9 . B
When α-carbons are deprotonated, the negative charge is resonance stabilized in part by the
electronegative carbonyl oxygen, which is electron-withdrawing. Alkyl groups are actually
electron-donating, which destabilizes carbanion intermediates; this invalidates statement II.10 . C
All of the answer choices are nitrogen-containing functional groups, but only enamines are
tautomers of imines. Imines contain a double bond between a carbon and a nitrogen; enamines
contain a double bond between two carbons as well as an amine.
11 . C
When succinaldehyde (or any aldehyde or ketone with α-hydrogens) is treated with a strong base
like lithium diisopropylamide (LDA), it forms the more nucleophilic enolate carbanion.
12 . C
Aldol condensations contain two main steps. In the first step, the α-carbon of an aldehyde or
ketone is deprotonated, generating the enolate carbanion. This carbanion can then attack
another aldehyde or ketone, generating the aldol. In the second step, the aldol is dehydrated,
forming a double bond. This double bond is between the α- and β-carbons, so the molecule is an
α,β-unsaturated carbonyl.
13 . D
Because benzaldehyde lacks an α-proton, it cannot be reacted with base to form the nucleophilic
enolate carbanion. Therefore, acetone will act as our nucleophile, and both choices (A) and (B)
can be eliminated. In order to perform this reaction, which is an aldol condensation, acetone will
be reacted with a strong base—not a strong acid—in order to extract the α-hydrogen and form
the enolate anion, which will act as a nucleophile.
14 . B
This is an example of an aldol condensation, but stopped after aldol formation (before
dehydration). After the aldol is formed using strong base, the reaction may be halted by the
addition of acid. Butanal in strong acid, described in choice (C), would be likely to deprotonate
without gaining the hydroxyl group. Methanal in diethyl ether would not be reactive because
diethyl ether is not a strong enough base to abstract the α-hydrogen, eliminating choice (A).
Reaction of the two aldehydes methanal and ethanal in catalytic base would form 3-
hydroxypropanal (which would dehydrate to form propenal), not 3-hydroxybutanal.