For example, how many mL of a 5% glucose solution would be needed to have 25 grams of glucose? 7-3

Copyright Prof. Lawrence Mailloux, 2021

(

25 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 1

)𝑥(

100 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 5 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒

) = 500 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Or to use the conversation factor the other way we could ask, how many grams of glucose are present in 400 mL of a 5% glucose solution? (

400 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1

5 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒

) 𝑥 (100 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) = 20 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒

Problem 1 – Perform the following calculations a) What is the concentration in (m/v)% when 2.25 grams of NaCl are dissolved in enough water to give a total volume of 250 mL? b) How many grams of NaCl would there me in 450 mL of the solution from part a? c) How many mL of the solution from part a would be needed to have 5.85 g of NaCl

When a liquid solute is dissolved in a liquid solvent, the concentration is usually given in units of volume/volume percent (v/v). Volume/volume percent refers to the number of mL of solute dissolved in 100 mL of solution. 𝑣 𝑣

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

% = (𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) 𝑥100

For example, if 40.0 mL of ethanol (the kind of alcohol in alcoholic beverages), were dissolved in enough water so that the total volume is 90.0 mL the resulting solutions concentration is 𝑣

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

40.0 𝑚𝐿

% = (𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) 𝑥100 = (90.0 𝑚𝐿 ) 𝑥 100 = 44.4% (𝑚⁄𝑣) 𝑣 We can use these percent concentrations as a conversion factor in calculations too. For example, how many mL of a 44.4% ethanol solution would be needed to have 125 mL of ethanol? (

125 𝑚𝐿 𝑒𝑡ℎ𝑎𝑛𝑜𝑙 1

100 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

) 𝑥 (44.4 𝑚𝐿 𝑒𝑡ℎ𝑎𝑛𝑜𝑙 ) = 282 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Or to use the conversation factor the other way we could ask, how many mL of ethanol are present in 400 mL of a 44.4% ethanol solution? (

400 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1

44.4 𝑚𝐿 𝑔𝑙𝑢𝑐𝑜𝑠𝑒

) 𝑥 (100 𝑚𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) = 178 𝑚𝐿 𝑒𝑡ℎ𝑎𝑛𝑜𝑙

7-4

Copyright Prof. Lawrence Mailloux, 2021

Problem 2 – Perform the following calculations a) What is the concentration in (v/v)% of a bottle of vodka made by dissolving 350 mL of ethanol in water to a final solution volume of 850 mL? b) What is the concentration in (v/v)% of a solution made by dissolving 45 mL of ethanol in 230 mL of water? c) How many mL of the solution from part a would be needed to have 275 mL of ethanol? d) How many mL of ethanol would be present in 1250 mL of the solution from part b?

When both the solvent and solute amounts are given in units of mass, the concentration is usually given in units of mass/mass percent (m/m). Mass/mass percent refers to the number of grams of solute per 100 grams of solution. 𝑚 𝑚

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

% = (𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) 𝑥100

For example, if 5.0 grams of benzoyl peroxide (a common acne medication) were dissolved in 45 grams of ointment (solvent) the resulting solution’s concentration is 𝑚 𝑚

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

5.0 𝑔 𝑠𝑜𝑙𝑢𝑛𝑒

% = (𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛) 𝑥100 = (5.0 𝑔 𝑠𝑜𝑙𝑢𝑡𝑒+45 𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡) 𝑥 100 = 10 % (𝑚⁄𝑚)

Notice that I divided by the total mass of the solution and not simply by the mass of the solvent. This is an easy mistake to make. Compare this example of part b of the previous problem. Be sure you are always dividing by the total mass (or volume) and not just the amount of the solvent. Just as was the case with (m/v) and (v/v) percentages, we can use (m/m) percentages as conversation factors too. For example, how many grams of a 10% benzoyl peroxide ointment would be needed to have 12 grams of benzoyl peroxide? (

12 𝑔 𝑏𝑒𝑛𝑧𝑜𝑦𝑙 𝑝𝑒𝑟𝑜𝑥𝑖𝑑𝑒 1

) 𝑥 (10.

100 𝑔 𝑜𝑖𝑛𝑡𝑚𝑒𝑛𝑡 𝑔 𝑏𝑒𝑛𝑧𝑜𝑦𝑙 𝑝𝑒𝑟𝑜𝑥𝑖𝑑𝑒

) = 120 𝑔 𝑜𝑖𝑛𝑡𝑚𝑒𝑛𝑡

Or to use the conversation factor the other way we could ask, how many grams of benzoyl peroxide are present in 400 g of out 10% ointment? 7-5

Copyright Prof. Lawrence Mailloux, 2021

(

400 𝑔 𝑜𝑖𝑛𝑡𝑚𝑒𝑛𝑡 1

10 𝑔 𝑏𝑒𝑛𝑧𝑜𝑦𝑙 𝑝𝑒𝑟𝑜𝑥𝑖𝑑𝑒

)𝑥 (

100 𝑔 𝑜𝑖𝑛𝑡𝑚𝑒𝑛𝑡

) = 40 𝑔 𝑏𝑒𝑛𝑧𝑜𝑦𝑙 𝑝𝑒𝑟𝑜𝑥𝑖𝑑𝑒

Problem 3 – Perform the following calculations a) Salicylic acid is used to fight acne. What would be the concentration (m/m)% of a solution made by dissolving 6.0 grams of salicylic acid in 294 grams of alcohol? b) How many grams of salicylic acid would be present in 50 grams of the above solution? c) How many grams the above solution would be needed to have 9.5 grams of salicylic acid?

Although percentages are the most common unit of concentration in medicine, in lab chemists often use a different unit of concentration, molarity. Molarity is abbreviated with a capital letter M and is defined as follows. 𝑴𝒐𝒍𝒂𝒓𝒊𝒕𝒚 (𝑴) =

𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝟏 𝑳 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏

𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆

= 𝟏𝟎𝟎𝟎 𝒎𝑳 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏

Bottles of chemicals in your CHM1033L class are often labeled using molarity. For example, a bottle of HCl labeled 12.1 M HCl would be read as “12.1 molar hydrochloric acid.” This means that inside a one liter bottle of this acid there would be 12.1 moles of HCl. Recall from Chapter Six that one mol of anything is equal to 6.02 x 1023 of that thing. So how many HCl molecules would there be in 500 mL of a 12.1 M HCl solution? (

500 𝑚𝐿 𝐻𝐶𝑙 1

12.1 𝑚𝑜𝑙 𝐻𝐶𝑙

6.02 𝑥1023 𝐻𝐶𝑙

) 𝑥 (1000 𝑚𝐿 𝐻𝐶𝑙) 𝑥 (

1 𝑚𝑜𝑙 𝐻𝐶𝑙

) = 3.64 𝑥 1024 𝐻𝐶𝑙 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠

Let’s consider another example. What is the molarity of a solution made by dissolving 0.9 grams of NaCl in water to a final volume of 100 mL? First we add up the mass of NaCl using a periodic table. Na = 23.0 g/mol and Cl = 35.5 g/mol therefore NaCl = 23.0 + 35.5 = 58.5 g/mol. Next we convert to moles (

0.9 𝑔 𝑁𝑎𝐶𝑙 1

1 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙

) 𝑥 (58.5 𝑔 𝑁𝑎𝐶𝑙) = 0.0154 𝑚𝑜𝑙𝑒 𝑁𝑎𝐶𝑙

Then we divide by the volume of the solution in liters (100 mL = 0.1 L)

7-6

Copyright Prof. Lawrence Mailloux, 2021

𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑀) =

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 1 𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

=

0.01538 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 0.1 𝐿 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

= 0.154 M NaCl

Problem 4 – Perform the following calculations a) What is the molarity of a solution made by dissolving 15.5 grams of KBr in water to a final volume of 255 mL? b) What would be the concentration (in molarity) of a solution made by dissolving 16.6 grams of sodium nitrate (NaNO3) in water to a final volume of 185 mL? c) How many KBr formula units are present in 175 mL of the solution from part a? d) How many moles of NaNO3 are present in 445 mL of the solution from part b?

Solution Concentration and Dilution Just like people purchase concentrated cleaning chemicals and then dilute them prior to use, chemists often need to dilute their solutions before using them. Chemists refer to the concentrated solutions they keep on hand in the lab as stock solutions. These solutions are often very concentrated so one bottle of stock solution can last a very long time. To know how much water they need to add, chemists use the dilution equation given below.

C1V1 = C2 V2

Where: C1 = starting concentration, C2 = final concentration V1 = starting volume , V2 = final volume.

One can use any units of concentration or volume they wish when using the dilution equation provided that the same units are used on both sides of the equation. Consider the following example. A 12.1 M bottle of concentrated HCl is available. How many mL of this concentrated stock solution would be needed to prepare 150.0 mL of 1.20 M HCl?

C1V1 = C2 V2

12.1 M (V1) = 1.20 M (150.0 mL)

V1 = 14.9 mL

This means that to prepare 150.0 mL of 1.20 M HCl one would need to place 14.9 mL of 12.1 M concentrated HCl into a container and add water (150.0 mL – 14.9 mL = 135.1 mL) until the total volume of the solution is 150.0 mL. Problem 5 – Perform the following dilution problems 7-7

Copyright Prof. Lawrence Mailloux, 2021

a) How would 450. mL of a 0.9% NaCl solution be prepared if the stock solution on hand is 35.0% NaCl? b) What would be the final concentration of a solution of KBr prepared by combining 25.0 mL of 10.0% KBr with 75.0 mL of water? c) When 37.0 mL of a concentrated solution of LiOH was dissolved in 145 mL of water the final concentration was 2.20 M. What was the initial concentration of the LiOH solution? d) How many mL of water would be needed to dilute 40.0 mL of a 6.0 M solution of glucose to a final concentration of 2.0 M?

Water as a Solvent Chemicals reactions often take place in solution because the act of dissolving a substance separates its particles on the molecular level. This molecular level mixing greatly increases the speed of the resulting chemical reaction. One very common solvent is water. In fact, water is such a common solvent that it is sometimes called the “universal solvent” because it is able to dissolve so many different chemicals. Furthermore, as nurses the chemical reactions you will be dealing with are those that occur within the human body. Because the human body is made largely of water, it is the solvent for most biochemical reactions and will be the solvent we will be concerning ourselves with in this class. Water is what is known as a polar molecule. This means that one side of the molecule has a slight negative charge, and the other side has a slight positive charge. Polar solvents like water are very good at dissolving other polar substances such as sugar (C6H12O6), or alcohol (C2H5OH) as well as most ionic compounds such as sodium chloride (NaCl), or silver nitrate (AgNO3) just to name a couple. To understand why water is such a good solvent let’s look at a water molecule more closely. Recall from Chapter Five that when two nonmetals bond together they share their electrons to make a covalent bond. This sharing of electrons need not be fair/equal however. How strongly an atom is attracted to the electrons in its covalent bonds in known as an atom’s electronegativity. The more electronegative an atom is, the harder it pulls on its electrons. Because elements in the upper right corner of the periodic table need to gain electrons to fill their outermost energy levels, these elements are also the most electronegative. For the purposes of 7-8

Copyright Prof. Lawrence Mailloux, 2021

this class we will concern ourselves only with the four most electronegative elements, nitrogen, oxygen, fluorine, and chlorine which are highlighted on the following periodic table, of which oxygen is by far the most important physiologically. Because the noble gases do not form chemical bonds, the concept of electronegativity does not apply to them.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

IA

IIA

IIIB

IVB

VB

VIB

VIIB

----

VIIIB

----

IB

IIB

IIIA

IVA

VA

VIA

VIIA

VIIIA

H

He

The four most electronegative elements Li

Be

Na

Mg

K

Ca

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Rb

Sr

Y

Zr

Nb

Mo

Tc

Ru

Rh

Pd

Cs

Ba

*

La

Hf

Ta

W

Re

Os

Ir

Fr

Ra

**

Ac

Rf

Db

Sg

Bh

Hs

Mt

B

C

N

O

F

Ne

Al

Si

P

S

Cl

Ar

Zn

Ga

Ge

As

Se

Br

Kr

Ag

Cd

In

Sn

Sb

Te

I

Xe

Pt

Au

Hg

Tl

Pb

Bi

Po

At

Rn

Ds

Rg

Cn

Nh

Fl

Mc

Lv

Ts

Og

Nitrogen, Oxygen, Fluorine, and Chlorine

* **

Ce

Pr

Nd

Pm

Sm

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

Lu

Th

Pa

U

Np

Pu

Am

Cm

Bk

Cf

Es

Fm

Md

No

Lr

Whenever any other element is bonded to one of these four atoms, the electrons in their bonds are pulled tightly by the more electronegative element. For this example, we will consider an oxygen-hydrogen bond because the H-O bond is found in many biologically important molecules, such as water. Because oxygen is more electronegative than hydrogen, the electrons shared between them are not shared evenly. Oxygen pulls much more strongly on the electrons than hydrogen. This causes the oxygen atom to have a slight negative charge and leaves behind a small positive charge on the hydrogen atom. We use a lowercase Greek letter delta (δ) to indicate 7-9

Copyright Prof. Lawrence Mailloux, 2021

a small, or partial, electric charge. Bonds like these are called polar bonds. They occur whenever any atom is bonded to one of the four very electronegative elements listed above. Using this notation a hydrogen-oxygen bond can be written as follows. δ+

H-O δOxygen -More electronegative -Has partial negative charge

Hydrogen -Less electronegative -Has partial positive charge

Overall water molecules are neutral, however because water molecules are shaped like the letter V, this makes the hydrogen side of the molecule have a slightly positive charge, while the oxygen end of the molecule has a slight negative charge. These slight positive and negative charges will be attracted to ions of the opposite charge. The diagrams below show water molecules surrounding and stabilizing sodium and chloride ions. This process is known as solvation, or when the solvent is water (as is often the case in introductory classes) hydration. Notice how the orientation of the water molecules is different depending upon the charge of the ion being hydrated.

These partial charges are critical for waters ability to dissolve ionic compounds like NaCl, and other polar substances. There is however, another class of chemicals that water can’t dissolve; non-polar chemicals. These chemicals usually do not have electronegative atoms (O, N, Cl, or F) in their formulas and therefore don’t have any polar bonds with charges for water to hold onto. Chemists have a saying, “like dissolves like.” This means that polar chemicals can dissolve other polar chemicals (like alcohol and water) while nonpolar chemicals can dissolve other 7-10

Copyright Prof. Lawrence Mailloux, 2021

nonpolar chemicals, for example grease and paint thinner. Polar chemicals and nonpolar chemicals, like oil and water, do not mix. You have likely encountered this concept in other classes. In biology you may have hear the terms hydrophilic (water loving / polar) and hydrophobic (water fearing / non-polar). In everyday language, we would call polar substances water based and non-polar substances as oil based. Paint can be water or oil based. Water based paint can be cleaned up with warm water, while oil based paint requires paint thinner. We will learn more about polar and nonpolar chemicals in the second half of this class when we study organic chemistry.

7-11

Copyright Prof. Lawrence Mailloux, 2021

Introduction to Osmosis If you have ever gotten pool or ocean water up your nose, or in your eyes, you know how much it stings. Despite this, people frequently use saline sprays and eyedrops to treat congestion, allergies, and dry eyes with no pain at all. The reason why saline does not burn, while pool or ocean water does has to do with a phenomenon called osmosis. Osmosis is the flow of a solvent (generally water) across a special membrane that allows only some substances (in this case water) to cross but not others. This membrane is called a semipermeable membrane (SPM). Osmosis occurs when two solution with different solute concentrations are separated on opposite sides of a SPM, depicted below with a vertical dashed line. When this happens, water will flow across the SPM in the direction needed to equalize the solute concentration. In the case of the diagram below, to the right. The resulting difference in water level generates what is known as an osmotic pressure. The pressures generated can be quite large, and as we will see, have significant biological consequences.

Water flows from side of low solute concentration to high solute concentration

Many tissues in our bodies, such as the membranes of red blood cells, behave as semipermeable membranes. This means that if a red blood cell is exposed to any solution who’s solute concentration is different than that inside a red blood cell an osmotic pressure will develop across the cell membrane. Two important solutions with solute concentrations equal to the inside of a red blood cell are 0.9% NaCl and 5% glucose. Any solution with this same concentration (such as saline) is said to be isotoinic. If a redblood cell is exposed to an isotonic solution, nothing happens. Saline does not burn. If the solution inside and out are not the same however things are different. Any solution with a solute concentration less than 0.9% NaCl, or 5% glucose, is called a hypotonic solution, while any solution with a concentration greater than this is called a hypertonic solution. If a red blood cell is placed into a hypotonic solution, water will rush into 7-12

Copyright Prof. Lawrence Mailloux, 2021

the red blood cell in an attempt to equalize the solute concentration. As a result, the red blood cell will swell (or even possibly burst) in a process called hemolysis, on the other hand if a red blood cell is placed into a hypertonic solution, water will rush out of the cell causing it to shrivel up in a process called crenation. Neither of these processes are desirable and this is why pool water (hypotonic) and ocean water (hypertonic) both burn. It is for this reason that fluids put into a human body must be isotonic. The diagram below summarizes what happens to a red blood cell when exposed to either a hypotonic or hypertonic solution.

Hypotonic/Hemolysis -Water flows into red blood cell. -Cell swells and/or pops

Hypertonic/Crenation

-Water flows out of red blood cell. -Cell shrivels up

Problem 6 – Fill in the blanks in the following narrative. A(n) _________ solution is any solution with a concentation of ____% NaCl or _____% glucose. Any solution with a solute concentration higher than this is called _______ and will cause a red blood cell to ______ in a process called _______. On the other hand, any solution with a lower solute concentration, will cause the red blood cell to _______ in a process called _______.

7-13

Copyright Prof. Lawrence Mailloux, 2021

Neti Pots and Lube? Medically Relevant Examples of Osmosis People with sinus problems sometimes wash their sinuses using a device called a Neti Pot, which looks like a small plastic teapot as shown below. First one fills the pot with warm sterile/distilled water (to prevent rare but serious infections it is important that only distilled water is used). Then to make the solution isotonic one dissolves premeasured salt packets that come with the Neti Pot. Once the solution is isotonic and at body temperature, it can be poured through the sinuses without any discomfort, washing out mucus, allergens, and other irritants.

Another unexpected, but interesting, investigation of osmotic pressure titled, “Is Wetter Better?” was published in the journal PLOS ONE1. The most common route of HIV transmission is sexual activity; specifically unprotected anal or vaginal intercourse. During sexual intercourse, micro-injury to vaginal or rectal tissue may occur. To decrease such injury, and to in increase pleasure, many people use personal lubricants during sexual intercourse. Examples of a few products are shown below.

1. C. S. Dezzutti et al., Is wetter better? An evaluation of over-the-counter personal lubricants for safety and anti-HIV-1 activity. PLOS One, 2012, 79 (11).

7-14

Copyright Prof. Lawrence Mailloux, 2021

Because damaged tissues are a less effective barrier to HIV transmission, it has been assumed that by helping prevent trauma, using lubricants might reduce the risk of HIV transmission. But are all lubes created equal? What if the lubricant itself is irritating to tissues? Could this impact HIV transmission? It is these questions that the researchers sought to answer. In their study, the researchers purchased a variety of different over the counter lubricants (Astroglide, Boy Butter H2O, Boy Butter Original, Elbow Grease, Gynol II , Good Clean Love, ID Glide, KY Jelly, PRÉ, Replens, Slippery Stuff, Sliquid Organic, and Wet Platinum). All of these products were subjected to a wide variety of laboratory tests to determine their osmolality, irritation to rectal/vaginal tissues, and potential impacts on HIV transmission. The results of the study* showed that aqueous (water based) lubricants that were isotonic did not cause damage to tissue samples and did not appear to increase susceptibility to HIV transmission. Lubricants that were hypertonic, or hypotonic, caused damage to tissue samples and increased tissue susceptibility to HIV transmission. Presumably the osmotic pressure generated by these nonisotonic products caused tissue damage and thus theoretically increase the potential for viral transmission. In summary, the researchers found that the “safest” lubricants were either aqueous and isotonic, and thus did not generate an osmotic pressure, or silicone oil based and thus not capable of generating an osmotic pressure as osmosis only applies to water based solutions. I put the word “safest” in quotes because all of these experiments were performed in a laboratory using tissue samples in test tubes. To perform such testing on real people actually having sex would be ethically impossible due to the risk of HIV infection.

*There is a lot of detail and information in this study. This is only a very brief summary of the this studies results. To see the entire study simply Google “Is wetter better plos,” and the study will be one of the first hits. PLOS is cool because it is an open access journal. Many journal articles are behind a pay wall.

7-15

Copyright Prof. Lawrence Mailloux, 2021

Osmosis vs. Dialysis In osmosis, solvent crosses the SPM, from the side of low solute concentration of the side of high solute concentration, thus equalizing the concentration of the two solutions. There is another process involving semipermeable membranes that you may have heard of before, dialysis. Dialysis is a process where small solute particles (like glucose or urea) pass through a semipermeable membrane that, unlike in osmosis, allows the passage of small solute particles instead of just solvent. Larger colloidal particles like starch or proteins are too large to pass through the membrane. It is this process that your kidneys use to clean your blood of water products. If a person’s kidneys are not working they can be hooked up to a machine that cleans the blood for the damaged kidneys. This process is known as hemodialysis. The frequency and duration of treatments depending upon the degree of kidney failure that patient suffers from. While lifesaving, a dialysis machine is not as good as a working set of kidneys. Many patients on dialysis are waiting for kidney transplants. The diagram below illustrates the basics of a dialysis process. Take a moment to compare the dialysis diagram below to the osmosis diagrams at the start of this section; then answer problem seven below.

Small solute particles flows from side of high solute concentration to low solute concentration. Large solute particles don’t cross SPM

Problem 7 – what is the main difference between Osmosis and Dialysis? In your answer be sure to mention who/what crosses the SPM (solvent or solute) and the direction in which it moves across the membrane.

7-16

Copyright Prof. Lawrence Mailloux, 2021

Acids and Bases What makes something an acid or base? Acids are chemicals the release H+ ions when dissolved in water; this causes the concentration of H+ ions in solution to increase. Chemists use square brackets [ ] to indicate concentration. Using this notation [H+] would mean “concentration of H+ ions.” You can identify an acid easily as they always have an H in the front of their formula. Below are a few examples of some common acids. We will cover naming of acids shortly. HCl – hydrochloric acid (stomach acid) H2CO3 – carbonic acid (important for maintaining pH of blood) HC2H3O2 – acetic acid (vinegar is dilute acetic acid, usually 5%) When dissolved in water these acids release their H+ ions. For example, when HCl is dissolved in water the following ionization reaction takes place. HCl → H+ + ClWhen the H+ ions are released into the water they stick to the first H2O molecule they encounter as the reaction below shows. As a result, some books/professors write H3O+ instead of H+. This is just a matter of notation, they mean the same thing. The H3O+ ion is known as the hydronium ion. H+ + H2O → H3O+ Some additional properties of acids of which you should be aware are: •

Acids are corrosive chemicals and should be handled with care.

•

Are neutralized by bases

•

Taste sour – for example citric acid in grapefruit, ascorbic acid (vitamin C), and aspirin (acetylsalicylic acid) are all sour.

•

Turn blue litmus paper red. You will work with litmus paper in CHM1033L.

7-17

Copyright Prof. Lawrence Mailloux, 2021

Bases do the opposite, bases are chemicals that decrease the [H+] ions in solution. Most bases do this by releasing hydroxide (OH-) ions. The OH- ions then react with the H+ ions from the acids to produce water as shown below. H+ + OH- → H2O Some additional properties of acids of which you should be aware are: •

Bases are corrosive chemicals and should be handled with care.

•

Are slippery (like soap or bleach)

•

Are neutralized by acids

•

Taste bitter

•

Turn red litmus paper blue. Remember “blue to base.” You will work with litmus paper in CHM1033L.

Bases are easy to spot as well, they usually have an OH at the end of their formula. There are no special rules for naming bases. Below are examples of some bases. NaOH – sodium hydroxide (a.k.a. lye, used to make soap in Fight Club) Mg(OH)2 – magnesium hydroxide (in milk of magnesia, a laxative an antacid) Because most bases release OH- ions some books define bases as chemicals that release OH- ions when dissolved in water. This definition however is a bit oversimplified, as OH- is not the only substance that can neutralize H+ ions, nor do all bases have an OH in their formula. Ammonia (a gas, NH3) is one common example. When ammonia gas is dissolved in water it reacts with the water to form ammonium hydroxide which makes the solution basic. NH3(g) + H2O(l) → NH4OH(aq) Because ammonia is a gas it is generally used as a solution dissolved in water. Chemists often use the terms ammonia, NH3, NH4OH, and ammonium hydroxide interchangeably. This can cause confusion for students in their CHM1033L lab class; they all refer to the same thing. You will use these chemicals in your lab class.

7-18

Copyright Prof. Lawrence Mailloux, 2021

Another couple of bases that don’t have hydroxide in their formula that we need to be familiar with are carbonates (CO32-) and bicarbonates (HCO3-). When carbonates or bicarbonates react with acid a decomposition reaction occurs and carbon dioxide gas is released. This is what happens when baking soda (sodium bicarbonate, NaHCO3) and vinegar (acetic acid, HC2H3O2) are mixed. The bubbles you see are carbon dioxide gas. HC2H3O2 + NaHCO3 → NaC2H3O3 + H2O + CO2 Carbonates and bicarbonates are important chemicals. The limestone in Florida is made of calcium carbonate (CaCO3). Slowly over the past few thousands of years, naturally acidic rain water has been slowly eating away at the rocks under our feet. In north and central Florida, which are farther above sea-level, this has resulted in the formation of massive underwater caves and springs. Florida has hundreds of these springs, many are state parks. Below at left is prof Mailloux swimming in Alexander Springs, on the right a dry cave within Santos Mountain Bike park near Ocala FL. The water clarity in these springs is amazing. In the photograph below, prof Mailloux is about 25 feet underwater. I had just swam under the large rock on the bottom right and was looking around before returning to the surface. The photograph was taken by my friend from the surface.

7-19

Copyright Prof. Lawrence Mailloux, 2021

You don’t need to drive as far as Ocala to see caves and other evidence of this erosion. Several small caves exist within Matheson Hammock park located just a few miles from Miami Dade College Kendall Campus. Below are two pictures of caves within Matheson Hammock park taken by Prof Larry Mailloux. In the photo at right you can see my friend Jose struggling to squeeze through a small opening.

In addition to forming beautiful caves and springs, bicarbonates play an essential role in stabilizing the chemistry of your blood, as we will learn about shortly.

7-20

Copyright Prof. Lawrence Mailloux, 2021

Acids Strength and the pH Scale Chemists use a scale known as the pH scale to describe how acidic a solution is. Because acids are chemicals that release [H+] ions when dissolved in water, the stronger an acid solution is, the higher the [H+] ions will be. Mathematically pH is defined as follows: pH = - log [H+] You may (vaguely) recall logarithms from your math class. We won’t be doing complex math with them so don’t worry. What is important for you to understand is because the scale is not linear, even a small change in pH will cause a big change in [H+] as explained shortly. The pH scale runs from 0-14 (negative and greater than 14 pH values are possible, but not important for this class). Neutral water has a pH of seven, any pH less than seven is acidic and any pH greater than seven is basic; with the acid or base strength increasing the father one gets from seven. The diagram below shows the pH values of some common substances. The only substance who’s pH you need to memorize is that for blood, who’s pH must stay close to 7.4.

6

7

7-21

8

9

10

11

12

13

14

14.0 Lye (cleaner)......... ……………….

5

11.0 - ammonia ………….

4

Most Basic

8.3 Baking soda……….….

3

4.2 – Orange Juice……….

2

2.8 – vinegar ……………….

1

1.2 – stomach acid ……..

0

Neutral

7.0 - pure water …………. 7.4 – Human Blood….…..

Most Acidic

Copyright Prof. Lawrence Mailloux, 2021 Because the pH scale is logarithmic, every change of one pH unit equals a 10-fold change in acidity. For example something with a pH of 3 is ten times more acidic than something with a pH of 4 and 100 times for acidic than something with a pH of 5. With this in mind take a look at the following table which shows how the concentration of [H+] varies with pH. pH

[H+]

Acidic/Neutral/Basic

0

1 x 100

Acidic

1

1 x 10-1

Acidic

2

1 x 10

-2

Acidic

3

1 x 10-3

Acidic

4

1 x 10

-4

Acidic

5

1 x 10-5

Acidic

6

1 x 10

-6

Acidic

7

1 x 10-7

Neutral

8

1 x 10-8

Basic

9

1 x 10

-9

Basic

10

1 x 10-10

Basic

11

1 x 10

-11

Basic

12

1 x 10-12

Basic

13

1 x 10

-13

Basic

14

1 x 10-14

Basic

Notice that whatever the value of the pH, it is equal to the negative exponent on top of the [H+]. For historical reasons the pH scale is the most commonly used scale to describe how acidic a solution is. There is a less frequently used scale that describes how basic a solution is. This scale is known as the pOH scale. The pOH scale is defined similarly to the pH scale: pOH = - log [OH-]

7-22

Copyright Prof. Lawrence Mailloux, 2021

The pH and pOH scales are very closely related. Study the table below to see how pH, pOH, [H+], and [OH-] are interrelated. pH

pOH

[H+]

[OH-]

Acidic/Neutral/Basic

0

14

1 x 100

1 x 10-14

Acidic

1

13

1 x 10

-1

-13

Acidic

2

12

1 x 10-2

1 x 10-12

Acidic

3

11

1 x 10

-3

-11

Acidic

4

10

1 x 10-4

1 x 10-10

Acidic

5

9

1 x 10-5

1 x 10-9

Acidic

6

8

1 x 10-6

1 x 10-8

Acidic

7

7

1 x 10-7

1 x 10-7

Neutral

8

6

1 x 10

-8

-6

Basic

9

5

1 x 10-9

1 x 10-5

Basic

10

4

1 x 10

-10

-4

Basic

11

3

1 x 10-11

1 x 10-3

Basic

12

2

1 x 10

-12

-2

Basic

13

1

1 x 10-13

1 x 10-1

Basic

14

0

1 x 10-14

1 x 100

Basic

1 x 10

1 x 10

1 x 10

1 x 10

1 x 10

You should have noticed that, just as with the pH scale, the value of the pOH is equal to the exponent on the [OH-]. Additionally, the sum of pH + pOH always adds to 14. You will be expected to know how the qualities in the table above interrelate. We will only concern ourselves with round number pH’s. For pH values that are not integers (like 7.4, human blood), one would need to use a calculator, however for this class we will keep things as simple as possible. Problem 8 – Fill in the following table pH

pOH

[H+]

[OH-]

Acidic/Neutral/Basic

1 x 1014

0 13

1 x 10-10

4 1 x 10-5

5

Acidic Acidic Neutral

8

6

1 x 10-6 7-23

Copyright Prof. Lawrence Mailloux, 2021

Naming Acids Naming acids is not difficult. Acids are divided into two categories, each with their own naming rules. Binary acids are acids that contain only two different elements. These acids are named as follows: “hydro” + first part of anion + “ic acid” For all practical purposes there are only four binary acids, so learning to name them is as simple as knowing the list below. HCl – hydrochloric acid HBr – hydrobromic acid HI – hydroiodic acid HF – hydrofluoric acid Oxyacids (a.k.a. ternary acids) are acids that contain a polyatomic ion. If the ion name ends in “ate” then you get an “ic acid,” if the polyatomic ions name ends in “ite” you get an ”ous acid.” Acids are neutral compounds. They have as many H’s in front as necessary to cancel out the negative charge of the anion. Below are the acids derived from the polyatomic ions you were instructed to memorize back in Chapter Five. NO3- - nitrate → HNO3 – nitric acid CO32- - carbonate → H2CO3 – carbonic acid C2H3O2- - acetate → HC2H3O2 – acetic acid SO42- - sulfate → H2SO4 – sulfuric acid PO43- - phosphate → H3PO4 – phosphoric acid None of the polyatomic acids you were instructed to memorize in Chapter Five ended in “ite.” Polyatomic ions ending in “ite” have one less oxygen than their “ate” cousins do and are less common. Here are some examples of polyatomic ions that end in “ite,” and their corresponding acids. You do NOT need to memorize these polyatomic ions, however if given the name and

7-24

Copyright Prof. Lawrence Mailloux, 2021

formula of a polyatomic ion you would be expected to be able to apply these rules to write the formula for, and name, its corresponding acid. NO2- - nitrite → HNO2 – nitrous acid SO32- - sulfite → H2SO3 – sulfurous acid Problem 9 – fill in the following table of acid names and formulas Formula

Name

H2CO3 Hydroiodic acid HC2H3O2 Nitric acid Hydrochloric acid HBr H3PO4 H2SO4 Hydrofluoric acid

Problem 10 – below are the formulas and names of some polyatomic ions that you probably have never heard of. Apply the rules explained above to fill in the following table by writing the formulas and names of their corresponding acids.

Formula of

Name of

Formula of

Name of

Polyatomic Ion

Polyatomic Ion

Corresponding Acid

Corresponding Acid

BO33-

Borate

CrO42-

Chromate

C2O42-

Oxalate

PO33-

Phosphite

7-25

Copyright Prof. Lawrence Mailloux, 2021

Neutralization – When Acids and Bases React When acids and bases react the H+ from the acid and the OH- from the base combine to form neutral water while the remaining anion from the acid and cation from the base form an ionic compound. Chemists refer to ionic compounds as “salts,” and we say acid + base → salt + water There are millions of salts, NaCl, the salt you put on your French fries, is only one common example. When referring to NaCl chemists often say “table salt,” to avoid confusion. For an example consider the reaction of HCl with NaOH. The H+ and OH- from the acid and base form water and the left over Na+ and Cl- form NaCl. HCl + NaOH → NaCl + H2O Here is another example (not yet balanced), the reactions of sulfuric acid with potassium hydroxide H2SO4 + KOH → K2SO4 + H2O

(unbalanced)

The water part is easy, but how do you figure out the formula of the salt? To figure out the correct formula of the salt you must do the crissy-crossy trick you learned in Chapter Five with the K+ and the SO42Na+

SO42-

K2SO4 To balance our reaction it is easiest to balance the acid and base parts first. The number of H+ from the acids must be equal to the number of OH- from the bases. In this case we have two H+ from the H2SO4 and only one OH- from the KOH, therefore we need to put a “2” in front of KOH. Additionally, each H+ that combines with an OH- forms one water. Thus we would have two H2O in our reaction, the K’s automatically balance themselves giving the final balanced reaction shown below. H2SO4 + 2 KOH → K2SO4 + 2 H2O

7-26

Copyright Prof. Lawrence Mailloux, 2021

Problem 11 – Complete and balance the following acid-base neutralization reactions. ______ H3PO4 + ______ Ca(OH)2 → ______ H2SO4 + ______ Ba(OH)2 → ______ H3BO3 + ______ Fe(OH)3 → ______ HBr + ______ Mg(OH)2 → ______ H3PO4 + ______ NaOH →

Acids, bases, and your blood. Recall that even a small change of pH means a big change in acidity. This means that even a small change in pH can have major consequences on a person’s health. Your body keeps the pH of its blood very close to 7.4. To do this your body uses what’s called a buffer. Buffers are solutions that are resistant to changes in pH. Your blood is buffered at a pH of 7.4. In order to do this all buffers need two components, a weak acids and “its salt” (we won’t consider buffers made from weak bases in this class). By “salt” we mean related ionic compound. Below are some examples of weak acids and related salts that could be used to make a buffer. The identity of the added cation (Li+, Na+, K+, etc.) does not matter. Weak Acid

Salt

(Reacts with added base)

(Reacts with added acid)

HF

NaF

HF

KF

HC2H3O2

LiC2H3O2

HNO2

NaNO2

H2CO3

NaHCO3

You will notice the last one is highlighted in red, this is the buffer system your blood uses to maintain its pH. By having two components to it, a buffer solution is able to neutralize the addition of reasonable amounts of acid or base and thus keeps the pH of the solution stable.

7-27

Copyright Prof. Lawrence Mailloux, 2021

Let’s consider a buffer made of H2CO3 and NaHCO3. If base (in this example NaOH) is added to the solution, the carbonic acid (H2CO3) will neutralize the added base and form more salt, (sodium bicarbonate, NaHCO3) and water. H2CO3 + NaOH → NaHCO3 + H2O If instead acid (in this example HCl) is added to the solution, the salt with neutralize the added acid, to from more carbonic acid. HCl + NaHCO3 → H2CO3 + NaCl Problem 12 – which of the following pairs of substances could be used to make a buffer? a. HCl / KCl b. HF / KCl c. HNO3 / LiNO3 d. H3PO4 / LiH2PO4 e. HBr / NaBr

Problem 13 – if a small amount of acid is added to an unbuffered solution the pH will a) decrease slightly b) decrease significantly c) stay the same d) increase slightly e) increase significantly Problem 14 – if a small amount of base is added to an unbuffered solution the pH will a) decrease slightly b) decrease significantly c) stay the same d) increase slightly e) increase significantly

7-28

Copyright Prof. Lawrence Mailloux, 2021

Problem 15 – if a small amount of acid is added to a buffered solution the pH will a) decrease slightly b) decrease significantly c) stay the same d) increase slightly e) increase significantly Problem 16 – if a small amount of base is added to a buffered solution the pH will a) decrease slightly b) decrease significantly c) stay the same d) increase slightly e) increase significantly

7-29

Copyright Prof. Lawrence Mailloux, 2021

Answers to In Chapter Problems 1. a. 0.9% NaCl b. 4.05 g NaCl c. 650 mL 2. a. 41.8 % b. 45 / (45 + 230) x 100 = 16.4%, NOT 45/230 = 19.6%. Divide by total volume c. 667 mL d. 205 mL 3. a. 6 / (6 + 294) x 100 = 2.00%, although the difference in the answer is small in this case, you should still be dividing by the total mass of 6 + 294 = 300 grams, and not just by 294 g, which would give 2.04%. b. 1.0 g c. 475 g 4. a. 0.510 M b. 1.06 M c. 5.37 x 1022 d. 0.472 mol 5. a. 11.6 mL. One would dilute 11.57 mL of the 35.0% concentrated solution with water until a total volume of 450 mL is reached b. 2.5% Be careful, make sure you divide by the total volume of 25 + 75 = 100 mL instead of only dividing by 75% which gives the common error of 3.33 percent. c. 10.8 M. As with part b, always divide by the total volume d. Final volume equals 120 mL, therefore 80 mL of water are needed

7-30

Copyright Prof. Lawrence Mailloux, 2021

6. A(n) _________ solution is any solution with a concentation of ____% NaCl or _____% glucose. Any solution with a solute concentration higher than this is called _______ and will cause a red blood cell to ______ in a process called _______. On the other hand, any solution with a lower solute concentration, will cause the red blood cell to _______ in a process called _______. 7. In osmosis solvent moves across the SPM in the direction from low solute concentration to high solute concentration. In dialysis small solute particles more across the SPM in the direction from high solute concentration to low solute concentration. 8. pH

pOH

[H+]

[OH-]

Acidic/Neutral/Basic

0

14

1 x 100

1 x 10-14

Acidic

1

13

1 x 10-1

1 x 10-13

Acidic

4

10

1 x 10-4

1 x 10-10

Acidic

5

9

1 x 10-5

1 x 10-9

Acidic

7

7

1 x 10-7

1 x 10-7

Neutral

8

6

1 x 10-8

1 x 10-6

Basic

9. Formula

Name

H2CO3

carbonic acid

HI

Hydroiodic acid

HC2H3O2

Acetic acid

HNO3

Nitric acid

HCl

Hydrochloric acid

HBr

Hydrobromic acid

H3PO4

Phosphoric acid

H2SO4

Sulfuric acid

HF

Hydrofluoric acid

7-31

Copyright Prof. Lawrence Mailloux, 2021

10. Formula of

Name of

Formula of

Name of

Polyatomic Ion

Polyatomic Ion

Corresponding Acid

Corresponding Acid

BO33-

Borate

H3BO3

Boric Acid

CrO42-

Chromate

H2CrO4

Chromic acid

C2O42-

Oxalate

H2Cr2O4

Oxalic acid

PO33-

Phosphite

H3PO3

Phosphorus acid

11. ___2__ H3PO4 + ___3__ Ca(OH)2 → ______ Ca3(PO4)2 + ___6__ H2O ______ H2SO4 + ______ Ba(OH)2 → ______ BaSO4 + ___2__ H2O ______ H3BO3 + ______ Fe(OH)3 → ______ FeBO3 + ___3__ H2O ___2__ HBr + ______ Mg(OH)2 → ______ MgBr2 + ___2__ H2O ______ H3PO4 + ___3__ NaOH →______ Na3PO4 + ___3__ H2O

12. a. HCl / KCl – no HCl is a strong acid b. HF / KCl – no, KCl is not a salf of HF. Would need KF instead. c. HNO3 / LiNO3 – no, HNO3 is a strong acid d. H3PO4 / LiH2PO4 - yes e. HBr / NaBr – no, HBr is a strong acid

13. a) decrease slightly b) decrease significantly c) stay the same d) increase slightly e) increase significantly

7-32

Copyright Prof. Lawrence Mailloux, 2021

14. a) decrease slightly b) decrease significantly c) stay the same d) increase slightly e) increase significantly

15. a) decrease slightly b) decrease significantly c) stay the same d) increase slightly e) increase significantly

16. a) decrease slightly b) decrease significantly c) stay the same d) increase slightly e) increase significantly

7-33

Copyright Prof. Lawrence Mailloux, 2021

Chapter 8 Drawing Structures of Covalent Compounds Bonding Patters In Chapter Five we learned that there are two kinds of compounds. Ionic compounds and covalent (molecular) compounds. Ionic compounds are made of ions that result from the transfer of electrons between metals and non-metals. These ions then arrange themselves into a repeating array of cations and anions that we call a crystal lattice. Covalent compounds on the other hand, are formed when two or more non-metals share electrons to give all of their atoms an octet of electrons. Unlike ionic compounds, whose structures are all very similar (crystal lattices) covalent compounds exist as distinct molecules, each with a complex and unique shape. As we will learn in this and later chapters, the three dimensional shape of these molecules determines their chemical properties. Every drug you have ever taken (legal or illegal) is covalent compound who biological/pharmaceutical properties are a direct consequence of its three dimensional shape. In this chapter, we will learn how to draw structures of simple covalent compounds. In subsequent chapters, we will learn how to apply these concepts to more complex organic compounds and how this related to the design and manufacture of pharmaceuticals. To draw structures of covalent compounds scientists use a notation that uses dots to represent valance electrons known as Lewis Dot Structures. Using this notation, we would arrive at the following symbols for the second row of the periodic table:

In these chapters we will be concerning ourselves with covalent/molecular compounds. Because covalent compounds are between all non-metals, we will ignore Li, and Be which are metals. Additionally, boron is a semimetal so to keep things simple we will ignore it as well. Lastly, Ne and the rest of the noble gases do not form compounds, therefore we can ignore these elements too. Thus of the elements in the second row, we only need concern ourselves with carbon, nitrogen, oxygen and fluorine as shown below.

8-1

Copyright Prof. Lawrence Mailloux, 2021

Most of the compounds we will be studying in the following chapters contain many hydrogen atoms. Hydrogen has only one valance electron giving the following dot structure.

Recall that the elements in the same column of the periodic table have the same number of valance electrons. This means the chlorine, bromine, and iodine also have seven valance electrons just like fluorine. Additionally, sulfur has six valance electrons like oxygen, and phosphorus, like nitrogen, has five. We will be working with these elements as well. These elements have the following dot symbols.

Covalent compounds are formed when nonmetals share electron to fill their outermost energy levels and achieve the electron configuration of the nearest noble gas. For elements other than hydrogen, non-metals share electron to fill their outermost energy level with eight electrons. We call this the octet rule. With hydrogen, the nearest noble gas is helium; because helium only has two valance electrons, hydrogen only needs two electrons to be happy. While there are rare exceptions to the above stated rules, we will not concern ourselves with them in this class. The number of valance electrons an element has determines the number of bonds that element forms. For each electron less than eight, the element must form one bond to fill its octet. Thus carbon with four valance electrons must form 8 - 4 = 4 bonds to have an octet. Nitrogen must forms 8 – 5 = 3 bonds to have an octet. Oxygen forms two bonds, while hydrogen and the halogens all form only one bond.

8-2

Copyright Prof. Lawrence Mailloux, 2021

When elements share one pair of electrons they form a single bond. When they share two pairs of electrons they form a double bond and when they share three pairs of electrons they form a triple bond. There is no such thing as a quadruple bond. Elements can use a combination of single, double, and/or triple bonds to achieve an octet. Carbon (and silicon below it) for example forms four bonds; this can take the form of four single bonds, two double bonds, one double bond and two single bonds, or one single bond and one triple bond. Below are the four bonding patters that you will see for carbon in this class.

Nitrogen (and phosphorus below it) forms three bonds with a lone pair of electrons. This can take the form of three single bonds and a lone pair, one double bond and one single bond with a lone pair, or one triple bond with a lone pair. This give the following bonding patters for nitrogen that you will encounter in this class.

Oxygen (and sulfur below it) forms two bonds with two lone pairs giving the following bonding patters.

Fluorine (along with the other halogens below it) and hydrogen form one single bond as shown below

Keep these bonding patters in mind as your proceed through the next section. If you draw a Lewis structure that does not follow these bonding patters it’s probably wrong! 8-3

Copyright Prof. Lawrence Mailloux, 2021

Steps for Drawing Lewis Structures To draw Lewis structures of covalent compounds follow the following steps. 1. Count the number of valance electrons in your compound. 2. Identify the central element. Hints: a. Usually written first in the compound b. Usually gives a more symmetrical structure c. Is never H or F as these two elements can only form one bond. 3. Put in the central element and draw single bongs to surrounding elements. 4. Give surrounding elements an octet of electrons (except H – only gets two) 5. Count the number of electrons you have used so far. If you have any left over, put them on the central atom. 6. If the central atom does not have an octet of electrons share electrons off surrounding atoms to make double or triple bonds. 7. Check to ensure that all elements have an octet.

8-4

Copyright Prof. Lawrence Mailloux, 2021

Example 1 - SiCl4. Step 1: Count the valance electrons: Si = 4 valance electrons,

Cl = 7 valance electrons each

4 + 7 x 4 = 32 valance electrons total Step 2: ID central element Si is central as it is written first. Also putting a single Si in the center and surrounding it with Cl atoms results in a symmetrical structure. Step 3: Put in central element and draw single bonds to surrounding elements

Step 4: Give surrounding atoms an octet. (Except H – only gets two)

Step 5: Count valance electrons – put any leftover electrons on the central atom. So far we have used a total of 32 electrons; 4 x 6 = 24 electrons from the three sets of lone pairs on each chlorine, plus 8 more from each of the four single bonds. Because we started with 24 electrons there are none left over to worry about. The structure is finished.

8-5

Copyright Prof. Lawrence Mailloux, 2021

Example 2 - AsF3. Step 1: Count the valance electrons: As = 5 valance electrons,

F = 7 valance electrons each

5 + 3 x 7 = 26 valance electrons total Step 2: ID central element As is written first indicating that it is the central atom. Also because F can only form one bond it cannot be the central atom.. Step 3: Put in the central element and draw single bonds to surrounding elements

Step 4: Give surrounding atoms an octet. (Except H – only gets two)

Step 5: Count valance electrons – put any leftover electrons on the central atom. So far we have used a total of 24 electrons; 6 x 3 = 18 electrons from the lone pairs, plus 6 electrons from the three single bonds. Because we started with 26 electrons we have 26 - 24 = 2 electrons leftover. Therefore, we have to put a lone pair on our central atom, giving us our final structure.

8-6

Copyright Prof. Lawrence Mailloux, 2021

Example 3 - H2S. Step 1: Count the valance electrons: H = 1 valance electron each, S = 6 valance electrons each 1 x 2 + 6 = 8 valance electrons total Step 2: ID central element Here H is written first however because H can only form one bond it can’t be the central atom. Also putting S in the center gives a more symmetrical structure. Step 3: Put in central elements and draw single bonds to surrounding elements

Step 4: Give surrounding atoms an octet. (Except for H – only gets two) Do not put an octet of electrons on hydrogen.. Remember that hydrogen emulates the electron configuration of He and therefore only wants two electrons, which are supplied by the single bonds. Step 5: Count valance electrons – put any leftover electrons on the central atom. So far we have used a total of 4 electrons, two in each of the two single bonds. This leaves 8 – 4 = 4 electrons leftover. We put our four leftover electrons as lone pairs on the sulfur atom giving the final structure of our compound.

8-7

Copyright Prof. Lawrence Mailloux, 2021

Example 4 - SiSe2. Step 1: Count the valance electrons: Si = 4 valance electrons,

Se = 6 valance electrons each

1 x 4 + 6 x 2 = 16 valance electrons total Step 2: ID central element Si is written first and putting it in the center will give a symmetrical structure Step 3: Put in central elements and draw single bonds to surrounding elements

Step 4: Give surrounding atoms an octet. (Except for H – only gets two)

Step 5: Count valance electrons – put any leftover electrons on the central atom. So far we have used a total of 16 electrons, 12 in each of the three lone pairs on the two Se atoms, plus 4 in the two single bonds. This leaves 16 – 16 = 0 electrons leftover. However, the structure is not correct yet. Si lacks an octet. Step 6: Central atom lacks octet – share off surrounding atoms to make multiple bonds.

Sharing electrons off the surrounding Se atoms forms double bonds and give Si an octet. It does not matter which electrons you chose to move. I have selected the outermost pairs.

8-8

Copyright Prof. Lawrence Mailloux, 2021

Example 5 - FCN. Step 1: Count the valance electrons: F = 7 valance electrons,

C = 4 valance electrons

N = 5 valance electrons

7 + 4 + 5 = 16 valance electrons total Step 2: ID central element C is written in the center as a hint to the compounds structure. Also, because carbon forms four bonds, it will fit nicely between F and N using these bonding patterns:

F–

–C≡

≡ N:

Step 3: Put in central elements and draw single bonds to surrounding elements

Step 4: Give surrounding atoms an octet. (Except for H – only gets two)

Step 5: Count valance electrons – put any leftover on central atom. So far we have used a total of 16 electrons, 2 x 6 =12 from the lone pairs on the surrounding F and N atoms plus four in the two single bonds leaving 16 – 16 = 0 left over. However, the structure is not correct yet. Carbon lacks an octet. Step 6: Central atom lacks octet – share off surrounding atoms to make multiple bonds. Sharing electrons off the surrounding fluorine and nitrogen atoms forms double bonds and gives C an octet. It does not matter which electrons you chose to move.

8-9

Copyright Prof. Lawrence Mailloux, 2021

Problem 1 – Fill in the following table – keep your bonding patters in mind Compound

Lewis Dot Structure

Compound

NH3 _________ Valance Electrons

CCl4 _________ Valance Electrons

H2O _________ Valance Electrons

CH4 _________ Valance Electrons

SF2 _________ Valance Electrons

AsCl3 _________ Valance Electrons

PCl3 _________ Valance Electrons

SiH4 _________ Valance Electrons

8-10

Lewis Dot Structure

Copyright Prof. Lawrence Mailloux, 2021

Problem 2 – Fill in the following table - keep your bonding patters in mind

Compound

Lewis Dot Structure

Compound

CH2S _________

BrCP _________

Valance Electrons

Valance Electrons

CCl2O _________

CS2 _________

Valance Electrons

Valance Electrons

SeI2 _________

H2Se _________

Valance Electrons

Valance Electrons

SiBr4 _________

CO2 _________

Valance Electrons

Valance Electrons

8-11

Lewis Dot Structure

Copyright Prof. Lawrence Mailloux, 2021

Problem 3 – Fill in the following table - keep your bonding patters in mind

Compound

Lewis Dot Structure

Compound

NF3 _________

HCN _________

Valance Electrons

Valance Electrons

CH2O _________

NI3 _________

Valance Electrons

Valance Electrons

PBr3 _________

SeF2 _________

Valance Electrons

Valance Electrons

CBr4 _________

HCP _________

Valance Electrons

Valance Electrons

8-12

Lewis Dot Structure

Copyright Prof. Lawrence Mailloux, 2021

Answers to In Chapter Problems

8-13

Copyright Prof. Lawrence Mailloux, 2021

8-14

Copyright Prof. Lawrence Mailloux, 2021

8-15

Copyright Prof. Lawrence Mailloux, 2021

Chapter 9 Drawing Structures of Organic Compounds Making a Carbon Backbone In the previous chapter we learned how to draw structures of covalent compounds. In this chapter we will focus our attention on drawing the structures of carbon based compounds known as organic compounds. Carbon is unique amongst the elements in that it is able to easily bond to itself in virtually an unlimited number of ways. By connecting multiple carbon atoms together a backbone of carbon atoms is formed; hydrogen atoms fill in the remaining bonds to ensure that every carbon has four bonds. Throughout this chapter, and future chapters, keep in mind that carbon always forms four bonds. If you draw a structure in which carbon has any number of bonds other than four, it is wrong. Recall from the prior chapter that carbon and hydrogen can only bond in the following ways.

In organic compounds the carbon atoms bond to one another forming a carbon backbone. To start out we will only consider carbon-carbon single bonds. The simplest compound is CH4 (methane) as shown below.

We can just as easily connect two carbon atoms together giving C2H6 (ethane).

9-1

Copyright Prof. Lawrence Mailloux, 2021

Or three carbons atoms can be connected giving C3H8; propane, the gas used in grills.

This process can repeat over and over again. Connecting eight carbon atoms in a row gives octane (C8H18) which is a major component of gasoline.

Just like an octopus has eight tentacles, octane has a carbon chain eight atoms long. Scientists name organic chemicals based upon the number of carbon atoms they have in their longest continuous chain of carbon atoms. The table below summarizes this information. Take time to commit this information to memory, as we will be making use of it throughout the rest of the semester.

Number Carbon Atoms 1 2 3 4 5 6 7 8 9 10

Condensed Structure CH4 CH3CH3 CH3CH2CH3 CH3CH2CH2CH3 CH3CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3

9-2

Name methane ethane propane butane pentane hexane heptane octane nonane decane

Copyright Prof. Lawrence Mailloux, 2021

Branches and Structural Isomers Carbon atoms are not limited to connecting themselves into straight chains. Whenever there are four or more carbon atoms in a compound there are multiple ways those atoms can be bonded together. Compounds with the same number and kind of atoms bonded in different ways are known as structural isomers. Below are the structures of the two possible isomers of C4H10. Chemists refer to the carbon atoms hanging off the main chain as braches. I have exaggerated the length of the carbon-carbon bond to the branch to make the structure easier to read.

Branch

Just because the two compounds above have the same formula (C4H10) does not mean they are necessarily similar chemicals. Different isomers are different chemicals with different structures, properties, and names. The three dimensional structure (that is the 3D shape) of a chemical determines its properties. Because the two isomers above have different structure, they will have different chemical properties. As the number of carbon atoms increases so too does the number of ways the carbon atoms can be connected. Do the two problems that follow to practice this concept. Problem 1 – There are three different structural isomers of C5H12. Fill in the following table below by drawing all three possible isomers.

9-3

Copyright Prof. Lawrence Mailloux, 2021

Problem 2 – There are five structural isomers of C6H14. Fill in the following table below by drawing all five possible isomers.

Using Bond Line Notation By this point you have likely noticed that writing every single C and H can get very tedious. Chemists call these structures skeletal structures. To save time scientists use an abbreviated notation known as bond line notation. In this notation, the carbons and hydrogen atoms are not shown. It is understood that everywhere there is a bend or end in the structure, there is a carbon atom with enough hydrogen atoms bonded to it to total four bonds. For example, consider butane, who’s structure is shown below using both notations:

Using this same notation octane would be written as

9-4

Copyright Prof. Lawrence Mailloux, 2021

When branches are present they are shown as follows

Below is a more complex example with multiple branches.

It is important to understand that it is the connectivity of the atoms that matters, not how they are arranged on paper. In the example above what matters is that on the third carbon there is a twocarbon branch and on the fifth carbon there is a one carbon branch. The fact that in bond line notation the branches are both on the bottom, while in the skeletal structure one branch is on top and the other on the bottom does not matter. In addition, left to right does not matter either. The two structure below represent the exact same chemical.

9-5

Copyright Prof. Lawrence Mailloux, 2021

Multiple bonds are easy to show using bond line notation. You must however keep in mind that carbon only forms four bonds. This means that the presence of double or triple bonds will result in less hydrogen atoms being attaching to those carbon atoms. In the example below you will notice that the second and third carbons containing the double bond only have one attached hydrogen. Remember carbon always forms four bonds.

Carbon atoms can also form rings. Below is cyclobutane, a four-sided ring.

Atoms other can carbon and hydrogen are known as heteroatoms. Heteroatoms are always written when using bond line notation. Look at the example below. Notice oxygen is forming two bonds, as was explained last chapter. Also, notice that the carbon with the oxygen on it has no hydrogen atoms attached as it already has four bonds.

While it take a little bit of time to get used to bond line notation is much faster and easier to write than skeletal notation. This is the notation we will be using most of the time in this class so it’s important you get used to it. Your exams will be written using this notation. Do the practice problems that follow to become familiar with this notation. We will begin using this notation in the next chapter. 9-6

Copyright Prof. Lawrence Mailloux, 2021

Problem 3 – Fill in the following table by writing either the skeletal or bond line formula as appropriate. When checking your answers in the back keep in mind it is the connectivity, not the orientation on paper, of the drawing that matters. Skeletal Notation

Bondline Notation

9-7

Copyright Prof. Lawrence Mailloux, 2021

Problem 4 – We will be learning about pharmaceuticals in this course. Every drug you have ever taken (legal or illegal) is an organic compound who’s biological effects are a direct consequence of its three dimensional structure. It is important that you understand what you are looking at when you see these structure. Fill in the following table by giving the skeletal structures of each of these drugs in the right hand column. Bond Line Notation

Skeletal Notation

AZT (Zidovudine) – first anti-HIV medication, introduced in 1987.

Emtricitabine – anti-HIV medication of the nucleoside analog reverse transcriptase inhibitor class (NRTI). One of the medications in Truvada (used to prevent HIV).

Efavirenz – anti-HIV medication of the non-nucleoside analog reverse transcriptase inhibitor class (NNRTI). We will learn more about these toward the end of the course.

9-8

Copyright Prof. Lawrence Mailloux, 2021

Answers to In Chapter Questions 1.

The three structural isomers of C5H12 are as follows

2.

The five structural isomers of C6H14 are as follows

9-9

Copyright Prof. Lawrence Mailloux, 2021

3. Remember it’s the connectivity that matters, not left/right/up/down on paper. If you are unsure that your answer is equivalent to the answers shown below, ask your instructor. Skeletal Notation

Bondline Notation

9-10

Copyright Prof. Lawrence Mailloux, 2021

4. The skeletal structures of these antiviral medications are as follows Bond Line Notation

Skeletal Notation

AZT (Zidovudine) – first anti-HIV medication, introduced in 1987.

Emtricitabine – anti-HIV medication of the nucleoside analog reverse transcriptase inhibitor class (NRTI). One of the medications in Truvada (used to prevent HIV).

Efavirenz – anti-HIV medication of the non-nucleoside analog reverse transcriptase inhibitor class (NNRTI). We will learn more about these toward the end of the course.

9-11

Copyright Prof. Lawrence Mailloux, 2021

Chapter 10 Alkanes, Alkenes, and Alkynes Hydrocarbons Compounds made of carbon and hydrogen atoms are known as hydrocarbons. Hydrocarbons are nonelectrolytes, non-polar, insoluble in water, and flammable. They tend to smell like “chemicals,” that is they have a gasoline/paint-thinner type smell. Chemists classify hydrocarbons based upon the presence or absence of double or triple bonds. Compounds that have the maximum number of hydrogen atoms possible for the number of carbon atoms in the compound are said to be saturated (we will learn how this concept applies to nutrition and saturated vs. unsaturated fats in Chapter 15). Such compounds will contain only carbon-carbon single bonds and are known as alkanes. The names of alkanes end in “ane,” and they have a ratio of carbon to hydrogen of CnH2n+2. Consider octane which is present in gasoline. As its name suggests, octane has eight carbon atoms, therefore it must contain 2 x 8+2 = 18 hydrogen atoms. The formula of octane is therefore C8H18 as shown below:

There are many different ways to bond eight carbons and 18 hydrogens, the example above is only the simplest one of many. As you learned in Chapter Nine, compounds with the same number and type of atoms (that is with the same formula) bonded together differently are known as isomers. Below are examples of two isomers of C8H18. It does not matter how you connect them; as long as you use only single bonds, you will always get a CnH2n+2 ratio of C to H.

10-1

Copyright Prof. Lawrence Mailloux, 2021

The above isomers of C8H18 are different chemicals. They have different chemical and physical properties. While it is true that chemicals with similar formulas and structures are often chemically similar, this is not always the case. Sometimes one isomer and another may be very similar, while other times they may be very different. In the case of pharmaceuticals, some of these difference can be quite drastic, it is possible for one isomer to be a safe and effective medication, while another isomer may be toxic. Because different isomers are chemically different, they have different names. Naming organic compounds is completely different than naming the ionic and covalent compounds we have studied thus far. Alkanes are the easiest organic compounds to name so we will learn about these first. It is important that you get a solid grasp of naming alkanes, because the naming or other types of organic compounds (such as alcohols) are based upon the same rules.

Naming Alkanes The names of alkanes are based upon the number of carbon atoms in the longest chain in the structure, which is called the parent chain. Below are the names for organic compounds with up to ten carbon atoms in their longest chain. Compounds can be longer than this, but we will not consider such compounds in this course. Number Carbon Atoms 1 2 3 4 5 6 7 8 9 10

Condensed Structure CH4 CH3CH3 CH3CH2CH3 CH3CH2CH2CH3 CH3CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3

10-2

Name methane ethane propane butane pentane hexane heptane octane nonane decane

Copyright Prof. Lawrence Mailloux, 2021

The first step in naming an alkane is to find the longest continuous chain. It is important to note that the longest chain will not always be readily apparent. The longest chain may take many turns and need not be written from left to right. Left, right, up and down are human concepts. All that matters in organic chemistry is the connectivity of the atoms. Consider the following examples below where the longest chain contains seven carbons as numbered below.

Hanging off of the parent chain are additional carbon and hydrogen atoms. Chemists call things hanging off the parent chain branches or side groups. If the branches are made of only carbon and hydrogen they are also called alkyl groups. The names of branches are based upon the number and kind of atoms attached. In this class we will be dealing only with the following groups which are the most common in organic chemistry. Branch/Alkyl Group

Structural Formula

methyl

-CH3

ethyl

CH2CH3

propyl

-CH2CH2CH3 (hanging from end carbon)

isopropyl -CHCH3 | CH3

(hanging from middle carbon) fluoro

-F

chloro

-Cl

bromo

-Br

iodo

-I

10-3

Copyright Prof. Lawrence Mailloux, 2021

Once the longest continuous chain has been found it is then numbered to give the branches the lowest possible numbers. The locations of the branches are then indicated in the name based upon the numbering. Numbers and letters are separated with dashes. Let’s consider our example compound again. The longest chain has seven carbon atoms. The chain was numbered from left to right to give the branch the lowest number.

The name of this compound would therefore be 3-methylheptane Had the compound been written the other way we would have numbered it from right to left as shown below.

Regardless of how it is written on paper, the –CH3 (methyl group) is still on the third carbon from the end and the longest chain is still seven carbon atoms long. The two structures below are identical. They are both equally correct ways to draw the structure of 3-methylheptane. is exactly the same as

10-4

Copyright Prof. Lawrence Mailloux, 2021

If a compound contains two or more of the same kind of branch we use the prefixes di, tri, or tetra to indicate the number of each branch. The location of each group is indicated by a number. Every branch gets a number, even if they are both bonded to the same carbon atom. Numbers are separated from each other using commas. In the example below the longest carbon chain is six carbons long. The chain was numbered from left to right to give the branches the lowest numbers. The methyl groups are on the second and third carbon. Therefore the structure below is 2,3-dimethylhexane.

In this next example, both bromine branches are on the third carbon from the end. Therefore, the following compound is 3,3-dibromoheptane. Remember each branch gets its own number thus 3-dibromoheptane is not correct.

If different branches are present, they are listed in alphabetical order. For example, the structure below would be named 3-bromo-3-fluoroheptane, and NOT 3-fluoro-3-bromoheptane.

10-5

Copyright Prof. Lawrence Mailloux, 2021

Prefixes do NOT count toward alphabetization, that is tribromo is alphabetized as a “b” not a “t” and dichloro is alphabetized as an “c” not a “d.” This means that “bromo” will always be listed before “chloro” regardless of how many bromine or chlorine atoms there are in a given compound. The example below would be named 2,2,3-tribromo-4-chloroheptane and NOT 4chloro-2,2,3-tribromoheptane.

I have been using skeletal notation (as opposed to bondline notation) for the past few pages to help you see how the atoms are connected as we learn this new topic. Your exams however will use bondline notation; therefore, the rest of this chapter will use bondline notation. Below are a couple examples using bondline notation. After studying these do the practice problems that follow. The rules for naming alkanes are the foundation for how all organic compounds are named. Therefore, you must master naming alkanes before moving forward. The structure of 3-ethyl-2-methyloctane is given below. Notice that ethyl was listed before methyl.

The structure of 6-ethyl-3,3-dimethylnonane is shown below. Notice that ethyl is first because the “di” prefix does not count toward alphabetization. Also the longest continuous chain in this compound was not written horizontally. Always look carefully to find the longest chain as it can take unusual turns.

10-6

Copyright Prof. Lawrence Mailloux, 2021

Problem 1 – Name the following alkanes

Naming Cycloalkanes Up until now all the examples we have looked at have been what are known as open chain compounds. It is however possible for the carbon atoms in a compound to bond to each other so as to form a ring. Chemists call such compounds cycloalkanes. Because of the carbon-carbon bond needed to close the ring, cycloalkane have two less hydrogen atoms than their open chain counter parts and have the formula CnHn. When naming cycloalkanes, the ring is always the parent chain. We will only work with rings with up to eight carbon atoms in them. Larger rings are possible. It is also possible for a compound to have more than one ring; however we won’t consider such compounds in this class. Below are the structures of the cycloalkanes we will be learning about in this class.

10-7

Copyright Prof. Lawrence Mailloux, 2021

Name

Structure

Formula

cyclopropane

C3H6

cyclobutane

C4H8

cyclopentane

C5H10

cyclohexane

C6H12

cycloheptane

C7H14

cyclooctane

C8H16

The rules for naming cycloalkanes depends upon the number of branches the compound has. If a ring only has one branch no number is needed as the location of the branch is understood to be position number one. Below are a few examples:

bromocyclobutane

methylcyclopentane

ethylcyclohexane

If a ring has two branches, the branch that comes first alphabetically is given the number one. The ring is then numbered in the direction that gives the second branch the lowest possible number. The names of the branches are listed in alphabetical order regardless of their numbering.

10-8

Copyright Prof. Lawrence Mailloux, 2021

For example, the structure below is 1-ethyl-3-methylcyclohexane:

And below we have the structure of 1-ethyl-2-isoproplycyclopentane

If three or more branches are present, the ring is numbered to give all the branches the lowest possible numbers. As always, the branches are listed in alphabetical order. Consider the structure below:

To give all the branches the lowest numbers possible the ethyl branch on top is given the number one. Then the ring is numbered clockwise as this will give chlorine the number two and the bromine the number four. The branches are then listed alphabetically giving the name 4-bromo2-chloro-1-ethylcyclohexane. Numbering the ring counterclockwise would be incorrect as this would give the chlorine atoms the five position instead of two.

4-bromo-2-chloro-1-ethylcyclohexane NOT 4-bromo-5-chloro-1-ethylcyclohexane 10-9

Copyright Prof. Lawrence Mailloux, 2021

Problem 2 – Name the following cycloalkanes

10-10

Copyright Prof. Lawrence Mailloux, 2021

Naming Alkenes If a compound has less than the maximum number of hydrogen atoms it is said to be unsaturated. Unsaturated compounds contain one or more multiple bonds. Compounds that contain one or more double bonds are known as alkenes, while compounds with one or more triple bonds are known as alkynes. The names of alkenes end in “ene,” while the names of alkynes end in “yne.” For simplicity, we will only deal with compound that contain one multiple bond. The formation of multiple bonds reduces the number of hydrogens in the compound. Alkenes have the general formula CnH2n while alkynes have the formula CnH2n-2. The rules for naming alkenes are similar to those for naming alkanes. The first step is to identify the parent chain, which for alkenes and alkynes is the longest continuous chain of carbons that contains the double or triple bond. The parent chain is then numbered to give the multiple bond the lowest number. Notice that, unlike with alkanes, it is the multiple bond that is given the lowest number and not the branches. This is because the multiple bond is the part of the molecule that controls the chemistry of alkenes and alkynes, and it is therefore more important than the branches. Below are the structures of the two possible isomers for butane. Functional groups are atoms in a compound bonded in a certain way that gives them predictable chemical reactivity. Double and triple bonds are examples of functional groups. We will learn about many more functional groups (such as alcohols) in later chapters. In each case, the functional group is always given the lowest possible number. For an example let us name the alkene shown below.

10-11

Copyright Prof. Lawrence Mailloux, 2021

First, we must find the parent chain which is the longest continuous chain that contains the double bond. We then number the chain to give the double bond the lowest number. In this case we number from right to left.

= 6-ethyl-5-methyl-3-octene

Because the double bond is located between carbons 3 and 4, the parent chain is 3-octene, that is the number indicates the carbon atom on which the double bond starts. The groups are then listed alphabetically giving 6-ethyl-5-methyl-3-octene.

Other than numbering to give the double bond the lowest number, instead of the branches, naming alkenes is exactly like naming alkanes. If multiples of the same kind of branch are present, we use prefixes. Consider the compound shown below.

First we find the parent chain, the longest continuous carbon chain that contains the double bond. Remember, as is the case with this example, the longest chain does not necessarily run horizontally. The chain is then numbered so the double bond gets the lowest number. Because three methyl groups are present we use the prefix “tri.”

= 3,4,5-trimethyl-3-heptene

10-12

Copyright Prof. Lawrence Mailloux, 2021

Problem 3 – Name the following alkenes

Naming Cycloalkenes Sometimes the carbon chain containing the double bond forms a ring. When this happens, the compound is known as a cycloalkene. Cycloalkenes are named similarly to cycloalkanes except that with cycloalkenes the ring is always numbered such that the double bond is between carbons one and two. The rest of the ring is then numbered such that the branches have the lowest possible numbers; the branches are listed alphabetically in the name. Consider the following example:

10-13

Copyright Prof. Lawrence Mailloux, 2021

The double bond must lie between carbons 1 and 2. This gives us two possible numbering options as shown below. Which do we pick?

Do we number it

or

?

In this case we chose the clockwise numbering as it puts the methyl groups at the 3 and 5 positions instead of the 4 and 6 positions. Therefore

is named 3,5-dimethylcyclohexene

If either direction gives the same numbers, go in the direction that gives the branch that occurs first alphabetically the lowest number. For example in the compound below we would number to give the bromine a lower number than chlorine because “b” comes before “c” in the alphabet.

is numbered

and NOT

Using the correct numbering we arrive at the correct name 3-bromo-6-chlorocyclohexen and NOT 6-bromo-3-chlorocyclohexene, a common wrong answer from numbering backwards.

10-14

Copyright Prof. Lawrence Mailloux, 2021

Problem 4 – Name the following cycloalkenes

Naming Alkynes Alkynes contain triple bonds and have the formula CnH2n-2. Their names end in “yne.” They are named exactly like alkenes. That is, they are numbered such that the triple bond has the lowest number and branches are listed alphabetically. The only difference is that the name will end in “yne,” to indicate that the compound is an alkyne.

10-15

Copyright Prof. Lawrence Mailloux, 2021

For example, consider the structure below. We number this compound from right to left to give the triple bond the lowest number. The branches are then listed in alphabetical order. The compound shown below is named 5-ethyl-4,4-dimethyl-3-heptyne. Remember the prefix “di” does not count towards alphabetization.

is named 5-ethyl-4,4-dimethyl-2-heptyne

Rings containing triple bonds (cycloalkynes) are very unstable and very uncommon. Therefore, we will not concern ourselves with them in this class. Problem 5 – Name the following alkynes

10-16

Copyright Prof. Lawrence Mailloux, 2021

Benzene and Aromatic Compounds A six carbon ring with alternating single and double bonds is known as benzene. For reasons that go far beyond the scope of this class, benzene rings are special and are chemically very different than alkenes. Because benzene rings are so commonly seen in organic compounds they can be drawn several different ways. Multiple equivalent structures of benzene are shown below.

is the same

as is the same as

Benzene rings occur frequently in organic compounds. For example, below are the structures of several different drugs, all of which contain a benzene ring. From left to right they are: methamphetamine, Efavirenz (Sustiva) an NNRTI, Raltegravir (an integrase inhibitor)

Small molecules containing benzene rings often have strong distinctive odors are sometime called aromatic compounds. We will not be doing any naming or reactions of benzene in this course. Most of the examples so far have been structure to name. You will also expected to be able to go from name to structure. Do problem six to practice going the other direction.

10-17

Copyright Prof. Lawrence Mailloux, 2021

Problem 6 – Draw structure for the following compounds 4-ethyl-2,2-dimethylnonane 3-bromo-2-chloro-2-methyl-3-ethylhexane 2,3-dimethyl-2-hexene 2,3,4-tribromo-1-octene 2-bromo-1-chloro-3-iodocyclohexane 3-bromo-2-chloro-4-iodocyclohexene 3-bromo-3-methyl-1-heptyne

Answers to in Chapter Problems 10-18

Copyright Prof. Lawrence Mailloux, 2021

Problem 1 – Name the following alkanes 2,4-dimethyloctane

3-ethyl-2-methylheptane

3-bromo-5-chloro-4-ethylheptane

2,3,3-trichlorohexane

3-isoproplyheptane

Problem 2 – Name the following cycloalkanes 10-19

Copyright Prof. Lawrence Mailloux, 2021

1,2-dimethylcyclohexane

1-ethyl-3-methylcyclohexane

3-ethyl-1,1-dimethylcyclopentane

1-bromo-2-isopropylcyclobutane

1-bromo-2-chloro-3-fluorocyclopropane

10-20

Copyright Prof. Lawrence Mailloux, 2021

Problem 3 – Name the following alkenes 4,5-dimethyl-2-heptene

5,6-dimethyl-2-heptene

4-ethyl-3-methyl-1-heptene

2-bromo-5-chloro-3-hexene

4-isoproply-3-methyl-2-octene

10-21

Copyright Prof. Lawrence Mailloux, 2021

Problem 4 – Name the following cycloalkenes

3-ethylcyclohexene

2,3-dimethylcyclohexene

3-chlorocyclobutene

4-isopropyl-3-methylcyclopentene

1,2,3-trimethylcyclopentene

10-22

Copyright Prof. Lawrence Mailloux, 2021

Problem 5 – Name the following alkynes

4-isopropyl-5-methyl-2-octyne

4,4,5-trimethyl-2-heptyne

4,5-dimethyl-2-heptyne

4-ethyl-5-methyl-1-hexyne

5-bromo-4-chloro-2-octyne

10-23

Copyright Prof. Lawrence Mailloux, 2021

Problem 6 – Draw structure for the following compounds 4-ethyl-2,2-dimethylnonane

3-bromo-2-chloro-2-methyl-3-ethylhexane

2,3-dimethyl-2-hexene

2,3,4-tribromo-1-octene

2-bromo-1-chloro-3-iodocyclohexane

3-bromo-2-chloro-4-iodocyclohexene

3-bromo-3-methyl-1-heptyne

10-24

Copyright Prof. Lawrence Mailloux, 2021

Chapter 11 Reactions of Alkenes, Alkenes, and Alkynes As mentioned previously all medications are organic compounds made up mainly of carbon and hydrogen with a few other atoms such as sulfur, nitrogen, oxygen, or halogens. Chemists that specialize in the making of carbon compounds are know as organic chemists. Just like a chef makes a meal by combining knowledge of techniques learned over many years with recipes; an organic chemist knows how to use chemical reactions to turn one compound into another. Furthermore, just as a chef has many basic recipes memorized, organic chemists learn dozens of fundamental reactions. By combining, these reactions scientists can build a wide range of organic compounds much like a Lego set can be used to make anything from a pirate ship to a castle. In this class you will learn about how many different types of compounds react. Keep in mind that while you will only be learning the basics of these reactions, in most cases scientists have a detailed understanding of how and why these reactions work. If you wish to understand more about the details of these reactions you would need to take several more chemistry courses. In this course we survey the basics of these reactions, as it is reactions like these that are used to make nearly every drug (legal or illegal) that you can think of. By learning these chapters, we are giving you an idea of the science behind how the pharmaceuticals you will dispense are made as well as how the chemical reactions in your body operate.

Functional Groups Chemists divide organic reaction up by functional group. A functional group is a grouping of atoms with predictable chemical reactivity. You will learn about roughly a dozen functional groups in this class. Double bonds and triple bonds are examples of functional groups.

11-1

Copyright Prof. Lawrence Mailloux, 2021

Reactions of Alkanes Because alkanes lack any specific functional groups, they do not undergo very many chemical reactions. One very important and common reaction that alkanes can undergo is combustion. In a combustion reaction the hydrocarbon reacts with O2 in the air to form carbon dioxide and water vapor. You have seen these reactions before, here is an example from Chapter Six for the combustion of butane, found in cigarette lighters. 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O To balance these reactions the only difference now is that instead of writing the problem statement using formulas like this C4H10 + O2 →

?

you will now be presented with the problem with the structure of the compound drawn out. + O2 →

?

To balance this equation all you need to do is count the number of carbons and hydrogens in the structure and then balance as normal. Remember that everywhere there is a bend or end in a structure there are hidden carbon atoms with enough hydrogens attached to give each carbon four bonds. Consider the following example for the combustion of octane, a component of gasoline; what are the products of the following reaction? +

O2 →

Examining the structure you should count that there are eight carbons atoms and 18 hydrogen atoms. Therefore, I simply rewrite the problem as C8H18 + O2 → CO2 + H2O and then balance as explained in Chapter Six. 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

11-2

Copyright Prof. Lawrence Mailloux, 2021

Problem 1 – Determine the formula for the following hydrocarbons.

=

=

=

Problem 2 – Complete and balance the following reactions.

______

+ _____ O2 →

______

+ _____ O2 →

______

+ _____ O2 →

11-3

Copyright Prof. Lawrence Mailloux, 2021

In addition to burning, alkanes can also under go reactions with halogens. These reactions are known as halogenation reactions and they consist of substituting a hydrogen atom on the alkane for a halogen atom. Simply mixing an alkane with a halogen will not result in a chemical reaction. In order to get the reaction started (recall activation energy from Chapter Six) energy in the form of either intense heat (Δ) or light (hν, UV) in needed. The abbreviation hν is commonly used in chemistry to indicate high energy light. In real life, the different halogens (F2, Cl2, Br2, and I2) do not all react exactly the same and some hydrogen atoms in alkanes are more reactive than others. These are complexities that we will ignore in this class. In all the problems we will consider small symmetrical molecules so it will not matter which hydrogen atom you chose to swap for a halogen. Let us consider an example. If ethane is exposed to chlorine in the presence of intense light a halogenation reaction will occur.

+ Cl2

𝑈𝑉

→

?

To determine the product of this reaction we simply swap any hydrogen atom for a chlorine atom. The other halogen atom bonds with the replaced hydrogen atom to form HCl as a byproduct.

+ Cl2

𝑈𝑉

+ H-Cl

→

Here is an example in bondline notation 𝑈𝑉

+ Br2 →

+ H-Br

11-4

Copyright Prof. Lawrence Mailloux, 2021

Problem 3 – give products for the following halogenation reactions

+ F2

∆

→ ?

𝑈𝑉

+ Cl2 →

ℎ𝑣

+ Br2 →

+ I2

?

?

∆

→ ?

Reactions of Alkenes In alkenes, it is the double bond functional group that is the site of chemical reactivity. Alkenes can undergo many reactions known as addition reactions in which new atoms are added to the double bond; we will learn about four such reaction in the class. One of the most important addition reactions an alkene can undergo is the addition of hydrogen to the double bond. These reactions, known as hydrogenation reactions involve the addition of H2 to a double bond. This has the effect of converting an alkene into an alkane, that is it converts an unsaturated compound into a saturated compound. You can think of it as a double bond eraser. These reactions are important in nutrition, we will learn more about hydrogenation reactions when we learn about saturated and unsaturated fats in Chapter 15. To make a hydrogenation reaction occur a transition metal catalyst is needed. A catalyst is a chemical that makes a reaction go faster by lowering the activation energy needed to get the reaction started. Only a very small amount of a catalyst is needed so that are not shown as part of the balanced chemical equation, but rather are indicated above the reaction arrow. Living systems (your body) use natural catalysts called enzymes that we will learn more about in Chapter 16. Common transition metal catalysts are platinum (Pt), nickel (Ni), and palladium (Pd).

11-5

Copyright Prof. Lawrence Mailloux, 2021

Let’s consider an example, below is the hydrogenation reaction of propene.

+ H2

𝑁𝑖

→

Or using bondline notation

+ H2

𝑁𝑖

→

Problem 4 – Give products for the following reactions 𝑃𝑡

+ H2

→ ?

+ H2

→ ?

+ H2

+ H2

𝑁𝑖

𝑃𝑑

→

?

𝑁𝑖

→ ?

Other addition reaction are also possible. When a halogen X2 (F2, Cl2, Br2, I2) is added to an alkene the halogen adds to the double bond. No catalyst, heat, or energy required. Such a reaction is known as a halogenation reaction. Below is an example of a halogenation reaction

+ Cl2

→

Or in bondline notation

+ Cl2

→

11-6

Copyright Prof. Lawrence Mailloux, 2021

Problem 5 – Give products for the following reactions

+ F2

→ ?

+ Cl2

→ ?

+ Br2

→ ?

+ I2

→ ?

So far the addition reaction we have considered involved adding two of the same atom to the double bond. In the last two addition reactions we will look at this will not be the case. The addition of both a hydrogen and a halogen atom to the double bond (H-F, H-Cl, H-Br, or H-I) is known as a hydrohalogenation reaction. As with halogenation reactions, no catalyst or extra energy is needed. Below is an example of a hydrohalogenation reaction.

+ H-Cl

→ ?

When determining the products of these types of reactions chemists follow a rule known as Markovnikov’s rule. To determine which carbon gets the hydrogen and which one gets the halogen. Markovnikov’s rule states that the hydrogen atom goes to the carbon atom that already has the most hydrogens to begin with; that is, the rich get richer.

11-7

Copyright Prof. Lawrence Mailloux, 2021

In the case of our example reaction the central carbon has less attached hydrogens and thus gets the chlorine atom

+ H-Cl

Carbon has one hydrogen -gets the Cl

→

Carbon has two attached hydrogens -gets the H

When using bondline notation the newly added hydrogen atom is understood and is not written.

+ H-Cl

→

If both carbon atoms have the same number of hydrogen atoms it does not matter which carbon gets the hydrogen and which gets the halogen. For example, in the reaction below, both carbon atoms making up the double bond have one attached hydrogen atom, therefore placing the halogen on either side of the double bond gives the correct product. In real life both products would be obtained as a mixture, therefore either answer would be accepted. both are

+ HBr

acceptable

→

answers

11-8

Copyright Prof. Lawrence Mailloux, 2021

Problem 6 – give products for the following reactions

+ HF

→ ?

+ HCl

→ ?

+ HBr

→ ?

+ HI

→ ?

The last alkene addition reaction we will consider involves the addition of water to the double bond in what is known as a hydration reaction. The products of hydration reactions are known as alcohols, which are the subject of the next chapter. To make it easier for you to understand these reaction I will sometimes write H2O as H-OH. Hydration reactions require an acid catalyst as indicated by H+ written above the arrow. Just like with hydrohalogenation reactions, the carbon atom with more hydrogens gets the new H and the other carbon gets the OH. In the example below, the central carbon has less hydrogens and therefore gets the OH.

+ H-OH

𝐻+

→

Or using bondline notation

+ H-OH

→

As with hydrohalogenation reactions, if both carbon atoms have the same number of attached hydrogens it does not matter who get the H and who gets the OH.

+ HOH

is exactly

𝐻+

→

the same as

11-9

Copyright Prof. Lawrence Mailloux, 2021

Problem 7 – give products for the following reactions 𝐻+

+ HOH

→

𝐻+

+ HOH

→

𝐻+

+ HOH

→

𝐻+

+ HOH

→

?

?

?

?

Reactions of Alkynes In alkynes the triple bond is the center of chemical reactivity. While alkynes can undergo many different reactions we will only concern ourselves with one, hydrogenation. Hydrogenation reactions of alkynes are exactly like hydrogenation reactions of alkenes except that more hydrogen is needed to reduce a triple bond to a single bond than a double bond, hence the two before the H2 in the reactions that follow. You can think of these reactions as triple bond erasers. The product of the hydrogenation of an alkyne is simple the corresponding alkane. For example

+ 2 H2

𝑁𝑖

→

Problem 8 – give products to the following reactions 𝑃𝑡

+ 2 H2

→ ?

+ 2 H2

→ ?

𝑁𝑖

11-10

Copyright Prof. Lawrence Mailloux, 2021

Answers to In Chapter Problems Problem 1 – Determine the formula for the following hydrocarbons.

= C7H16

= C10H22

= C7H14

Problem 2 – Complete and balance the following reactions.

______

+ __11__ O2 →

7 CO2 + 8 H2O

___2__

+ __31__ O2 →

20 CO2 + 22 H2O

___2__

+ __21__ O2 →

14 CO2 + 14 H2O

11-11

Copyright Prof. Lawrence Mailloux, 2021

Problem 3 – give products for the following halogenation reactions

+ F2

∆

+ H-F

→

𝑈𝑉

+ H-Cl

+ Cl2 →

ℎ𝑣

+ H-Br

+ Br2 →

+ I2

∆

+ H-I

→

Problem 4 – Give products for the following reactions 𝑃𝑡

+ H2

→

+ H2

→

+ H2

+ H2

𝑁𝑖

𝑃𝑑

→

𝑁𝑖

→

11-12

Copyright Prof. Lawrence Mailloux, 2021

Problem 5 – Give products for the following reactions

+ F2

→

+ Cl2

→

+ Br2

→

+ I2

→

Problem 6 – give products for the following reactions

+ HF

→

+ HCl

→

+ HBr →

+ HI

→

11-13

Copyright Prof. Lawrence Mailloux, 2021

Problem 7 – give products for the following reactions

+

+ HOH 𝐻 →

+

+ HOH 𝐻 →

+

+ HOH 𝐻 →

+

+ HOH 𝐻 →

Problem 8 – give products to the following reactions 𝑃𝑡

+ 2 H2

→

+ 2 H2

→

𝑁𝑖

11-14

Copyright Prof Lawrence Mailloux, 2021

Chapter 12 Reactions of Alcohols The Alcohol Functional Group In Chapter 11 we saw that by adding water to a carbon-carbon double bond (hydration) we can attach an –OH (hydroxyl group) to an organic molecule. Molecules that have a hydroxyl group bonded to one of its carbon atoms (C-OH) are called alcohols. An alcohol is an example of a functional group - a part of a molecule with specific chemical reactivity. The double and triple bonds in alkenes and alkynes from the prior chapter are also functional groups. This means that a hydration reaction turns an alkene into an alcohol. The alcohol can then be further reacted upon resulting in additional functional groups that we will learn about later in this chapter. This is what gives organic chemistry its power. By carefully combining different chemical reactions in sequence chemists can modify the functional groups of a molecule and custom build a molecule with a specific desired structure. The three dimensional structure of a molecules determines its chemical properties. Pharmaceuticals are a very important application of this fact. The biological activity of a molecule is determined by its three dimensional structure. Every drug/medication you have ever taken, legal or illegal, is an organic chemical who’s effects are a result of its three dimensional structure. Organic chemistry is NOT memorization, although in a survey course like this it may seem like all memorization. Chemists do not just memorize chemical reactions. What they do is learn a large number of reactions and then creatively connect them together to build a final desired molecule. This is just like cooking. Chefs do have to memorize a large number of kitchen procedures and facts, however what makes someone a chef is the ability to creatively combine and apply these pieces of information together into a good tasting meal. In this chapter we will learn about the chemical reactions that alcohols undergo, however before we can do so, we must first learn how to name and classify alcohols. Luckily, this is not difficult if you have learned the naming covered in Chapter 10.

12-1

Copyright Prof Lawrence Mailloux, 2021

Naming Alcohols Naming alcohols is very similar to naming alkenes, except instead of the C=C being the focus of attention, the –OH group is the functional group of interest. To tell the reader that the compound is an alcohol, the suffix is changed to “ol.” Consider the alkene below:

Here I numbered the chain from right to left to give the C=C the lowest number. In this case the double bond starts on the second carbon, thus the parent chain is 2-octene. This numbering puts the bromine on the 7th carbon and chlorine on the 6th carbon. Listing the groups in ABC order gives the name: 7-bromo-6-chloro-2-octene. Now consider the following alcohol below.

Here we do exactly the same thing except now we number to give the –OH (that is the hydroxyl functional group) the lowest number. Because the –OH lies on the 2nd carbon the parent chain is octanol, where the suffix “ol” indicates the compound is an alcohol. The groups are then listed in ABC order as usual giving the name 7-bromo-6-chloro-2-octanol. If the –OH is bonded to a carbon in a ring we have a cyclic alcohol. Cyclic alcohols are named exactly like cycloalkenes, except instead of starting by at the double bond, we start at the cabon where the –OH is bonded to. Then we number in the direction that gives and branches the lowest numbers just like with cycloalkenes in Chapter 10.

12-2

Copyright Prof Lawrence Mailloux, 2021

Consider the following examples below. Just as was the case with cycloalkenes, we will always start numbering at the functional group (in the case of alcohols, the –OH) so we do NOT put a “1” in the name. Notice how similar the two names are.

3-methylcyclohexene

3-methylcyclohexanol

All organic chemistry naming follows the same basic rules covered thus far, which are: 1. Find longest continuous chain that contains the functional group of interest 2. Number to give the functional group the lowest number a. Alkanes have no functional groups, number to give branches the lowest number. b. All other compounds we number to give the functional group the lowest number c. Remember there is no “1” in names when i. The functional group is attached to a ring ii. The functional group is always at the end of a chain as with aldehydes and carboxylic acids covered later in this chapter 3. List branches in ABC order when writing name 4. Use ABC order to break ties 5. Change parent chain suffix to indicate the functional group present a. Alkanes – “ane” b. Alkenes – “ene” c. Alkynes – “yne” d. Alcohol – “ol” e. Aldehyde – “al” (covered later in this chapter) f. Ketone – “one” (covered later in this chapter) g. Carboxylic acids – “oic acid” (covered later in this chapter) Do the practice problems on the next page.

12-3

Copyright Prof Lawrence Mailloux, 2021

Problem 1 – Name the following alcohols

12-4

Copyright Prof Lawrence Mailloux, 2021

Classification of alcohols The outcome of a chemical reaction depends upon the location of the –OH functional group on the carbon backbone. This is related to the number of carbon atoms directly attached to the carbon atom to which the –OH is attached. Using this system alcohols fall into one of three categories: primary (1°), secondary (2°), and tertiary (3°). In a primary (1°) alcohol the –OH lies at the end of a carbon chain. This means the carbon directly attached to the OH is directly bonded to one other carbon atom. Below are some examples of primary alcohols.

In a secondary (2°) alcohol the –OH lies in the middle of the carbon chain. This means the carbon directly attached to the OH is directly bonded to two other carbon atoms. Below are some examples of secondary alcohols.

In a tertiary (3°) alcohol the –OH lies at in the middle of the carbon chain and there is also a branch on the carbon with the –OH. This means the carbon directly attached to the OH is directly bonded to three other carbon atoms. Below are some examples of tertiary alcohols.

Problem 2 – Classify the alcohols in problem one as primary, secondary, or tertiary. Now that we can name and classify alcohols, let’s learn about the kinds of reactions they can undergo.

12-5

Copyright Prof Lawrence Mailloux, 2021

Reactions of Alcohols Dehydration Reactions – Turning Alcohols into Alkenes Alcohols can under go a variety of different reactions. The first we will consider is the reverse of the reaction we ended the prior chapter with, hydration. Not surprisingly, the name of the reverse reaction is called dehydration, and it involves the loss of a water molecule. When an alcohol molecule is heated in the presence of acid, the bond to the –OH group, and the bond to an adjacent H atom are both broken. The released H and OH combine to make water, leaving behind a double bond as shown below.

∆/𝐻 +

+

→

H-OH

Let’s consider some examples. Sometimes it does not matter which side of the –OH carbon that the double bond forms on as the products are exactly the same thing. ∆/𝐻 +

→

+

H-OH

+

H-OH

Is exactly the same as… ∆/𝐻 +

→

12-6

Copyright Prof Lawrence Mailloux, 2021

Below is a similar example involving a cyclic alcohol.

∆/𝐻 +

→

+

H-OH

+

H-OH

Is exactly the same as…

∆/𝐻 +

→

In the prior examples it did not matter where the double bond ended up. This is not always the case. In the reactions below, it does matter which side of the –OH carbon the double bond ends up on as the two products are not the same. For an example consider the oxidation of 2-methyl-3pentanol. The question then is does the reaction proceed like this

∆/𝐻 +

+

→

H-OH

Or is it like this…

∆/𝐻 +

+

→

H-OH

If the two reactions above were performed one would likely get a mixture of both products, with the first product being the major product making up most of the product mixture and the second product only making up a small amount of the product mixture. When a situation like the above occurs the alkene with double bond on the carbon with the branch is the major product. A chemist would say, “the more highly substituted alkene” is the major product. On the next page is an example using a cyclic alcohol.

12-7

Copyright Prof Lawrence Mailloux, 2021

When 2-methylcyclohexanol dehydrates the major product (best/correct answer) is

∆/𝐻 +

→

+

H-OH

+

H-OH

And the minor product is

∆/𝐻 +

→

Chemists have learned how to steer such reactions towards one outcome or another, however this level of detail is not necessary in this course. To keep things simple just give the major product as that would be the main chemical produced. Lastly remember carbon only forms four bonds. In the reactions below only one product is possible as the other product would give carbon more than four bonds. This is because the neighboring carbon needs to lose a hydrogen to allow for the double bond to form. If one of the neighboring carbons does not have a hydrogen to lose than the double bond cannot form on that side of the –OH carbon. Consider the following example. When 2,2-dimethyl-3-pentanol dehydrates the reaction that takes place is

∆/𝐻 +

→

+

H-OH

+

H-OH

And not

∆/𝐻 +

→

As there is no H to be removed on the second carbon from the left. Additionally, carbon can only form a maximum of four bonds making the second product impossible.

12-8

Copyright Prof Lawrence Mailloux, 2021

The same rules apply to cyclic alcohols. When 3,3-dimethycyclohexanol dehydrates the reaction is that takes place is

∆/𝐻 +

→

+

H-OH

+

H-OH

And not

∆/𝐻 +

→

As doing so would give an impossible product.

12-9

Copyright Prof Lawrence Mailloux, 2021

Problem 3 – Give the products of the following reactions ∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

12-10

Copyright Prof Lawrence Mailloux, 2021

Oxidation and Reduction (Redox) Reactions We are now going to learn about a totally different class of chemical reactions called oxidation/reduction reactions. Historically people observed that when some substances react with oxygen in the air; not surprisingly these reactions came to be known as oxidation reactions. You are already familiar with many oxidation reactions in your daily life: •

Iron rusting

•

Lettuce turning brown

•

Apple darkening after being bit into

•

Guacamole turning dark

•

Fruit juice turning into vinegar after sitting for a long time.

These reactions were observed first because they are so common. Eventually scientists learned that oxidation reactions were much more than simply reaction with oxygen. It turned out that these reactions are actually quite complicated, and that not just oxygen can make something “oxidize.” The name however has stuck. Chemicals that cause oxidation are called oxidizing agents. Some oxidizing agents include: •

Oxygen itself (O2)

•

Sodium hypochlorite (NaClO) – bleach

•

Oxyclean®

•

Hydrogen peroxide (H2O2)

•

Chromic acid (H2CrO4) – will use in CHM1033L

Oxidation reactions are even taking place when your body burns food to generate the calories needed to power your daily activities. These and other oxidation-reduction reactions are quite complicated and we will only cover the basics here.

12-11

Copyright Prof Lawrence Mailloux, 2021

To begin we will learn about three new functional groups that contain a carbon double bonded to an oxygen (C=O) which is known as a carbonyl group. The carbonyl is formed from the alcohols –OH group. The first two (aldehydes and carboxylic acids) always come at the end of a carbon chain and result from oxidation of primary alcohols.

Aldehyde

Carboxylic Acid

In the third functional group, ketones, the carbonyl occurs in the middle of the carbon chain. Ketones come from secondary alcohols. Below is a generic structure of a ketone. Tertiary alcohols as we will learn do not oxidize.

Ketone Below are some examples of aldehydes in bond line notation

Below are some examples of carboxylic acids in bond line notation

Below are some examples of ketones in bond line notation

12-12

Copyright Prof Lawrence Mailloux, 2021

Oxidation of Primary Alcohols Whenever an organic compound oxidizes the following things happen: •

Two hydrogen atoms are lost

•

The number of bonds to the oxygen atom increases (carbonyl, C=O, forms)

•

Sometimes another atom of oxygen is added entirely (vigorous conditions)

We use the symbol [O] over the arrow to denote oxidation. Let’s see what happens when we do this to a primary alcohol. Consider the following example with ethanol, the alcohol in beverages. Below is the structure of ethanol.

First we remove two hydrogen atoms , giving the following intermediate structure

Next we increase the number of bonds to the oxygen atom from one to two. This gives carbon back its required four bonds, while also giving oxygen its required two bonds and forming the carbonyl. What we are left with is an aldehyde. Mild oxidation of primary alcohols gives aldehydes.

In bond line notation this reaction would look like this [𝑂]/ 𝑚𝑖𝑙𝑑

→

12-13

Copyright Prof Lawrence Mailloux, 2021

For another example, consider the oxidation of 3-ethyl-4-methyl-1-pentanol shown below.

[𝑂]/ 𝑚𝑖𝑙𝑑

→

Notice the carbon skeleton remains unchanged and our product is an aldehyde. To name this compound we number the chain starting at the aldehyde group and list branches in alphabetical order as usual. To indicate that the compound is an aldehyde we change the suffix to “al.” Because an aldehyde is always at the end of a chain, it will always be at the one position. Because of this, the number “1” does NOT appear in the name. The aldehyde shown above is 3ethyl-4-methylpentanal. Problem 4 – Name the following aldehydes

12-14

Copyright Prof Lawrence Mailloux, 2021

Aldehydes are not the end of the story. If the oxidation of a primary alcohol is allowed to go on long enough, or under strong enough (vigorous) conditions, then the aldehyde that is formed will oxidize further. Another oxygen atom will be added between the carbonyl and the H atom on the end of the chain to give a carboxylic acid. Vigorous oxidation of primary alcohols gives carboxylic acids.

[𝑂]/ 𝑣𝑖𝑔𝑒𝑟𝑜𝑢𝑠

→

We name carboxylic acids exactly like aldehydes except the suffix is changed to “oic acid.” As with naming aldehydes, carboxylic acids can only be at the end of a chain; therefore we don’t put a one in the name. The product of the reaction above is therefore named 3-ethyl-4methylpentanoic acid. Problem 5 – Name the following carboxylic acid

12-15

Copyright Prof Lawrence Mailloux, 2021

In this book we will use the words “mild” and “vigorous” to indicate what kind of conditions were present. If you are starting with a primary alcohol and see the word “mild,” then the product will be an aldehyde, if you see the word “vigorous” than the product will be a carboxylic acid. Lastly, if we start with an aldehyde the product will be the corresponding carboxylic acid. In this case the words “mild” or “vigorous” will not be used as carboxylic acid is the only possible product. For example, 3-ethyl-4-methylpentanal oxidizes to 3-ethyl-4-methylpentanoic acid as shown below.

[𝑂]

→

12-16

Copyright Prof Lawrence Mailloux, 2021

Oxidation of Secondary Alcohols to Ketones Now that we have finished learning about oxidation of primary alcohols, let’s consider what would happen if we did the same thing to a secondary alcohol. Consider the following secondary alcohol, 2-propanol, better known as isopropyl, or rubbing, alcohol.

First we remove two H atoms giving the following intermediate structure

Then we will increase the number of bonds to oxygen, giving carbon its four bonds back. The result is a ketone, named 2-propanone, better known as acetone (a kind of nail polish remover).

In bond line notation the oxidation of 2-propanol would look like this [𝑂]

→

For another example consider the oxidation of 4,5-dimethyl-2-hexanol shown below. The carbon skeleton is unchanged. We simply turn the alcohol into the corresponding ketone.

[𝑂]

→

12-17

Copyright Prof Lawrence Mailloux, 2021

To name this ketone we number in the direction that gives the carbonyl the smallest number, in this case starting on the right as shown below. The groups are then listed using prefixes in ABC order as usual. The suffix “one” indicates that this compound is a ketone. Below is the structure of 4,5-dimethyl-2-hexanone.

If the secondary alcohol was a cyclic alcohol, then the product will be a cyclic ketone. For example consider the oxidation of 2-methylcyclohexanol oxidizes shown below.

[𝑂]

→

To name a cyclic ketone, we always start numbering at the carbonyl. Because the carbonyl is always in the one position, the number one does NOT appear in the name. We number in the direction that gives the branches the lowest numbers and list the branches alphabetically in the name. The suffix is changed to “one.” For the reaction above the product would be numbered as shown below and named 2-methylcyclohexanone.

12-18

Copyright Prof Lawrence Mailloux, 2021

You may have noticed that I did not use the words “mild” or “vigorous” in any of the proceeding examples. That’s because ketone is the end of the line for these oxidation reactions if we try to oxidize a ketone further we will get no reaction. For example neither 4,5-dimethyl-2-hexanone or 2-methylcyclohexanone, shown below, will oxidize.

[𝑂]

→

[𝑂]

→

no reaction

no reaction

Problem 6 – Name the following aldehydes

12-19

Copyright Prof Lawrence Mailloux, 2021

Tertiary Alcohols Don’t Oxidize With a tertiary alcohol, the carbon with the –OH is in the middle of the carbon chain and there is also a branch on the –OH carbon. Carbon-carbon bonds are quite strong and don’t break easily. Furthermore, there are no hydrogens on the carbon with the hydroxyl groups to remove. Therefore, if we were to try and oxidize a ketone by following our usual steps we would arrive at the following impossible structure where carbon has five bonds:

→

→

For the reaction above to occur, carbon would need to form five bonds. As you should be well aware by now, carbon only forms four bonds. As a result tertiary alcohols do not oxidize. Below are two examples of tertiary alcohols 2-methyl-2-pentaol and 1-methylcyclohexanol.

If one were to try an oxidize them nothing would happen as shown below. [𝑂]

→

[𝑂]

→

no reaction

no reactions

12-20

Copyright Prof Lawrence Mailloux, 2021

Reductions of Aldehydes and Ketones into Alcohols The opposite of an oxidation reaction is a reduction reaction. Reduction reactions are indicated by [H] above the arrow. When a reduction reaction takes place, the products are the exact reverse of the corresponding oxidation reaction. Two H’s are added to the structure, and the double bond to oxygen (C=O, carbonyl) is reduced to a single bond with a hydroxyl group making an alcohol. Below are some examples of reduction reactions An aldehyde reduces to a primary alcohol [𝐻]

→

A ketone reduces to a secondary alcohol.

[𝐻]

→

I can’t give you any everyday examples of reduction reactions because our atmosphere, being full of oxygen, is an oxidizing (therefore not reducing) environment. Just because we don’t see reduction reactions happening in our daily lives does not mean they don’t happen. Reduction reactions are commonly performed in laboratories all over the world. In addition, both oxidation and reduction reactions are happening right now in your body without you even realizing it. Because this is only an introductory survey course we will not concern ourselves with reduction reactions in this class.

12-21

Copyright Prof Lawrence Mailloux, 2021

Problem 7 – Give products for the following redox reactions. If no reaction occurs write “no reaction” [𝑂]/𝑚𝑖𝑙𝑑

→

[𝑂]/𝑚𝑖𝑙𝑑

→

[𝑂]/𝑚𝑖𝑙𝑑

→

[𝑂]/𝑚𝑖𝑙𝑑

→

[𝑂]/𝑚𝑖𝑙𝑑

→

[𝑂]/𝑚𝑖𝑙𝑑

→

[𝑂]/𝑣𝑖𝑔𝑒𝑟𝑜𝑢𝑠

→

[𝑂]/𝑚𝑖𝑙𝑑

→

[𝑂]/𝑣𝑖𝑔𝑒𝑟𝑜𝑢𝑠

→

12-22

Copyright Prof Lawrence Mailloux, 2021

Problem 8 – Give products for the following redox reactions. If no reaction occurs write “no reaction” [𝑂]

→

[𝑂]

→

[𝑂]

→

[𝑂]

→

12-23

Copyright Prof Lawrence Mailloux, 2021

Answers to In Chapter Questions Problems 1 and 2 – Naming and classification of alcohols 4-methyl-2-pentanol, secondary 2,4-dimethyl-2-pentanol, tertiary

5-bromo-6-chloro-2-heptanol, secondary

2-methylcyclohexanol, secondary

1-methylcyclohexanol, tertiary

2-bromo-5-chlorocyclopentanol, secondary

2,3,5-tribromocyclopentanol, secondary

5-ethyl-4-methyl-3-heptanol, secondary

5-ethyl-3-methyl-3-heptanol, tertiary

12-24

Copyright Prof Lawrence Mailloux, 2021

Problem 3 – Give the products of the following reactions ∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

∆/𝐻 +

→

12-25

+

H-OH

+

H-OH

+

H-OH

+

H-OH

+

H-OH

+

H-OH

+

H-OH

+

H-OH

Copyright Prof Lawrence Mailloux, 2021

Problem 4 – Name the following aldehydes butanal

2,3-dimethylbutanal

4-ethyloctanal

4-bromo-3-chloropentanal

2,3,4-trimethylhexanal

12-26

Copyright Prof Lawrence Mailloux, 2021

Problem 5 – Name the following carboxylic acid 3,4-dimethylhexanoic acid

4-ethyl-3-methylhexanoic acid

3,4,4-trimethylhexanolic acid

Problem 6 – Name the following aldehydes 2-hexanone

4,5-dimethyl-2-hexanone

3-chloro-2-methyl-2-octanone

2-ethyl-3,6-dimethylcyclohexanone

2-bromo-5-chlorocyclopentanone

12-27

Copyright Prof Lawrence Mailloux, 2021

Problem 7 – Give products for the following redox reactions. If no reaction occurs write “no reaction” [𝑂]/𝑚𝑖𝑙𝑑

→

[𝑂]

→

[𝑂]

no reaction

→

[𝑂]

→

[𝑂]

no reaction

→

[𝑂]

→

[𝑂]/𝑣𝑖𝑔𝑒𝑟𝑜𝑢𝑠

→

[𝑂]/𝑚𝑖𝑙𝑑

→

[𝑂]/𝑣𝑖𝑔𝑒𝑟𝑜𝑢𝑠

→

12-28

Copyright Prof Lawrence Mailloux, 2021

Problem 8 – Give products for the following redox reactions. If no reaction occurs write “no reaction” [𝑂]

→

[𝑂]

no reaction

→

[𝑂]

→

[𝑂]

no reaction

→

12-29

Copyright Prof. Lawrence Mailloux, 2021

Chapter 13 Reactions of Carboxylic Acids In the prior chapter we learned that the vigorous oxidation of a primary alcohol (or oxidation of an aldehyde) yields a carboxylic acid as shown in the example below.

In this chapter, we will learn about reactions of carboxylic acids to make even more functional groups. We will then apply this information towards the end of the textbook when we learn about applications of these compounds to biochemistry, the HIV virus, and how modern medications allow most HIV positive people to live healthy normal lives. In this chapter we will encounter three new functional groups: esters, amines, and amides. The structures of these three functional groups are shown below.

Ester

Amine

Amide

We will now consider some reactions of carboxylic acids, and introduce these three new functional groups as we encounter them.

13-1

Copyright Prof. Lawrence Mailloux, 2021

Ionization The simplest reaction that a carboxylic acid can undergo is to ionize. Back in Chapter 7 we learned that acids are chemicals the release H+ ions when dissolved in water. This means acids are electrolytes as you saw with the lightbulb demo in experiment six of CHM1033L. Below is the ionization of acetic acid, (the acid in vinegar, HC2H3O2) that occurs when placed in water. The acidic hydrogens that ionize are always written at the start of the formula. Acetic acid has one acidic hydrogen, underlined above. It is the release of these ions that makes the solution conductive allowing the light bulb to light up. When the H+ ion is lost by an acid it does not leave alone. Rather the H+ that is lost is taken up by a nearby water molecule making H3O+, which is called the hydronium ion. HC2H3O2(l) + H2O(l) ↔ H3O+(aq) + C2H3O2-(aq) Showing all of the carbons and hydrogens we can see the water taking the acidic hydrogen off the carboxylic acid group making H3O+ (hydronium ion) and C2H3O2- (acetate anion). The arrows below show the movement of electrons.

The above reaction is often simplified and written as follows, it’s understood H+ means H3O+. + 𝐻2 𝑂

HC2H3O2(l) ↔

H+(aq) + C2H3O2-(aq)

Using bond line notation this would be written as follows:

Here is another example, also in bond line notation.

13-2

Copyright Prof. Lawrence Mailloux, 2021

Problem 1 – Give products of the following ionization reactions. 𝐻2 𝑂

+

↔

𝐻2 𝑂

+

↔

𝐻2 𝑂

+

↔

Ionization and Neutralization reactions Another simple reaction that a carboxylic acid can undergo is to be neutralized by a base. Earlier in the course we learned that the reaction of an acid with a base produces a salt and water. Consider following example where the neutralization of acetic acid with sodium hydroxide makes sodium acetate (a salt) and water. HC2H3O2 + NaOH → NaC2H3O2 + H2O In this reaction, the OH on the NaOH steals an H from the HC2H3O2 molecules forming NaC2H3O2 and water. Below is this reaction with all the carbons and hydrogens shown.

In bond line notation this reaction would be written as follow.

13-3

Copyright Prof. Lawrence Mailloux, 2021

Here are a couple more examples using bond line notation. The base can be written above the arrow as in the prior example, or it can be written to the left of arrow.

Problem 2 – Give products of the following neutralization reactions. 𝑁𝑎𝑂𝐻

+

→

𝐾𝑂𝐻

+

→

𝐿𝑖𝑂𝐻

+

→

13-4

Copyright Prof. Lawrence Mailloux, 2021

Naming and Classification of Amines Amines are organic chemicals that have an N-H bond. They are the organic cousins of ammonia, NH3. Like alcohols, they can be classified as primary, secondary, or tertiary by the number of carbons directly attached to the N atom. The small simple amines we will be learning about are very easy to name. One simply lists the branches off the N-atom in alphabetical order. Examples of the structures and names of each type of amine are given below.

Ammonia

Methylamine

Ethylmethylamine

Primary (1°)

Secondary (2°)

Ethyldimethylamine Tertiary (3°)

Problem 3 – Name and classify the following amines.

One feature that all amines have in common is the lone pair of electrons on the nitrogen atom. These electrons allow ammonia, and amines, to steal hydrogen atoms from acids. That is amines, like ammonia, are basic. Below is the reaction that occurs when NH3 is dissolved in water. NH3(g) + H2O(l) → NH4OH(aq)

13-5

Copyright Prof. Lawrence Mailloux, 2021

Showing all the carbons and hydrogens we can see how the nitrogen uses its lone pair of electrons to take an H off of a nearby water molecule to make ammonium hydroxide, a weak base.

The only part of the molecule that reacted was the long pair on nitrogen. The surrounding hydrogen atoms did not do anything. Therefore, if we replace one or more of the hydrogens on NH3 with carbon groups the resulting amines will also be basic as they will still have a lone pair of electron on the nitrogen atom. Below is the ionization reaction of methylamine showing all the carbons and hydrogens.

Like all bases, amines react with acids to form a salt. In the case of amines, the product is known as an amine salt. The reactions below illustrate this fact.

Neutral amine Poor H2O solubility

Amine Salt Good H2O Solubility

Why should be care about amine salts? The answer is solubility. Amines are oily chemicals that have a nasty fish like smell. Because they are oily, they do not dissolve in water very well. Amine salts on the other hand are ionic compounds, and ionic compounds do dissolve well in water. Because the human body is made largely of water, any medication that you take must have some level of water solubility for it to be able to do its job within your body. To improve the water solubility of medications, they are commonly sold as their amine salts. Because these salts are made by reacting the amine with HCl (HBr can also be used), the product name often has •HCl written after it. Many medications are supplied as their amine salts. One such medicine 13-6

Copyright Prof. Lawrence Mailloux, 2021

is diphenhydramine, better known by its brand name Benadryl®. The structure of diphenhydramine is given below, with its amine group indicated.

Being a neutral amine, diphenhydramine is not very soluble in water. To improve its water solubility, it is sold as its amine salt which is made by reacting diphenhydramine with hydrochloric acid to give diphenhydramine•HCl as the product.

Diphenhydramine•HCl

Diphenhydramine Neutral amine / poor H2O solubility

Amine salt / good H2O solubility

Looking at a package of Benadryl® you will notice the HCl written after the drug name.

13-7

Copyright Prof. Lawrence Mailloux, 2021

Many other over the counter and prescription medications are sold as HCl salts. Look at your medication bottles in your home and you may find HCl (or •HCl) written on them. This means the medication if being provided as its HCl salt. Problem 4 – Give the amine salt products of the following reactions. 𝐻𝐶𝑙

→

𝐻𝐶𝑙

→

𝐻𝐶𝑙

→

Problem 5 – Velafaxine (Effexor®)and Loperamide (Immodium AD ®) are two medications that are taken in their amine salt form. The neutral amine structures of these compounds are given below. Give the structure of these medications in their amine salt form.

𝐻𝐶𝑙

→

𝐻𝐶𝑙

→

13-8

Copyright Prof. Lawrence Mailloux, 2021

Condensation Reactions Condensation reactions are reactions where two compounds bond together and in the process spit out a water molecule. This happens whenever an alcohol, or a primary or secondary amine, reacts with a carboxylic acid. The H from the alcohol or amine, combines with the OH from the carboxylic acid to form water. The rest of the molecules bond together to make two new functional groups, esters and amides. When the product is an ester, these reactions are sometimes called esterification reactions, which simply means reaction that makes esters. Amide forming reactions don’t have a special name.

Making Esters: Carboxylic Acid + Alcohol → Ester + Water When a carboxylic acid reacts with an alcohol, the products an ester and water

Carboxylic Acid

Alcohol

Ester

Water

For specific examples, consider the following reactions

Unlike amines which have a nasty fish smell, esters smell pleasant. The smell of many fruits, as well as natural and artificial fragrances and flavors, are esters. In CHM1033L you will make and smell esters in Experiment 11. You will only need to be able to give the products of such reactions, you don’t need to remember the specific smells.

13-9

Copyright Prof. Lawrence Mailloux, 2021

Problem 6 – Give the products of the following esterification reactions. 𝐻+

+

+

→

𝐻+

+

+

→

𝐻+

+

+

→

𝐻+

+

+

→

Making Amides: Carboxylic Acid + Amine → Amide + Water When a carboxylic acid reacts with a 1° or 2° amine, the products are an amide and water

Carboxylic Acid

1°/2° Amine

`

Amide

For a specific examples consider the following reactions.

13-10

Water

Copyright Prof. Lawrence Mailloux, 2021

Problem 7 – Give the amide products of the following condensation reactions. 𝛥

+

→

+

→

+

→

+

→

𝛥

𝛥

𝛥

+

+

+

+

Esters and amides are important functional groups. We will learn more about their importance in biochemistry in the upcoming chapters. For now, we will learn some of the reactions they can undergo.

13-11

Copyright Prof. Lawrence Mailloux, 2021

Hydrolysis Reactions Breaking Down Amides with Water Just as esters and amides are formed by the elimination to water (condensation), under the right conditions, adding water to an ester or amide can break it back apart into the carboxylic acid and alcohol/amine from which they were made. The term used to describe reactions that involve breaking something down with water are known as hydrolysis reactions. Hydrolysis reactions are simply the reverse of condensation reactions. The human body is largely made of water. As a result, many chemical reactions that take place in your body are hydrolysis reactions. When you eat food, your body uses hydrolysis reactions to break the food down. Let’s look at the hydrolysis of amides to make amines and carboxylic acids first. Consider the generic example below:

Amide

Water

Carboxylic Acid

For a specific examples, consider the following reactions.

13-12

1°/2° Amine

Copyright Prof. Lawrence Mailloux, 2021

Problem 8 – Give the products of the following hydrolysis reactions. +

H2O

+

H2O

+

H2O

+

H2O

𝛥/𝐻 +

+

→

𝛥/𝐻 +

+

→

𝛥/𝐻 +

+

→

𝛥/𝐻 +

+

→

13-13

Copyright Prof. Lawrence Mailloux, 2021

Breaking Down Esters with Water With esters the situation is a bit more complicated. Depending upon if the reaction is carried out under acidic or basic conditions, the form of the carboxylic acid product will change slightly. Under acidic conditions, the hydrolysis of an ester is simply the reverse of the condensation reaction that formed the ester.

Ester

Water

Carboxylic Acid

For a specific examples, consider the following reactions

13-14

Alcohol

Copyright Prof. Lawrence Mailloux, 2021

Problem 9 – Give the products of the following hydrolysis reactions under acidic conditions. +

H2O

+

H2O

+

H2O

+

H2O

𝛥/𝐻 +

+

→

𝛥/𝐻 +

+

→

𝛥/𝐻 +

+

→

𝛥/𝐻 +

+

→

If the hydrolysis reaction takes place under basic conditions the carboxylic acid that forms will be immediately neutralized by all the base present. As a result, the product of the base hydrolysis of an ester is an alcohol and the salt of the carboxylic acid. Generically the reaction would look like this:

Ester

Water

Carboxylic Acid

13-15

Alcohol

Copyright Prof. Lawrence Mailloux, 2021

For a specific examples, consider the following reactions

Soap is made by hydrolyzing esters found in fats and oils as vividly portrayed in the movie Fight Club when the Narrator (played by Edward Norton) and his alter ego Tyler Durden break into a liposuction clinic to steal human fat to make soap. For this reason, base hydrolysis of an ester is sometimes called a saponification (soap making) reaction. We will learn more about this in the Chapter on lipids (Chapter 15)

Problem 10 – Give the products of the following hydrolysis reactions under basic conditions. +

H2O

+

H2O

+

H2O

+

H2O

𝑁𝑎𝑂𝐻

→

𝐿𝑖𝑂𝐻

→

𝐾𝑂𝐻

→

𝑁𝑎𝑂𝐻

→

13-16

+

+

+

+

Copyright Prof. Lawrence Mailloux, 2021

Answers to in Chapter Problems Problem 1 – Give products of the following ionization reactions. 𝐻2 𝑂

↔

𝐻2 𝑂

↔

𝐻2 𝑂

↔

+

H+

+

H+

+

H+

Problem 2 – Give products of the following neutralization reactions. 𝑁𝑎𝑂𝐻

→

𝐾𝑂𝐻

→

𝐿𝑖𝑂𝐻

→

13-17

+

H2O

+

H2O

+

H2O

Copyright Prof. Lawrence Mailloux, 2021

Problem 3 – Name and classify the following amines. Diethylmethylamine (tertiary)

Ethylpropylamine (secondary)

Diisopropylamine (secondary)

Ethylamine (Primary)

Problem 4 – Give the amine salt products of the following reactions. 𝐻𝐶𝑙

→

𝐻𝐶𝑙

→

𝐻𝐶𝑙

→

13-18

Copyright Prof. Lawrence Mailloux, 2021

Problem 5 – Velafaxine (Effexor®)and Loperamide (Immodium AD ®) are two medications that are taken in their amine salt form. The neutral amine structures of these compounds are given below. Give the structure of these medications in their amine salt form.

𝐻𝐶𝑙

→

𝐻𝐶𝑙

→

Problem 6 – Give the products of the following esterification reactions. +

+

+

+

𝐻+

→

𝐻+

→

𝐻+

→

𝐻+

→

13-19

+

H2O

+

H2O

+

H2O

+

H2O

Copyright Prof. Lawrence Mailloux, 2021

Problem 7 – Give the amide products of the following condensation reactions. 𝛥

+

→

+

→

+

→

+

→

𝛥

𝛥

𝛥

Problem 8 – Give the products of the following hydrolysis reactions. +

H2O

+

H2O

+

H2O

+

H2O

𝛥/𝐻 +

+

→

𝛥/𝐻 +

+

→

𝛥/𝐻 +

+

→

𝛥/𝐻 +

+

→

13-20

+

H2O

+

H2O

+

H2O

+

H2O

Copyright Prof. Lawrence Mailloux, 2021

Problem 9 – Give the products of the following hydrolysis reactions under acidic conditions. +

H2O

+

H2O

+

H2O

+

H2O

𝛥/𝐻 +

→

𝛥/𝐻 +

→

𝛥/𝐻 +

→

𝛥/𝐻 +

→

+

+

+

+

Problem 10 – Give the products of the following hydrolysis reactions under basic conditions. +

H2O

+

H2O

+

H2O

+

H2O

𝑁𝑎𝑂𝐻

→

𝐿𝑖𝑂𝐻

→

𝐾𝑂𝐻

→

𝑁𝑎𝑂𝐻

→

13-21

+

+

+

+

Copyright Prof. Lawrence Mailloux, 2021

Chapter 14 Carbohydrates Carbohydrates and Monosaccharides Carbohydrates are compounds made of carbon, hydrogen, and oxygen. There are many different kinds of carbohydrates, we will learn about a few of these in this chapter. All carbohydrates are made of simple sugar ring units known as monosaccharides. The most well-known monosaccharide carbohydrate is glucose (aka. dextrose or blood sugar) which is made by plants from water and carbon dioxide via photosynthesis; the balanced reaction for which is shown below. 6 CO2 + 6 H20 → C6H12O6 + 6 O2 Structurally carbohydrates are polyhydroxylaldoses or polyhydroxylketoses. Try saying, or even pronouncing, those ten times fast! Let’s us break these crazy words down. “Poly” means many, and “hydroxyl” is an OH group. Lastly aldoses are carbohydrates with an aldehyde functional group in them and ketoses are carbohydrates with a ketone in them. Therefore, this simply means carbohydrates are carbon compounds with many OH groups and either an aldehyde or ketone functional group in them. Scientific jargon may be unwieldly, but it’s usually systematic and can often be broken into pieces. In addition to being classified as aldoses or ketoses, carbohydrates are also classified according to the number of carbons they contain as the table below summarizes. Number of Carbons

General Name

Aldose

Ketose

3

Triose

Aldotriose

Ketotrisoe

4

Tetrose

Aldotetrose

Ketotetrose

5

Pentose

Aldopentose

Ketopentose

6

Hexose

Aldohexose

Ketohexose

7

Heptose

Aldoheptose

Ketoheptose

14-1

Copyright Prof. Lawrence Mailloux, 2021

For our first example, consider the structure of D-glucose (sugars are asymmetrical molecules can exist in both left (L) and right handed (D) forms however only the D form occurs naturally) shown below. CHO H

OH

HO

H

H

OH

H

OH CH 2OH

Carbohydrate structures of this type are called Fischer projections, The CHO at the top is an abbreviation for an aldehyde functional group. D-glucose is an example of an aldohexose. Below is the structure of D-fructose, a ketohexose, and an important monosaccharide that we will learn about later in this chapter. CH 2OH O HO

H

H

OH

H

OH CH 2OH

Problem 1 – Below is the structure of D-galactose. Along with D-glucose, and D-fructose, Dgalactose is one of the three important monosaccharides that we will be learning about in this chapter. Classify this carbohydrates according to the table discussed above. CHO H

OH

HO

H

HO

H

H

OH CH 2OH

14-2

Copyright Prof. Lawrence Mailloux, 2021

Before we go further let me state that you do NOT need to memorize exact structures of any of these monosaccharides. You may be asked things about their structures, such as similarities or differences between them as is often asked in CHM1033L, but not exact structures. Problem 2 – Classify the following carbohydrates according to the table discussed above. CHO

CH 2OH

HO H

O

HO

H

H

HO

H

H

OH

OH

H

OH

H

OH

CHO

CH2OH

CH 2OH

CH 2OH

Forming Rings, Fischer and Haworth Projections We live in an incredible universe. Even something as simple seeming as sugar is actually quite complex when studied in detail. Up until now we have been seeing carbohydrates as open chain molecules in their Fischer projections. However, sugars have more complex shapes than the prior structures show. Under the correct conditions a hydroxyl (OH) group can react with the carbonyl group in either an aldehyde or a ketone. When this happens, a new bond is formed between the alcohol oxygen and the carbon from the carbonyl. The hydrogen that was initially on the alcohol OH moves up to the oxygen from the carbonyl. OH

O C

C

R

+

H

O

C

C

C

R

O

C

There is no requirement that the carbonyl and OH group be on different molecules. If a molecule is long and flexible enough it is possible for an OH on one end of the molecule to react with a C=O on the other end, closing the molecule up into a ring. With carbohydrates this happens with monosaccharides with five or more carbons, that is pentoses, hexoses, and heptoses spend most

14-3

Copyright Prof. Lawrence Mailloux, 2021

of their time as closed rings. The closed form of a monosaccharide is known as its Haworth projection, while the open chain form is called a Fischer projection. These reactions are reversible, as indicated by the ↔ arrow. This means that monosaccharides can close into rings and then open back up over and over again. Real monosaccharides spend most of their time as closed rings, therefore we will use Haworth projections frequently in the rest of this chapter. Below are the structures of the three most important monosaccharides (glucose, galactose, and fructose) in both their Fischer and Haworth forms. In the far-right column is a simplified cartoon like abbreviation for these monosaccharides that I will use later in this chapter or when I write on the board. We will learn more about these three monosaccharides throughout this chapter. Monosaccharide

Fischer Projection

Haworth Projection

Cartoon Form

CHO H

H OH

OH

HO

H

H

D-Glucose

O

HO

H

OH

H

HO H

H

OH

Gl

OH

H

OH

CH 2OH CHO H HO

OH OH

OH

H

H

D-Galactose

O

Ga

H HO

H

H

HO H

H

OH

OH

H

OH

CH 2OH CH 2OH O HO

D-Fructose

CH2OH

H

H

OH

H

OH

CH2OH O

H

OH

H

OH HO

CH 2OH

14-4

F

H

Copyright Prof. Lawrence Mailloux, 2021

When a monosaccharide such as glucose, galactose, or fructose closes to make a ring the OH on the right-hand side can end up either on the top or bottom of the ring. These two different forms of a sugar ring are known as anomers; the carbon with the OH is called an anomeric carbon. When the OH on the anomeric carbon is down the molecule is designated as alpha (α), when the OH is up, the molecule is designated as beta (β). The α and β forms of glucose are shown below. H OH

H OH H

H

O

O

HO

HO H H

OH

HO

H

HO

H

OH

OH

H

OH

α-D-glucose

H

β-D-glucose

The importance of alpha vs beta will become apparent in the section on Disaccharides below. Before we move onto the disaccharides there are a few things you are expected to know about the monosaccharides glucose, fructose, and galactose first. Glucose (C6H12O6) is the most important carbohydrate of all. As was mentioned at the start of the chapter glucose is blood sugar. Healthy blood glucose levels are important. What constitutes healthy blood sugar levels depends upon how the test is administered with regards to when the person ate last. For example, according to Mayo Clinic, after an overnight fast blood sugar levels less than 100 mg/dL (5.6 mmol/L) is normal. Numbers from 100 to 125 mg/dL (5.6 to 5.9 mmol/L) is prediabetic, and >126 mg/dL (7 mmol/L) is considered diabetic.2 You don’t need to memorize these specific numbers as you will learn more about these topics in your nursing courses from more qualified instructors. Fructose also has the formula C6H12O6. Because many fruits, juices, and honey are rich in fructose, it is also known as “fruit sugar.” Unlike glucose that is immediately ready for use, fructose is first metabolized by the liver into glucose before your body can burn it.

2. Mayo Clinic Patient Care and Health Information, Diabetes. https://www.mayoclinic.org/diseases-conditions/diabetes/diagnosis-treatment/drc-20371451 (Accessed 2/17/2021)

14-5

Copyright Prof. Lawrence Mailloux, 2021

The last of our three monosaccharides Galactose also has the formula C6H12O6. It is found as one of the two monosaccharides in lactose which itself if found in milk. Hence galactose is sometimes called “milk sugar.” Like fructose, your body first metabolizes galactose into glucose before it is used. Although all three monosaccharides have the same formula, in this book when you see C6H12O6 written assume it refers to glucose unless otherwise stated as it is the most important of the monosaccharides.

Disaccharides When two monosaccharides bond together they split out a water molecule and form what is known as a disaccharide. Taken together mono and disaccharides make up what are known as simple carbohydrates. In this class we will learn about three important disaccharides: sucrose, maltose, and lactose.

Sucrose Sucrose is a very common carbohydrate, however you probably know it better by its more common name, table sugar. Chemically sucrose is made by the combination of an α-D-glucose and a β-D-fructose as shown in the following reaction. Amazingly the sugar you put on your cereal is really this giant chemical! H OH

CH2OH

CH2OH O

H

O H

HO H

H

+

HO

H

D-glucose

→ OH

OH

HO H

OH

α-1,2-glycosidic linkage

H

OH

D-fructose

Sucrose

Like all rings, there are rules for numbering sugar rings. You do not need to worry about the details of these rules, only the results of them discussed here. In the case of sucrose, the glucose and fructose are bonded together at the 1 and 2 positions via an α-1,2-glycosidic linkage.

14-6

Copyright Prof. Lawrence Mailloux, 2021

This reaction is reversible. Sucrose can be easily hydrolyzed to release a glucose and a fructose. Hydrolysis means “breaking with water.” However, if one just sits sucrose in water the reaction is so slow that it basically does not happen. To makes things go faster a catalyst is needed. Catalysts are chemicals the make reactions go faster but are not consumed like a normal reactant. Usually only a small amount of catalyst is needed to speed up a reaction. In CHM1033L you will use heat and acid (H+) as a catalyst to speed up the hydrolysis of sucrose when you do your carbohydrates experiment. In addition to acid, your body also uses special enzymes to speed up this reaction as sucrose is broken down and used by your body for energy. Enzymes are nature’s catalysts and are important for speeding up many chemical reactions inside living organisms, digestion is just one common example. We will learn more about enzymes in Chapter 16 on proteins. Below is the reaction for the hydrolysis of sucrose, it is simply the reverse of the prior reaction that joined the glucose and fructose together in the first place.

H OH

CH2OH H

H2 O →

O

H

+

HO

OH

H

HO H H

Sucrose

CH2OH O

H

OH OH

D-glucose

14-7

OH HO

H

D-fructose

Copyright Prof. Lawrence Mailloux, 2021

Maltose Another disaccharide is maltose. Maltose is made from two α-D-glucoses bonded together. Due to the numbering rules of sugars the bond ends up between the 1 and 4 carbons on the two rings. This downward pointing bond that holds these two glucoses together is an α-1,4-glycosidic bond. The reaction of two glucoses to make a maltose is shown below. α-1,4-glycosidic linkage H OH

H OH H

O

+

HO H

HO H

OH

H

H

D-glucose

→ H

HO H

OH

O

HO

H

OH OH

D-glucose

maltose

And as with sucrose, maltose can also be hydrolyzed to give two glucoses as shown below H OH

H OH

+ H2O →

H

+

O

HO H

HO H

D-glucose

OH

O H

HO H

OH

H

maltose

H HO

OH

H

OH

D-glucose

Lactose The last disaccharide we are interested in is lactose. Lactose is made from a β-D-galactose and an α-D-glucose and is found in milk (human milk is about 8% lactose, cow milk about 5%). Per the numbering of the sugar rings we again arrive upon a 1,4 glycosidic bond holding the two rings together. In this case however the upward pointing β-D-galactose results in an upward pointing β-1,4-glycosidic bond. The reason we care about this is because not all people are able to digest lactose. As we have learned in prior chapters, the 3D structure of a compound is critical to its biochemical properties. Thus, having a different shape from its alpha cousins, this beta linkage requires its own enzyme, lactase, to be broken down in the body. Although many people continue to produce lactase throughout their entire life (adult consumption of milk makes humans unique amongst animals) some people do not make sufficient lactase to allow them to digest lactose and are said to be lactose intolerant. The lactose sitting undigested in the body 14-8

Copyright Prof. Lawrence Mailloux, 2021

producing cramps and diarrhea. Below is the reaction of D-glucose and D-galactose to make lactose.

β-1,4-glycosidic linkage

→

+

D-glucose

D-galactose

Lactose

As with other monosaccharides, lactose can also be hydrolyzed to release the monosaccharides from which it was made

+ H2O →

Lactose

+

D-glucose

D-galactose

Polysaccharides Polymers are large molecules made from many small repeating units called “mers.” There are many different polymers, both natural and manmade in existence. Examples of polymers include: plastic, rubber, latex, proteins, DNA and some carbohydrates, the topic of this section. Very long Carbohydrates are polymers made by connecting anywhere from hundreds to thousands of glucose rings together into giant chains. These long chain carbohydrates polymers are known as complex carbohydrate, in contrast to mono and disaccharides which are called simple carbohydrates. In this class we will learn about three important complex carbohydrates: starch, glycogen, and cellulose.

14-9

Copyright Prof. Lawrence Mailloux, 2021

Starch Starch is the way in which plants store their energy. During photosynthesis glucose is made from carbon dioxide, water, and sunlight. The energy from the sunlight is stored in the chemical bonds in the glucose rings that are stored bonded together into two large polysaccharides. The two polysaccharides that make up starch, amylose, and amylopectin, are quite similar. A few of their similarities and differences are described below. Amylose is made up of anywhere from 250 to about 4000 glucose rings connected together. Amylose has a straight chain structure due to having only α-1,4-glycosidic linkages. The structure of a short segment of amylose is shown below:

Amylopectin has a very similar structure to amylose, however every 20 to 25 sugar rings a α1,6-glycosidic linkage sneaks in. This causes amylopectin to have a branched structure.

occasional α-1,6-glycosidic linkage causes branches

Partial hydrolysis of starch breaks the amylose and amylopectin down into smaller and smaller chains of glucose rings called oligosaccharides (the prefix “oligo,” means few). Complete hydrolysis breaks it all the way down to nothing but a large number of glucose rings.

14-10

Copyright Prof. Lawrence Mailloux, 2021

Glycogen Just like plants, animals also need to store excess glucose (energy) for later use. Animals do this using a polysaccharide known as glycogen. Glycogen is an animal’s reserve of quick energy, as opposed to fat which is for long term energy storage. Human can hold about 1,500 Calories in quick reserves this way. After exhausting these supplies, one will feel tired. Runners call this “hitting the wall,” and cyclists refer to it as “bonking.” With adequate training one can increase the size of their glycogen stores. Additionally, one can temporarily increase their body’s reserves by eating starchy meals in the time leading up to competition through a process known as “carb loading.” Structurally glycogen is very similar to amylopectin, except that it has α-1,6-glycosidic linkages, even more frequently (every 10 or so units). This causes glycogen to have be even more branched than amylose. As is the case with starch, complete hydrolysis of glycogen produces glucose.

Cellulose Plants use the sugars they make for more than just energy storage. Cellulose is a polysaccharide made from glucose rings that are bonded together by β-1,4-glycosidic linkages as shown below.

β-1,6-glycosidic linkage make cellulose indigestible

As we have seen with lactose, breaking beta linkages requires different enzymes than breaking alpha linkages. This means that cellulose is very resistant to being broken down making it a perfect building material for plants. In fact even termites alone can’t even break down cellulose; it’s actually bacteria in the stomach of termites that break down cellulose. Fiber is made of cellulose and because we are unable to digest it, fiber does not contribute Calories (recall subtracting the fiber in Chapter 3). 14-11

Copyright Prof. Lawrence Mailloux, 2021

This concludes the directly lecture portion of this chapter. The section following the next set of practice problems contains more information on carbohydrates intended to help you with your CH1033L carbohydrates experiment. Problem 3 – What monosaccharides make up sucrose? Problem 4 – What monosaccharides make up maltose? Problem 5 – What monosaccharides make up lactose? Problem 6 – In sucrose, the two monosaccharides are held together by _________ linkages. Problem 7 – In maltose, the two monosaccharides are held together by _________ linkages. Problem 8 – In lactose, the two monosaccharides are held together by _________ linkages. Problem 9 – Some people are unable to digest the disaccharide __________ because they lack enough functioning _________ enzyme. These people are said to be __________. Problem 10 – Hydrolysis of the disaccharide sucrose gives what monosaccharide(s)? Problem 11 – Hydrolysis of the disaccharide maltose gives what monosaccharide(s)? Problem 12 – Hydrolysis of the disaccharide lactose gives what monosaccharide(s)? Problem 13 – Complete hydrolysis of starch gives what monosaccharide(s)?

14-12

Copyright Prof. Lawrence Mailloux, 2021

Laboratory Tests for Carbohydrates in CHM1033L Benedict’s Test for Reducing Sugars In CHM1033L you will test for carbohydrates in lab using two different tests, Benedict’s test and the iodine test. Benedict’s is a solution containing copper (II) ions (Cu2+), and like all Cu2+ containing solutions, Benedict’s solution is a clear blue color. When one of these copper (II) ions bumps into an aldehyde group on certain sugars (known as reducing sugars) an oxidation reduction reaction takes place. The aldehyde is oxidized, just like with the oxidation reactions back in Chapter 12. At the same time the blue copper (II) ion is reduced to a cloudy brick red colored copper (I) ion in the form of a copper(I) oxide (Cu2O) precipitate. In general: 𝛥

Cu2+(aq) + Reducing Sugar → Cu2O(s) This reaction is similar to the oxidation reactions you performed in Experiment Nine where you oxidized alcohols using orange Cr6+ (in form of chromic acid, H2Cr2O4). Except now instead of looking for green Cr3+ to indicate the reaction was successful, you will be looking for a cloudy red Cu2O precipitate. Just like how not all alcohols reacted with chromic acid during Experiment Nine, not all sugars will react with benedicts. Only what are known as reducing sugars will react with Benedicts. The following sugars are reducing sugars: 1. All monosaccharides a. Glucose b. Fructose c. Galactose 2. All disaccharides except sucrose a. Maltose b. Lactose c. Sucrose (not a reducing sugar must hydrolyze first)

14-13

Copyright Prof. Lawrence Mailloux, 2021

Sucrose and polysaccharides (like starch) will not react with Benedicts unless they are first hydrolyzed. When sucrose is hydrolyzed it breaks up to release glucose and fructose, both of which are reducing sugars. In the case of starch, hydrolysis produces free glucoses which then react with benedicts. These reactions are depicted below using the cartoon like notation introduced previously. When sucrose is treated with Benedicts, nothing happens

G

F

l

𝐵𝑒𝑛𝑒𝑑𝑖𝑐𝑡𝑠 (𝐶𝑢2+)

→

no reaction, stays clear blue

However, if the sucrose if first hydrolyzed the glucose and fructose monosaccharides are released. These will then react with Benedicts.

G

F

l

𝛥,𝐻+

G

→

𝐵𝑒𝑛𝑒𝑑𝑖𝑐𝑡𝑠 (𝐶𝑢2+)

F

→

l

Cloudy red, Cu2O

Likewise, starch alone will not react with Benedict’s. Below is a short section of a starch polymer molecule. When treated with benedicts nothing happens as the glucoses are not free to react.

G

G

G

l

l

l

𝐵𝑒𝑛𝑒𝑑𝑖𝑐𝑡𝑠 (𝐶𝑢2+)

→

no reaction, stays clear blue

After hydrolysis the glucoses are released…

G

G

l

l

G

l

𝛥,𝐻+

→

14-14

G

G

G

l

l

l

Copyright Prof. Lawrence Mailloux, 2021

…and the liberated glucoses are now free to react with Benedict’s

G

G

G

l

l

l

𝐵𝑒𝑛𝑒𝑑𝑖𝑐𝑡𝑠 (𝐶𝑢2+)

→

Cloudy red, Cu2O

The pictures below show what a negative (left) and positive (right) benedicts test looks like.

Iodine Test The presence of starch in a solution is very easy to detect. When a solution of iodine (which is initially a clear golden amber) is exposed to starch it turn black almost instantly. No heating, or hydrolysis required. 𝐼𝑜𝑑𝑖𝑛𝑒 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 G

G

G

l

l

l

→

14-15

Black

Copyright Prof. Lawrence Mailloux, 2021

The pictures below show what a negative (left) and positive (right) starch test looks like.

Problem 14 – What does Benedict’s reagent test for? What does a positive Benedicts test look like? What does a negative test look like? Problem 15 – What disaccharide does not react with Benedict’s? What can be done to make it react? Problem 16 – Fill in the following narrative: The disaccharide _________ and the polysaccharide __________ do not react with Benedict’s reagent. This means they are not ________ sugars. Although these two compounds do not react with Benedict’s their hydrolysis products will. When the disaccharide _________ is hydrolyzed it is broken down into the monosaccharides _______ and ______ which are both reducing sugars. Or in the case of the polysaccharide _______ complete hydrolysis breaks the complex carbohydrate into many _______ monosaccharides which reducing sugars which are than free to react with Benedict’s. Problem 17 – What does iodine reagent test for? What does a positive iodine test look like? What does a negative iodine test look like?

14-16

Copyright Prof. Lawrence Mailloux, 2021

Answers to In Chapter Problems 1. Aldohexose 2. Ketopentose, aldopentose, aldotetrose 3. Glucose + fructose 4. Two glucoses 5. Glucose + galactose 6. α-1,2 7. α-1,4 8. β-1,4 9. Lactose, lactase, lactose intolerant 10. Glucose + fructose 11. Two glucoses 12. Glucose + galactose 13. Glucose 14. Reducing sugars (all monosaccharide and all disaccharide except sucrose), a positive result is indicated by the solution turning from clear blue to cloudy red. 15. Sucrose, hydrolysis 16. Sucrose, starch, reducing, sucrose, glucose, fructose, starch, glucoses 17. Iodine tests for starch, a positive result is indicated by the solution turning from clear amber to black (too dark to really worry about clear vs cloudy)

14-17

Copyright Prof Lawrence Mailloux, 2021

Chapter 15 Lipids Lipids are a class of nonpolar (hydrophobic, fat soluble) molecules found in many living organisms. In this class we will divide lipids into two categories, those with ester functional groups (fats, waxes, and oils) and those without esters (steroids). Both perform many important functions within your body as described below.

Steroids - Lipids without Esters Steroids are fat-soluble molecules that have a steroid nucleus. The structure of a steroid nucleus is shown below. Any molecule with this basic structure will be a steroid.

All steroids contain this basic core shape. Below are some examples of steroids you may have heard of. From left to right they are testosterone, progesterone (birth control), prednisone (prescription anti-inflammatory medication), and hydrocortisone (OTC anti-itch cream).

Another substance you might have heard of (but probably don’t think of as a steroid) is cholesterol. Below is the structure of cholesterol, as you can see from its structure, it is clearly a steroid.

15-1

Copyright Prof Lawrence Mailloux, 2021

Cholesterol, is a natural steroid that is necessary for health in moderation. Although cholesterol can be obtained from our diet, most cholesterol is made by your body. Cholesterol performs a variety of function in the human body. It is involved in synthesis of vitamin D, the production of hormones, and as a component of the lipid bilayer of cells. Steroids are a great example of how the three-dimensional structure of a molecule determines its biological properties. Drugs/medications (difference is politics) are organized into classes based upon their three dimensional shape. For example, the three molecules shown below are all examples of opiates. Their three dimensional shape allows them to bind to opiate receptors in the brain. This means that all of these molecules have the same basic properties common to all opiates, and are all strong pain killers. They also have the same risks and side effects. All of them can cause itching and constipation and all can be habit forming.

Codeine

Morphine

Heroin

Likewise the following molecules are both amphetamines.

Adderall

Methamphetamine

Like all amphetamines, both Adderall and methamphetamine are powerful stimulants. They both increase concentration and decrease appetite. Both can be addictive, and both have the potential to be abused.

15-2

Copyright Prof Lawrence Mailloux, 2021

This is something that doctors, and pharmacists, learn about in great detail. This is important. For example, the two molecules below are antibiotics of the beta-lactam class.

Penicillin

Amoxicillin

Because they have similar structures this means that they both kill bacteria in the same way. Therefore, if an infection did not respond to penicillin, or if a patient was allergic to penicillin, a doctor would likely not proscribe amoxicillin next. Instead, they would proscribe an antibiotic of a different class such vancomycin or tetracycline, which clearly have different structures, and work differently.

Tetracycline

Vancomycin Later in Chapter 18 we will look at some different classes of antiviral medication that are used to treat HIV. As with antibiotics, antiviral medications biological activity is a direct consequence of their three dimensional structure.

15-3

Copyright Prof Lawrence Mailloux, 2021

Waxes, Fats, and Oils - Lipids with Esters The second category of lipids we will consider are those that have ester functional groups in their structure. These are the lipids that make up most fats, waxes, and oils, such as the olive oil you cook with. Recall that esters are formed when an alcohol and a carboxylic acid bond together. In the process a water molecule is released.

Fats, waxes, and oils are formed in a similar way. The only difference is that the alcohol in question (glycerol/glycerin) has three -OH groups on it and therefore each can bond with up to three different carboxylic acids molecules. Also, the carboxylic acid molecules involved are quite long and are often called long chain fatty acids. When one fatty acid chain is present the product is called a monoglyceride. The reaction below shows the formation of a monoglyceride: O

OH

HO OH

OH

O O

OH

+ H2O

OH

When two fatty acid chains are attached, the product is called a diglyceride:

When three fatty acid chains are attached, the product is called a triglyceride:

15-4

Copyright Prof Lawrence Mailloux, 2021

Believe it or not you have eaten all three of these molecules many times in your life. Both mono and diglycerides are used a food emulsifiers to keep oils from separating in products such as Jiff peanut butter. Fats (butter, lard, bacon grease), oils (olive, canola, peanut), and waxes (beeswax) are all composed a triglycerides. Look at the nutrition label of the foods in your kitchen to see if any of them contain mono or diglycerides.

Types of Fats, Saturated, Unsaturated, and Trans When discussing lipids people usually refer to solids as fats and liquids as oils. Generally, most solid fats (butter, lard, bacon grease...) come from animals while most lipids derived from plants (olive, canola, peanut…) are liquids. This is because animal products tend to be higher in saturated fats (which are solids), while plant products tend to be higher in unsaturated fats (which are liquids). There are exceptions to this of course, coconuts for example are high in saturated fat and thus coconut oil is a solid at room temperature, although it will melt on a hot day. As you may have guessed from the name, saturated fats are lipids in which all the carbon atoms are bonded to the maximum number of hydrogen atoms. That is, they are alkanes and contain only carbon-carbon single bonds. Unsaturated fats on the other hand contain one or more carbon-carbon double bonds. Like all alkenes the double bonds in unsaturated fats can either be arranged as cis or trans, giving rise to cis fats and trans fats. Nearly all naturally occurring unsaturated fats are of the cis variety, so the “cis” part is generally implied. Unlike naturally occurring cis fats, trans fats are synthetically made and are solids at room temperature.

15-5

Copyright Prof Lawrence Mailloux, 2021

To understand why it is that saturated (and trans) fats are solids while cis unsaturated fats are liquids we must consider their structures. Recall from chapter three that in solids the particles are tightly packed in a regular way, while in liquids the particles are able to flow past each other and are not arranged in an orderly way.

Solid

Liquid

Gas

The same is true for lipids. In saturated and trans unsaturated fats the hydrocarbon tails on the lipid molecules stack together nicely like spoons in a drawer. Thus, saturated fats and trans fats are solids at room temperature.

Saturated fat

Unsaturated trans fat

When the double bonds are arranged in a cis orientation the carbon chains are bent and do not pack well. This causes most cis fats (naturally occurring unsaturated fats) to be liquids at room temperature.

Unsaturated cis fat

15-6

Copyright Prof Lawrence Mailloux, 2021

Hydrogenation of Fats and Formation of Trans Fats Recall from chapter 11 that alkenes can be converted to alkanes by adding H2 in the presence of an appropriate catalyst.

In chapter 11 we were working with small molecules, however these same reactions can be performed on any alkene. In the case of lipids one can either fully hydrogenate or partially hydrogenate. In fully hydrogenated lipids, all of the double bonds are converted to single bonds. This transforms an unsaturated fat into a saturated fat. Partial hydrogenation on the other hand only converts some of the double bonds into single bonds. Of the double bonds that remain, some get converted from their natural cis configuration to a trans configuration. That is partial hydrogenation creates trans fats.

Cis unsaturated fat

H2 / Metal Catalyst

O

O

O

O

O

O

O

O

O

O O

O

Fully hydrogenation makes saturated fat

Partial hydrogenation makes trans fat

15-7

Copyright Prof Lawrence Mailloux, 2021

This means that any food product that says “partially hydrogenated…” will contain trans fat. Companies do this because trans fats are solids and have a long shelf life which is important for processed foods that need to last for a long time without refrigeration. This is important to know because trans fats are bad for your health and should be consumed as infrequently as possible. Because people are becoming aware that trans fats are bad for you, the serving size on products are sometimes adjusted such that there is a small enough amount of trans fat present to allow the manufacture to round down to 0 grams trans fat on their label. One well-known food that contains trans fats is powdered nondairy coffee creamer. Because most Americans use more than “one serving” worth of most foods products when they eat this can add up over time. If you put nondairy creamer into your coffee everyday, over time you will eat quite a bit of trans fat. Real milk is healthier, of course it requires refrigeration

15-8

Copyright Prof Lawrence Mailloux, 2021

Saponification – Making Soap Recall from chapter 13 that when an ester is subjected to hydrolysis under basic conditions that the products are an alcohol and a carboxylate salt.

Ester

Carboxylate Salt

Alcohol

This reaction can be performed on any ester, including lipids. When a lipid is subject to base hydrolysis the product is an alcohol with three OH groups on it and three separate long chain carboxylate salt molecules. These long chain salt molecules are soap.

A generic lipid molecule H2O / NaOH

+

Glycerin/glycerol (an alcohol)

salt of a long chain fatty acid (aka. soap)

15-9

Copyright Prof Lawrence Mailloux, 2021

To understand how soap works, let’s take a closer look at a soap molecule. We are only interested in the organic part, the sodium ion is not important here.

A soap molecule, the left side is polar while the long tail is nonpolar Recall that oxygen is a very electronegative molecule which causes it to form polar bonds. This combined with the negative charge on the far left oxygen cause this end of the soap molecule to be polar and attract water. The carbon chain on the other hand is made of only carbon-carbon and carbon-hydrogen bonds which are nonpolar. Thus, the long tail on the right is nonpolar and attracts other non-polar (i.e., oily) substances. To illustrate this soap molecules are often drawn like little tadpoles with the head being polar and the tail nonpolar.

Polar head

------------------------ nonpolar tail --------------------------→

When we use soap, the polar head interacts with the water molecules while the non polar tail interacts with the oily non polar molecules, such as those present on our skin or in oily stains. This allows the soap to act as an intermediary between the water and the oil allowing it to be washed down the drain. The following image (from defeatdd.org) illustrates how this works.

15-10

Copyright Prof Lawrence Mailloux, 2021

Glycerphospholipids and Lipid Bilayers Many cells, and even some viruses like HIV, are surrounded by a lipid bilayer. You likely have seen images of lipid bilayers in your high school or college biology books. Below is a common representation of a lipid bilayer. Notice that the polar heads hangout with the polar heads while the nonpolar tails hangout with the nonpolar tails.

The molecules that make up the lipid bilayer are called glycerphospholipids. Glycerphospholipids have a similar structure to soap, however they each have two nonpolar tails. Below is the structure of a glycerphospholipid.

Because each glycerphospholipid has two nonpolar tails, the tadpole like diagram for them have two tails.

15-11

Copyright Prof Lawrence Mailloux, 2021

End of Chapter Problems 1. Lipids are broken into two categories based upon the presence or absence of esters in their structures. Those without esters are known as ________ and can be recognized by the presence of a(n) ____________ in their structures. Lipids with esters make up the ________ and _______ in our diets. 2. ____________ is an example of a natural steroid that is necessary to vitamin D synthesis. 3. Which of the following molecules would be a steroid?

4. The biological activity of drugs/medications depends upon their ____________. 5. The molecule below is an example of a(n)___________ O O

OH

OH

6. The molecule below is an example of a(n) _____________

7. The molecules in questions five and six are commonly found in food and act as ____________ 8. The molecule below is an example of a(n) ___________.

9. Lipids derived from animals tend to be higher in ____________ and thus are __________ at room temperature while lipids derived from plants tend to be higher is ___________ and thus are _______ at room temperature.

15-12

Copyright Prof Lawrence Mailloux, 2021

10. We usually refer to solid lipids as _______ and liquid lipids as ________. 11. The molecule below is an example of a(n) ___________ and would be more common in animal products. It would likely be a(n) ________ at room temperature

12. The molecule below is an example of a(n)___________ and would be more common in plant products. It would likely be a(n) ________ at room temperature.

13. The molecule below is an example of a(n) __________. A synthetic fat, tt would likely be a(n)_________ at room temperature. O O

O

O

O O

14. When an unsaturated fat is exposed to H2 in the presence of an appropriate metal catalyst a(n) ____________ reaction occurs. These reactions can be performed two ways. ____________ produces only saturated fat while ____________ produces some transfat. 15. Hydrolysis of an ester under _______ conditions makes soap. These reactions are called __________ reactions. 16. On a soap molecule the _______ is polar and interacts with water while the ________ is nonpolar and interacts with _________. 17. Lipid bilayers are made of __________. Their diagrams look similar to those for soap molecules however they have ___________ instead of one.

15-13

Copyright Prof Lawrence Mailloux, 2021

Answers to End of Chapter Problems 1. Steroids, steroid nucleus, fats, oils 2. Cholesterol 3. The molecule shown below is betamethasone, a prescription steroid.

4. Three dimensional shape 5. Monoglyceride 6. Diglyceride 7. Emulsifiers 8. Saturated fat 9. Saturated fat, solid, unsaturated fat, liquid 10. Fats, oils 11. Saturated, solid 12. Unsaturated, liquid 13. Transfat, solid 14. Hydrogenation, full hydrogenation, partial hydrogenation 15. Basic, saponification 16. Head, tail 17. Glycerphospholipids, two

15-14

Copyright Prof. Lawrence Mailloux, 2021

Chapter 16 Proteins and Enzymes The 20 Amino Acids Proteins, like carbohydrates, are polymers. However, unlike carbohydrates where the repeating molecules are glucose sugar rings, proteins are made of large numbers of amino acids bonded together. All amino acids have an amine group on one side and a carboxylic acid group on the other side. Below is the structure of a generic amino acid.

There are 20 different amino acids in nature. Of these 20 amino acids, our bodies can make ten of them, while the other ten of them must be obtained from our diet. The ten amino acids that must be obtained from our diet are called essential amino acids. Foods that supply all of the essential amino acids (like beans and rice) are called complete proteins. These 20 amino acids differ by the identity of the side chain attached at the R position (in chemistry the letter R represents a generic side chain). The amino acids are classified according to the kind of side chain that is attached. Different books classify the amino acids in different ways. In this book we will break them down into two categories: polar, and non-polar, with the polar ones further broken down into neutral and charged side chains. Acidic side chains have a negative charge and basic side chains have a positive charge. On the next page are the 20 naturally occurring amino acids along with their three letter abbreviations arranged according to the above scheme. You do NOT need to memorize these structures; they will be given to you on exams if needed.

16-1

Copyright Prof. Lawrence Mailloux, 2021

Nonpolar (Hydrophobic) Glycine (Gly, G)

Alanine (Ala, A)

Valine (Val, V)

Tryptophan (Trp, W)

Phenylalanine (Phe, F)

Proline (Pro, P)

Serine (Ser, S)

Threonine (Thr, T)

Leucine (Leu, L)

Isoleucine (Ill, I)

Polar (Hydrophilic) and Neutral Cysteine Tyrosine Asparagine (Cys, C) (Tyr, Y) (Asn, N)

Polar (Hydrophilic) / Negatively Charged Aspartic Acid Glutamic Acid (Asp, D) (Glu, E)

Methionine (Met, M)

Glutamine (Gln, Q)

Polar (Hydrophilic) / Positively Charged Lysine Arginine Histidine (Lys, K) (Arg, R) (His, H)

16-2

Copyright Prof. Lawrence Mailloux, 2021

Amino Acids to Proteins Recall from Chapter 13 that when a carboxylic acid reacts with a primary or secondary amine the products are an amide and water.

Carboxylic Acid

Amine

Amide

Water

Because amino acids have a carboxylic acid group on one side, and an amine group on the other, they are able to bond to each other repeatedly. When two amino acids bond together, the resulting structure is called a dipeptide. Three amino acids bonded together yields a tripeptide; and many amino acids bonded together gives a polypeptide, or a protein. For example, the reaction below shows the formation of a dipeptide formed when alanine bonds to valine.

alanine

valine

a dipeptide

Because the resulting dipeptide still has an amine on one end and a carboxylic acid on the other it can bond to another amino acid (for example tyrosine) to make a tripeptide.

a dipeptide

tyrosine

a tripeptide

This process repeats over and over again until the final protein is complete. Real proteins can be hundreds, or even thousands, of amino acids long. Hydrolysis of a protein breaks it down into its amino acids. When you eat protein your body breaks it down into its amino acids which your body then makes use of. 16-3

Copyright Prof. Lawrence Mailloux, 2021

Protein Structure and Function With 20 different amino acids to choose from, there are virtually an unlimited number of possible combinations of amino acids. This rich variety of possible structures is what allows proteins to perform all kinds of different functions inside a living organism. Below are some of the different roles that amino acids play in your body. 1. Reaction catalysis – virtually all biochemical reactions are catalyzed by proteins called enzymes. Enzymes are nature’s catalysts. Below are some examples of enzymes. a. Hydrolase – catalyzes breaking of bonds by water b. Lactase – breaks down lactose c. Lipase – breaks down lipids d. Amylase – breaks down starch e. Reverse transcriptase – converts viral RNA to viral DNA (Chapter 18) f. Integrase – inserts viral DNA into host genome (Chapter 18) g. Protease – finishes maturation of HIV virus particles (Chapter 18) 2. Oxygen transport – oxygen is carried around your body by hemoglobin, a protein found in red blood cells. Along with an Fe atom, the hemoglobin in your blood carries oxygen molecules and transports them to the muscles in your body. 3. Movement of muscles – proteins molecules called actin and myosin allow for expansion and contraction of muscles. 4. Immune system – specially shaped proteins called antibodies bind to protein markers (antigens) on the surface of invading bacteria and viruses. (See Chapter 18 for details) 5. Hormones – many hormones, such as the insulin your body uses to break down sugar, are proteins. 6. Structural components – hair, whites of eyes, skin, and muscles are all made of protein. As with all molecules, the three dimensional structure of a protein determines its biological activity. Protein structure is broken down into four levels of organization: primary, secondary, tertiary, and quaternary.

16-4

Copyright Prof. Lawrence Mailloux, 2021

Primary Structure The primary structure of an amino acid is the sequence of amino acids in the polymer chain. In the tripeptide example on the first page of this chapter, the primary sequence is alanine-valinetyrosine. Real proteins are much longer and their primary sequence can be hundreds, or even thousands, of amino acids long. Genes in an organisms DNA determines the primary structure of a protein. We will learn more about this in Chapter 17.

Secondary Structure The secondary structure of a protein is caused by hydrogen bonding that occurs at regular intervals along the backbone of a protein. Hydrogen bonding is a special kind of bond that occurs between a hydrogen atom bonded to an electronegative element (oxygen and nitrogen being the most important biologically) and another electronegative element. In proteins this occurs between a hydrogen atom that is bonded to an oxygen or a nitrogen atom and a nearby oxygen atom. The dotted lines in the image below represent a hydrogen bond occurring between the backbones of two amino acids (the squiggly lines indicate that the protein chains continue in both directions for many amino acids).

16-5

Copyright Prof. Lawrence Mailloux, 2021

Hydrogen bonding results in the formation of two different kinds of secondary structural subunits. When the hydrogen bonding is from nearby parts of the protein chain the backbone will coil like an old school phone cord forming an alpha helix. When the hydrogen bonding is from distant parts of the protein, the backbone will fold into a pleated flat region called a beta sheet. The image below (obtained from Wikipedia) is of reverse transcriptase, an important viral enzyme found in HIV that we will learn about in Chapter 18. The coils are alpha helices while the flat arrows represent beta sheets.

Protein backbone

Beta Sheet

Alpha Helix

16-6

Copyright Prof. Lawrence Mailloux, 2021

Tertiary Structure The tertiary structure of proteins is caused by interactions between side chains. Below are the four different kinds of interactions we will consider in this class along with examples of each. 1. Salt Bridges. Side chains with opposite charges will attract each other, while two amino acids with side chains of like charge will repel one another. a. A negatively charged aspartic acid amino acid will repel another negatively charged amino acid such as glutamic acid. b. A negatively charged aspartic acid amino acid will attract a positively charged amino acid such as lysine. 2. Hydrophobic/hydrophilic interactions. Chemists have an expression “like dissolves like.” This means than nonpolar amino acid side chains will attract other nonpolar amino acids side chains, while polar side chains will attract other polar side chains. a. A nonpolar amino acid such as alanine will be attracted to other nonpolar amino acids such as valine, or leucine. b. A polar amino acid such as serine will be attracted to another polar amino acid such as another serine, or a different polar amino acid, such as threonine. 3. Disulfide linkage. Under the correct conditions the –S-H groups on two nearby cysteine amino can bond together to form a sulfur-sulfur bond known as a disulfide linkage. a. –S-H + -S-H → -S-Sb. Perms work by breaking disulfide linkages and then reforming them 4. Hydrogen bonding between side chains on amino acids pulls those amino acids closer together. It is important that you not confuse this with hydrogen bonding that occurs along the backbone of the protein which gives rise of the secondary structural subunits of alpha helices and beta sheets.

Quaternary Structure Some proteins are complete after they obtain their tertiary structure. Others are made up of multiple tertiary subunits that bundle together to give the final quaternary structure of the protein. 16-7

Copyright Prof. Lawrence Mailloux, 2021

Denaturing Proteins Its unique three-dimensional shape determines a protein’s biochemical properties. Many chemical and physical processes can destroy a proteins three-dimensional shape, a process called denaturing. One a protein has been denatured its biological activity is destroyed. Some ways to denature proteins include: 1. Heating a protein denatures it. Most proteins have evolved to operate at body temperature. If the temperature is raised too high the protein will denature. 2. Alcohol and other solvents alter the secondary and tertiary interactions destroying the proteins three-dimensional shape. This is why alcohol and other solvents kill bacterial and viruses – it denatures proteins necessary for the microorganisms to live. 3. pH changes. Changing pH of a solution affects salt bridges and hydrogen bonding altering the shape of the proteins. 4. Heavy metal ions like Ag+, Pb2+, Hg2+, etc. disrupt disulfide linkages and salt bridges altering the proteins structure. 5. Reducing agents break disulfide linkages. The number of disulfide linkages determines the difference between straight and curly hair. More disulfide linkages means curlier hair. Perms work by breaking disulfide linkages and then reforming them after the hair has been reshaped.

Many of the above interactions happen in the kitchen. When you cook eggs you are denaturing the egg whites. Citric acid, found in lemons and limes, are used to make ceviche. Jell-O is made of proteins. Addition of acid or alcohol disrupts the three dimensional structure of the proteins. This is why adding citrus fruit or alcohol (as in Jell-O shots) causes the gelatin to not set as firmly.

16-8

Copyright Prof. Lawrence Mailloux, 2021

Enzymes – Natures Catalysts Enzymes and Activation Energy Recall from Chapter Six that in order for a chemical reaction to take place the starting materials must have sufficient energy to overcome the activation energy barrier (Ea). In Chapter six we

Increasing Energy →

considered the example of a hot pack which had the following energy diagram.

Reactants Fe, O2

Ea

Heat Released

Product(s) Fe2O3

Reaction Progress →

The higher the energy of activation, the longer a chemical reaction will take to occur. Most biochemical reactions are very complex and will only take place in a reasonable amount of time if a catalyst (enzyme) is present. Enzymes work by providing an alternative, lower energy, pathway for the reaction to follow. Enzymes lower the activation energy. The diagram below shows the energy diagram for a generic biochemical reaction. The solid line represents the Ea for the uncatalyzed reaction while the dotted line represents the Ez of the catalyzed reaction. In a

Increasing Energy →

biochemical, reaction the starting materials are called substrates.

Substrate(s)

Ea

Product(s)

Reaction Progress 16-9 →

Copyright Prof. Lawrence Mailloux, 2021

Lock and Key Theory and Induced Fit To understand how enzymes work, consider the following diagram. The shape below represents the disaccharide lactose with the red square representing the glucose ring and the green circle representing the galactose ring.

Substrate

When someone who is not lactose intolerant consumes lactose the body uses the enzyme lactase to break the disaccharide down into separate glucose and galactose monosaccharides. The enzyme has a location called the active site, which is shaped specifically to hold the substrate in question, in this case lactose.

When the enzyme encounters a substrate molecule the substrate will settle into the active site just like a key fits into a lock (substrate is the name given to the molecules that enzymes help react). This forms an unstable intermediate called an enzyme substrate complex. The enzyme acts as a template and lines up the atoms that need to react with each other. Then the enzyme puts strain on the chemical bonds in the intermediate causing old bonds to break and new bonds to form. The products are released (in this case separate glucose and galactose molecules) and the enzyme is left unchanged and ready to break down more substrate molecules.

16-10

Copyright Prof. Lawrence Mailloux, 2021

Enzyme Inhibition Many drugs/medications work by inhibiting enzymes. There are two ways an enzyme and be inhibited, competitively and noncompetitively. In competitive inhibition a blocker molecule with a similar shape similar to the target substrate enters the active site. This is like breaking a key off in a lock. So long as the blocker is present the active site will be blocked and no reaction will take place. This is displayed in the following figure.

In a noncompetitive inhibitor, a molecule binds at a location away from the active site. This causes a change in the shape of the active site disabling the enzyme. Active Site Shape Changed

16-11

Copyright Prof. Lawrence Mailloux, 2021

End of Chapter Problems Problem 1 – Draw structures of the following tripeptides a) Serine-threonine-cysteine b) Methionine-aspartic acid-glutamine c) Valine-valine-tryptophan Problem 2 – List six functions that proteins perform Problem 3 – What reactions do the following enzymes catalyze? a) Hydrolase b) Lactase c) Lipase d) Amylase e) Reverse transcriptase f) Integrase g) Protease Problem 4 – How are amino acids categorized? Why are these categories important? Problem 5 – What determines the primary structure of a protein? Problem 6 – What causes the secondary structure of a protein? Problem 7 – Hydrogen bonding from nearby locations on the protein backbone causes the formation of _________ while more distant hydrogen bonding causes the formation of ______. Both of these are examples of _______ structure. Problem 8 – Hydrogen bonding between side chains is one of the interactions that gives rise to protein’s _______ structure. Problem 9 – Not all proteins exhibit this level of structure. Problem 10 – The sequence and identity of amino acids is the ________ structure of a protein. Problem 11 – What is a salt bridge? What kind of structure does this interaction cause?

16-12

Copyright Prof. Lawrence Mailloux, 2021

Problem 12 – Like dissolves like. This means that ______ amino acids will be attracted to other ______ amino acids while ______ amino acids will be attracted to other _______ amino acids. Problem 13 – Give an example of a polar amino acid. What other amino acid would be attracted to this amino acid? Problem 14 – Give an example of a nonpolar amino acid. What other amino acids would be attracted to this amino acid? Problem 15 – What two amino acids would one expect to be attracted to the amino acid lysine? Problem 16 – Heat, heavy metals, acid, solvents, and reducing agents are all ways to _____ a protein? Problem 17 – Disulfide linkages occur between two _____ amino acids. This interaction helps give rise to a proteins ______ structure. Problem 18 – Most enzymes work best as _____ temperature. Excess heat can cause the protein to _______. Problem 19 – A ______ is the molecule that an enzyme acts upon. The _______ fits into the ______ on an enzyme like a key fits in a lock. Problem 20 – Enzymes work by lowering the ______ of a reaction. Problem 21 – A ______ inhibitor has a shape similar to the ______ of a given enzyme; it works by ________. Problem 22 – A _______ inhibitor binds at a location away from the _______ which causes it to change _______, disabling the enzyme.

16-13

Copyright Prof. Lawrence Mailloux, 2021

Answers to End of Chapter Problems 1. Serine-threonine-cysteine

Methionine-aspartic acid-glutamine

Valine-valine-tryptophan

16-14

Copyright Prof. Lawrence Mailloux, 2021

2. Reactions catalysis, oxygen transport, movement of muscles, immune system antibodies, hormones like insulin, structural components like hair and muscles. 3. These kinds of enzymes catalyze the following kinds of reactions: a. Hydrolase – breaking bonds with water b. Lactase – digestion of lactose c. Lipase – breaks down lipids d. Amylase – breaks down starch e. Reverse transcriptase – conversion of viral RNA to viral DNA f. Integrase – insertion of viral DNA into host genome g. Protease – finishes maturation of HIV virus particles 4. Amino acids are categories as by their side chains as polar or nonpolar. Polar amino acids further broken down into negatively charged and positively charged side chains. This is important because opposite charges attract (like charges repel) and like dissolves like. These attractions and repulsions are responsible for a proteins tertiary structure. 5. The primary structure of a protein is the sequence of amino acids specified by genes in your DNA. 6. Secondary structure is caused by hydrogen bonds that occurs along the protein backbone. 7. Alpha helices, beta sheets, secondary 8. Tertiary 9. Quaternary 10. Primary 11. Salt bridge refers to the attraction between positively and negatively charged amino acids. These interactions give rise to a proteins tertiary structure. 12. polar, polar, nonpolar, nonpolar 13. Serine is an example of a polar amino acid, it would be attracted to any other polar amino acid 14. Glycine is an example of a nonpolar amino acid, it would be attracted to any other nonpolar amino acid. 15. Lysine has a positive charge, thus it would be attracted to negatively charged amino acids such as aspartic acid or glutamic acid. 16. Denature 16-15

Copyright Prof. Lawrence Mailloux, 2021

17. Cysteine, tertiary 18. Body, denature 19. Substrate, substrate, active site 20. Activation energy (Ea) 21. Competitive, substrate, binding to and blocking the active site. 22. Noncompetitive, active site, shape.

16-16

Copyright Prof. Lawrence Mailloux, 2021

Chapter 17 DNA, RNA, and the Genetic Code Nucleic Acids Living organisms use nucleic acids to store and transmit genetic material. There are two kinds of nucleic acids, DNA (deoxyribonucleic acid) and RNA (ribonucleic acid). To understand how our bodies make use of DNA and RNA we must first learn about their structures.

Deoxyribonucleic Acid (DNA) DNA is the instructions needed to make a living organism. Every cell in your body (excluding sex cells) contains the same set of DNA molecules. Nerve cells, skin cells, and liver cells, all have the same DNA, yet each is different because it only uses the part of those instructions (genes) that are necessary for that type of cell. Skin cells for example would activate genes needed to grow air, while nerve cells would not. Like carbohydrates and proteins, DNA is also a polymer. In the case of DNA the repeating units are small molecules known as nucleosides. There are four different nucleosides found in DNA: adenine, thymine, guanine, and cytosine. These are abbreviated using the letters A, T, G, and C. Below are their structures:

Adenine (A)

Thymine (T)

Guanine (G)

17-1

Cytosine (C)

Copyright Prof. Lawrence Mailloux, 2021

These nucleosides are bonded to a small sugar molecule named deoxyribose. Attached to the sugar molecules is a phosphate (PO43-) group. Together these make up what are called nucleotides. Below are the structures of nucleotides made from each of the four different nucleosides A, T, G, and C:

To make a polymer the phosphate on one nucleotide’s sugar bonds to the bottom of the sugar ring of another nucleotide. This process repeats itself thousand and thousands of times, DNA molecules are very long. The structure of four nucleotides bonded together is shown below.

17-2

Copyright Prof. Lawrence Mailloux, 2021

Thousands of nucleotides hooked together as shown in the image above gives the structure of one strand of DNA. The DNA inside a living organism however is double stranded, that is it consists of two strands of DNA bonded together into a spiral pattern known as a double helix. The two strands of DNA are held together by hydrogen bonds between nucleosides of on each strand. Specifically an adenine (A) on one strand always bonds to a thymine (T) on the adjacent strand, while a guanine (G) on one strand always bonds to a cytosine (C) on another strand. We call strands of genetic material that are partnered A to T and G to C complementary. This is important; be sure to remember that A bonds to T and G bonds to C. The image below, obtained from Wikipedia, shows the structure of short sequence of double stranded DNA.

Such a sequence of DNA can be written much more compactly by simply abbreviating using the letters A, T, G, and C. TGT GGA AGT (DNA) ACA CCT TCA (Complementary DNA strand)

17-3

Copyright Prof. Lawrence Mailloux, 2021

Problem 1 – give the complementary DNA strand for the following sequences of DNA a) ATCGCAAGGGCC b) GGCTATCGCAAT c) CTAGCGGAGTAT

Problem 2 – A segment of DNA is determined to contain 15% A. Given this information, determine the percent of the other nucleosides. ______ % T ______ % G _________ % C

Ribonucleic Acid (RNA) RNA has a structure similar to DNA except for three small, but important, differences. The first difference is that RNA does not use thymine (T), instead it uses a very similar nucleoside called uracil (U). This means that RNA uses the letters A, U, G, and C, instead of the A, T, G, and C that DNA uses. Below are the structures of uracil and thymine.

Thymine (T, in DNA)

Uracil (U, found in RNA)

The second difference is in the sugar that RNA uses. DNA uses the sugar deoxyribose in its sugar phosphate backbone, while RNA uses a slightly different sugar, ribose. The difference between deoxyribose and ribose is that deoxyribose has an H where ribose has an OH. The “deoxy” in the name refers to the lack of this OH group. The images below compare the structures of ribose and deoxyribose.

Deoxyribose (in DNA)

Ribose (in RNA)

17-4

Copyright Prof. Lawrence Mailloux, 2021

The last difference between RNA and DNA is that while DNA is double stranded, RNA is single stranded. The image below shows an example of a short segment of single stranded RNA.

17-5

Copyright Prof. Lawrence Mailloux, 2021

The Central Dogma of Biochemistry Inside the nucleus of all of the cells in your body is your DNA. This DNA contains all the instructions necessary to make you. These instructions are called genes. A gene is a specific sequence of A, T, G, and C’s in your DNA that codes for a particular protein. This code is known as the universal genetic code and, with only very few exceptions, is the same for all life on this planet. This means that you, your cat, and an onion all use the same genetic code! Breaking this code was one of the greatest scientific achievements of the 20th century. The code is read in three base pair groups called codons. With four base pairs (ATGC) available, this gives a total of 4 x 4 x 4 = 64 possible combinations, with which to code for the 20 naturally occurring amino acids. Thus some amino acids are coded for by multiple codons. Additionally there are codons that signal for the process to start and stop. Below is the universal genetic code. You do NOT need to memorize this.

The central dogma of biochemistry states that genetic information flows from DNA to RNA to proteins. Or put into words “DNA makes RNA which makes proteins.” DNA → RNA → Proteins Let’s consider an example. A skin cell wants to grow some hair. For a skin cell to grow hair (a protein), it would need to activate (express) the genes that code for hair proteins. Let’s consider a short segment of DNA. For this example, we will only consider a segment of three codons (nine base pairs) long. This will only code for three amino acids. Real genes are hundreds or thousands of codons long as real proteins are made of far more than three amino acids. However to keep the

17-6

Copyright Prof. Lawrence Mailloux, 2021

example simple we will only use three codons. Below is our hypothetical three codon long double stranded DNA sequence for a hypothetical hair gene. TAA-CGC-CAC

(DNA coding for desired protein)

ATT-GCG-GTG

(Complementary DNA strand)

DNA is double stranded, however only one of the two strands is actually read at a given time. The two strands of DNA open up like a zipper. Then one of the two strands is chosen to be read. Don’t worry about how your body decides which strand to read. We will use the top strand from the segment above in our example. TAA-CGC-CAC

(DNA – single unzipped strand)

Now that the DNA is unzipped, your body must make a copy of the gene it wishes to express. Your body does this by making a complementary strand of RNA called messanger RNA (mRNA). Recall that A bonds to T and that G bonds to C. However because RNA does not use tymine (T) the A’s in the DNA will be paired up with uracil (U) instead. This process of making a complementary mRNA strand from a strand of DNA is called transcription. This gives the following sequence for our complementary mRNA strand. TAA-CGC-CAC ↓Transcription – occurs in nucleus AUU-GCG-GUG

(complementary mRNA – put into genetic code wheel)

Problem 3 – give the complementary mRNA strand for the following sequences of DNA a) TGGCGCGAA b) GAGCCGAGA c) CGAGCCCAC Transcription takes place in the nucleus. Once the mRNA strand is complete, it leaves the nucleus and enters the cytoplasm, the part of the cell outside the nucleus. Once outside the nucleus it arrives at an organelle called a ribosome. Ribosomes (which are themselves proteins) act as factories for making proteins based upon instructions brought in by the messenger RNA.

17-7

Copyright Prof. Lawrence Mailloux, 2021

The process of making proteins from the instructions carried by the mRNA is called translation. Translation takes places at the ribosomes. At the ribosomes the mRNA binds to another kind of RNA called transfer RNA (tRNA). Transfer RNA are small, hairpin shaped loops of RNA with amino acids stuck to them. Each of the 64 different three letter codon possibilities given in the universal genetic code has its own tRNA molecule. On the bottom of the tRNA molecule is an anticodon which binds to the mRNA in a complementary fashion (A to U and G to C). Below is the structure of a t-RNA molecule carrying an isoleucine amino acid, note the anticodon at the bottom.

If the anticodon is a match for the codon in the mRNA the tRNA will add its amino acid to the growing protein chain. For our three codon long example below: (mRNA)

AUU-GCG-GUG

The three amino acids coded for are AUU = isoleucine

GCG = alanine

GUG = valine

The three codons above will bind to the following t-RNA with the complementary anticodons UAA, CGC, and CAC:

17-8

Copyright Prof. Lawrence Mailloux, 2021

When the anticodons on the tRNA molecules binds to the codons in the mRNA as shown below AUU-GCG-GUG

(mRNA)

UAA CGC CAC

(complementary tRNA anticodons)

the t-RNA molecules release their amino acids which join the growing protein chain. For our example, the three amino acids are isoleucine, alanine, and valine. The structure of this tripeptide is shown below.

Remember, real proteins are much longer than three amino acids. In reality, the above process would continue many times until a STOP codon is reached. Problem 4 – What amino acid (or stop) is coded for by the following codons? a) GGU b) UAA c) CUG d) UGG 17-9

Copyright Prof. Lawrence Mailloux, 2021

Problem 5 – What tRNA anticodons would bond to the mRNA codons given below? a) UAG b) UGG c) AUC d) CCA Problem 6 – A protein chain will continue until it reaches a stop codon. What are those codons? Problem 7 – Which of the following DNA sequences would code for a tripeptide? a) GGCAGACATTAA b) CGATATATTACT c) ACTGGACCCGAC Problem 8 – What amino acids would be coded for by the following sequences of DNA? a) CGGAGACCAATT b) AAAGCGTATATC c) GGCTTCGAGACT For problems 9 to 14 fill in the blanks Problem 9 – The conversion of DNA to RNA is called _______ and occurs in the ______? Problem 10 – t-RNA link up with mRNA and make proteins. This occurs at the _______? Problem 11 – The conversion of DNA to mRNA is called ________ while the conversion of mRNA to tRNA is called _______? Problem 12 – DNA is read ____ base pairs at a time. Each set of ___ base pairs is called a ____? Problem 13 – tRNA molecules have ______ on them which bond to the _____ in the mRNA. Problem 14 – While DNA and RNA are very similar, they do have important differences fill in the blanks regarding these differences. a) DNA is ______ stranded while RNA is _____ stranded. b) DNA uses the sugar _____ while RNA uses the sugar ______. c) DNA uses the letters ___, ___, ___, and ___, while RNA uses ___, ___, ___, and ___. 17-10

Copyright Prof. Lawrence Mailloux, 2021

Answers to in Chapter Problems 1. a) TAGCGTTCCCGG b) CCGATAGCGTTA c) GATCGCCTCATA 2. 15% T, 30% G, 30% C 3. a) ACCGCGCUU b) CUCGGCUCU c) GCUCGGGUG 4. a) GGU = glycine b) UAA = stop c) CUG = leucine d) UGA = tryptophan 5. a) AUC b) ACC c) UAG d) GGT 6. There are three mRNA stop codons: UAA, UAG, and UGA.

17-11

Copyright Prof. Lawrence Mailloux, 2021

7. a) GGC-AGA-CAT-TAA CCG-UCU-GUA-AUU

(DNA to be copied) (mRNA for genetic code)

Codes for Proline-serine-valine-isolucine, codes for four amino acids, not a tripeptide b) CGA-TAT-ATT-ACT GCU-AUA-UAA-UGA

(DNA to be copied) (mRNA for genetic code)

Codes for Alanine-isoleucine-stop-stop, codes for only two amino acids, not a tripeptide c) ACT-GGA-CCC-ATC UGA-CCU-GGG-UAG

(DNA to be copied) (mRNA for genetic code)

Codes for stop-proline-glycine-stop, only codes for two amino acids, not a tripeptide d) GAA-TTC-CGA-ACT CUU-AAG-GCU-UGA

(DNA to be copied) (mRNA for genetic code)

Codes for leucine-lysine-alanine-stop, codes for a tripeptide 8. a) CGG-AGA-CCA-ATT GCC-UCU-GGU-UAA

(DNA to be copied) (mRNA for genetic code)

Codes for alanine-serine-glycine-stop b) AAA-GCG-TAT-ATC UUU-CGC-AUA-UAG

(DNA to be copied) (mRNA for genetic code)

Codes for Phenylalanine-arginine-isoleucine-stop c) GGC-TTC-GAG-ACT CCG-AAG-CUC-UGA

(DNA to be copied) (mRNA for genetic code)

Codes for proline-lysine-leucine-stop 9. Transcription, nucleus 10. Ribosomes 11. Transcription, translation 12. Three, three, codon 17-12

Copyright Prof. Lawrence Mailloux, 2021

13. Anticodons, codons 14. a) Double, single b) Deoxyribose, ribose c) A, T, G, C in DNA and A, U, G, C in RNA

17-13

Copyright Prof. Lawrence Mailloux, 2021

Chapter 18 DNA, RNA, and Antiviral Medications Professor Mailloux is NOT a physician, he is a chemist. This book is NOT intended to give medical advice. It is meant to illustrate how the chemistry you learn in CHM1033 is applicable to the real world. Proper HIV treatment requires a coordinated working relationship between a patient and a doctor who specializes in treating HIV. If you or someone you know thinks they may have HIV don’t panic and go get tested right away. To find testing simply Google “free HIV testing near me.” Once at the testing center qualified personal will be there to help you should you test positive. Home HIV test kits are also available at pharmacies like CVS and Walgreens for about $40. They have detailed instructions and phone numbers to call for help.

Structure of Nucleosides and Nucleotides Recall from the prior chapter that DNA contains nucleosides A, T, G, and C, while RNA uses A, U, G, and C. The structures of the five nucleosides are shown below.

Adenine (A)

Thymine (T, in DNA)

Guanine (G)

Cytosine (C) Uracil (U, in RNA)

In DNA and RNA the nucleosides are bonded to sugar rings with an attached phosphate. These molecules are called nucleotides. Below are the structures of nucleotides found in DNA for each of the four different nucleosides A, T, G, and C:

18-1

Copyright Prof. Lawrence Mailloux, 2021

In RNA the nucleotides are very similar in structure except U is used instead of T and the sugar ring is ribose, instead of deoxyribose. Below is the nucleotide containing uracil as found in RNA.

Although you do not need to memorize these exact structures, you do need to know their general shape as this will be important for understanding how antiviral drugs work. Take to moment to take a mental snap shot the general structure of the above molecules. We will come back to them later in this chapter. First, we need to cover some basics about HIV.

How HIV Is Spread Despite being a very deadly virus, HIV is actually fairly hard to catch and easy to kill outside the body. HIV is not spread by casual everyday activities such as sharing utensils, hugging, kissing, or sharing a bathroom. There are three ways to catch HIV: blood, birth, and sex. Let’s take a look at each of these three routes of infection in more detail.

Blood Unbroken skin provides a very effective barrier against HIV. If blood from an HIV infected person lands on healthy, unbroken skin, the virus can’t enter the body. For an infection to take hold the infected blood must get inside one’s body. This can occur in a variety of ways. If infected blood were to land on skin with cuts or scrapes the virus could get in this way. To prevent this from happening health care workers follow universal precautions and always wear gloves when touching a patient, or handling samples of bodily fluids. Another way this could happen is through needles. In a health care setting, the most common way this could happen is by an accidental needle stick by a contaminated needle. To prevent this needles are shielded and dirty needles are always disposed of in a sharps container. If despite these steps a needle stick were to occur, prompt treatment with antiviral medications can greatly reduce the risk of contracting HIV. Taking antiviral medications after exposure to reduce the risk of infection is called post exposure prophylaxis (PEP). The sooner PEP is started the more effective it is. Per the Centers for Disease Control and Prevention (CDC), this must be within 72 hours at the latest. 18-2

Copyright Prof. Lawrence Mailloux, 2021

In the 1980’s when the HIV pandemic was just starting a number of people caught HIV through transfusions of contaminated blood. This is no longer a problem as all blood is tested and the risk of catching HIV from a blood transfusion is now essentially zero. The most common way HIV (and other diseases) are spread via blood is through sharing of dirty needles when injecting intravenous drugs like heroin. To help prevent this many localities (including Miami Dade) offer needle exchange programs where drug users can exchange dirty needles for clean ones. Many years of research have shown these programs to be safe and effective at reducing the spread of disease. Additionally years of research has shown that these programs do not increase illegal drug use and do not increase crime.

Birth Without treatment an HIV positive pregnant woman has an about 23% chance of passing HIV onto their baby. The details of this are beyond the scope of this book, but with a combination of medications, and other measures (caesarean deliveries and use of formula instead of breastfeeding), this number can be greatly reduced, to about 2%1. In 2015 the World Health Organization validated the Cuba was the first country in the world to essentially eliminate mother-to-child transmission of HIV.

Sex The most common mode of transmission of HIV is through unprotected vaginal or anal intercourse; with the receptive partner at the greater risk. This is because the mucus membranes of the vagina and anus are not an effective barrier against disease. Condoms can greatly reduce this risk when used correctly. Although this risk is not zero, and it is impossible to determine exact numbers, oral sex is not a common route of transmission for HIV, and the CDC lists this risk of transmission from oral sex as low. For more details, see the CDC’s website: https://www.cdc.gov/hiv/risk/estimates/riskbehaviors.html Regardless of how it happens, HIV must enter the body to cause an infection. Let’s now take a look at what happens when HIV gets inside your body.

1.

Fan, H. Y., Conner, R. F., Villarreal, L. P. AIDS Science and Society 7 th ed. Jones and Bartlett Learning, 2014

18-3

Copyright Prof. Lawrence Mailloux, 2021

The HIV Life Cycle HIV virus particles are very small, roughly 100 nm in diameter, and spherical shaped. On the outside of the virus particle is a lipid membrane with protein spikes sticking out of it. Inside the lipid membrane is a capsid made of protein. Inside this capsid are the viruse’s genetic material (RNA) and three enzymes, reverse transcriptase, integrase, and protease. Below is a diagram of the HIV virus obtained from the National Institutes of Health (NIH).

Notice that HIV does not use DNA to carry its genetic information. HIV belongs to a class of viruses known as RNA viruses (Covid-19 is also an RNA virus). Keep this in mind as it will be important later. Viruses can not replicate on their own. For any virus to replicate it must first enter a host cell. In the case of HIV the target cells are white blood cells. Then once inside it needs to hijack the cell and trick the cell into using its cellular machinery to make copies of the virus. This is done in a series of steps. If any of these steps is interrupted, the virus will not be able to replicate. Different HIV medications target different parts of the HIV life cycle. There are more than two dozen HIV medications available, with each class targeting a specific step in the virus’s life cycle. Covering all of these is way beyond the scope of this class. We will only look at a couple of these steps in detail.

18-4

Copyright Prof. Lawrence Mailloux, 2021

When an HIV virus particle encounters a white blood cell, it uses its spike proteins to bind to receptors on the cells surface. Once the spike binds to the receptor the lipid membrane of the virus fuses with the cell membrane. The capsid enters the cell and releases the viruses RNA along with an enzyme called reverse transcriptase. HIV is an RNA virus, it must first convert its viral RNA to viral DNA. Because converting RNA to DNA is the opposite direction of transcription, (see chapter 17) this process is called reverse transcription. The virus uses its reverse transcriptase enzyme to turn its viral RNA into viral DNA. Medications called reverse transcriptase inhibitors block this step. We will learn more about these kinds of medications later in this chapter. After the viral DNA is ready, another enzyme, integrase, is used to integrate the viral DNA into the host cell’s DNA. Now the cell has been hijacked and will start making more HIV virus particles. Those particles then assemble into new immature virus particles which bud from the cell surface. One final viral enzyme, protease, finishes the maturation process and the new virus particles drift away an infect other white blood cells. The diagram on the next page (also from the NIH) summarizes this process.

18-5

Copyright Prof. Lawrence Mailloux, 2021

18-6

Copyright Prof. Lawrence Mailloux, 2021

Typical Untreated HIV Timeline Everybody is different, and while a small percentage of lucky people are naturally resistant to HIV, the vast majority of people are not. Although the time line can vary greatly, most people who catch HIV will develop AIDS within ten years, some much sooner. The diagram below summarizes a typical HIV timeline, actual times vary greatly from person to person.

When the virus first enters the body the immune system is totally unprepared. As a result the viral load (the number of virus particles in a given volume of blood) will rapidly increase (red line) while the CD4 cell count (a kind of immune cell, blue line) drops quickly. During this time some people will experience mild flu like symptoms, others will feel fine. Because the symptoms (fever, fatigue, body aches…) are common for so many ailments most people will just assume they have a cold. Unfortunately, due to the high viral load people at this stage are very contagious. Additionally, because the person has yet to develop antibodies they will still test negative for HIV. It can take anywhere from three to six months for a person to develop antibodies against the virus. During this window period a person will continue to test negative. After the window period passes the patient will begin to develop antibodies (green line) and will now test positive for HIV. As the body produces antibodies the CD4 count will rebound partly, and the viral load will drop (but not to zero). Although still contagious, at this point an HIV positive person will feel healthy and may unknowingly spread the virus to others.

18-7

Copyright Prof. Lawrence Mailloux, 2021

Over the ensuing months and years, the immune system will battle the virus (flat region). Unfortunately, this is a losing battle. For reasons that are poorly understood some people’s immune systems are able to hold off the virus longer than others; however after about ten years essentially everyone’s immune system will no longer be able to keep up with the virus and the CD4 count will begin to drop and the viral load will increase. As the immune system continues to weaken the patient will become susceptible to opportunistic infections that are not a problem for people with a healthy immune system. At this point, a person has developed AIDS (acquired immunodeficiency syndrome) and will not be able to fight off even simple infection. Eventually they will catch something their weakened immune system can’t defend against and death will occur. When the HIV pandemic first started in the early 1980’s HIV was a death sentence. Doctors and loved ones were powerless to help as their patients and friends progressed through the above stages. Fortunately, this is no longer the case. With proper treatment HIV positive people can live healthy normal lives and don’t need to fear infecting their partners. Modern medicine as transformed HIV from a death sentence to a manageable chronic disease much like diabetes or hypertension. In the rest of this chapter we will learn how a couple of these modern medicines work.

18-8

Copyright Prof. Lawrence Mailloux, 2021

HIV Public Health Campaign In Miami Miami Dade County has one of the highest rate of HIV in the nation. This is unacceptable because humanity already has the knowledge necessary to end this pandemic, but to accomplish this goal, this information needs to be known to the general public. To that end a public health campaign was undertaken in Miami. You may have noticed signs and billboards at the bus and train stations in Miami about HIV. These billboards were part of the inspiration for this book. The picture below was taken at the South Miami Metrorail station.

As future health care professionals, you should have a deeper understanding than the average person about each of the three points listed above. Let us consider each one in more detail.

18-9

Copyright Prof. Lawrence Mailloux, 2021

Routine Testing for HIV and STD’s Getting tested for HIV is quick and easy. Simply Google “free HIV testing near me,” or if you prefer you may test yourself at home. Home test kits are available at most major drug stores like CVS and Walgreens. They cost about $40 and take less than a half hour. When performed correctly according to the instructions they are very accurate. Below is a picture of the box for a home HIV test kit.

Inside the kit you will find a test strip, testing solution, detailed instructions, and even a black bag to allow you to throw the box away in the trash discreetly should privacy be a concern for you. This book is NOT an instruction manual for using these tests, it is just to give you an idea of what to expect. You must follow the instructions exactly to ensure accurate results.

18-10

Copyright Prof. Lawrence Mailloux, 2021

This particular product tests oral fluid (some tests use a drop of blood) for presence of antibodies. It is very important to understand that these tests do NOT test for the actual virus. Some people’s bodies start making antibodies sooner than others. Although for most people it is within three months, in rare cases it can take up to six months for some people’s immune system to start making antibodies. During this window period an HIV positive person may still test negative. This person can still be contagious. This fact must be taken into account when getting tested. Antibodies are proteins made by the immune system that have a specific three dimensional shape that is custom made by the immune system to attack the virus. On the surface of the virus are proteins with their own three dimensional shape. These viral proteins are called antigens. To test for antibodies using the above kit the mouth is swabbed using the device pictured below and allowed to sit in the test solution to develop.

After the proscribed time has passed the test is ready to be read. To ensure the test is working correctly a control is built in and a line should appear next to the letter C, which stands for control. If this line does not appear something is wrong either with the test or how it was performed. Either way you will need to get a new test kit. If antibodies are present they will bind to the antigen. This will cause a second line to appear next to the letter T, which stands for test. This is a positive test result. If no antibodies are present, no line will appear; this is a negative test result. If a positive test result is obtained a second more sensitive test will need to be performed by a doctor to confirm the result.

18-11

Copyright Prof. Lawrence Mailloux, 2021

PrEP and Condoms Can Reduce Your Risk of Getting HIV When used correctly condoms are very effective at reducing the risk of catching HIV, pregnancy, and other STD’s. To further reduce the risk of catching HIV PrEP is an option. PrEP (PreExposure Prophylaxis) is taking antiviral medications prior to exposure to reduce the risk of catching HIV. When taken exactly as directed PrEP is quite effective. Exact numbers are difficult to obtain and depend upon mode of transmission. The CDC’s website summarizes the results of several studies. Failure to take PrEP regularly and as directed greatly reduces its effectiveness.2 It is also important to keep in mind that while PrEP can be effective at reducing the risk of contracting HIV, it provides no protections against other STD’s or pregnancy’ thus condoms are still important. As of the writing of this book in 2021 there are two FDA approved medications for PrEP, Truvada (also available as a generic) and Descovy. A doctor will determine which is best. These medications work in similar ways, but for brevity this book only covers Truvada in detail. Truvada is a once a day pill that contains two medications, Emtricitabine and tenofovir disoproxil fumarate. Both of these medications work by preventing the virus from converting its viral RNA to viral DNA. The structures of these medications are shown below.

Tenofovir

Emtricitabine

Take a moment to look back at the structures on the first page of this chapter. You will notice that they are very similar in shape to the nucleotides used to make up DNA/RNA. This similarity in shape is key to how they work

2.

Centers for Disease Control and Prevention. Effectiveness of Prevention Strategies to Reduce the Risk of Acquiring or Transmitting HIV. https://www.cdc.gov/hiv/risk/estimates/preventionstrategies.html (Accessed December 3, 2021)

18-12

Copyright Prof. Lawrence Mailloux, 2021

Recall that HIV uses an enzyme called reversetranscriptase to turn its viral RNA into DNA, a key step in the viruses replication. The normal substrate for reversetranscriptase are nucleotides. When a nucleotide bonds to the active site of the reversetranscripate enzyme, the virus adds it to its growing DNA strand. The example below shows a letter C being added by the enzyme.

Emtricitabine and tenofovir act as competitive inhibitors of the reverse transcriptase enzyme. Because of their similar shape to nucleotides these molecules are able to enter the active site and block it preventing further DNA growth stopping the virus from replicating. The image below shows Emtricitabine blocking the active site.

Medications like Emtricitabine and Tenofovir that work by mimicking the shape of nucleosides they are called Nucleoside analog Reverse Transcriptase Inhibitors (NRTI).

Treatment is Prevention While the two medications in Truvada are sufficient to reduce the risk of contracting HIV, they alone are NOT sufficient to treat HIV. If an HIV positive person takes only Truvada the virus will quickly develop resistant to these medications. Therefore, people taking Truvada still need to get tested for HIV regularly so that if they are unlucky enough to catch HIV they can be correctly medicated.

18-13

Copyright Prof. Lawrence Mailloux, 2021

Treatment of HIV is accomplished with Antiretroviral Therapy (ART). ART consists of three or more medications from at least two different classes. By using more than one kind of medication the risk of resistance is greatly reduced as the virus would have to develop mutations against more than one kind of drug. With dozens of medications available, there are multiple combinations of drugs that can be used. The exact set of medications used must be determined by a qualified physician. For this chapter we will consider just one such medication, Altipla. I chose Altipla as it makes a good example for this course, I am in no way endorsing any particular medications. Like Truvada, Altipla contains Emricitabine and Tenofovir. In addition to these NRTI’s, Altipla contains a third medication, Efavirenz. The structure of Efavirenz is shown below.

Take a moment to compare the structure of Efavirenz to the structures on the first page of this chapter. You will notice that they are nothing alike. Efavirenz acts as a noncompetitive inhibitor of the reverse transcriptase enzyme and is an example of a Nonnucleoside analog Reverse Transcriptase Inhibitor (NNRTI) a different class of medication. As shown in the image below, when Efavirenz binds to a site away from the active site it causes a conformational change in the enzyme that alters the shape of the active site, disabling the enzyme.

The goal of ART is to lower a patient’s viral load below levels that not are detectable with normal tests. This is called being undetectable. Being undetectable does NOT mean the virus is gone, just that the amount present is very low. If one stops taking their medications, the viral load go back up again. Additionally, failure to take medications regularly increases the risk of the virus having a chance to develop resistance, which is bad. When someone is infected with a resistant strain of HIV, their options for medications will be more limited; possibly requiring the use of medications with more side effects. 18-14

Copyright Prof. Lawrence Mailloux, 2021

Being undetectable is the goal of ART because a person who achieves and maintains an undetectable viral load will be healthy and will not have to worry about passing HIV along to their sexual partners. The fact that undetectable persons are not at risk of spreading the virus is encapsulated in the saying Undetectable Equals Untransmittable, or U = U for short.3 In summary HIV is no longer the death sentence it used to be. By using combinations of different kinds of medications it is now possible to suppress the level of virus in the body to the point that the patient no longer has to worry about spreading the virus to their sexual partners. Patients who take their medications as directed, see their doctor regularly, and keep up on their lab tests can live healthy normal lives. This is why “treatment is prevention.” Thank you for taking the time to take this course and read my book. I hope you found it interesting and informative.

3.

Centers for Disease Control and PreventionHIV Treatment as Prevention. https://www.cdc.gov/hiv/risk/art/index.html (Accessed December 3, 2021)

18-15

Copyright Prof. Lawrence Mailloux, 2021

End of Chapter Questions 1. How can you find free HIV testing near you? 2. About how much does a home HIV test kit cost and where can you get them? 3. What is in a home HIV test kit? 4. What are the three ways HIV can be spread? 5. What is PEP and what does it stand for? When might it be needed? 6. The sooner PEP is started the better. According to the CDC, by when should PEP be started? 7. Although the risk can’t be reduced to zero, modern treatment can greatly reduce the risk of a pregnant woman passing HIV onto their child. What country was the first in the world to be recognized by the United Nations as having eliminated mother-to-child HIV transmission as a public health risk? 8. What sexual acts pose the greatest risk of catching HIV? 9. The HIV virus is ______ shaped. On the outside of the virus particle is a membrane made of ______ with protein spikes sticking out of it. Inside the membrane is a _______ made of _____. Inside the _____ is the viral ______ and a few ______. 10. To enter a cell the HIV virus first binds to ______ on the surface of a white blood cell using its protein spikes. Once the virus successfully binds the membrane of the virus and the cell ______ allowing the ______ to enter the cell. Once inside the _____ breaks down and releases ______ and ______. The virus then uses its ______ enzyme to convert its viral _____ into viral ____. Next the virus uses its ______ enzyme to stick its DNA into the host cells DNA. The cell will begin to produce immature virus particles that _____ from the surface of the cell. Lastly the virus uses its ______ enzyme producing active virus particles that drift away to infect other cells. 11. What reaction does the reverse transcriptase enzyme catalyze? 12. What enzyme is used to stick the viral DNA into the host cells DNA? 13. What enzyme is used to finish the maturation process releasing infectious virus particles? 14. Although it varies greatly from person to person, most people who catch HIV will develop AIDS within ______ years. 15. The amount of virus in a person’s blood is called a(n) _______.

18-16

Copyright Prof. Lawrence Mailloux, 2021

16. It can take up to ________ months for a person to develop antibodies. This period is known as the ________. During this time a person will test negative despite having the virus and being contagious. 17. What does HIV stand for? 18. What does AIDS stand for? 19. What is the difference between HIV and AIDS? 20. As of the writing of this book in 2021, what county has one of the highest rates of HIV in the nation? 21. What do HIV tests actually test for? 22. What are antibodies? 23. What are antigens? 24. A home HIV test stick has the letters C and T on them. What do these stand for? What is the purpose of the C part? 25. What does PeEP stand for? 26. As of the writing of this book what are the two medications on the market for PrEP? 27. Emtricitabine and Tenofovir belong to the _____ class of medications. Both work by blocking the active site of the _________ enzyme. These medications are both examples of a(n)________ . 28. What does ART stand for? What does it consist of? 29. Efavirenz belongs to the ______ class of medication. It works by binding to a location on the ______ enzyme away from the active site. This causes the shape of the active site to change. This makes Efavirenz an example of a(n) ________. 30. What does it mean to be undetectable? 31. Is a person that is undetectable cured of HIV? 32. What does U = U stand for? What does it mean?

18-17

Copyright Prof. Lawrence Mailloux, 2021

Answers to End of Chapter Questions 1. Google “free HIV testing near me.” 2. Home test kits cost about $40 and are available at pharmacies like CVS and Walgreens. 3. Instructions, test swab, test solution, discrete disposal bag. 4. Blood, birth, and sex. 5. Post Exposure Prophylaxis is taking antiviral medications after an exposure event like a needle stick or sexual assault. 6. Per the CDC PEP needs to be started within 72 hours. 7. Cuba 8. Unprotected anal and vaginal sex with the receptive partner at the greatest risk. 9. Spherical, lipids, capsid, protein, capsid, RNA, enzymes 10. Receptors, fuse, capsid, capsid, RNA and enzymes. 11. Reverse transcriptase inhibits the conversion of viral RNA to viral DNA 12. Integrase enzyme inserts the viral DNA into the host cells genome. 13. Protease enzyme finishes the maturation process producing infectious virus particles. 14. Ten years 15. Viral load 16. Three to six months, window period 17. HIV stands for human immunodeficiency virus 18. AIDS stands for acquired immune deficiency syndrome 19. AIDS is the late stage of infections with the HIV virus when the immune system is too weak to fight off infections. 20. Miami Dade County 21. HIV tests test for antibodies, not the actual virus. 22. Antibodies are specially shaped proteins made by the immune system that target antigens. 23. Antigens are specially shaped proteins on the surface of viruses that are targeted by antibodies. 24. The letter C stand for control and the letter T stand for test. The control is line indicates the test is working correctly. 25. PrEP stands for pre-exposure prophylaxis. 26. Truvada and Descovy 18-18

Copyright Prof. Lawrence Mailloux, 2021

27. NRTI, reverse transcriptase, competitive enzyme inhibitors. 28. ART stand for Antiretroviral therapy. It consists of a combination of at least three medications from at least two different classes. 29. NNRTI, reverse transcriptase, noncompetitive enzyme inhibitor 30. Undetectable means ones viral load is so low that it is not picked up on routine tests. 31. Undetectable does NOT mean cured. As of 2021 there is no cure for HIV. If one stops taking their medications the viral load will increase again. 32. Undetectable equals Untransmittable. This means that HIV positive people who achieve and maintain undetectable status, see doctor regularly, and keep up with their lab tests, do not need to worry about giving HIV to their sexual partners.

18-19