Passage VIII Questions 47 - 53) Wolff-Kishner Reduction versus Clemmensen Reduction
55. Choice Cis correct. The Grignard reagent (methyl magnesium bromide) adds on one carbon (a methyl group), so
choice A and choice B, the choices are narrowed down to either choice C or choice D based on the four-carbon
electrophile requirement. The best way to get a secondary alcohol as the product is to react H3CMgBr with butanal, which will yield the product with the alcohol group on carbon number 2. Pick C for an altogether great
experience and correct answer. Choice D will form 2-methyl-2-butanol.
56. Choice D is correct. The reaction of a Grignard reagent with a carbonyl yields an alcohol. Stereoisomers are formed when the alkyl groups are all different on the alcohol carbon. For this to be true, the Grignard nucleophile must add an alkyl substituent different from the alkyl groups present on the carbonyl. Choice A is eliminated, because the ketone has an ethyl and methyl group attached. The Grignard reagent adds an ethyl group. The alcohol that is formed is achiral (3-methyl-3-pentanol). Choice B is eliminated, because the ketone is symmetric (has two ethyl groups), so the product must be symmetric and thus achiral. An ester when reacted with excess Grignard reagent yields an achiral tertiary alcohol (it adds the same R-group twice), thus choice C is eliminated. Choice D is best, because the final alcohol has an ethyl group, methyl group, and hydrogen attached. This makes the alcohol asymmetric and thus chiral. Tiie product mixture is two enantiomers.
57. Choice B is correct. Aldehydes are carbonyls with at least one hydrogen and often one carbon attached to the carbonyl carbon. The addition of a Grignard reagent places a second carbon on the electrophilic aldehyde site and thus reduces the aldehyde to a secondary alcohol. Pick B for the inner peace that a correct answer can bring.
HoC
RMgBi
OMgBr H3C OH
H30^
cAh
58. Choice B is correct. An ester has both a leaving group and a carbonyl with which to react. The first R-group to attach converts the ester into a ketone by displacing the alkoxy leaving group. The ketone that is formed as the intermediate product can react with a second equivalent of the Grignard reagent and thus add a second R-group.
The final product is a tertiary alcohol with two identical R-groups. The conclusion is that a Grignard reagent reacts twice (not just once) with an ester. Pick B to tell the world "I know Grignard!"
59. Choice D is correct. The Grignard reagent is a strong base, so an aprotic solvent such as ether is chosen. This makes choice A a valid statement. Because the carbonyl compound is converted into an alcohol in a Grignard reaction, a bond to oxygen has been lost. Tliis means that the carbon has been reduced, making choice Bvalid. A Grignard reagent can be used to form a secondary or tertiary alcohol, as shown in the passage. To form a primary alcohol, a Grignard reagent can be added to either formaldehyde or the epoxide ethylene oxide. This makes choice C valid. Because the Grignard reagent is a strong base, it can react with the weakest of acids to form an alkane. This is why the reaction must be carried under anhydrous, aprotic conditions. It is thus not true that the Grignard reagent should be added under acidic conditions to obtain the highest yield. Pick D and move on with another correct answer under your belt.
60. Choice C is correct. Propanoic acid is the conjugate acid of propanoate, which when bonded to carbonyl group forms an anhydride. This question is asking for the relative reactivity of an anhydride. Acid chlorides are more reactive than an anhydride, so choice A is eliminated. Amides are less reactive than an anhydride, so choice B is eliminated. Esters are less reactive than an anhydride, so choice C is true. The difference in reactivity between anhydrides will not be distinguishable by pKa values that only differ by 0.1. There is a difference in steric hindrance to consider. Choice D may or may not be true, but regardless, choice C is definitely true, thus choice D is assumed to be false within the context of this question.
Passage X (Questions 61 - 67) Aldol Condensation and Alpha Hydrogens
61. Choice B is correct. The most acidic hydrogen on the molecule is the H bonded to the alpha carbon (marked on the molecule as carbon b). The best answer is choice B. The aldehyde proton is not as acidic as the alpha proton, because the lone pair that would form on the carbonyl carbon is not stabilized by resonance, like the alpha
carbon.
Copyright © by The Berkeley Review® 81 CARBONYLS & ALCOHOLS EXPLANATIONS
62. Choice D is correct. When counting carbons starting from the carbonyl, the molecule is a 4-carbon chain with a methyl group bonded to the number 2 carbon. The compound has an aldehyde functional group, so the correct
IUPAC name is 2-methylbutanal, answer choice D.
63. Choice A is correct. An aldehyde with excess alcohol in the presence of an acid will lose a water molecule, and go on to form an acetal. It is an equilibrium reaction, but the excess alcohol will push the reaction to the acetal (product side of the reaction). A hemiacetal is formed when the reaction is carried out under basic conditions.
Choice A is the best answer.
64. Choice A is correct. The reaction isan aldol condensation. The acid workup causes elimination and thus ensures the formation of the a,fi-unsaturated ketone from the fi-hydroxyketone intermediate product. Choice D is eliminated, because it has seven carbons, which is not possible. Choice C is not an a,fi-unsaturated ketone, so it can be eliminated. Choice B, when broken apart by a retro-aldol reaction, would yield fragments of two and four carbons, not three and three carbons. Choice Bis eliminated. The only cc,fi-unsaturated ketone composed of two
three carbon units is choice A.
65. Choice D is correct. This isjust one ofthose things that you need to learn. The conversion ofa ketone into anenol or an enol into a ketone is referred to as tautomerization. Select D as your answer. Tautomerization is involved
in glycolysis in the conversion of glucose-6-phosphate into fructose-6-phosphate, which is catalyzed by
isomerase.
66.
67.
Choice Cis correct. This is acase of recognizing the weakest base. Only one of the answer choices is not astrong enough base to deprotonate an alpha hydrogen, so it must be the weakest base. Carbonate, C032", is not avery strong base, because the electrons are stabilized by resonance involving the double bond (C=0 Jt-bond). It is therefore the weakest base out of the choices listed, and thus is the one not strong enough to deprotonate the H on the alpha carbon. Choice Cis your choice for abrighter tomorrow, assuming of course that the brightness of
tomorrow depends on your answer choice for this question.
Choice C is correct. Only a ketone and an aldehyde can undergo aldol condensation reactions. An acid would
rather lose the proton from its hydroxyl group than lose the less acidic alpha proton to undergo the condensation
reaction. Choice C, the carboxylic acid, cannot undergo aldol condensation.
Passage XI (Questions 68 - 73) Claisen Condensation Reaction
68. Choice Bis correct. The most acidic proton on a fi-ketoester is the alpha hydrogen that when lost results in a carbanion conjugated to the most 7t-bonds. Choices Cand Dare eliminated, because neither is an alpha hydrogen. Choice Ais only conjugated to the keto carbonyl, while choice Bis conjugated to both the keto carbonyl and the ester carbonyl. The most acidic proton is proton b, thus you should pick choice B.
69. Choice C is correct. As stated in the passage, the lowest yield is found with the fi-ketoester anion intermediate that has no alpha hydrogen. For this to occur, the starting ester must have only one alpha hydrogen. Of the choices, only choice Chas just one alpha hydrogen. You should pick C. The reaction is shown
below.
O
H,C H
OEt 'OEt.
HOEt"
HoC
Nothing protonates the alkoxide, thus the equilibrium favors the reverse reaction.
O O
H,C
H3C
OEt + "OEt
CH,
No alpha hydrogen, therefore the alkoxide leaving
group cannot deprotonate the final fi-ketoester.
70. Choice Ais correct. AClaisen reaction is not possible with esters that have no alpha hydrogen. Of the choices only choice Ahas no alpha hydrogen. This makes choice Athe best answer. As apoint of interest, this molecule
isnot known toexist as a stable compound at room temperature.
Copyright © by The Berkeley Review® 82 CARBONYLS & ALCOHOLS EXPLANATIONS
71. Choice D is correct. The product of an ester reacting with itself by way ofa Claisen condensation reaction is recognizable from the carbon chain length on the carbonyl portion of the ester. In the answer choices, the two alkyl groups can behighlighted to recognize the alkyl groups that must be present on the two esters that react, considering that only the alkoxide leaving group is lost. The alkyl groups are highlighted for choice A are:
H3CCH2COCH2CO2CH2CH3, therefore choice A is formed from the condensation of two different esters:
H3CCH2CO2CH2CH3 and H3CCO2CH2CH3. This same analysis of the product for choices B, C, and D will lead to the conclusion that choice D is the correct answer. In choice D, the starting ester is (C6H5)CH2C02CH2CH3 which leads to (C6H5)CH2COCH(C6H5)C02CH2CH3.
JU
Choice A: O O
A ♦ A
H3CH2C OEt H3C OEt
NaOEt
HOEt H3CH2C
Choice B: O O
A ♦ A
H3C OEt H3CH2C OEt
NaOE£
HOEt H3C
Choice C: O O
A ♦ A
HgCgHjC OEt H3C OEt
NaOEt
HOEt HsCgHjC
Choice D: O O *'
OEt
OEt HgC^C A OEt HOEt
NaOEt
O O
OEt CH,
.JUOEt
O O
H5^6H2^A H5C6H2C OEt
C6Hs
72. Choice A is correct. The starting molecule possesses two ester sites so an intramolecular Claisen reaction is possible. An intramolecular Claisen reaction results in a cyclic fi-ketoester. This eliminates choices B and D.
The chain contains six carbons, so choice C can be eliminated, because it requires seven carbons excluding the alkoxy group to form a six-carbon ring. The best answer is choice A. The reactionis shown below.
O O O
EtO
OEt NaOE
HOEtW EtO
OEt
EtO O
73. Choice D is correct. The two esters can either react with each another or they can react with themselves. This results in four structural isomers being formed. Of the structural isomers, two have one stereocenter, therefore two of the four structural isomers have two possible stereoisomers (enantiomers). The total number of isomersis
therefore six. Choice D is correct. The isomers are drawn below:
O
A + H3C
H,C OEt
Copyright © byThe Berkeley Review®
O
A
O O
H,CX^K.'— OEt HoCHoC
3 AA 3 ^ BA
O O
A^A
NaOEt
O O
AB =
CH3
O o
O O
C OEt HOEt A^A
A
H H B
H,C OEt H3CH2C
Bl £ CHo
O O
X^X,
BB I
CH3 —3
83 CARBONYLS & ALCOHOLS EXPLANATIONS OEt
OEt
OEt HgCHfeCT ^T OEt
CH,
Passage XII (Questions 74 - 80) Transesterification Reaction
74. Choice A is correct. The ratio of the integrals for each signal in the proton NMR of spectrum I is given as 3 : 2 : 3.
This implies that the compound has either eight total hydrogens or some multiple of eight total hydrogens. Of the choices, choice B (methyl ethanoate) has six hydrogens and choice D (ethyl methanoate) has six hydrogens. This narrows the question down to either choice A (methyl propanoate) or choice C (ethyl ethanoate). The key feature in the spectrum is the singlet far down field caused by a methyl bonded to the oxygen. This peak implies that the compound must have a methoxy group on the carbonyl carbon. Pick A for optimal satisfaction. The structures are drawn below:
a b
H3CH2C
O
OCH,
Methyl Propanoate a: Triplet (3H) b: Quartet (2H) c: Singlet (3H)
O
H3C OCHo
Methyl Ethanoate a: Singlet (3H) b: Singlet (3H)
o
H3C"
b c
OCH2CH3 Ethyl Ethanoate a: Singlet (3H) b: Quartet (2H) c: Triplet (3H)
The bolded hydrogen will be found the furthest down field.
O
a / ^ \ * b c
H OCH2CH3
Ethyl Methanoate a: Singlet (1H) b: Quartet (2H) c: Triplet (3H)