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25 Must Kn ow Reactions - DESTROYER DYNAMITE SERIES, by Dr. Jim Romano
1.거\/ 거〉
(CH 3b CO-N é!+ / \ . . / (CH3bCOH
E2
/""、/
E2
NaOC2 H5 C2H5 0H
Zaitsev Elimination
A
2.
*Othεr basεs
~
4.
..
Hoffman EliminatioÞl
such as NaOH , NaNH2 , NaOCH 3 could also
NaCN
r
DM8 。
havε bεεn
elnployed.
~
8N2
•/
81\11
λ/
8N 1 with rearrangement
CN
↓ι I
H2。
Aγ
CH 30H,
,
。
CH3
4γ
까、/
H2。
E•
A
*3 0 n1ay also do SNl , but hεating (rεf1ux) favors
λ..../ 7.
Br2 hv
싸ι
thε
El
Radical 8ubstitution
Br
*Br2 favors 3 0 position
?
니
CH30H
L
·j
째내
해
8 끄μ
cH3 0 -
H
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Epoxide opening under basic conditons
25
Know Reactions - DESTROYER
H3 0 +
10.
/
H3 0+
CH3CH2CN
, by Dr. Jim Romano
Epoxíde opening under acidíc condítons
CH 30H
9. 끄μ
SERIES
‘
닙H
Nitrile Hydrolysis
CH3CH2COOH
A
11. CH 3CH 20H
s 。 CI? 。r
Mg
CH 3CH 3CI
CH 3CH 21\IIgCI
PCl 3
\1) CO 2
2)H 3 0+ \i
CH 3CH 2 COOH 1)LiAI~ 2)
2)H 3 0+
@@arH3
H 2。
CH 3CH 2 CH 2 0H (Acid Reduction)
Grígnard Preparation and Reactions
12. Mg CH 3CH 2 0H 열흙 CH 3CH 2CI-•
CH 3 CH 2 Mg다
PCI 3
μ“
。
P
야
카-
jη-
샌〕
5
C
태
Grignard Preparation and Rεactions with oxidations
\H
κι
CH 3
며싸
CH 2
。y
‘
CH 3 - 우 -H
태 대
H3 0+
------*‘
마
K2 Cr2 7 。
ν7 t /，껴
。
11 CH 3 -C-CH 2CH 3
/
CH 3CH 2 CH 2 CH 2 0H
。H
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25 Must Kn ow Reactions - DESTROYER DYNAMITE SERIES, by Dr. Jim Romano
+
까
c”
∞
q )
‘r
끽
A
꺼
/P
대 대
암
CH 3 CH 2 'C-CH 2 'C 、
랜
。
13. 4
4
)
。H
Dεcarboxylation
of a
ß-
Kεto
Acid
14. 。
H3。+
CH 3-C- O-C 2 Hs
1
CH~- c'l。
느
.;)
HO-C2 Hs
-\뻐
KOH 。
//
CH 3 -C、
-
’
"Q -K-r +
HOC2 Hs
Ester Hydrolysis under Acidic
때d
Basic conditions
15.
렷
‘
DMS 。
햇
+
。
q」〉
lH
꽤 매 | 매 l
해
메
+
-3 Pl (1
Wittig Rεaction
16. 센 우 H3
센 우 H3
HBr þ
CH3c=q CH 3
CH3 C-?-CH3 해 Br
Markovnikov Addition
1) BH3 , THF
|
끼「〕
l
B
℃|깨
C
바캔 〕
킷〕
대
H| C
2) H2 0 2 , OH-
비
며
FH3
CH 3 우 -F-CH3 。
Anti-Markovnikov Addition
CHγC 、
H rI
Anti-Markovnikov Addition
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]
、。
+
PH3
。 =C、
。 zonolysis
‘
CH
3
,
SERIES by Dr. Jim Romano
25 Must Know Reactions - DESTROYER
17. 1) NaNH -2 CH 3 'C 三 C-H
느
2)C 2 Hs l
-
/Na or Li
/'
/ H 3 C、
CH 3 'C 드 C-C 2 H s
H2' Li ndlars catalyst
\
In
NH 3 、
/H
H、
션 C=C. H3닝 、C 2 H S
C=C. 나 C 2 Hs
Alkyne reactions: Substituion and Reductions
18.
비
H2 0 , H2S0 4 CH 3 'C 프 C-H
HgS0 4
-
비
Tautomeric Shift
CH 3 유 -?-H
IÞ
CH3F=c-H
。fI
。H
eno1 .. unstable
NaBH
Alkyne Hydration with Reduction
비 CH3 p-C헤 3 。
H
19.
H H
H-C-C-H H-C-OH
(1
”
(1 |
ICH 3
”C
대
H2。
。
3
」n
CH 3'C-H
H 。 H
。
。
+
H2。
Aldol condensation followed by dehydration
7
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25 Must Kn ow Reactions - DESTROYER DYNAMITE SERIES, by Dr. Jim Romano
(
ι
)
”
ζυ
j
니
/
。
껴4
(-
ζ니
]
”c
니”
대 q
c”
。
。
20.
+
CHI3
Haloform Reaction
21. Br2
。 비 CH 3 .CH 2 .C-N 、
H 느
CH 3 CH 2 .N 、
。 H- ， H20
H
+
CO?
H
Hoffman Rearrangment H3 0+ 。
/’
。
CH 3CH 2.C:
CH 3CH 2. C 、。 . . . " -K+ _1/+
+ .j..
+
NH L1 +
。H
NH 3
(Acidic hydroysis of an amide)
(Basic hydrolysis of an amide)
22. H
+
\c/μN A 11
/ C \ r.1\1 H Diels Alder Reaction
(1
C
Claisen Condensation
24.
거v
CH 3NH 2
+
H 2。
Imine Formation
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빽닝
싹 얘
평짧
싹
{
때났매
84. During tbe chlorination of benzenesulfonic acid, wbicb structure represents a major resonance form?
85. Complete the product: κ、1 CH3CH2CH2Cl Br2 11 ’ ---닝 ~/ AIC1 3 hv
T\_1
ι
PT{、f1 1 l")
휩 |간H
써 r-C\A
c) - CNC22-HHsC5H3
띤〕
g
찮ld
C
87. Complete the sequence:
o
때 CH3CH2CH2~-다 즈땐정 짧 1) Sn HCl AIC1 3
HCl
H2S04 2) OH-
ξ〉설강〈죄 융tl a)
仁〉뀔징장Cl
b) Cl
)
웹굉
집
?
c)
d)
섰r뀔정효 @
Cl
b)
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34
90. Which position would Br‘ 2 and light or 어BS preferentially attack?
/
"CH2 CH 2 CH 2 CH 3 、
H a) 1
b)2 c) 3 d)4 e) 5
9 1. Complete the product: Peroxide
H
c=éH
남
Tnlti ' l f r. r
--.:
C)O d)O
e) A andB 41
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ORGA힘IC
CHEMISTRY - PROBLEMS
122. 1, 3-Dimethylcydohexane has two chiral carbons. How many possible stereoisomers can be written? a) 2 b)3 c)4 d)5 ε) 1 123. Which is true about the SN2 reaction? a) In thε transition statε， the carbon involved in thε rεaction changεs frOlTI sp 3 to Sp2 hybridizεd state rate ofrεaction depends on both nucleophilε and halidε c) 10 halides react faster than 3 0 halides d) Polar aprotic solvents arε often εlTIployed in thε reaction ε) All ofthεse
b)Thε
124. Which of these statements are true? a) Strong bases such as C 2 H s O下Ja↓ and CH 30-Na+ favor E 2 rεactions when 0 rεacted with 3 halides b) A highly polar solvεnt has a high diεlectric constant c)Thε propagation steps of f1 uoridatiol1 are h팽11yendothεrmlc d)A and B ε) All arε correct 125. Which statement about Pyrrole is true? a) It is an unsaturated aromatic amine b) It is a poor basε c) It is f1 at (planar) d) Two are corrεct e) All are correct 126. Which of the below compounds will react with cydohexanol to produce the following compound? 「
、ì
0 11
、\/'、、O/C\CH3 a) Acetyl chloride acid , H 3 0 c)Acεtic anhydride d)A and C e) All ofthese
-성)Acεtic
T
127. Which ofthe following will have bonds that vibrate at the highest frequency?
a)샤~
c) ø'\~
b)~\Z、/\ d)/、/\-.../
e)O
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128. An unknown compound gave a strong broad band at 1720 cm- 1 in the IR spectra, while the proton 하웹R showed only a triplet and a quartet. ，원lich is a iikely rnolecuie for the unknown'? n 0 ;-、 ’ P‘ , / CH CH CH c) CH CH ë-CH CH d) HO\ ) e) More than onε a) 3 T 2 3 3자 2 3 、
o용
thesε
129. Which bond is the shortest? F 亡
’
x
이
I
、l'
‘。 -CH
\딩
1
치|
[]
a b c d ε
사잉이 이 여 야
a
130. Which cornpound reacts fastest in an S아1 t)lle of reaction '?
f
a)l)
Fl
Cl
b) 대 c) 숨 Cl
d)때 e없 13 1. Which halide reacts fastest in the SNl reaction? a) Methyliodidε b) Isopropylchloridε c) t-butylbron1idε d) 2-chlorobutane e) t-butyliodidε 132. Which statement(s) are true? a) DNA lacks a 3 ’ 。 H group , thus is said to contain a deoxy sugar b)As the % ofadεnmε and thymine i11crεases so doεs the melting point of the sample c) Sugar and phosphatε groups arε on thε outsidε of‘ thε hεlix d) The nitrogenous basεs arε stackεd in parallε1 planεs outsidε the hεlix e) More than 011ε ofthεse ar;ε corrεct
43
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ORGANIC CHEMISTRY - PROBLEMS 133. Which compound below wiU absorb UV light at the longest wavelength? a)Bεnzene b)Anthracεne
c) Naphthalεnε d) 1 ， 3-butadienε ε) 1 ， 3 ， 5-octatrienε
134. Which gives the weakest IR signal for its functional group? H~C‘
a)
~H
b) ￡ c) 人〈
yy/C=C\ H CH3
e)
d) H3Ct=c:CH3 H H
CH3C프 C-CHrCHr CH3
135. Complete the product:
CHrCHr CH2-몽=몽-H
CH '1 0H 1) BH3 , THE 봐 (1 mole) 2) H20 2, OH- CH 2C1 2 H+ H
H H
c) CHr CHr CHr C\-CH3 O
떠 CH 3 -CH r CHr끽-F-o-CH3
H 0
OH
OH
/ CHrCHrCHrCH꽉-H
b)
e) None of these
CH:
ÇH 3 CH 3 "CH 2 "CH 2 "N-H
e) None of these
295. A researcher determined the % oxygen in five compounds. If each compound had four carbons, which would have the greatest % of oxygen by mass? a) b) c) d) e)
Aldehyde Ester Kεtone
Ether Alcohol
296. Which compound is most acidic? 에
d)
c)
a)
띠 ÞO
e)
0 0
297. The following halides underwent E2 elimination. Which halide would give the greatest number of products?
a) 것L\
c)
λ、/"\
d) 책k
b) 뽑r
298. (S) -lactic acid has a specmc rotation of +3.80 , while (R) -lactic acid has a specific rotation of 0 ‘ 3.8 • Which instrument would one use to identify the two compounds? t
A
ri
짧
m·ι
야
m
ε
S
m
앉따
-m
m
덩
Q
ρι
싸 ω π]
이 띠 리
’
t
따빠
E
M mS pi % K
때찌싸 따
하
사띠
N
mm
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84
-SOLUTIONS l. B This is a Sin1ple acid-base reaction. The base [potassimTI t- butoxide] simply abstracts a proton fron1 the acid [diisopropyl aminε]. Not，εAminεs arε usual1 y bases , but hεrε it acts as an acid! Thε final products arε tbutylalcohol and potassiun1 diisopropylmTIinε
2.D A and B ar，ε quickly εliminatεdbεcause thε two solvents cannot bε miscible in εach othεr. To extract an acid , we must have a basic solution, i.ε. pH = 10. At pH = 10 , we can deprotonatε the acid and bring it from the ethεr layεr， (C2H5)20 , into thε aquεous ， H 20 layεr. The ethyl bεnzεne will bε lεftbεhind and dissolvεsmcεly into thε εthεrlayεr.
3.A have 3 groups
Wε
hεre
; the OH group gives up a proton most easi1y to form the co ‘ jugate basε.
4.E For bases: -CH 3 > -NH 2 > -OH > Br- Thus A,B ,C are incorrec t. D is incorrect since thε Ocanrεmovε a small amount of electron dεnsity away from thε nitrogεn. Choicε E is corrεct， with less oxygεns ， the elεctrons arε morε availablε to thε N in order to capturε anH+. 5.D This rεaction has
thε
1TIercuriniunl ion as the
Ìntεrmεdiatε.
This cyclic
Ìntεrmεdiatε
1S
Sεεn
in choicε D.
싫Xj Br蘇Xj 따Mj 이4
/U니”\
ex.
·n
쨌
시대
’””m
ι 앤 바|‘
M mW …”
빼 적낭
(Any other structures would represent duplications)
r줬o
7. B Silver con1pounds such as Ag(NH3)2+ (tollens reagent) , or Ag 20 aldehydes into acids. 8. A Con1pound A is (trans) - 1, 4 -
diεthylcyclohexanε.
arε
n1ild oxidizing agεnts used to oxidize
For B , let us draw in thε hydrogens:
H C ，.， H익-~\
-
νr 、객-C
2 s H
H Onε ethyl is below a hydrogen, the Compound B Ìs also (trans) - 1, 4 - díεthylcyclohεxanε.
85
othεr εthyl
is above a hydrogεn. This is trans.
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ORGANIC CHEMISTRY -SOLUTIONS 9.A This is vεry tricky. This molecule has zεro degrees ofunsaturation, thus no doublε bond, ring, or triple bond can bε prεsεnt. Amides contain a double bond thus would not be seen. Let us find out why we have 0 degrees ofunsaturation. For each N , subtract one H , and recall , oxygen has no bearing, thus think ofthis compound as C 2oH 42 . Now comparε it to the nεarest alkanε: C 2o H 42 • glvεn C20 H 42 • nearest alkane The diffiεrence in hydrogεns is zero , thus thε molecu1e is complεtel y saturated. 10. B
꾀{I낭 당
Thε
ethyl directs thε br01nine para, followed by side chain oxidation.
11. B This is simply an oxidation followed by an addition of a Grignard. H
l
nr
H f
、
?•
I
H
H
,
、
2)
m
H
맛
mm
~~~_) ~~~L
H l
?ι
..-,n
m l
「~
1) r=t-CH2MgCl CH ~， C1 CHiCHì-C-H ---.H 2 2 H 0+
늦
9
pcc
CH~CH"CH"OH j ι ι
H
에
-「」
-찌〕
-rl· -”*]
71
-
에
rI
∞il--띠 ---
c
더-멈
-Fi--
]
센 -π ’
싸 --H
CryHζ R
i
[This added Grignard is called an Allylic Grignard] 12. D This is a tough about -6.
onε!
The H coming off a carbonyl
oxygεn
lS
εxtremely
acidic.
Thε
pKa for such a proton is
13. D Spεctroscopy is an idεntification techniquε not a sεparation tεchniquε. A mixture is first dissolved in εthanol and cooled. The para iS01nεr will crystallize out of solution, the ortho isomer will no t. The crystals are isolated, and the mothεr liquor is run through a COlUlTIll to purify the ortho isomεr.
14. D In thε first conlpound, thε carbon and N atoms are sp hybridizεd. Thε nonbonding ε1εctrons ofN occupy its second sp orbita1. In thε sεcond compound , thε carbon and N atoms arε Sp2 hybridized. Thε nonbonding electrons of the N occupy its sεcond Sp2 orbital.
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-SOLUTIO아S
15. B Condεnsation. Relllovε
CH需-O-CH 3 짧§꽁 +
갱 λ녕 밥쇄
This is a mixεd- Claisen shown bεlow:
alpha H from the second εstεr and thε 。CH 3 fron1
=
+
thζ
first as
CH 3 0H
This product has 1 chiral carbon( seε star) Thus 2 1 = 2 sterεOlsomεrs
16. B During recrystallization a solid is dissolved into a minimum amount of solv앉1t with thε goal to crystallizε out a more pure solid. The ideal solvent has thε solid fully soluble at high tεmperaturε and barεly solublε at low tempεraturε. This makεs choice B incorrect. Cooling should be donε slowly in ordεr to avoid gεtting any in1puritiεs trappεd in thε crystal. Long crystals indicate high purity since in1puritiεs will inhibit growth of thε crystal. The ideal solvent surely doεsn't bind to thε lattice structure sincε this would inhibit crystal growth, thus we want a low affinity for thε solid crystal. 17. B nvzisthε
mass to charge ratio. Essεntially， it is the molεcular mass of a compound missing a single εlεctron. This spεcie [radical cation] then is subjectεd to fragnlentation. Thε molεculε has a formula of C SH ll N0 2 . (Count this carefully) The mass is 153. Thus wε saythε n1/z ratio of thε molecular ion is 153. Notε ifa compound has an odd nl01ecular weight it mεans an odd nUlllbεr ofnitrogεns are prεsεnt. This is called the Nitrogen Rule. For practice , what is thε m1z ratio for the lllOlεcular ion of
Ck
CI A
/CI
CI)γlCI CI
This is C 6C1 6 12 x 6 = 72 for carbons 35 .4 5 x 6 = 213 for chlorinεs 72 + 213 = 285 for nvz
87
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ORGANIC CHEMISTRY -SOLUTIONS 18. D The carbon-carbon bonds in choice d are made up of sp-' -sp.) orbitals thus
arε thε
longes t.
Consider b:
비
H-C~C 드 C-H
버\ Sp3 - sp orbitals Consider c: H
H‘
세
‘
니
/C=C
~C: "，
H\
I-Í
II
\C/ 니 니
Sp3 - sp2 orbitals
Consider e:
H、
r.l r.l " " C=C-:-C-H
니
|꺼
Sp2 - sp3 orbitals
--?4
“
n3
rε-write thε
PH2CH2CH2 CH3- 던 ÇH 2 。
product and
gεt
유
19. C Thε trick is to
·n
-없
”M
뼈
mf
댐
”「‘끼/-
”삐
c mn o1mv
the 0 ’ s closε as show below:
K
Now Join
o=C
CH 3 20. C Y ou should be familiar with each of the above separation tεchniquεs. An azε。trope is a constant boil ing mixturε that Îs made up of substances in a cεrtain ratio that simple distillarion cannot sεparate. Thε mixture can eithεrhavε a boiling point higher or lowεr than eithεr cOlllponent. Ethanol - H2 0 is an examplε ofan azεotropε. Stεam distillation is vεry useflll to distill high boiling point substancεs and is vεη usεf띠 for the isolation of oils , waxes , and COlllplεx fats.
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PDFANDELEι'TRONIC FORMATS ARE
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2 1. C This reaεtion is a free-radical reaction. Remove the 3 0 benzylic y1랴 ds the most stable radical as shown: :、 ?H3 H3C~
hydrogεn
and substitute with a Br ... its rellloval
r~-CH2CH3
A very stable 3 0 benzylic radical is producεd when the benzylic H is re1lloved. 22. B lllust draw a chair and place the larger of thε two groups in the equatorial position. A mεthyl group has 4 atoms , Br has 1 atom; thus a mεthyl group occupies morε spacε. Now , renlεmbεr that wε need thε trans conformer so ...
Wε
Br
서
H 3 C-하-커-H Thε IIIεthyl
group is placεd equatorial, whilε the bronlinε is axia l.
23.B 。H
ν~CH3 CI'
씨5
ι」
\γ
3
4 Thε
C with the OH is #1.
Phεnols havε
an OH
attachεd
to a benzene rÍng
24. B Recall single bondεd carbons arε Sp3 , double-bonded carbons are Sp2 , and triplε-bonded are sp. 25. C A is a 10 B is a 3 0 C is a 2 0 D is a 3 0 E is a 4 0
anline amine allline amine amine (ammonium salt)
26. B This is the
enεdiol rεarrangεmεnt. Thε
thε εnedio l.
H
alpha hydrogen is rε1110ved to fonll an εnolatε ， Finally proton transDεrs as shown finÎsh the job. '/.0:
\,..-7
r, H
’
HÖ낀펴냥--oH HO -l- H
protonation yields
;Ö: 、c
C-OH
--•
r--‘
H
H-+-OH
OH
H-백
OH
CH 2 0H
thεn
CH 2 0H D-fructose
업lolate
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ORGANIC CHEMISTRY -SOLUTIONS 27. E This is a bit tricky. Choicεs B and D arε εlünÌnated quickly sincε the groups are eclipsεd， with choice B the most unstablζ. C is wrong it has 2 gauchε intεractions between brominεs and brominε and the mεthyl group. Choicε A is eliminatεd since CH 3 is a fairly largε group and this will raise thε energy. Choicε E has one Br - Br gauche interaction which is not nearly as bad as Br - CH3 gauche. 28. A In a solvolysis rεaction， the solvεnt acts as the nuclεophilε. Herε wε havε a 30 halide reacting with a polar solvεnt
in an SNl
CH':t CH3-t I
•
rεaction.
C..， H익 OH
~---，
CH y ':tj .L.J.
느
CH3 - 수 O-C H
’
2 s
CH3
CH
(t-butyl ethyl ether)
29. C Alcohols havε hydrogεn bonding which increases the intermolεcular forcεs of attraction which givε rise to a high boiling point. Choicεs C and D are alcohols. Now, recall that branching lowεrs thε boiling point because a branched molεcule has less surface arεa. Choicε C is corrεct sincε it is an unbranched alcoho1. 30. E This pair represεnts tautomεrs ， which are sinlply constitutional isomers that Usually, the more stable keto form prεdOlllinates.
까v
--•ÞOO--
ket。
co-εxist
in rapid ζquilibriulll.
까、/ enol
31.A A base will usε its 10nε εIεctron pair to capturε a proton. Electron withdrawing groups lllakε the compounds less basic. Choicεs B , C, E all have electron withdrawing groups which dεcreasε basicity. Thε εlεctrons are lllorε 100sεly he1d in choice A than choicε D , since choicε A has electrons in an sp-' orbital while choicε Dhas its elεctron pair in thε Sp2 orbital which is more tightly held. (Do you think electron pair is tightly held in an sp orbital).
CH3C프N: is very basic? No! Thε
32. D us draw thε molecule out: H ÇH 3
Lεt
C2 H 5' 、~H This is bεlow
thε trans confonllation. Thε εthyl is the larger of thε two groups and is placed equatoria1. an H , methyl1llust be above an H to get thε trans iS Olller.
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90
Sincε
ethyl is
써 F 낭 IF
X F
34. D Thε correct sεquεnce is as follows:
대 {끼》 「
o 35. C
COOH
m-bromobenzoic acíd
Simp1ε
distillation is based on molεculεs that have different boiling points. If the boiling points arε about 50 0 C apart, simplε distillation is a fine techniquε. The first c Olnpoul1d is to1uenε， BP = 111 oC and the second conlpound is diεthy1 ethεr， BP = 35 0 C. SiInple distillation will work here , with the ethεr distilling firs t.
36.D
37. C Thε sεquencε
is as follows:
。H
영
H “ C- H
0
‘
느명+
K2 Cr2。7 γ 기 H3 0+
생
P(Phh-CH2 악기 \ι/
\/
(Wittig product)
38. C 。
I')
excess
셨/ 擬7 느 거k/ 2 。H-=s H 0 ""
，/'、-.../
*1 2 ,
εxcεss ，
:1
'-'1
1
OH- is the ha1oform rεaction. This
。
q,
Hq 。+
V 。
B>C. Now, E is a 10 halidε as wε11 as D , but Br is a bεtter leaving group than Cl. Reca11 I>Br>Cl>F. Thus thε corrεct answer is C. 94. B This problεlTI
illustratεs
(~찮
야~
)3 P:
the synthesis of an ylide. The reaction is named the Wittig rεaction.
훌þ，.
THF /'““‘、‘，/"
o
(Smp뚫 $짧 2 reactl당 n)
n
당tt람서 c헤:) -c-O 섬
I'mιlut‘lB
95. A Elution order is depεndent on thε polarity of thε n101ecule. Let us considεr elution ordεr of solutes in colmTIll chrOlTIatography, alkanes>alkenes>ethers>estεrs>aldehydes and ketonεs>alcohols>acids. Choice C would be last to εlute sincε it is thε most polar. Choice A is an ester , and is thε least polar of thε molεculεs prεsentεd. Since the εstεr is 1εast polar, it wiU εlute first.
ηι
101
\/앓 〔시 댐
methylpentanoic acid
Cli----H μμ
5- hydroxy- 3-
H
”/-
〔닝fI
H
>
힘ν냉
、 αJt
-
FL
대
ζ、‘ g、흥~I"'\ 。
kf
(앗 ‘總
....
l
“따
..
가
~ HO- 갑 H 2 -CH 2 -Ç-Ç -c 、 .1. .1 、。 H
離 繼
r
~
l nπ 씨
96. D This is sin1ply a lactone bεing hydrolyzεd， followed by oxidation. $H3 ￠싹3 H .0
/
3- methylpentanedioic acid
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ORGANIC CHEMISTRY -SOLUTIONS 97. E Recall nitro groups , CN groups , halogεns ， S03H groups , and εlεctron-withdrawing groups increasε acidity. III is most acidic. II is 1εast since it lacks an aromatic group that hεlps to incrεasε acidity. 1 is lllore acidic than IV , slncε IV has a CH3 group [εlectron-donating] and will decrease acidity. 98. D This anlidε， propanamide is vεryunrεactive in nu c1 eophilic acyl substitution rεactions， but doεsundεrgo acid and basε hydrolysis as shown below.
S
。H-
。
H
ll
l
CH 3CH 2C -N -H
CH3CH2-C-0- + NH3 “
。
’
H3 0 '"
---•þo-
11 、
。H
startùlK compollnd Thε anlidε
has resonancε， thus no
frεε
+NH 4
+
CH3CH2 - C
rotation will occur betwεεn the C-N bond. Let us εxmllmε
thε resonancε
structure:
S부 CH 3CH 2C -N-H
?
+ 2H
CH 3CH 2C =N:'
τ~....
H
The molecule has double bond character between thε C and N , thus rotation is not permitted.
99.B D2， likε H 2 simply adds across thε double bond in a syn fashion. Thε D 2 moleculε must COlllε in syn fronl the bottom face bεcausε thε addition from the top is too stεrically hindered. Choice B c1εarly shows thε produc t. ‘
100. C HI04 is Pεriodic acid. It will
cleavε
OH groups on adjacεnt carbons and oxidize thelll ìnto
aldehydεs orkεtones.
때
OH
?텍 OH
fI
H iH
H3C-Ç-CHrCH2-Ç-t Ç-C2H5
4 ·
OH 0 H 3C-Ç-CH2-CH2-t-H H
Cut, then oxidize 101. B 1720 cm-l suggests a
O t.- - - .- .- -.-.-
N‘애~orma 따lly an OH group gives a broad band at about 3300 Clll-
O
1
.
。
An Anhydride: R -C-O-C-R would givε a pair ofbands around 1700 cm- 1• Notε:
AC드N
(Nitrilε) ~
2200 cm- 1
A )C=C(' (Alkene) ~
1650 cm- 1
A-C드C- (Alkyne) ~
2200 cm- 1
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0
+ H-~-C2H5
102. E A positivε Tollεn ’ s tεst can suggεst εithεr an aldehydε or an alpha -hydroxy kεtone. A positive tεst with 2 ,4DNP suggests a carbonyl group. B is an alpha-hydroxy kεtone ， whilε C is an aldehyde. Choice A would givε a positive test with 2 ,4 -DNP , but not with the Tollεns reagent. 103. B This con1pound would elεctronεgativε 。xygen
giviε
10 C-13 signals since it has 10 diffìεrent typεs of carbons. C-3 howεvεr has two at0111S dirεctly bonded to it; hence , the signal would appear n10st downfiεld.
vεry
104. B When in acid. attack the 1110rε substituted carbon. When in basζ attack the least substitutεd carbon. Cut
、
/LJ~" ‘ H판숫\:""CH~ 파
파\
ι
C2H5-0~H;) TT十 ‘
n
(HJ
I C 2H s I
y
녁L-J
、ν
1)"
Hf-F-CH3
H H
More substituted carbon
105. B This estεr is
phεnyl-2-mεthylbutanoate.
(Esters usually
106. B This is N-[3-chlorophenyl] propanan1ide, a 2 0 OH CH3-f-해- H ------ 10 Amide
o
CH~
o
CH1
smε11
good and are
usεd inpεrfUlnεs
and flavorings)
amidε.
CH3-f-하-다 ------ 20 Alnide CH3-t-하-CHx ------ 3O Amidet A cyclic
a111idε
is a Lactam e.g.
많。 니”
l
107. A Aromatic compounds must be cyclic , planar, with εvεry atom of the aromatic ring having a p-orbital , whilε obεying Huckel ’ s Rule of 4n + 2 π electrons.
A ~ Aron1atic B ~ Non-aromatic. It has an sp'、 carbon , thereforε lacks a p-orbital C ~ Anti-aromatic. It has the Sa111ε critεria as aro111atic , not 4n + 2 π electrons , but 4n π 랴εctrons. D ~ Non-ar0111atic. The Sp3 carbon does not have an elnpty p-orbital , thus the elεctrons cannot delocalize.
103
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영3뽑 품H3 ;C품j3jg짧f 정
fffir ROOR
109.B C1
짧많
110. B This is
thε
영
* HBr, ROOR adds the H and Br Anti -Markovnikov
OH
C2HSO-K+ C2H sOH
〈γCH3 1) BH3, T많 2) H20 2, OH-
V
타폈
‘…
l、 )'"Cμ3 + ..bnantlom
Diels-Alder Reaction, a 4 + 2 cycloaddition. This mεans that 4 carbons con1e fr01TI thε conjugated 2 carbons come from thε diεnophile. It is a onε-stεp process , occurring through a sing1e transition
diεnε ， whilε
state. [C oncεrtedMεchanism]
ψ
+
C總
11 1. A Sin1ply rεmove thε alpha hydrogens and replacε them with dεuteriums.
Choicε
A shows thε
corrεct structurε.
112. E Rand S reuεr to configuration, which is the arrangεn1ent of atoms in space at each chiral center. + and - refer to rotation, which must bε εxpεnmεntally detenninεd. Somε R moleculεs arε + son1e are -. Thε SalTIe goεs for S lTIolecules. + Rotation = dextrorotatory whilε - Rotation = levorotatory, thus nonε ofthesε are corrεct. (If you have an R enantiomer, its mirror image wi l1 bε an S.) 113. A This is a tough problem. Vinylic and aldehydic protons arε not very acidic , thus D and E arε quickly eliminatεd. C is not nearly as acidic as the two a1pha protons A and B. Howevεr， since A is εaSlεr for a base to rεmove ， it wi l1 bε morε acidic than B. Remembεr， an acid- basε reaction occurs quickly , thus A is less sterically hindered, and it will be removed faster. 114. A Propynε
is donating a proton to -:NH2 . A proton donor is a Bronstεad-Lowry acid. : NH 2 is thε BronsteadLowry base sincε it is accepting a proton. Notε thε arrow movemε， nt and makε surε you undεrs따ndthis forn1a1ism.
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115. B This is
thε
Tollens
Rεagent，
which
V\쇠
116. C This is a tough proble111. Lεt ’ s do the111 out H 까 Br') 짝 rR A ( \(~、H CH쩍~ 악-F-H L、 .) ι、 / Br
(
B
C
g
Br2 CH 2Br2
/\~\
앓5 crt쏠
D
o
Br,., r
CH 2Br2
+
Br Br.、~\
Br캘Hs
B뽑
Br CH 2Br2
흙Br
Not OpticalIy Activc
Will have a11 enantio111er only , it has only onε chiral carbon More tha11 one chiral center exists here , thus diast!εrε0111εrs arε possiblε
For εxa111plε would be diastereomers
Will have an enantiomεr on1y
117. A Proton A is from
thε
aldehyde group and will
appεar
around 9-10 ppm.
105
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ORGANIC CHEMISTRY -SOLUTIONS 118. C This rεaction is thε acetoacetic εster
synthεsis; do not worry about the n없nε ， just undεrstand thε chεnlistry. H 0 "' ~ ~ ~ ~~ 0 H 0 11 -,1 - 1 ) NaOC 1i< μ L 2 CH 3-C-C-C-O-C 2 H s -~-.. CH 3 G- ç-C-O-C 2 H s 2) CH 3 CH 2 Br H
o
T
ÇH 2
CH 3 Of1 (ζstεr
o
H
hydrolysis)
0
CH:r C-C-C-O-
C2H SOH
+
CH 2
9
H
|
CHrC{f-H 를
CH?
o
H
0
11
1
11
CH: r C-C-C-OH
(p rotonation)
1ε carboxylation 0 1'
‘
a +C02
A ß keto acíd
CH 3
CH2 CH 3
N otice in the first step how thε alkylatεd via an Sl‘'Ì2 rεaction.
doublε
alpha proton was
rεmovεd
by N aOC 2 H 5 , a strong base and was
119. B Rεcall Sp3 > Sp2 > sp in length. Look at the adjacent carbons on both sides of the respective bond:
I • sF-sp2
• 3• 4• 5• 2
sp2-sp2 Sp2 -Sp3 Sp3_ Sp 3 Sp3_ Sp 3
Clearly bond 2 had the
grεatest
amount of s character
120. D 30 alcohols dehydratε fastes t. A and E are 10 alcohols , B and C are 2 0 , only D is 3 0
CH1
샤μ 2-1ne뼈-2-but때0] OH
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•••
12 1. E Small rings such as cyclopropanε and alkynεs also hydrogenatε:
will open whεn
trεatεd
with H 2 / P t.
Alkenεs
as well as
H')
/\/
D
cyclobutanε
판
+H2
’
-- ~
휴
122. B Trap! ÇH i f"'*、
\/
、 CH
3
Notewε have 2 chiral carbons. So let us find thε nun1ber of thεoretical stereoisomεrs using the 2 N rulε， where N
= # of chiral carbons. 2 2 = 4 , but bεwarε of any possible rnεso compounds. Lεt ’ s sεε whatwε havε: IH3
!H3
탠γH3 털YH3
fH3
I
H3
III is rneso , thus would not havε an εnantiomer *Rεcall a lneso compound has no optical activity. It has chiral carbons , but thεre exists a pl때ε of symrnetry. Somε eXaInplεs ofn1εso compounds includε: ÇH3 ÇOOH
|I~’".’".’"'’".’ ...
Cl
Cl
'-、、/~ 흩CH
H-• -OH ! COOH 3
123. E us rεview the SN2 rεaction: a) The stereochemical result is 100 % invεrS lOn b) Rate = k[halide] [nucleophilε] c) Backsidε attack of the nuclεophi1e on thε halide. The Sp3 carbon goes to Sp2 then back to Sp3 d) Genεrally works best in solvεnts such as DMSO , acetone , HMPT which are polar aprotic ε) Works best with strong nucleophiles such as CN0 0 0 f) Methyl> 1 > 2 > 3 in rεactivity ___ H òB B c앵g_ H꽁 페페:/ + H"…1… x l…샌 1 c--'좌 - • t따R 에-꽉딩 따 :뉴-“………-“……-“…-“…. Lεt
•
1 ..
i- 、강~
Å
-1
Pεntacoordínatε
Transitíon State 124. D Strong basεs such as CH 30-Na+ ’ C 2H 5 0-Na' ’ and NaNH 2 favor the E 2 reaction , εspεcially if on a 2 0 or 30 halidε. A solvent with a high polarity such as H 20 has a high dielectric constant. If a solvent has a diεlectric constant ovεr 15 , wε rεfer to it as polar. Fluorinε is the most reactive rnεmber ofthε halogεn family. Rεcall F 2 > Ch > Br2 > h in radical rεactÎons. F 2 has a very high εxothennicity in thε propagating stεps ， thus rεacts very quickly. h has a vεry endothennic first propagating step , thus rεacts slowes t. 107
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ORGANIC
CHEMIST월Y
-SOLUTIONS
125. E { .. )
T;J H
is Pyrrole. It is an unsaturatεdhetεrocyclic aromatic amine Iht wwrmomm1u 뼈 1
126. E Let’ s do thεm out: a)
CH3파Cl
-D 노 CH3-@-썩=>
+ HO
b) CH3-@-OH +
+HCl
H잊] 노 CH3-@-달) +H2。
이 CH3-&。꿇H3 + H달) 검」 cH3-@-썩=:>+ 때3-@-OH 127. C Conjugation decrεasεs thε frequεncy ， thus choìce B is wrong. or singlε bonds , thus wiU havε thε greatεst frequency. 128. C
Trìplε
bonds have a greater
strεngth
than double
o
-~- group while a triplεt-quaπet indicates an ethyl group 1720 cnf 1 indicates a O
CHTCH2-~-CHTCH3 EiIlet
l
Q Ulirtet
(Note: choicε B would givε a singlet for thε C-l methyl group in addition to the triplet-quartet, and D has an 1 group which gives an IR signal around 3300 cm- )
。H
129. A ， bond has the grεatεst anlount of double bond character. B and D are long The shortεst have rεsonancε and are not true double bonds.
singlε
130. E F act: SN 1. Think 2 stεps ... Step #1 is slow. We form 감le carbocation; Here , s Î1nply sεε which one forms thε most stab1e carbocation!
thε
m
·wm‘ri ,
바 아 orril빼 bl)leζ야 e anti aromatic hm빼 십 i엄s horr
A
--’- ·% 4
·mι
+
is
뼈
옮 ir 2 띠
뼈
iS
+
+
띤
〔닝
[>
。
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108
thε
second step is
bonds. C and E
attack.
13 1. E wε rεvlewεd thε
S1‘.J2 reaction. Let us now rεVlεwthε SN 1 reaction: a) Reactivity is faster for 30 halidεs~ 30 >2 0 >>1 0 b) Rate = k[halidεL nucleophile indepεndεnt c) Sterεochεn1ical outcomε is partial invεrsion and partial racen1ization d) Accelεratεd by polar protic solvents such as H 2 0 and CH 30H , which can transltlon statε e) The slow step is the formation of thε carbocati 0n H H
--r、
hεlp
stabilize the carbocation
•==
Methylenεcyclohεxanε
152. B We must know a few facts bεfore we start. If you do an elimination on a cyclohexanε ring, draw a chair conformation firs t. Sεcondly， rlεcall that thε H and thε halogen (or Tos group) must both bε in thε axial position. Also , make sure that if a t-butyl group is prεsent that it is in thε equatorial position. By placing a group in thε εquatorial position, wε lllinlIlllZε 1,3 -diaxial intεractions. Only choicε B satisfiεs these critεria. Lεt us have a look: WhyB? 1/\ ÇH3 I This lll0 여 lε 링 cu 띠 괴 1 le 엽 i s a (ci 엽비 잉 S )-I-t 얘 t쉰 but)양떠 찌 y 4ι1-4.’-꽤’-’ FH3 H 臨잃 f뻐 inth 此 짧없때i려 p ε 뼈 떠 osS잎it뼈 뼈 10 on， 臨 ha 잃sn 뼈 리19 c 빼 뻐 h bo 야r디 n빼 ng 짧H짧'’S없 없X a 었 없i따’ x」 tH3 ι F-CH3 h and a t-butyl group equatoria This is able now 1. to CH3 r 4-t-butylcyclohεxenε H undenw εlimination to vicld * Thεεncircled atOl11S can depart readily r、
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153. B Hεrε wε
must first do CH3 Cl~H
thε
Rl S configuration.
C1뉘효-H
CHj Now sünply convert the Newlnan into a Fischer by doing as 1 present: CH i CH~ H r C1 "back" has been
C1γh
Cl'\--,../’ H CH:) RS vs. RR reprεsent
-t:-ì1
(Thε
swi雄d)
H diastεreomers
154. B Draw the moleculε out: /'、/‘、/ This wi lI be a conj ugated diεnε. It does th응 1, 2-addition at -40 oC but is under kinetic control , and at 50 oC we do the 1, 4-addition , but it will bε undεr thεnnodynanlic contro l. 155. E
H N
0 짧d '(섣)
Pyrrole
Furan are aromatic
C is non-arOlnatic , D is non-aromatic also , sincε both lack a p-orbital. 156. A This is a 30 halide reacting with a weak basε. Thε 2-mεthyl-2-butanol product is a rεsult of an SN 1 reaction. In an SNl rεaction (uninl01ecular) , thε ratε does not depend on thε nudeophile , only the halidε. Ratε = k [halidε] So if halidε triples , while nudeophile doubles , ratε will only triplεsmcε nudεophile plays no rolε in thε kinetics ‘
157.E SOCh and PCb will replacε thε OH group and crζatε the lεaving group , thus NaCl and C1 2 would not work.
halidε.
HCl also would work.
Thζ
OH group is a poor
158. B A , D , E all contain meta-directors that withdraw electrons and destabilize the transition statε during thε reaction. C and B both arε OIP activators , but thε OR group is morε activating than just an R group.
113
Copyright ((;)2006-2 ρ 15 ORGO}vL4N, LLC PDF AND ELECTRONIC FORMAT~S’ AREILLEGAL
ORGANIC CHEMISTRY -SOLUTIONS 159. A
펀
‘파얀
Tu1
H
‘ ‘
t
까liι
던￠쇄
‘ 웰 끼μ*
i
CH격- c;“L낸f 一---..
H
펴
.....
뿔 & 냐 ·H
묘1') J..&
TT
칸잃
짧- 캡H 2 꾀 H 2 -때 繼r
劉
HH ..J._.l.
CH3" 검늪검-H H훌 S()4
H2 S04 H되던l ’ ‘ ”까 펀H홀- 진-칸 됨 - lal.ROrl없nc 딘.tUI t
H
뭘따
H
1
뀔H3-갤-c낸11 따H Ul1펴table Erol
J
160. D
V [c 뚫k
(protected carbonyl , now an O/p director)
-
HOCH2CH20H
+ H+
FeBr3
혔k
H3 0+
‘
161. A In acid, attack the morε substituted carbon, wε
whilε
in
basε ，
are in basε.
C ·M
꺼@
1μ
,
’
- H3C-r--F-H CH3 CH3
H3
따.
이4
??
F2ff5JJj
o \o
·--n ℃
1
‘
(뭘/〉 [
c--H/ Ml
‘‘‘‘
깜
nn, /C
없
끼‘，‘
C、
H
O
m
162. B Alkεnεs
wiU add in a 1. 2 - addition mode.
~、
C1 2
/녕기 CI
Y
Z
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114
attack the least
， substitutεd
carbon. In this
εxanlplε ，
ε
163. A This is the rεduction of a Nitrile. Nitrile rεduction forms 1 anlinεs. LiAIH4 with H:z O (it would be a ruce way to blow yourselfup).
Hεrε thε ammε
164. D NaBH4 is a reducing agεnt that wiU nitriles , or acids. Lεt us look:
but unlike LiA IH 4 it doesn ’ t reduce
0
쩌 m
파← 꽤
짧l~/|대l 때머 대
o
Il--
‘
니
LiAlH 4, Et20
2)H2 0
7-?
•
1I
때 CH3-짧HrC1 +
Thε
or
kεtonεs，
use
estεrs ，
OH
벌
N샘H4i ..-
COOH
CH 30H
CH?
띠
~ 'COOH
CH?
COOCH 3
AlCI3
Nεver
COOCH 3
혔
CH3
lJ
H3 t-butylbe
10 halide rearranged! Think of it likε this:
ÇH 3 \I ~
H3C-t-Jt4c1
il- il Thε
rεducε aldεhydεs
is l-Butananline.
ÇH 3 짝 - H3C-F-F-H
H
carbocation formεd is now attackεdbythε
/、、
/
115
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ORGANIC CHEMISTRY -SOLUTIONS 166. A u1fonation is a rεvεrsib1e procεss.
$03 H
(j.1 \、 ι~
+ H 2 S04 - Þ ~ 、1 +H2S04~OL-~ J
J
、γ
\、 ß
$03H
r상l
hζ따
~~ Hεre
‘
( )
H 1 0+ \/ is a quick “ Road Map" of aromatic chemistry:
젊 K2따씬까
싫 魔
따
넓 E; 뚫단 젊 2
f
2
Cl
꿨표‘
J 피」
7、‘
116
ILLEGAL
」 LlO
c
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fli
H 0+
CH
PDFANDELEι'TRONIC FORMATS ARE
1카」
QQ
야써
e
짜
mn”
ηm
따
ilι
,, l
1
써뼈 뼈
μ ”
G g
+ CH4
않
H3C-C=~-MgBr
ClflO H3 B
H
기」
돗~CH3 MgBr -
‘키
‘、 H 3 C-C프 C-
+
「L
f
ι W ι 넉n 넉며 「 ~
167. C
f
H
(1j
-SOLUTIONS 168.B This
11101εculε
has
CH"OH
l'-' LlO~
ahεn1IaCεtal
1:1
group as shown with an arrow at C -1 : This is a D-sugar since CH20H group is above the plane of the ring
(anomerÍ c carbon) at c-l
H
If a he111iacetal group is prεsent， the 11101eculε can open up to its free test, mutarotate , and be classifiεd as arεducing sugar
aldεhydε
group ,
glVε
a+
Toll앉lS
and
Bεnεdicts
* If the OH group at C-I bεta
is down , we call it thε aψha form. If the OH group at C-I is fonn. Alpha and beta forms arε rεlated as anomεrs.
pushεd
up , we call it the
169.B This is a con1l11Only asked question. Sucrosε is fructosε to forn1 thε below con1pound:
madε bythε
joining of 1 molecule of glucosζ with 1 moleculε of
-Li nkage at glucose
二:\{H HO I
Ò
,OH r、
/ν\\
l
Of왜←--
ß-Lillkage at f
‘--냥’ CH，OH OH
Sucrose Sucrose doεs not have a hemiacetal group , thus is a non-reducing sugar. Thε linkagε bεtwεεn thε two molecules is called a glucosidic linkagε (Acetal group). A Non-Reducing sugar will not mutarotatε. It will give a negativε Tollεns and Benedicts Test.
170.B
껴3/ 3으말ik/ (CH3)3CO-K~... ~CH 3 COOOH 갓~ 댈으플 H-묻-짧H3 개잔utHel'e
(CH3)COH
HT
Ó찌
3
페;낮뀐3.; 17 1. D PCC is a nlild oxidizing agεnt that oxidizes 10 alcohols into 。、/H
aldεhydes
and 2 0 alcohols into kεtonεs.
PCC
ì OH
L
r1 2Ln2un
CH2 C12
0
117
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ORGANIC CHEMISTRY -SOLUTIONS 172. E Let ’ s do thεmout:
-p
-+
>
」
o 훨 6 뚫환
λ끼
빽니
189. C
때댔
(Cyc1 ohexylmagnesium bromidε)
+CH30H
(Cyc1 ohexanoic acid)
A
” :,‘ …
t
]뀔…
κ…
f
O
…
I) CH3MgCl
ÞO-
2)H30+
OH H~C-갓-CH~
*always assume 2 moles ofGrignard ifyou havε an ester
tH3 (t-butylalcohol)
19 1. C This is a 20 halide , base is wεak， solvent is polar aprotic_ 100% inversion of configuration and thε ratε law is: Ratε = k [halide][ nucleophilε] Thε final product n1ust bε S: H_ R.-NaN'1 HA ，N~ 져차ν Acetone ;>ζν
R
Thesε arε thε
conditions that suggest SN2 , thus
wε gεt
S
192. D Branchingnεxt
to a doublε bond often will indicatε a possiblε shift. CH찌 H H I CH3 H H Hvdride shiH to ield CH찌 HH -Ç-Ç=C-H + +.6、 Slow - CHx-t-t-ty ylelu a::;C 많 H3 쉰쉰쉰끌끌3'γγ’-3-’ H H
파파
값
파낭 파
mo않 S빼le c總1on
k
|파 파
H 20:1 CH~
rH3 CH3-Ç-CH-2CH-3
ÓH
2-methyl-2-butanol COPJ ’right ([;;2006-2015 ORGOlvfAN, LLC
122
CH3-t-kH2-CH3 +:O-H “
Tj
H씨~\
\'-
/t1
193. B
o
An
0
εster has this group: R-간-O-R. The R-칸-
derivζd from thε acid
~ Es않r+wat벼
| Acid + Alcoho1 Leγs
portion is
do an εxample:
H十
9 ------
R-ë낙OHj
9
+:.. __H• OCHr CH3- -• R-C-O-CH2 CH3 + H 20
t
-γ
194. A At a pH = 1.6 , all groups will be in their protonatεd forn1s. H
+H3N-F-COOH ÇH2
\
γ '1 、/'
r pKa ~2.3
I thus at pH = 2 .3 wε 10sε 50 % ofthεsε hydrogens
thus at pH = 9.6 we lose 50 % ofthεsε hydrogens
* 댐Thεn this n101εculε lS nεutral i.e. at its isoelεctric point, wε sεε: H
+H3N-f-cooÇH2 ι갓\
ιJ 195. C This is the
[
+
and call it a Zwitterion
Diεls-Alder rεactÌon.
NRf:쏠
heat
띠fr§
Note: reaction producεs the trans isomer here , since thε
diεnophilε
was also trans. Thus , thε reaction is said to
bε stεreospecific.
123
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ORGANIC CHEMISTRY -SOLUTIONS 196. B This is thε
Claisεnrεaction. Wε havε
an εstεr with an alpha hydrogεn in basε. Proceed as follows:
[Assumε 2molεs]
o
HO
CH3-CH2-f-쉰-C 2熊 + CH3-j 것 -O-C 2H s 뭔!@ @
j
1_
H 0
，-、
(g) CH3-t-f-o-C2H5 tFrom C=O ÇH2 CH 1
ì
~ From @
+ C2HsOH
197. E C펜~:
CH2C12
CA---F-H
OH페 O
i
CH3-t녁섣텔 다
'-'-'--'-5
0 ” 。-9
ι꺼 f
).、‘
-ζU
\3
Aγ
”4-
mm 3
Aldol Product
꺼n
~CC
--
&B
O /1l
-
o
f”
n 범
lll
---.:.,
석---?
.")
묘u
CH 때r OH
0
H-C-C-H
At C-3 wε sεε a “ doublε alpha" proton. This wiU be thε most acidic. By removal of thε double alpha proton, wε yield a rεsonancε stabilizεd Enolate anion as shown: o 0
I^서낭〉
••
셋갔」、/ n
Rεmoval
of the alpha hydrogen C-l or C-5 will yield either from an enolatε ， but with only two resonancε fonns , not threε.
III 거、~
199. D This is a Markovnikov addition, but sincε therε is branching adjacent to the double bond, bε suspicious of a shift! ÇH 3 ÇH 끽 ij ij ÇH3~ ~ dξ CH3-&-t-t-H ζ • CH;-t-t-t-H j j + H-:-Br 3 ωlit C:
H，파닫 몽도풍-H ζy~ 파
l
'-'.L.L
파냥 파
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바뼈
124
k
때
- i
:다r:
CH3-&-CH2댔
-SOLUTIONS 200 ‘ A Look for 4
difúεrεnt
“ paths". Chiral carbons will bε starrεd ...
*
CH;
Wε have 8 chiral carbons. [If askεd for thε numbεr of stereoisomers 2 or 256 ] 8
20 1. C FεBr3 is thε
Lεwis
Acid catalyst usεd.
때+ Br2 織 202.B The morε
conjugatεd thε
system, the more
stablε.
@聽때& 빠
203.B A Low Vapor Prεssure = High Boiling Point. Compounds with hydrogen bonding such as alcohols show high boiling points and low vapor prεssurεs.
m
H H
There are only 2 chiral carbons , The lniddle C is not chiral , because the same groups appear above and below it.
… H
205. B
。
R
R-t-CI + H-하-H Acyl halide
10 Alnine
OR
R-t-하-H + HCl Amide
206. C This is a simple rεduction of an alkεnε using a Pt or Pd catalys t.
~
H2/'"\/\/ Pt
207.B The carboxy acid has thε greatest aInount of hydrogen-bonding watεr to thε grεat，εst extent.
ability~
thus wi I1 be lnost likely to dissolve in
208. C A pipet is calibrated to deliver a spεcific volumε of a liquid whεn filled to thε mark and burεt will dεliver known volumεs of solution, but during a titration. 125
allowεd
to drain. The
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ORGANIC CHEMISTRY -SOLUTIONS 209.E Let us do
thε
Rl S CP--.......
OH
\
↓‘’H이
A)
R。
@CHf
R
(R)-2-butanol
써\]
@
/3
Carboxy Acids>Alcohols>Ketones and Aldεhydes> Alkynεs>Alkεnεs>Alkanes (Most Acidic) (Least Acidic) 25 1. B This is saponification. An estεr (actually a cyclic 。
or lactonε)
。
Cut
κ〉’18
1) KOH
lγ 거
2) H3 0+
11
εstεr
‘
-1 \
n
I /I"'\ LJ ‘
rOH
~麻
252.B Ha10gεns
such as Br or C1 wi I1
dεactivate
the ring , but wi I1 direct ortho/para. 141
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ORGANIC CHEMISTRY -SOLUTIONS 253.D In column chromatography, a mixturε of compounds is placεd atthε top of a solid support. The nlotion of solute and solvεnt through thε solid phase is called elution. The nlore polar a compound is , the slowεr it will travε1， and thε longer it would take to elute. Thε silica gel is polar, and attracts polar molεculεs ， while the nonpolar nl01εcules do not strongly attract , thus will elute firs t. Here , only cyclohexanε is non-polar, thus will εlute firs 1. All thε others arε polar molεculεs that you should εasily be able to draw.
254.B The insoluble Ílnpurity, namεly sand must first be removed. lt is easily donε by filtration. Now , thε two renlaining c Olnpounds are hεxanoic acid and toluεnε. Basic εxtraction is done to separate the acid into the aqueouslayεr. Toluenε is all that remains in the ethεrlayεr， and can be relnoved by εvaporation of the ethεr. 255.B An aldεhydε
proton would bε sεεn around 9ppm, thus choice C is eliminatiε d. Therε is no signal around 1113ppm thus , d and ε are eliminated. A COOH proton comes in vεry downfield bεtween 11 and 13 ppm. Choice A is wrong since no triplεt-quartet is noted for thε εthyl group. Choice B is correct the signal is split at around 7 ppln since an εlectronεgativε atom is dirεctly attached to the benzenε nng. 256.A Thε kεY
to this problem is to rεalize that the N O 2 group , likε the CN , S03H , COOH groups are εlεctronwithdrawing. An electron -withdrawing group will destabilize thε carbocation and increase the εnεrgy 1εading to the transition statε. This is whya meta director such as N0 2 is said to bε deactivating. Choice A is clεarly the 1εast stable sincε an elεctron -withdrawing group residεs on an already electron-deficient carbon. Relnelnber; avoid placing a deactivating group dirεctl y on a carbon bεaring a positive charge! 257.A A carbonyl group appears around 1720 cm- 1 thus B , D , E arε all easily εliminated. An alcohol gives a broad signal around 3300 cm- 1 and is choicε A. Choice C is an ethεr and doεs not give a signal in thε 3300 cm- 1 rε glOn.
258.A Rεcall
that al합1 groups such as lUεthyl releasε electrons by hyperconjugation. Examinε choice A carεfully.ln Choice A , you will see that the methyl group is dirεctlyattachεd to thε εlectron deficient carbon (i.ε. thε one bεaring the positivε charge). A group such as lnethyl can release electrons and stabilizε thε carbocation sincε the positiv，ε charge is rεadily dispεrsed in thε transition state. Choicεs C, and D arε all valid resonancε hybrids, thεy do not havε thε stabi1ity , as that seen in hybrid A. B is wrong since a positivε charge resides on a carbon with four bonds. 259.C This is a 3 0 halide being hεatεd with a strong base. Recall , a 2 0 or 3 0 halidε hεated with strong basεs such as KOH , NaOH, NaNH 2 , CH 30Na, C 2 H 5 0Na yield E2 elimination products. Sincε the basε is small , wε fonn the 1110re substitutεd alkenε， thε Zaitsev product. (Choicε B is a minor product, but would havε bεen the ln며 or product if a stεrically hinderεd base such as (CH 3 )3 CONa was employεd.
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142
260.B All ofthε rεsonance forms arε correctly writtεn and valid stnlcturεs. Howεvεr， choicε B is n10st stable sincε each atom has a con1plete octe t. 1n addition , an “ extra" bond is created in resonance hybrid B. Resonanc응 hybrid B makes a largε contribution to thε ovεrall structure. 26 1. ε
Saturated fatty acids arε found in grεat abundance in anin1al fat. Saturatεd fa얀yacid tails pack very tightly together, giving a triglyceride with a high melting point, and solid or semi-solid appearance. 1f a double bond is prεsεnt， wε havε alowεrmε 1ting point unsaturatεd fatty acid that tεnds to be fluid- likε. A doublε bond producεsa “ kink" in the molεcule and prεvεnts it fr Oll l tightly packing. As a rεsult， unsaturatεd fatty acids havε fewer intermolecular attractions and havε lowεrn1εlting points than saturated fattyacids. Fish and plant oils are abundant in unsaturated fatty acids.
262.C
CH 3 CH 2 CHi)H
-
K2Cr20 7 H •
ι0 CH 3CH2-C
\
OH
。11 CH 3CH2-C-Cl
o H Br, CH 3CH2NH2
를
OH-
11
.
CH3CH2-C-~-H
"
+COì
The last step 、iVas the Hoffmann rearrangεmεnt wherε a primary amide lost a C
143
== 0 and became an anlÌnε.
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ORGANIC CHEMISTRY -SOLUTIONS 263.E This is a mixed AldoL Sinlply rεmovε the alpha hydrogεn from that aldehyde which contains the alpha hydrogen then attach the aldehyde 1acking the alpha hydrogen. Let us try:
CH 3 CH2
eri
$
얹
”oc
o
-H20
11|
껴3
A
‘때 、 、、 Ll
dεcidε
'" Cl
R"-
+
CH，， -CH， γνν‘-’
-
H?O
11
H-C
l -:-
264.C First, let us
잉
職
%
H‘ ---1J
l“
c m?----• T---nu l c ; f
1o
뺑
?n m
짧
h
며
、‘
l
’
qμ
/
ε
、‘
•
샤따
([)? ·mm
M’kU mM
빠
M
@
ifthe nl01εcule is E or Z.
/ C==C / \ \
CH3
*denotes group of higher priority
H
-
;
1j
3
l
• • |
-
컨 T
C H~--
|
π.A
α 2
1
돼|니
This is clεar1y a Z. Next, draw out the molecu1ε and find the longest continuous carbon chain bearing thε double bond as shown:
‘ t
5
H ‘ /-. -H i
-?4
||
/0
H
H
C
Ju
?4
h
i
m 1n“V
n‘
p
뾰
4.
야
m
ζJ
껴/i
·m
/t‘ ·、
·E
잉
돼
CH--
빼
|| ?J
265.C Choices A , B and D all involvε shifts that result in a morε stablε carbocation. However, in choice C a 2。 carbocation goεs to a 10 carbocation; this shift is εnεrgεtically unfavorable and will not likεly occur. 266.C This is thε Baeyer-Villigεr reaction. An aldehyde or ketone is converted into an ester. If given a ketone , insεrt thε o closest to the most branchεd portion ofthε molecule closest to thε carbonyl group. If givεn an aldehyde , lnsεrt the 0 closεst to the hydrogen atOlTI.
Copy’right fructose>1품>NaCl b) NaCl>MgO>NaCl>HF c) MgO>NaCl>HF>fructosε d) MgO>NaCl>fructosε>HF ε)
HF>MgO>NaCl>참uctose
110. Which of the following reactions
짧ustrate
homolysis?
IBr •I" +Br" b) IBr •1" + Br+ IBr ---* r+ + Br d) IBr • IBr+ +e e) B and C 11 1. 10 amps are passed through a solution ofNiS0 4 for one hour. How many grams ofnickel would be electroplated? (1 mole e- = 96,500 C) a) b)
d)
(1 )(59)
96500 (36 , 000)(59) (96 , 500)(2) (36 , 000)(2) (59)(96, 500) (36, 000) (59)(96, 500) (96 , 500)(2) (36, 000)(59) 177
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GENERAL CHEMISTRY - PROBLEMS 112. Consider the following reaction:
=
2 Fe+2(aq) + Cu+2 (aq)
2 Fe+3(aq) + Cu (S)
Given the follow data for reduction potentials: 느 Fe+ 2 E O = 0.77 volts 1 e-+ Fe+3 느
2 e- + Cu+2
E O= 0.34 vo1ts
Cu
Which is a false statement? Thε E for this rεaction is 1.11 vo1ts O When calcu1ating E , the coefficients arε not takεn into account O
a) b) c) d) e)
At εqui1ibrium thε cell vo1tage is zero Cu+2 is the oxidizing agεnt Two ofthε above
113. Calculate the heat of combustion of C 2H 4 given the following data at 1 atm and 25 oC. Compound MI O (KJ Imole) -286 H2 0 (l) -242 H2 0 (g) -394 CO 2 (g) +52 C2~ (g) a) b) c) d) e)
-1412 KJ -1756 KJ -1324 KJ -3560 KJ -1570 KJ
114. CaCh has a weight of Illg/mole. How many grams of CaCh. 2H 20 would be required to make a 1M solution? 1
1ll
m) mf
孤
---------li 1li
mf
팽 r
,,
1li1li i
m
뼈
1
「/‘
이 띠 리
때)J% η--까
---14 , ---「/」
떠
N
m m
•
115. Consider th ction A + B C~ the followin 1! kinetic data was obtained: Trial [A] Rate ( moles Il'sec) [B] 9.1 X 10-6 .20 .40 8.0 X 10-5 2 .40 .60 9.1 x 10-6 3 .20 .60 Which is the value of the rate constant? a) b) c) d)
2 .2 x 10-4 M- 1 Sεc2.2 X 10-4 M-2 Sεc- 1 3.2 X 10-6 M-2 sec- 1 3.2 X 10-6 M- 1 Sεc
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178
116.
εonsider
the following reaction:dr 4 H\aq) + O2 (g) + 4e2 H2 0 'þ
(1)
If sodium hydroxide is added, and assuming 02(g) is at 1 atm and [H+] = 1.0 M , which statement(s) is true? a) b) c) d) e)
Thε E valuε would not changε smcε base is not part of this The E valuε would incrεase. The rεaction quotiεnt would takε H 20 (l) into accoun t. Two ofthεsε arε true. None are correc t.
systεm.
117. Calculate the pOH of a buffer solution that contains 0.10 M benzoic acid and 0.010 M potassium benzoate if the Ka of the acid is 6.5 x 10-5 • 0 oo 6 4 Q/5 II 「/ Q。
사띠 비 이 따 라
1)2
118. 찌lhich statement(s) is true? a) b) c) d) ε)
A rεaction that has a +MI and a -i1 S is spontanεous at all ten1pεratures. If the MI and i1 S arε both positive , a reaction will be spontaneous only at low tεlTIpεratures A -i1G usually impliεs that a reaction will occur quickly C4 H lO would have a highεr εntropy than C 3Hs. Two ofthε abovε.
119. A gas is compressed from 40 liters to 16liters at a constant pressure of 4.0 atm. If 10 KJ of energy is released as heat, find the work done for this process. a) b) c) d) e)
-96 1. atm +96 1. atm -10 1. atm -640 l. atn1 No work, since P is constan t.
120. Consider the below reaction: 2A+B JÞ 3C AG=-150앙 KJ Which statement(s) is false? a) b) c) d) ε)
Thε rεaction will1TIOVe rapidly from left to right The rεaction is spontanεous as writtεn The reaction is thermodynmnÌ cally favorable Thε εquilibrium mixture will consist almost exclusivεlyofC Two of the above
12 1. Which statement(s) is false? a) b) c) d) e)
All spontaneous rεactions result in an increase in enthalpy ofthε umvεrsε All spontaneous rεactions rεsult in an incrεasε 111 εntropy of thε umverse When a liquid vaporizεS ， AH and i1 S arε both positive q and w arε not statε functions Two ofthε above 179
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GENERAL CHEMISTRY - PROBLEMS 122. A 0.010M solution of an unknown compound Y has an absorbance of 0.5 at 360 nm. What concentration is a solution of Y which gives an absorbance reading of 0.25 at the same 360 nm and path length? a) b) c) d)
5 xl0- 3 M 1x10-2 M lx10-3 M 2x10-2 M 5x10-2 M
123. What is the best indicator that a chemical reaction is spontaneous? a) b) c) d) e)
Enthalpy Free Enεrgy Entropy Potential Enεrgy Thεrmal Energy ln 찌Tater.
124. Brine is a solution that usually consists of a) Bromine b) Sodium chloridε c) Pεtroleum d) Lithiun1 iodide ε) Barium sulfate
125. Consider the following particles; proton, electron, neutron, alpha particle. How do they compare in mass? a) b) c) d) e)
proton > elζctron > n응utron > alpha particle alpha particle > proton > neutron > e1ectron alpha particlε > neutron > proton > εlεctron neutron > alpha particle > proton > el응ctron nεutron > proton > alpha particle > eIεctron 14
126. The Ksp ofFe(OHh at 25 oC is 1.6xl0- • 찌That is the solubility of Fe(OH)2 in 0.025 센'1 FeClz?
d)
5.2 x 10-8 4 x10-7 4 X 10-9 6.8 x10- 11
e)
1. 0 x 10- 11
a) b) c)
3
127. A solution ofLiCI has a density of 1.13 g/cm at 25 oC. If the solution is 20 is the molarity? 7.6 M 0 .2 5 M 5.3 M 11 .2 M ε) 1. 5 M Copyright rt:J2006-2015 a) b) c) d)
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%
by weight LiCI, which
GENERAL
-PROBLE민1: S
128. Calculate the oxidation number on N and P in Kl'T 0 2 and P0 4-3 respectively. a) b) c)
d) e)
+3 , +5 +5 , +3 +2 , +3 +4 , -5 +3 , -5
129. Sodium sulfate reacts with barium iodide. Which reaction below depicts the net ionic equation? a) Na 2S03(aq) + Ba1 2(aq)
---+0-
BaS0 3(s) + 2 Na1(aq)
b) 2 Na+(aq) + S04치aq) + Ba+2(aq) + 2I-(aq) ---용 BaS04/~\ 4(s)
+2 '''-
Na-;-(~_\ + L 'W (aq) ,
2I- (aq)
':'i
c) Ba+2(a q) + S03칸a띠용 BaS0 3(s) d) Ba+2(aq) + S04-2(a~ BaSOis) e) 2 Na\aq) + 21-(aqτ--흥 2Na1(s)
130. How much heat is ~~~ HCI0 3 > HCI0 2 > HCIO in acid strength 1n a closed containεr， thε vapor pressure wílI dεpend on thε surface area of thε liquíd. Hydrogen bonding wi1l cause the solid- liquid line to be skewed to the left in a phase diagram ofP vs T. 1n a body-centered cubic unit cell , Wε have 2 atoms. B and C
140. Consider an NaN0 3 solution at 00 C. Calculate the osmotic pressure of a .20M solution assuming 100 q끼ó dissociation. a) b) c) d) ε)
5.0 atm 760 n1m Hg 9.0 atm 12 atm 2 .4 atm
14 1. 찌1hat is the net ionic equation if HBr reacted with LiOH? a)
Li+ (aq) + OH (aq) + H+ (aq) + Br- (aq) • H 2 0 (l) + LiBr(aq)
b)
LiOH(aq) • Li+(aq) +0표-(aq)
c)
Li +(aq) + Br - (aq) • LíBr(aq)
d)
H+ (aq) + OH (aq) • H 2 0 (l)
e)
HBr(aq) • H+ (aq) + Br- (aq)
•
•
142. Consider the foHowing: ‘ 2 C (g) 2A (g) + B (g) 、 .50 atrn of A and .20 atrn of B are placed in a flask at 300 K. At equilibriurn, the total pressure is found to be .60 aün. εalculate the Kp for the reaction. a) b) c) d) e)
15.0 4 .4 6.6 8.1 1. 0 X 10-4
[
내， i
c c c
1ll
、
씨/-
μ4f τI ‘
「}
n
mm 4
않때
%-
7
&‘
r
7、
&M
+ +
mN
난
、 m
m+’ m
M mw
n 냥
사띠 비 이 따 리
값) 없 ω
143. \Vhich reactions shown below wiIl not yield a precipitate?
144. 684 g of C 12H 22 0 11 and 360 g H 20 are rnixed. Calculate the vapor pressure of water over the solution, if the vapor pressure of pure H 2 0 at 20 oC is .023 atm. a) b) c) d) ε)
.036 atm .021 atlTI .014 atm .002 atm .011 atm 183
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GENERAL CHEMISTRY - PROBLEMS 145. Which salt would have its solubility most affected by pH change? a) b) ‘ c) d)
CuCl AgBr BaF2 AgI
ε)
Nonε
of the
solubilitiεs
will be significantly af:6εcted
146. A gas mixture consists of equal masses of Ne and Ar. What is the partial pressure of Ne if the total pressure was 300 mmHg? a) b) c) d) e)
200 nunHg 50 nunHg 100 nunH g 300 nm퍼g Nonε of these choices are
corrεct
147. Which of these have the greatest ionization energy? a) b) c) d)
Ar Na+ Na Mg
ε)
Kr
148. Consider liquid X at 35 0 Cand 200 C. Which of the following is true about kinetic energy, E K and potential energy, E p ? a) b) c) d) e)
EK [35 0 C] EK [35 0 C] EK [35 0 C] EK [35 0 C] EK [35 0 C]
> Ed20 0 C] < EK [20 0 C] > EK [20 0 C] 0 ~ Ed20 C] > EK [20 0 C]
; ; ; ; ;
Ep [35 0 C] > E p [35 0 C] = Ep [35 0 C] ~ Ep [35 0 C] > E p [35 0 C]
[N; ] at εquilibrium d) A .10 M solution ofHN 3 has a pH of 5.2 ε) Two of the above 195. An ionic solid gave a blue color. Which solid is most likely? a) b) c) d) e)
BaS04 AgCl CUS04 KI Li 2 S04
196. A solution of Na2HP04 is a) Acidic b) Basic c) Nεutral d) Cannot be determined ε) Unablε to dissociate Given for H 3P04: Kl =7.5xlO-3 K? =6.0xlO- 8
K3 = l >< I0-l2 197. A catalyst wiU change? a) Thε reaction lnechanism b) The energy of activation c) Thε spεcific rate constant d) Two ofthεse. ε) All ofthε above. 198.
、ìVhich
element does not have a d orbital vacancy?
a) Sc b) Ti c) V d) Fε ε)
Zn
193
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GENERAL CHEMISTRY - PROBLEMS 199. A 1 M solution of which of the following has the highest pH? a) b) c) d) e)
KOH N값이 0 3
CaCh B(OH)3 CH3NH2
200. Which set of quantum numbers may apply to the below figure?
혔
0 n=l , n=l , n=2 , n=3 , n=O ,
a) b) c) d) ε)
t =0 (: = 1
C= 0 t= 1 C= 3
·따 않
떠
N
O
m
m·nm pi
Pi
m m
mm po
·따
뻐 때 1
시”띠 리
T F C ·m B S
빼 밟”빼 뺑
20 1. A solid, liquid, and gas can all coexist at the?
202. A gas is heated from 20 oC to 60 oC at constant volume. Which is true? a) Thε pressurε mcrεasξs by a factor of 2. b) The prεssurε dεcreases by Ji c) The prεssure increases by less than 3. d) The pressure increases by a factor of 3. ε) The pressure remains the same. 203. Which molecule would be most polar in the gas phase? a) b) c) d) e)
I2 NaBr LiF NaF CsF
204. Which molecule has the greatest bond energy? a) b) c) d) e)
N2 O2 Ch I2 HI
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205. Which boiling point trend is correct? a) HI>HCl>HBr>HF b) HI>HF>HBr>Hεl c) HF>HBr>HC1>HI d) HF>HI>HBr>HCl ε) HI>HBr>HF>HCl 206. Arrange the following electromagnetic radiation in order of increasing frequency: violet, radio waves. a) b) c) d) ε)
visible, ultra-
Radio < IR < Visible < Ultraviolet IR < Radio < Visiblε < Ultravíolet Radio < Visiblε < IR < Ultraviolεt Ultraviolet < IR < Radio>Kb
a) The ratε linuting stεp is step 2 b) C1 is the intermediate c) C10 is the cata1yst d) Stlεp 1 has thε grεater εnεr잃T of activation ε) All of these arε true 239. How many lone electron pairs are found in AsBr3? a) 8 b) 10 c) 12 d) 6 e) 2 240. What mass ofwater is needed to dissolve 240 g ofNaCI to produce a 0.20 m aqueous solution? g。 。ζ σb oζ
---14k ζ!)αU
IK
14--)1K 까/
띠 랴 따 비 서딩
’I nu1K
1-
k O{)
201
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GENERAL CHEMISTRY - PROBLEMS 241. A 500 ml gas sample collected over H 20 at 740mmHg and 25 0 C is studied for analysis. What is the volume of the dry gas at STP if the vapor pressure of H 20 at 25 0 C is 24mmHg? 298 760 x-740+24 273 5OO x 740 - 24 x-273 b) 25 760 740+24 298 x-500x 273 760 d) 500 × 740-24 × 273 298 760 ε) None ofthese arε correct a)
500x
‘
떠← 값냉 셈꾀 선었
4 3 1
n」
힘
LU
E 돼 파찌
H H
d-H E」
g」
242. Consider the below reaction: C 2H 6 +Br2 • C 2H sBr+HBr KJ/molξ
c-c C-H
345 414
t디 -Br
강양 3
What is the net energy released in this reaction? a) 40 없 b) 52 kJ c) 80 없 d) 90 kJ ε) 140 kJ 243. Calculate the ~ H for the process Rh,_, (s) +O •
~2(g)
•Rh O ~
~~-2(s)
Fr0111 the following information: 2Rh,(s)_.+0 • - 2(g) •
,
2RhO ,., - (s) -~---
2RhO (5) ,. , +0 • - 2(g) • , 2Rh O - 2(s)、 -~--
L\.H
= -8 00kJ
ð.H =-300kJ
a) -550 kJ b) -250 월 c) -300 kJ d) -200 뻐 e) Nonε ofthese
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244. Consider the fo야Ilo~‘w Acid Ka x 10- 10 6 HCN 1.8 x • 0-5 CH 3 COOH 4.5 X 10 -4 HN02 10 C6H50H 1.3 X 10Which sequence lists the correct order of base strength? a) CN- > CH3 COO- > C 6 H 50-> NO; b) CN- >
NO~
> CH 3COO-> C 6 H 5 0-
c) N0 2 - > C 6 H50-> CH 3 COO->CNd) C 6 H 5 0-> CN- > CH3 COO-> NO; ε)
C 6 H 50
>CH 3 COO >N0 2
>CN-
245. Consider the following table:
PROTONATED COLOR DEPROTONATE'D COLOR INDICATOR PK.α A 4.4 Yellow Blue Pink B 5.0 Orange 7.상 C Yellow Purple R 8.0 Red Blue 5 An aqueous solution has a hydroxide concentration of 1 x 10- M would show what colors? ’
a) Yεllow with indicator A b) Blue with indicator B c) Yellow with indicator C d) Rεd with indicator D e) Two of the abovε arε possiblε 246. Consider the foHowing: 다잉二$
ran~e
1.2 - 2.8
red
methyl red
4.2 - 6.0
red
phenolphtha1 ein
8. 3 -10.0
red
alizarin yellow
10.1-12.4
yellow
V
“
1hu
What indicator would be employed to determine the pH of an a) b) c) d) e)
VJ
NH 4ε 1
째뼈뼈
thymol blue
Vι
Effective DH
짜않짜봐씨앉‘따
Indicator
it
solution?
thynlo1 blue it would bε rεd thymol bluε it would be green methyl red : it would bε orange phenolphthalεin it would bε yellow alizarin red it would bε blue
203
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GENERAL CHEMISTRY - PROBLEMS 247. Which statement is t.rue? a) The melting point of an impure compound is usually lowεr than that of a purε solid. b) Thε surfacε arεa of a liquid will afDεct the vapor pressurε. c) A compound with a vεry small specific heat will givε nsε to a large ~ T. d) AandB ε) A and C 248. Consider the following: CaS03(s) ζ~ CaO (5)
+ S02(g)
If this reaction is endothermic, which change will cause an increase in S02? a) Rεmoving CaO(s) b) Adding CaS03(s) c) Dεcreasing thε volum，ε ofthε container d) Increasing the telnpεrature ε) Morε than one of the abovε
쨌
×
nU4
1lli
‘
、，/
(‘“‘ 1 3- m n 3 × “””
ιι
m-많 &ω
mm ω g따 뾰 M
249. Consider the following: K-7 a-
1·lA
v
Which reaction shown below would occur to the greatest extent? a) HF + CH3 CH2 COO H )þ CH3CH2COOF + H 20
+ OBr
b)
HF
c)
HOBr
d)
CH3CH2COOH
e) HF
)Þ
+ F-
느
HOBr HF
+ CH3CH2COO-
+ OBr
느
+ F)þ
+ F-
HF
+ CH3CH2COO
CH3CH2COOH
+ F
250. 30.0 ml of 0.20 M barium hydroxide, Ba(OH)2 is required to neutralize 25 ml of citric acid , H 3C 6H s0 7 • What is the molarity of the H 3C 6 H s0 7 solution? a) b) c) d) ε)
0 .1 67 M 0.550 M 0 .2 5 M 0.75 M 0.084 M
251. Ammonium nitrate will dissolve in water at 25 0 C , and a decrease in temperature is noted. Which of the following is true? a) -~H~ +~ S b) -~H; -~S c) +~H; +~S d) +~H; -~S e) +~H; ~S = 0 Copyright ({;)2θ06-2015 ORGO~"ftvL4N，LLC PDF AND ELECTRONIC FORMATS ARE ILLEGAL
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252. Consider the dissolving of ammonium phosphate in water. How many moles of ions are present at high dilution? a) b) c) d) ε)
3 4 6 x 10ι3
4(6x 1023 ) 12
253. Which gas wiU have the greatest density at latnl and 25 0 C? a) C 4H lO b) O 2 c) Ch d) F 2 ε) PH3
254.
Pb-2θ8
a) b) c) d) e)
is formed from Th - 232. How many alpha and beta decays occur respectively?
6, 2 6, 4 6, 6 4, 6 8, 2
2용5. εonsider:
16H;A~\ 2AtJllO-;{ ,'u\ + 역 Zll;~，.\ + 2 생Íl;;n\ (Aq) + T ""lUitV4(Aq) T 5Z11 ~'LJit(s) • - r ~'LJIt(Aq) T ""lU (Aq) + 8H.. O If
If 200 ml of .02M KMn04 reacts with zinc, appronmately how many mg of zinc are needed? a) b) c) d) ε)
4001ng 6501ng 300 mg 7501ng 150 mg
256. Which is not a possible ground state configuration? a) a 3d6 sublevel with 6 unpaired elεctrons b) a 3d2 sublεvel with 2 unpairεd elεctrons ， c) a 5d3 sublevel with 3 unpairεdelεctrons 2 d) a 7s sublεvε1 with 0 unpairεd electrons ε) All are possible 257. Which statement(s) is false? a) A 3d orbital can hold a lnaxÌlnum of 10 elεctrons b) A 3p orbital has 3 spatial oriεntations c) Thε n=3 shell can contain 18 electrons d) A and B ε) B and C
205
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GENERAL CHEMISTRY - PROBLEMS 258. Which statement is false? If a rεaction is εxothermic ， the systeln evolves heat to thε surroundings During an exothεrmic reaction, a chemical substancε increases in potential εnεr잃7 During an endothεrmlc rεaction， the tεmpεraturε of the surroundings is lowεred B and C A andB
a) b) c) d) ε)
259. Which gas graph, assuming ideal behavior, is false?
a)
c)
pι T
At constant V
vι P T
d)
b)
wι
PL
At constant T
v
T
e) None of 않lese
260. Consider the decomposition of dinitrogen tetroxide: N 2 0 4(g)• 2N02(g) o O Thε Lì H was found to bε 55KJ and Lì S was found to be 175JIK. this reaction become spontaneous? 55 a) ~~ _ K .1 75 b)
175 -~~
55
K
c)
12 H 55) \ /、 ~K .175
d)
55 K 175
ε)
_~:_
Nonε ofthεsε. Thε rεaction
cannot bξ spontaneous
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Abovε
what lnininlum tempεraturε would
26 1. Consider the foHowing data:
What is a possible normal boiling point for compound B? a) 38 0 C
b) 100 0 C c) 85 0 C d) 90 0 C 0 ε) 50 C 262. At 0 oC a gas is dissolved in H 2 0 at a mole fraction of2 x 10-18 • What is the molality of this solution given the molecular 찌reight of the gas is 32 g/mole. a) 5 x lO-4m b) 1 x lO-I4 1n c) 1 x lO-I6 m d) lxlO- 12 m ε) 4xlO-7 m
263. Rank the 1'oHowing in order 0 1' increasing 1'requency: radio waves, ultra violet, infrared, and visible. a)
Ultraviolεt
< radio
wavεs
。 ~H>O ， ~S < 0 ~H= 0 , ~S > 0
270. At what temperature wiH a fixed amount of gas with a volume of of 200 750 Torr occupy a volume of 210 eat a pressure of 610 Torr?
e at 10 C and 0
(610)(210)(283) (750)(200) ‘. t
、
L ”U
/
(610)(210)(1 0) (750)(200)
c)
d)
(210)(283) (010)(750)(200) (200)(283) (610)(750)(210) Nonε ofthεse
27 1.
εonsider
2혈 u
+
the foHowing nudear equation:
!n~?
+
1끓 Ba
+
;:Kr
How many neutrons are needed to balance this equation? 사띠
--4
이 띠 어
272.
Ov
N
3
2 4
찌lhich
a)
”
b)
electron configuration represents an atom in the excited state? 2
이 리
ls ls22s2 2pl 2 2 6 2 3 ls 2s 2p 3s 3p 2 1 l ls 2s 2p ls
1
209
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GENERAL CHEMISTRY - PROBLEMS 273. Consider the following: N 2 + 3H 2 P 2NH 3 8 moles of N 2 and 8 moles of Hz are placed in a 2liter flask and allowed to come to equilibrium. At equilibrium, 2 moles of NH 3 are formed. Calculate the Keq. (8)2
a)
Keq =
/_', -:_, '" (8)(8)j (1)2
b)
(8)
c)
Kεq = (8)(8)2
d)
Kεq = (4)(2)3
e)
None ofthese
(4)2
274. Consider the following:
〔J끓월:jj=쇄
꾀l
Find the total pressure when the valve is open. (Assume constant temperature and the volume between the chambers is negligible) a) b) c) d) e)
2.0 atm 3 .4 atnl 6.0 atm 10 atm 8 atm
275. HF was added to a solution ofKOH. What could be the pH at the equivalence point? a) b) c) d) ε)
3 5 7 9 AorB
276. Which pair represents isotopes? 29339Np and 29349nl U b)
Ig 0
and
~~Cr and 24 d) l9lo 76 US and More than onε
17
80
2546Fe 17971Os ofthεse
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GENERAL 277. Consider compound AB. Element A has a low ionization energy, while element B has a high electron affmity. \Vhich type of bond is most likely? a) b) c) d) e)
Coordinate covalent Dispersion force Metallic bond Ionic bond Covalent bond
278. How does the speed of He compare to that of Ar? a) b) c) d) e)
4x as fast 2x as fast 3x as fast 1. 5x as fast 0 .2 5x as fast
279. Which statement is false? a) Entropy dεcrεasεs during the freezing of watεr b) Ifthε ~H of a reaction is nεgativε ， and the ~S is also negative , the rεaction will be spontaneous at low tεmpεraturε
c) As stεanl is condεnsεd ， d) If the reaction has e) More than one of these
thε ~S
m
is positivζ and + ~S thε reaction is
spontanε。us
at all telnpεratures
280. Hydrogen is approximately six times as fast as an unknown gas. Which gas is most likely?
0 ., CL,
b)
F) BH~
e) C vapor 281. Consider the following chamber: 폈。꺼 ~、、、 [
던따‘
}
‘\민~=--/서 If the total pressure is 1400 torr at 25 0 C and the mixture contained 1.2 moles CO 2 , 3.0 moles CO and 1. 0 moles of Ne, what is the partial pressure of Ne? a)
1400[딛] 5.2 (5 .2 )(1.2) 1400
c)
d)
백 5.2] 1. 2 1400 5.2 (1 .2)(1 400)(3.0) 5.2
211
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GENERAL CHEMISTRY - PROBLEMS 282. Consider the reaction below: 3COCL2{g}， +B~O '''2 3(‘ • , 2 혔CI 、 ~
-~~'3(g)
+3CO ‘ .
-~~2
Calculate the enthal JlYchange for this reaction given the below data: m[없 1m이셔 Compound
a) b) c) d) e)
-1270
B 2 0 3 (s)
-220
COC1 2 (g)
-394
CO 2 (g)
-404
BC1 3(g)
-60 KJ -3900 KJ +355 KJ -115 KJ +115 KJ
283. A compound is 40 0/0 carbon, 6.6 0/0 hydrogen, and 53.3 0/0 oxygen. What is the empirical formula?
a) CHO b) CH 2 0 c) CH 3 0 d) CH 2 0 3 e) None ofthεsε 284. A 5.0 liter container contains C 2 H6 , S 0 3 and N 2 at 25 0 C. If the container develops a pinhole, what describes the relationship between partial pressures of these gases after 2.5 hours? a) P 112 > P C2H6 > P SÛ3 b) P SÛ3 > P N2 > P C2H6 c) P C2H6 > P N2 > P SÛ3 d) P SÛ3 > P C2H6 > P N2 e)
PN2
> PsO3 > PC2H6
285. What is the
%
of oxygen in magnesium phosphate?
4x16 x100 263 b)8x 16 x100 263 c)12 x 16 x100 263 d)2x 16 xlOO 263 ε) Nonε ofthese a)
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286. Consider the below diagram: L
Pressurξ
(atm)
273
J73
----m
Temp (K)
The critical point and boiling point is located at which points respectively? a) L ~ P b) L ; Z c) Y ; P d) Z ; P ε)
x
p
287. When sodium reacts with excess water, the products are:
a) Na20 b) Na20+NaOH c) Na20+NaOH+H2 d) NaOH+H 2 e) NaOH 288. 、rVhich Lewis structure is correct?
rζ- Cl:
: Cl : Br-Cl
: Cl : : Cl ν
l
: Cl :
/
써써
n l‘ 、
1
: Br-Cl -CI-Cl:
: Br-Cl : Cl
e) None ofthese
289. Consider XeF2 • 찌That is the formal charge on Xe?
+1
a) b) c) d)
-1
ε)
-2
+2
0
213
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GENERAL CHEMISTRY - PROBLEMS 290. Consider CIF... 5. Which structure is correct? F
낸、F α「
a)
Fl F/
T:'
/~
. ··Fi
+CI-F /
\
F
F
F
mμ 「
해
F
I F / :CI-F /
F
\
, ht
낸
F/
|
‘F
F
N one of these 291. Consider an experiment in which O 2 gas is to be dissolved in water. Under which conditions would O 2 be most soluble? 사띠
N High prεssurε ~ low tempεrature 사”이 려
Low pressurε ; high tempεraturε High prεssure ~ high temperatur응 Lowpressurε
lowtεmperature
Tεmperaturε andprεssure havε
no
aff;εct
on O2 water solubility.
292. 찌!hat is the ratio of CO 2 to O 2 during the combustion of C 2H 2 ? 띠 에 서ν띠 며
112 3/2 5/4 4/5 3/1
293. Consider the below reaction:
A+B • C
Which of the following would not affect the reaction rate of this irreversible reaction? 사띠 씨”이 띠 리
Decrεasing
A B Adding a catalyst Relnoving C Incrεasing
Incrεasing Tεmperature
294. Which statement is true? a) The actual 111ass of an atomic nuclεus εquals the SUl11 ofthε massεs of its protons and nεutrons b) Thε actual 111ass of an atomic nucleus is a little greatεr than the SU111 of the masses of its proto11s and nεutrons
c) The actual mass of an atOlnic nucleus is a littlε smaller than the sum of the masses of its protons and neutrons d) Thε actual mass of an atornic 11ucleus is much greatεr than the smn of the lnasses of its protons and 11εutro11s
e) N one of these
arε corrεct
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295. If a Iiquid were heated , which property(s) would increase? a) V iscosity b) Surfacε tεnSlOn c) Vapor prεssure d) A and C ε) B and C 296. Which substance has the greatest vapor pressure at 25。 a) b) c) d) ε)
Acεtic
ε?
Acid
Ethanε Octanε Acεtone
Propanol
297. Consider N0 2 • Estimate the ONO bond angle. a) 120。 b) Ovεr 120。 c) Under 120。 d) 180。 ε)
109.5 。
298. Consider P20 칸 . How many valence electrons does this specie contain? a) b) c) d) e)
52 48 56 58 62
299. Consider radioactive nuclide X. If 88 0/0 of a pure sample dec윈TS in 24 hours, what is the half-life? a) b) c) d)
mi
.m”
m. m
없
k
o m---Lu eqr·
i
ku n “
」
4
4
’’’’」
1
4 ‘
4
%lii
Frl」 1ll 」
’
1/-
4
이→
-------」1
「 ιJ
e
-me
l
」
l ’ l
g+2 E
，，，
MLg
「/-「、
g+2
r·i
짧
+2
t
야
γ--~
1，，，‘，‘」 1
g --------O
뻐L 뻐
해A
리
Mm
u 1l
αuJ↑
””
+2
영
띠
K
%
이
이
K
생
야
K
생
떠
K
WA
뻐 -ML
h
v
앙
따
찌
ma”
「때 「빠m -L 「 뼈싸뀐 L
ε)
3 hours 8 hours 14 hours 16 hours 4 hours
215
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GENERAL CHEMISTRY - PROBLEMS 301. Consider the following reaction of hydrazine, N 2H4: N 2H 4 (1) 느 N 2 (g) + 2H2 (g) Ifthe L1 H a) b) c) d) ε)
=
-
13 Kcal; this reaction is:
N ot spontaneous at any temperaturε Spontaneous at all tempεratures Spontaneous only at low temperaturεs Spontaneous only at high tempεratures Cannot tε11 without the ~G
302. In what pair are both species polar? a)
AIC1 3
and
PC1 3
c)
H20
and
BC13
b)
CS 2
and
N0 2
d)
S02
and
SBr2
e)
이one
ofthese
303. Which molecule is likely to form a dimer? a) BF;
b) NO; c) N0 2
d) NO; ε) Nonε ofthεse
304. Consider the OCN- anion. How many resonance forms can be drawn? a) 1
b) 2 c) 3 d) 4 ε) None 305. In which compound does sulfur have the highest oxidation number? a) H2SO3 b) S02
c) SBr2 d) Na 2 SO 4 e) H')S
306. A glowing object is at 1832 0 F. What is the Kelvin temperature? a) b) c) d) ε)
1250 1273 2105 1400 1559
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216
307. An unknown mineral has a density of 5.0 g/cm is the mass of a 9 cm -' sample? a) b) c) d) ε)
j •
If a 2.0 cm -' sample has a mass of 10 grams, what
30 g 45 g 50 g 60 g 18 g
308. The total energy of the nucleus is less than the combined energy of the separated protons and neutrons. This energy is called: a) Ionization Energy b) B inding Energy c) Potential Energy d) Mεchanical Enεrgy ε) Intema1 Energy 309. Consider the titration between HBr and KOH. Which indicator would be best for such a titration'! Red (Pk IN = 8.1)
a)
Crεsol
b)
Mεthyl
c)
Mεthyl Yεllow
d)
Phεno1phthal망n
ε)
Bromothyn101 Bluε (Pk IN = 7.1)
Red (Pk IN = 5.5) (Pk IN = 3.3) (Pk IN = 9.3)
310. An indicator is red in the undissociated form and yeHow in the dissociated form. If a solution were at a very low pH, what color would the indicator appear? a) b) c) d) e)
Red Yellow Orange Purple Unable to
tεlL
more infonnation neεdεd
31 1. If Cs metal was exposed to light of a certain minimum frequency caHed the threshold frequency, electrons are ejected from the surface of the metal. What is the correct term for this phenomenon? a) b) c) d) e)
Quantum thεory Photoelectric Effect Bohr Effect Photo fluorescent Effìεct Tunneling Ef:6εct
312. Consider the foHowing reaction: 4 Al + 302 • 2Al 20
3
If 2.7 grams of Al react with excess O 2, 3.θg of Ah03 is produced. What is the
a) b) c) d) e)
%
yield?
20% 40% 60% 50% 80% 217
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GENERAL CHEMISTRY - PROBLEMS 313. How many ml of 0.120M KOH must be added to 60 ml of 0.20M HF to produce a solution with a pH of 3.3? (pKa of HF is 3.3) a) b) c) d) e)
50 ml 10 ml 15 ml 20 ml 12 ml
314. Consider the following reactions: S(s) + 02(g) • SOz(g) Find AH given the following data : S(S) + 302 (g) • S03(g)
빠 =-400월
2S0 2(g) + 02(g) • 2S0 3(g) AH = -200KJ
a) b) c) d) ε)
-300KJ -400KJ -600KJ -200KJ +200KJ
315. Which statement is false? a) b) c) d) e)
CH3F has London dispersion forces between molecules and dipole-dipole attractions CH3COOH has hydrogen bonding as the dominant intermolecular forcε London dispersion forces exist in all molecular solids and increase as the number of electrons Íncrease S02 possesses both dipole-dipole attractions and London dispersion forces The strontium cation is larger than the sulfur anion since negative ions gain electrons
316. Consider the following decomposition reaction 2KCI0 3 • 2KCI+30 2
If 10.0g of KCI0 3 decomposes at 250 C and 755 mmHg, what volume of O 2 is produced? 핫) 1)(2양용}
76입
b)
d)
‘
(!23) 246 f ‘양앓였셨3쟁힘
755
강쟁월
7쳤”
e)N one 0 f these
317. 1.0 gram of a hydrocarbon undergoes complete combustion to yield 3.1 grams of CO 2 and 1.3 grams ofwater. What % of carbon is contained in the hydrocarbon? a) b) c) d) e)
82% 60% 40% 95% 33%
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ARE lLLE’GAι
318. Consider the follow
4x 18 284
a)
~x100
b)
~x100
284 4x 18 c) ~x100 212 1Ox284
d)
100
4x 18
ε) ~x100
350
319. Consider the following simple apparatus in which a semi permeable membrane separates an aqueous sugar solution from pure water: A
B 녀t
mu
서써
----
”‘ 떠떼
씨總《
A
B
e
om
n· 야
---- e an
없
앉
4--o m
뼈
mn
-않
-·Ii
빼
e
없
e ---i
ku
jrl
‘
•
2 2
?
J//2
1ll
-이 3
,
1 ]
J/ 2
, U ,
1I !I
32 1. Consider the equation: Cr20~- +
+
•
Which coefficient 찌!ould be needed to balance a) 9 b) 7 c) 4 d) 5 ε)
않
않
·뼈 않
rι
씬
aL KU
앉
뿔
빠
·띠
뻐
띠 라 따 비 서니
따 ιι
‘
·UU
m· μ쇄
뻐뿔
짧
Which of the following will occur? a) Thε so1ution volunlε mcrεasεs in side B b) The concentration of sidε B increasεs c) The concεntration of side B dεcrεases d) The concentration of B rεmains constant e) A and C only Au kU c l f J1 Of q m a nal 4 0 nU U2 , ’ 4·4·4 1 1t U2
Cr+ +ρ +1 3 3
?
14
322. How many isomers [constitutional] can be drawn for ε4 H I0 0 ? a) b) c) d) e)
6 7
5 8 10 219
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GENERAL CHEMISTRY -
PROBLE깨1S
323. Arrange the foUowing in order of correct boiling point: H 2 S , H 2 0 , H 2 Se, H 2 Te a) H 2 Te > H 2 Se > H 2 S > H 2 0 b) H 2 Te > H 2 S > H 2 Se > H 2 0 c) H 2 0> H 2 S > H 2 Se > H 2 Te d) H 2 0>
H2了e
> H 2 Se > H 2 S
e) H 2 Se > H 2 S > H 2 0 > H 2 Te 324. Consider the Titration curve below:
pH §혁uìval흉ncepo썰t ...........
cm" 싹c ‘
ad양g경
The curve depicts which of the following? a) Strong acid - strong base titration b) 짜fεak base - strong acid titration c) Wεak acid-wεak base titration d) Wεak acid - strong base titration ε) Two of the abovε are possiblε 325. Organize in decreasing order with respect to atomic a) b) c) d) e)
rad피:
Br- > i> F K+ >Ca냐 >At 3 At3 > Na+> 0-K+> Ca++> Br 0- > 0+> 0-
326. A barometer is to be made using H 2 0 instead of mercury. What 까rould be the height of the column of water in this barometer if the density of merCUlγ is 13.6 times greater than the density of water? a) (1 3.6)(760) b) 760113.6 c) 760-13.6 d) 760+13.6 ε) (760)/1 3.6(22 .4)
Copyright 1. When calculating thε EO, wε do not take into account the coεfficients. E O= 0 at εquilibrium， and Cu -'-2 is bεing reduced to Cu, so it would bε thε oxidizing agεnt. 113. A First write out thε
balancεd εquation:
C 2H4(g) + 302(g) Notε:
þ
2 CO 2(g) + 2H 20 (l)
In con1bustion , H 20 is liquid (25 OC) and .ð.H o f of0 2 = 0
L
L
~= ~ f 0 products ~H f 0 reactants ~= ψ-394) + 2(-286)] - [+52] ~ = [- 788 + - 572] - 52 ~=-1360-52
~
=-1412 kJ /mol 245
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GENERAL CHEMISTRY SOLUTIONS 114. C Wenεεd
the mass of CaC12 • 2H 2 0 . Thus 111 + 36 or 147 grmTIs. Don ’t forget to add
thε
lTIaSS of the two
、;vaters.
115. A Let us keep A constant to
(뾰
e、raluatε
B. Look at trials 1 and 3. A is constant. Now proceed as follows:
6
91 x 10=> 1 5 =1 => x = 0 = ) .40 ) 9.1 x 10-6 Thus , wε sεezεro ordεr kinεtics with rεspεct to B. Now, hold B constant bεtw응εn trials 1 and 2 to
evaluatε
A.
5
(뾰) =~---.-A---6 oX 10.20)
9.1x10-
Rate = k
[A]
3 = 9 => x = 2 , now pick any trial , let us use Trial 2. To εvaluate k
=>
2
5
8 X 10- = k[.60] 2 8x10-S =k[3.6x10-1]
k -=
( k remains constant even if concentrations change)
5
8 X 10-
(3.6x10σ-1
1 1 k= 2.2 x 10σ-4M- 닝 Sεec-
Note 감 thε units ofk
Lεt’ s say the rate law was rate = k[ A ]2[B] 3 this is a 5 order reaction , so the units would have been th
M-4sec- 1
116. E To do this , let us
writε
out the Nernst Equation:
|E=E O -몇 !OgQI
EÜ= Standard ε1εctrodε potεntial E= 표MF at a new concentration n=Thε number of εlεctr“ons given in the half-reaction n=4 Q=Rεaction Quotient, a value that takεs thε S없nε form as thε Keq. Ohviously , basε (sodium hydroxide) wil1 remove thε H+ , thus thε acid concentration changes , thε E will changε ， No pure liquids and no solids are usεd， thus H 2 0 is excluded.
Q=
r
함+ [0 2 ]
E=E O - 포싼 lOQ: n
_
1
'-'[H十 r
[0 2 ]
Mathematically, we sεe as [H+] decrεases， so doεs the E. Hεrε ’ s a quick trick: .0591 log L _ [A] E = E 0 - .:..:.:..:..: L~ ~J n
'-' [B]
If [A] increases E would decrease If [B] increases E would increasε
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246
117. B A buffer solution may be a mixture of a weak acid and its salt. CH 3COOH and exan1ples.
CH3COONa~
HF and KF are
A buffjεr can also consist of a weak basε and its salt such as NH 3 and NH 4 Cl or CH 3NH2 and CH 3 NH딘Cr. For buffers , usε thε
Hendεrson -Hasselbach εquation:
p H = P Ka + log
[Sa1t ] .-L
~-~- .J,
[Acid] If [Salt] = [Acid] , the log tζrmbεcomεs zero. So pH = pKa: this mεans that you arε at what is callεd thε “ halfpoint" in thε titration. For εxamplε， you have acetic acid , whεn yourεach thε pKa i.e. pH = pKa we ge t:
εquivalεnce
CH3COOH
특--ξ;
CH3coo-
50%
50%
Now back to our problen1. First get the pKa: .+plμ= -log Ka pKa = -log 6.5 x 10-:1 pKa;::::; 4 .2 Practicε how to estin1atε logs! You know that 1 x 10-:1 has - log of 5 , so any nUlnber ovεr 1 x 10-:1 would give an answεr less than 5. n
--- ×
flt
-“
J 1linu
4 -----껴/
?ι
4·7/ × --iouS
4
× ---nu
4
j
‘
4*· 껴4
생생생
•.
g
야 1
Fr m
[Salt] pH=pka+1og---[Acid] .010 pH=4 .2 + log .-.-~.10 1 X 10-2 pH=4 .2 +10g ~_1 1 x 101
pH =4.2 + log 1 X 10-1 pH=4 .2 -1=3 .2 Thus 14-3 .2 =pOH
~
pOH = 10.8
118. D ~G =Llli- T~S that is spontaneous has a K > 1, + EO and a -.ð. G. A -.ð. G doεs not imply anything about speed , so C is wrong. If -.ð. H and +.ð. S arε presεnt， a rεaction will proceed spontanεously at all tεlnperatures ， whilε ifthζ .ð.H is + and .ð. S is -, it will not be spontι neous undεr any tempεrature. If +.ð. H and +.ð. S ， thε rεaction is only spontanεous at high tεmpεrature. If -.ð. H and -.ð. S, thε reaction is spontanεous only at low tεmperature . .ð. S, thε Entropy is a lneasure of the randomness in a system. Gas > Li안uid> Solid. Sincε C 4 H lO has n10rε atoms and therefore a greater molecular c0111plεxity than the sn1aller C3H8 , thε .ð. S would be higher for C4 H lO . Arεaction
247
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Lμ ιU
L
144
째쇠 」」
I
갱짜 깎 거
H Qx B V w= 1i w -------g6 /-‘2 w w m
120. A A -i1 G mεansthε rεaction is spontaneous , thus will be thεnnodynamically favorablε. virtually all of product C, since ð. G is so negative. i1G ÏInpliεs nothing about spεεd.
Thε mixturε
will
bε
12 1. A All spontaneous reactions result in an increase in thε entropy, not enthalpy of thε Universε. This is the Sεcond Law ofThεnnodynaInÎcs. If a liquid vaporizes , we need heat (+MI) and we are increasing the disordεr (+i1S). A state function such as ð.H , i1G , i1E , does not depεnd on a particular path. Statε functions dεpεnd only on the initial and final statε. q and w are not statε functions sinc응 thεy do depend on path. 122. A amount of radiation absorbed by a samplε ε = aconstant
Thε
lS glVεn
by thε Bεεr-LambεrtLaw: A= εbc
b = path 1εngth
c = concentratlOn and concεntration are dirεctly rεlatεd sincε our variablεs are on opposite sidεs Since our absorbancε halvεd， so doεs thε concentration.
Absorbancε
이
nU
mw
，、，)
-2-
×
OU4
ofthε εquation.
M
123. B Gibbs Fr∞ Energy is the best indicator for spontaneity. A reaction with a -ð. G 쁘뾰! be spontaneous. Entropy and εnthalpy nεεd not be positive or negative for a spontaneous reaction to occur. 124. B Brine has been used for 1nany purposes such as a preservative in foods to inhibit bactεrial growth. Brine is “ salt water" consisting of mainly sodium chloridε， NaC1. 125. C Alpha particlεs are ~Hι+ and 따e 1nost massi、ζ εlectrons thε lεast alpha particlε >nεutron > proton > electron 126. B This is the common ion effect; procζed as follows:
느 Fe+2(aq) + 2 OH-(aq)
Fe(O돼2(S)
Ksp = [Fe+ 2][OH-f Lεt x= [Fε+1] Lεt 2x = [OH-] BuL wε have an additional source ofFe+2 from the FεCh FeC12 lÞ Fe+2 + 2 ct .025
.025
2(.025)
So [Fε+2] actually is x + .025 Copyright CQ2θ06-20150RGOlvfAN，LLC PDF AND ELECTRONIC FORMATS ARE ILLEGAL
248
Ksp = [Fε+2] [OH-f 1. 6 X 10- 14 = [x + .025] [2x] 2 Now assun1e x density of solid T
This anomaly is due to H-bonding. Note how the solid -liquid line is skewed to the left. A body-centered cubic unit cell has 2 atoms , while a face-centered unit cell has 4 atoms.
140. C =MRTi π = Osmotic pressure M = Molarity T = Kelvin temperature i=Van ’ t Hoff facto r. This is the # of particles that will theoretically dissociate at infinite dilution. Here i = 2 becausε， NaN03 Na+ + NO i Jr
•
π = (.2 0)(.0821)(273)(2) π = 9atm
14 1. D LiOH is a strong basε ， and HBr is a strong acid, thus both arε complεtεly dissociated.
상Z(aq) + OH (aq) + H+ (aq) + 짧""'(aq) • 상