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The Organic Chemistry Bible Everything you need to know for GAMSAT Organic Chemistry www.AceGAMSAT.com The information contained in this guide is for informational purposes only. The publication of such Third Party Materials does not constitute a guarantee of any information, instruction, opinion, products or services contained within the Third Party Material. Publication of such Third Party Material is simply a recommendation and an expression of our own opinion of that material. No part of this publication shall be reproduced, transmitted, or sold in whole or in part in any form, without the prior written consent of the author. All trademarks and registered trademarks appearing in this guide are the property of their respective owners. Users of this guide are advised to do their own due diligence when it comes to making decisions and all information, products, services that have been provided should be independently verified by your own qualified professionals. By utilising this guide, you agree that the company AceGAMSAT is not responsible for the success or failure relating to any information presented in this guide. ©2018 AceGAMSAT. All Rights Reserved. AceGAMSAT is not affiliated with ACER in any way.
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About GAMSAT The GAMSAT (Graduate Australia Medical School Admissions Test) has been developed by the Australian Council for Educational Research (ACER). This test selects for students that have the greatest capacity to advance their studies in medicine, dentistry, and pharmacy. The GAMSAT was developed for Australian universities and its component of use for admissions has spread to institutions in the UK and Ireland. GAMSAT evaluates the nature and extent of abilities and skills gained through prior experience and learning, including the mastery and use of concepts in basic science as well as the acquisition of more general skills in problem solving, critical thinking and writing. Candidates whose first degree is in a non-scientific field of study can still sit GAMSAT and succeed in an application for admission to one of the graduate-entry programs. A science degree is not always a prerequisite and institutions encourage applications from candidates who have achieved academic excellence in the humanities and social sciences. However, it must be stressed that success in GAMSAT is unlikely without knowledge and ability in the biological and physical sciences.
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Table of Contents BASIC AND COMMON FUNCTIONAL GROUPS
1
ORGANIC CHEMISTRY BASICS
1
MOLECULAR REPRESENTATIONS
2
3D BOND-LINE STRUCTURES IDENTIFYING LONE PAIRS OF ELECTRONS RESONANCE HYBRID ORBITALS
5 6 7 8
ALKANES
10
NAMING ALKANES NAMING SUBSTITUENTS CONSTITUTIONAL ISOMERS OF ALKANES NEWMAN PROJECTIONS CYCLOHEXANE AND CHAIR CONFORMATIONS
11 13 14 15 16
STEREOCHEMISTRY
18
ISOMERISM IMPORTANT CONCEPTS IN STEREOISOMERISM DESIGNATING R AND S RELATIONSHIPS BETWEEN ENANTIOMERS AND DIASTEREOMERS MESO COMPOUNDS FISCHER PROJECTIONS
SUBSTITUTION REACTIONS
18 20 23 25 27 28
30
ALKYL HALIDES THE SN2 MECHANISM THE SN1 MECHANISM
30 31 33
ELIMINATION REACTIONS: STRUCTURE AND PREPARATION OF ALKENES NOMENCLATURE OF ALKENES STEREOISOMERISM IN ALKENES ELIMINATION REACTIONS E2 REACTIONS E1 REACTIONS
38 39 40 43 44 47
IV
ADDITION REACTIONS
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HYDROHALOGENATION HYDRATION HYDROGENATION HALOGENATION
49 50 51 51
ALKYNES
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RADICAL REACTIONS
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ALCOHOLS AND PHENOLS
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PHYSICAL PROPERTIES OF ALCOHOLS ACIDITY OF ALCOHOLS AND PHENOLS PREPARATION OF ALCOHOLS USING GRIGNARD REAGENTS
SPECTROSCOPY
56 56 59
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INFRARED SPECTROSCOPY NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY
AROMATIC RINGS
62 63
69
NOMENCLATURE OF AROMATIC RINGS
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AROMATIC SUBSTITUTION REACTIONS ELECTROPHILIC AROMATIC SUBSTITUTION SULFONATION NITRATION ACTIVATING GROUPS DEACTIVATING GROUPS
72 72 72 73 73 74
ALDEHYDES AND KETONES
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INTRODUCTION NOMENCLATURE OF ALDEHYDES NOMENCLATURE OF KETONES PREPARATION OF ALDEHYDES AND KETONES IMPORTANT REACTIONS OF ALDEHYDES AND KETONES
CARBOXYLIC ACIDS
76 76 77 78 80
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NOMENCLATURE STRUCTURE OF CARBOXYLIC ACIDS ACIDITY OF CARBOXYLIC ACIDS PREPARATION OF CARBOXYLIC ACIDS REACTIONS OF CARBOXYLIC ACIDS
83 84 84 85 86
V
AMINES
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CLASSIFICATION NOMENCLATURE PROPERTIES OF AMINES PREPARATION OF AMINES IMPORTANT REACTIONS OF AMINES
88 88 89 90 90
AMINO ACIDS, PEPTIDES, AND PROTEINS INTRODUCTION STRUCTURE AND PROPERTIES OF AMINO ACIDS ACID-BASE PROPERTIES PEPTIDE SYNTHESIS PROTEIN STRUCTURE PROTEIN FUNCTION
91 91 92 94 95 97 98
APPENDIX
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Basics of Organic Chemistry Carbon, hydrogen, oxygen, and nitrogen are the most common atoms found in organic chemistry. Carbon (C) can form four single bonds, two double bonds, or one triple bond with a single bond. Oxygen (O) can form two single bonds or a double bond and has two pairs of unshared electrons. Hydrogen (H) can only form one single bond. Nitrogen (N) may form 3 single bonds. Depending on the compound, it can also form double and triple bonds.
Fig. 1 - Common molecules in organic chemistry.
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Molecular Representations Video: Representing Structures of Organic Molecules
Molecules can be drawn using a variety of different styles. It is important to be able to understand the different styles, as they are likely to show up in the GAMSAT exam. Lewis structures are useful as they clearly show all atoms and bonds in the molecule. Each bond represents the sharing of two electrons between respective atoms.
Fig. 2 - Lewis structures of water, ammonia, and methane (top to bottom).
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Partially condensed structures do not show the bonds between C and H. Atoms are drawn beside each other. Three hydrogens bonded to carbon are shown as a methyl group (CH3).
Fig. 3 -Partially condensed structure of 2-propanol.
Condensed structures do not show any single bonds. The structural arrangement of atoms is still shown.
Fig. 4 - Condensed structure of 2-propanol.
The molecular formula does not provide information about the structural arrangement of atoms. It only shows the number of each type of atom present. Looking at a formula, it can be seen that there can be many arrangements. Many previous GAMSAT questions have asked candidates to determine the number of different arrangements (constitutional isomers) that can be produced from a certain molecular formula.
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Example: C3H7OH
Isopropanol
Propanol
Ethyl methyl ether
Fig. 5 - Different constitutional isomers of C3H7OH.
Bond-line structures are simple and easy to read. They are drawn in a zigzag-like fashion where the endpoints and corners denote a carbon atom. The hydrogen atoms that are bonded to the carbon are not shown. It is assumed there are enough hydrogen atoms so that each carbon atom has 4 bonds. The end points will have 3 hydrogen atoms.
Fig. 6 – Bond-line structure of hexane.
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This hexane has 6 carbons (count the corners and endpoints). Understanding bond-line structures is important in GAMSAT questions that involve counting the number of chiral centres (see stereochemistry chapter) or atoms in complex compounds such as cholesterol. Double bonds and triple bonds are shown with two and three bonds, respectively.
Fig. 7 - Double and triple bonded compounds.
3D Bond-Line Structures Wedges represent a group coming out of the page (towards you), and a dash represents a group going behind the page (away from you).
Fig. 8 - 3D bond-line structures.
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Identifying Lone Pairs of Electrons
There are several patterns that must be recognised for oxygen and nitrogen:
Oxygen A positive charge on the oxygen atom indicates 3 bonds and one lone pair of electrons; a negative charge on the oxygen atom indicates one bond and three lone pairs. No charge indicates two bonds and two lone pairs.
Fig. 9 - Lone pairs of electrons on oxygen.
Nitrogen A positive charge indicates 4 bonds and no lone pairs of electrons; a negative charge indicates two bonds and two lone pairs. No charge indicates three bonds and one lone pair.
Fig. 10 - Lone pairs of electrons on nitrogen.
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Resonance
Video: Resonance
Resonance structures are used to show the spread of positive charge. They show a combination of different structures in a linear fashion:
Fig. 11 - Resonance structures showing the spread of a positive charge.
In terms of the GAMSAT, questions usually arise that require the candidate to identify the most significant resonance structures out of several for a certain compound. There are rules that can be used to determine the significance of resonance structures:
1. Structures with minimal charges are more significant than structures with several charges. 2. Structures that have a full octet of electrons on their atoms are more significant than those that do not have a full octet. 3. If two carbon atoms in a structure have opposite charges, this structure is generally insignificant.
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Fig. 12 - The significance of difference resonance structures.
In summary, look for the structure that has minimal charges, a full octet of atoms, and does not show carbon atoms with opposing charges.
Hybrid Orbitals Video: Sp3 Hybridized Orbitals and Sigma Bonds Video: Pi bonds and sp2 Hybridized Orbitals
The concept of hybrid orbitals tends to confuse many students. For the purpose of the GAMSAT, detailed information is not necessary. Previous GAMSAT questions have asked to choose the compound that contains an sp2 hybridised carbon—easy marks! The following information is sufficient to tackle questions related to hybrid orbitals: An sp3 orbital carbon will be connected to 4 atoms/groups. The angle between atoms/groups is 109.5 degrees. An sp2 orbital carbon will be connected to 3 atoms/groups. The angle between atoms/groups is 120 degrees. An sp orbital carbon will be connected to 2 atoms/groups. The angle between atoms/groups is 180 degrees.
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Fig. 13 - Hybridisation of the carbon atom.
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Alkanes Alkanes are compounds that contain only carbon and hydrogen atoms. They are saturated hydrocarbons.
Fig. 14 - Common alkanes and their structures.
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Naming Alkanes Video: Naming Simple Alkanes Video: Naming Alkanes with Alkyl Groups
In the GAMSAT, it is highly likely to determine the product of a reaction in the provided stimulus and select its correct name from multiple-choice answers, which is where the skill of naming compounds comes into use.
When identifying the name of a compound:
1. Select the longest chain (parent chain) and count its carbons. If there are two chains of equal length, choose the chain with the greater number of substituents. 2. The number of carbons will determine the prefix; for alkanes, the name will end in -ane.
Fig. 15 - Naming alkanes.
It is crucial to memorise the following 10 prefixes: www.acegamsat.com
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Table 1 - Prefixes for naming alkanes.
If there are carbon atoms bonded in a cyclical structure, these are termed cycloalkanes. The prefix cyclo- indicates the presence of a ring structure. If there are hydrocarbon atoms bonded in a cyclical structure, these are termed cycloalkanes.
Fig. 16 - Common cycloalkanes and their structures.
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Naming Substituents
The substituents are named after the parent chain has been identified. -yl is added to the end of the name of each group—one carbon is methyl; two carbons, ethyl; three, propyl; and so forth.
Fig. 17 - Common substituents and abbreviations.
When assembling the name of the alkane, the carbon atoms in the parent chain are to be numbered. These numbers are used to identify the location of each substituent.
How can it be known which end of the parent chain to label as number 1? If one substituent is present, it is assigned the lowest number possible.
When there are many substituents, the first substituent receives the lowest number. If there is a tie, the second substituent should be as low as possible. If still tied, then the lowest number should be assigned in alphabetical order.
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If more than one substituent is present in a compound, a prefix is used to indicate how many times that substituent appears: Di- = 2 Tri- = 3 Tetra- = 4 Penta- = 5 Hexa- = 6
Commas are used to separate two numbers from each other, and hyphens are used to separate numbers from letters. Example:
1. Identify the parent chain: 8 carbon atoms 2. Identify and name the substituents: 2 methyl groups and 1 ethyl group 3. Number the parent chain and assign locations of substituents: Methyl groups at C2 and C3 and an ethyl group at C4. There are two methyl groups—prefix will be di-. 4. Arrange substituents alphabetically (ignoring prefixes): E has alphabetic priority to M. 4-ethyl-2,3-dimethyloctane
Constitutional Isomers of Alkanes
In previous GAMSAT exams, candidates have been asked to identify the number of constitutional isomers of a given alkane. For example, hexane (C6H14) has five constitutional isomers.
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Fig. 18 - Constitutional isomers of C6H14.
These isomers have the same number but a different arrangement of atoms. Be aware that some may look different at first sight but will still have the same arrangement of atoms.
Newman Projections Video: Newman Projections Video: Newman Projections 2
Students should know that the eclipsed conformation is the highest in energy (less stable), and staggered conformation is the lowest in energy (more stable).
Fig. 19 - Staggered versus eclipsed conformation. www.acegamsat.com
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Cyclohexane and Chair Conformations
Video: Chair and Boat Shapes for Cyclohexane
Cyclohexane can be found in a chair or boat conformation. At any moment, 99% of cyclohexane molecules are found in a chair conformation (lower energy = more stable).
Fig. 20 - Chair versus boat conformation
Chair conformation
The hydrogen atoms that are in the same plane as the ring are equatorial. The hydrogen atoms that are perpendicular to the ring are axial.
Fig. 21 - Axial and equatorial bonds in chair conformations.
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There are two conformations for each chair conformation. These come about via a chair “flip.”
Fig. 22 - The chair flip.
In the GAMSAT, students may be required to quickly determine the more stable chair conformation. The conformation with the substituent in the equatorial position is more stable because of less steric interactions. In the case above, the second conformation is more stable and will be favoured in equilibrium.
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Stereochemistry Isomerism Isomers have the same type and number of atoms. Students should know the following two types – constitutional isomers and stereoisomers. Constitutional isomers have the same molecular formula but a different connectivity of atoms.
Table 2 - Constitutional isomers.
Stereoisomers have the same molecular formula but a different spatial orientation of atoms.
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Fig. 23 - Categories of isomers.
In alkenes and cyclic compounds, geometric isomerism occurs through cis and trans compounds. When the substituents are on the same side of the ring or double bond, it is designated cis. When they are on opposite sides, it is designated trans.
Fig. 24 - Cis and trans isomers of 2-butene.
The trans isomer is more stable than the cis isomer. This is because the substituents are further apart and there is less steric repulsion.
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Important Concepts in Stereoisomerism Chirality Video: Introduction to Chirality Video: Chiral Examples 1 Video: Chiral Examples 2
Objects that are not superimposable on their mirror images are called chiral objects. Objects and molecules can be classified as chiral or achiral.
Fig. 25 - Examples of chiral and achiral objects.
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Fig. 26 - Examples of chiral and achiral molecules.
Chiral centres
A chiral molecule contains a carbon atom that is connected to four different groups. This carbon that is connected to four different groups is called the chiral centre.
Fig. 27 - Molecule with a chiral centre.
In the GAMSAT, it is likely that a large complex compound is given and students are to count the number of chiral centres.
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Fig. 28 – A complex compound containing a chiral centre.
There are two different ways to arrange four groups around a chiral centre. These two arrangements will be nonsuperimposable mirror images of one another. These two compounds only differ from each other in spatial arrangement and are thus called stereoisomers.
Fig. 29 - Stereoisomers.
Enantiomers
When a compound is chiral, it will have one nonsuperimposable mirror image. This is called an enantiomer.
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Fig. 30 - Enantiomers of alanine.
Designating R and S Video: Cahn-Ingold-Prelog System for Naming Enantiomers Video: R,S (Cahn-Ingold-Prelog) Naming System 2 The R/S system is used to provide information about the absolute configuration of a molecule. R/S configuration is assigned for each chiral centre in a compound. 1.) Assign priorities to each of the 4 groups connected to the chiral centre according to their atomic number. The atom with the highest atomic number is given higher priority and the lowest atomic number is given lowest priority.
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2.) If necessary, rotate the molecule so that the fourth priority is on the dash (facing backwards). Rotating the molecule may be difficult to visualise for some people so use this quick and simple method for step 2: In order to get the 4th group facing backwards (on the dash), swap the numbers on two adjacent groups around and then swap the other two numbers with each other.
3.) Count the other three atoms (1, 2, 3) and then determine if they are moving in a clockwise (CW) or counterclockwise (CCW) direction while counting from 1-3.
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If clockwise direction – R If counter clockwise direction – S Sometimes, the atoms connected to the chiral centre are the same. If so, look for the first point of difference to determine the group that is given highest priority. A double bond is considered two separate bonds. For example, a carbon double bonded to an oxygen will be treated as if the carbon is connected to two oxygen atoms. This also applies for triple bonds. The R or S is placed at the front of the name of the compound [e.g., (2R,3S)-3methyl-2-pentanol]. This has a chiral centre at carbon two and three. At carbon two, it is R configuration and L configuration at carbon three. In the GAMSAT, students may be asked to identify the correct compound based on a given name, or they may be given the compound structure and need to choose the correct name out of the multiple-choice options.
Relationships between Enantiomers and Diastereomers
Video: Stereoisomers, Enantiomers, Diastereomers, Constitutional Isomers, and Meso Compounds
Enantiomers are stereoisomers that are mirror images of one another. They have similar physical properties to each other.
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Diastereomers are stereoisomers that are not mirror images of one another. Cis/trans isomers are diastereoisomers. They have different physical properties from each other.
Fig. 31 - Examples of enantiomers and diastereomers.
To find the maximum number of stereoisomers: Max number of stereoisomers = 2n Where n = number of chiral centres
Example: Cholesterol has 8 chiral centres. 28 = 256 stereoisomers.
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Fig. 32 - Structure of cholesterol.
Any compound with 1 chiral centre must be a chiral compound. If a compound has 2 chiral centres, it can either be chiral or achiral. If the compound has a plane of symmetry, it will be achiral. If the compound lacks a plane of symmetry, it is most likely chiral (for the scope of GAMSAT, assume that it is always chiral).
Meso Compounds As previously mentioned, a compound is achiral if it has chiral centres and a plane of symmetry. These compounds are called meso compounds. If a group of stereoisomers contains a meso compound, there will be less than 2n stereoisomers. The following structure has two chiral centres and therefore 4 stereoisomers. In other words, two pairs of enantiomers are expected.
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The first pair of enantiomers meets our expectations. However, the second pair is actually just one compound. It has reflectional symmetry and is therefore a meso compound.
Fischer Projections This drawing style is commonly used with compounds that have multiple chiral centres. The following drawings are known as Fischer projections:
Fig. 33 - Fischer projections with different numbers of chiral centres.
The horizontal lines are coming out of the page (towards you) and the vertical lines are going into the page (away from you). Tip: Think of the compound as trying to hug you. The arms are the horizontal lines coming toward you.
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Substitution Reactions In a substitution reaction, one group is exchanged for another. This occurs when a suitable electrophile is treated with a nucleophile. The electrophile is known as the substrate, and it must contain a good leaving group. Common leaving groups include Br, I, and Cl.
Fig. 34 - Substitution reaction.
Alkyl Halides
Halogenated compounds are used as electrophiles in substitution reactions. Carbon atoms in the organic halide have special names. The carbon atom connected directly to the halogen is termed the . are connected to carbon.
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There are two mechanisms that can occur during the substitution reaction: 1. The nucleophilic attack and loss of a leaving group occur at the same time. This is known as a concerted process. 2. The leaving group is lost first and then the nucleophile attacks. This is known as a stepwise process.
The SN2 Mechanism Video: SN2 Reactions Video: SN2 Stereochemistry
Rate = k [substrate] [nucleophile] The rate equation above is second order because the rate is linearly dependent on the concentration of two different compounds. As a result, it is termed SN2 (substitution nucleophilic bimolecular). An SN2 reaction produces a concerted process. Remember: When an SN2 substitution reaction occurs, there is an inversion of configuration.
Fig. 36 - SN2 reaction showing an inversion of configuration.
This occurs because the nucleophile (Nu) can only attack from the side opposite to the leaving group (X). For an SN2 reaction, a methyl alkyl halide is most reactive. Reactivity decreases when moving from primary to secondary to tertiary alkyl halides.
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Fig. 37 - Change in rate with different types of substrates.
Steric hindrance from the extra side chains is responsible for the decrease in reactivity and reaction rate.
Fig. 38 - Steric effects.
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The SN1 Mechanism Video: SN1 Reactions
The following reaction contains a tertiary substrate. As previously mentioned, the SN2 reaction will experience steric hindrance and will be very slow. As a result, an SN1 reaction is likely to occur.
For an SN1 reaction, the rate equation is: Rate = k [substrate] (Br- is the nucleophile in this example) The SN1 reaction rate is only dependent on the concentration of the substrate. A change in nucleophile concentration will not affect the reaction rate.
Fig 39 - SN1 reaction and reaction rate.
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For the SN1 reaction, the leaving group (X) leaves first (rate determining step) and then the nucleophile (NU) attacks the carbocation formed. This mechanism is consistent with a stepwise process.
Fig. 40 - SN1 reactions can result in inversion or retention of configuration.
The trend for SN1 reactions is the reverse of SN2 reactions. Tertiary substrates are most reactive and methyl substrates are least reactive. When the leaving group leaves, a carbocation is formed and tertiary carbocations are more stable than secondary > primary > methyl alkyl halide. When the leaving group is eliminated, the carbocation formed is a planar compound. The nucleophile thus has an equal chance of attacking from either side of the carbocation. As a result, some products show retention of configuration and some show an inversion of configuration. For example, if the starting compounds have an R configuration, the products will show both R and S configuration (racemic).
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Fig. 41 - Example of SN1 reaction.
Remember: SN1: Tertiary substrates undergo substitution by the SN1 mechanism. SN2: Methyl and primary substrates undergo substitution by the SN2 mechanism. Secondary substrates can proceed by either SN1 or SN2. Look therefore at the nucleophiles to help identify the reaction mechanisms. Strong nucleophiles favour SN2 reactions and weak nucleophiles favour SN1 reactions. The following table should be memorised:
Table 3 - Common nucleophiles.
Also look at solvent effects when deciding what mechanism a secondary substrate will abide by. Protic solvents contain hydrogen connected to an electronegative atom, and polar aprotic solvents contain no hydrogen atoms connected to an electronegative atom.
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Protic solvents favour SN1 reactions while polar aprotic solvents favour SN2.
Fig. 42 - Protic and aprotic solvents.
Video: Solvent Effects on SN1 and SN2 Reactions
Overall mechanisms for SN1 and SN2 reactions (these will help with understanding):
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Table 4 – Comparing SN1 and SN2 reactions.
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Elimination Reactions: Structure and Preparation of Alkenes
Fig. 43 - Common alkenes and their structures.
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In an elimination reaction, a proton is removed fro -position together with a -position. After this elimination process, a C = C double bond is formed and is called an alkene.
Video: Naming Alkenes Examples Nomenclature of Alkenes 1. Identify the parent chain (replace the suffix -ane with -ene to indicate a double bond). Choose the longest chain that contains the double bond. 2. Identify the substituents. 3. Assign a location to each substituent (the double bond should receive the lowest number possible). A single number indicates the position of the double bond (e.g., if the double bond is between C3 and C4, only position 3 is used to designate location). 4. Alphabetically arrange the substituents.
Fig. 44 - Naming an alkene.
Longest chain including double bond = 7C – heptene Double bond is between carbons 2 and 3 – locant 2 assigned There are two methyl groups on carbon 5 and there is one methyl group on carbon 6 Putting these together makes: 5,5,6 – Trimethyl-2-heptene OR 5,5,6 – Trimethylhept-2-ene
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Stereoisomerism in Alkenes Cis and Trans Designations
Video: Cis-Trans and E-Z Naming Scheme for Alkenes
A double bond is composed of a sigma and a pi bond . The sigma 2 bond results in overlapping sp hybridized orbitals, and the pi bond results in overlapping p orbitals.
Fig. 45 - Sigma and pi bonds in ethylene.
Double bonds do not freely rotate at room temperature so the different stereoisomer must be observed. Cis/trans orientation is used to indicate the relative arrangement of similar groups.
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Cis - similar groups are on the same side of the double bond. Trans - similar groups are on the opposite side of the double bond.
Fig. 46 - Trans and cis configurations of 2-Butene.
E and Z Designations
Video: Entgegen-Zusammen Naming Scheme for Alkenes Examples
E and Z are used when an alkene possesses non-similar groups. Each substituent at the double bond is assigned a priority as follows: 1. Atoms of higher atomic number have higher priority. 2. If the first atoms are similar, higher priority is assigned to the group that has an atom with a higher atomic number at the first point of difference. 3. If atoms are all the same type, the group that contains a higher number of atoms will be given higher priority.
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Fig. 47 - E and Z configuration.
If the two groups with higher priority are on the same side of the double bond = Z configuration If the two groups with higher priority are on the opposite side of the double bond = E configuration *Easy way to remember: E is for enemy, which are in opposition to each other in battle.
Fig. 48 - E and Z configuration. www.acegamsat.com
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Stability of Alkenes
The trans alkene will generally be more stable than the cis alkene. This is because the cis alkene will produce greater steric strain, as the higher priority groups are closer to each other than when in trans configuration. Alkenes are more stable as they become more substituted. This is due to the delocalisation of electron density that produces a stabilising effect.
Elimination Reactions All elimination reactions involve the transfer of a proton and the loss of a leaving group. There are two possible mechanisms in elimination reactions: Concerted process: A base takes a proton and the leaving group leaves at the same time – E2 Stepwise process: The leaving group leaves first and then the base removes a proton –E1
Fig. 49 - Types of elimination reactions – E1 and E2.
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E2 Reactions
Video: E2 Reactions
The following is a representation of the E2 pathway. It is a concerted process. The base -carbon, a double bond forms, and the leaving group is eliminated.
Fig. 50 - Representation of pathway for E2 reaction.
Rate = K [Substrate][Base] The rate is linearly dependent on both the substrate and the base. As a result, it is a bimolecular elimination reaction (also termed E2).
Substrate Effect
The rate of reactivity increases from primary to secondary to tertiary substrates. This is because in the transition state, the C=C bond is forming and the double bond will be more substituted in tertiary substrates. A more substituted double bond will be more stable and will have a lower activation energy. Lower activation energy explains the observation that tertiary substrates react most rapidly in E2 reactions.
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Regioselectivity of E2 Reactions
The elimination reaction can sometimes produce more than one product. Hydrogen can -carbons. As a result, two regiochemical outcomes are produced:
Fig. 51 - Regioselectivity of E2 reactions.
The more substituted product is called the Zaitsev product. The less substituted product is called the Hofmann product.
Fig. 52 - Zaitsev and Hofmann products.
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position. It is a sterically unhindered base and thus it follows the general rule that the Zaitsev is the major product and Hofmann is the minor product. However, when a sterically hindered base is used, the Hofmann product is the major product and the Zaitsev is the minor product.
Fig. 53 - Reaction using sterically hindered bases.
In the GAMSAT, remember that the more substituted product (Zaitsev) would be the major product, unless it specifically mentions that a sterically hindered base is present.
Stereoselectivity of E2 Reactions
Having looked at regioselectivity, stereoselectivity will now be examined. The following compound forms only one product. This is because if an H position, the same product will result.
When this compound (3-bromopentane) undergoes an E2 reaction, two possible stereoisomeric alkenes are produced:
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Both cis and trans stereoisomers are produced. The major product is the trans stereoisomer and thus the reaction is said to be stereoselective (favours one stereoisomer over another).
E1 Reactions Video: E1 Reactions An E1 reaction occurs in a stepwise process. The leaving group is first removed and then the b carbon, forming the alkene.
Rate = k [substrate] Therefore, the rate only depends on the concentration of substrate (not the base). As a result, it is a unimolecular reaction. It is a unimolecular elimination reaction and is thus termed an E1 reaction.
Substrate Effect
The substrate effect has a similar trend to SN1 reactions in that tertiary substrates are more reactive. Tertiary carbocations are more stable than secondary > primary. Tertiary carbocations will thus have the lowest activation energy, allowing it to react more rapidly during the E1 reaction.
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Regioselectivity
For E1 reactions, the Zaitsev product (more substituted) is the major product. The minor product is the Hofmann product.
Stereoselectivity
In the resulting alkene, the trans isomer is the major product and the cis is the minor product.
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Addition Reactions Addition reactions involve the addition of two groups across the double bond of an alkene. This reaction is simply the reverse of an elimination reaction.
Hydrohalogenation
Hydrohalogenation involves the addition of hydrogen and a halogen across the double bond of an alkene. Two important rules include: 1. Markovnikov’s Rule: The hydrogen will bind to the carbon that is least substituted (already has the greater number of hydrogens). Observed for addition reactions involving HCl, HI, and HBr.
2. Anti-Markovnikov’s Rule: The hydrogen will bind to the carbon that is most substituted (fewer hydrogen atoms). If trace amounts of peroxides (ROOR) are present, then HBr will be added in an anti-Markovnikov fashion.
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The following reaction involves the electrophilic addition of H-Br to 2-butene:
Fig. 54 - Addition reaction of 2-bromo-2-butene.
The alkene removes a proton from the H-Br. A carbocation and a bromide ion are formed. The bromide ion then attacks the carbocation.
Hydration Hydration reactions involve adding water (H and OH or H2O) across the double bond of an alkene. The presence of an acid (H+) is important for catalysing these reactions. The following reaction occurs in a Markovnikov fashion:
Fig. 55 - Hydration reaction.
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For the scope of the GAMSAT, only concentrate on simple hydration reactions that occur in a Markovnikov fashion (shown above).
Hydrogenation
Hydrogenation generally requires the presence of a metal catalyst, which allows for the addition of molecular hydrogen (H2) across a double bond.
Fig. 56 - Hydrogenation of an alkene using a metal catalyst.
Halogenation
Halogenation involves the addition of Br2 or Cl2 across an alkene. The following shows the chlorination of ethylene to produce dichloroethane.
Fig. 57 - Halogenation of an alkene.
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Alkynes Alkynes are unsaturated hydrocarbons that contain a carbon-
Fig. 58 - Common alkynes and their structures.
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The following compound is acetylene.
Its triple bond contains two pi bonds and one sigma bond.
Fig. 59 - Sigma and pi bonds of an alkyne - acetylene.
Alkynes are named using the same steps when naming an alkene. However, the suffix yne is used to indicate the presence of the carbon-carbon triple bond instead of the -ene. The following shows the structure of 5,5,6-trimethyl-2-heptyne:
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Radical Reactions Video: Free Radical Reactions
In radical reactions, a single-barbed arrow is used to show the motion of one electron.
Fig. 60 - Motion of one electron in radical reactions.
There are three categories involving the fate of radicals.
Fig. 61 - Fate of radicals.
Initiation: Radicals are created. Termination: Two radicals form a bond with each other. Propagation: The location of the unpaired electron is moved from place to place.
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Alcohols and Phenols Video: Alcohols
Alcohols are compounds that contain a hydroxyl group (-OH).
Fig. 62 - Ethanol
The naming follows the same discrete steps as alkanes, but instead the suffix -e is replaced with -ol to indicate the presence of the -OH group.
Fig. 63 – Categories of alcohols.
When choosing the parent, identify the longest chain that includes the carbon connected to the -OH group. The -OH group will also receive lowest priority, despite the presence of double bonds or other alkyl substituents.
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Phenol is a unique type of alcohol. It has an -OH group attached directly to a phenyl ring. A phenol can specifically be used to identify hydroxybenzene.
Physical Properties of Alcohols Video: Alcohol Properties
The boiling point of an alcohol is much higher than its alkane counterpart. This is due to the hydrogen-bonding interactions that occur between the alcohol molecules. The -OH groups are responsible for these hydrogen-bonding interactions.
Table 5 – Differences in boiling points of alcohols and alkanes.
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Acidity of the Hydroxyl Group The acidity of a compound can be determined by looking at the stability of its conjugate base. The more stable the conjugate base, the greater the acidity.
Fig. 64 - Acidity differences in acids and their conjugate bases.
The conjugate base of an alcohol has a negative charge on the oxygen atom and is called an alkoxide ion.
Alcohols are more acidic than alkanes because the negative charge on the oxygen is more stable than the negative charge on a carbon.
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Predicting the Acidity of Alcohols and Phenols
Previously in the GAMSAT, students were required to predict which alcohol (or phenol), out of a number of alcohols, is more acidic than the others. The following are factors that affect the acidity of alcohols and phenols: 1. Resonance is an important factor affecting the acidity of alcohols. If resonance is present, the acidity is greater.
Fig. 65 - Examples of alcohols with different acidity.
2. Induction is another factor that can affect the acidity of alcohols. Electronegative atoms have the ability to attract negative charge towards them, which leads to the stabilisation of conjugate bases. The electron-withdrawing substituents (electronegative) can increase the acidity of a nearby atom.
Fig. 66 - Induction affecting the acidity of alcohols.
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Overall, remember these two important factors that affect acidity: - The electronegativity of the element (more electronegative = more acidic). - The distance between the electronegative element and the negative charge on the O in the conjugate base.
Preparation of Alcohols using Grignard Reagents A Grignard reagent is prepared by the reaction between an alkyl halide (RX) and magnesium (Mg). It has the form of R-Mg-X.
Fig. 67 - Grignard reaction.
The C-Mg bond is of great importance. Carbon and magnesium have such a great difference in electronegativity that the bond between them can be treated as separate charges.
The carbon nucleophile (R-) can then attack a range of electrophiles, including the carbonyl group of ketones and aldehydes.
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Fig. 68 - Steps in the Grignard reaction.
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Spectroscopy In previous years, students were required to identify the colour of light that has the longest wavelength or has the highest frequency out of a selected few wavelengths. An easy way to remember is through the colours of the rainbow, ROYGBIV (red, orange, yellow, green, blue, indigo, violet). The wavelength decreases and the frequency increases from R to V.
Fig. 69 - Electromagnetic spectrum.
An important equation to remember:
v (frequency) – number of wavelengths that pass a certain point in space per unit of time c – speed of light – the distance between adjacent peaks in an oscillating field
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Spectroscopy is used in organic chemistry for structure determination. In the GAMSAT, students are likely to receive questions that require the individual to identify the number of “peaks” that would show up for a compound. In other cases, students have been required to identify the desired compound, given the absorption spectrum.
Infrared (IR) Spectroscopy
The bonds in compounds will absorb IR radiation. Different bonds will absorb at different characteristic frequencies, which determines the type of bonds present in a compound. The process of IR spectroscopy involves irradiating the compound with all frequencies of IR radiation and then detecting which frequencies are absorbed. For example, a compound containing a C=O bond will absorb a frequency of IR radiation that is characteristic of the C=O bond. A plot is then constructed that shows the frequencies that were absorbed by the sample. This makes IR spectroscopy a useful method to identify the presence of functional groups in a compound. The following is an example of an absorption spectrum:
Fig. 70 - Absorption spectrum.
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For the purposes of the GAMSAT, students should be familiar with the above graph and the following general concepts: - Single bonds appear on the right side of the spectrum