The correct answer is A.
This question requires us to translate acid strength into Ka.
Recall that a smaller Ka corresponds to a weaker acid;
do not confuse Ka with pKa!
We can thus translate the question into: "Which substituted benzoic acid will
be the weakest acid?" Recall next that electron-withdrawing groups increase
acid strength since they help to stabilize the anion formed upon dissociation,
so a further translation is, "Which of the following is the least electron-withdrawing
as a substituent?" Of the choices offered, B, C, and D are electron-withdrawing
groups, while choices A and E are electron-donating. Of the electron-donating
groups listed, NH2 is more so than is CH3 due
to the lone pair of electrons on the nitrogen atom. (In electrophilic aromatic
substitutions, the amine group is a strongly activating ortho/para director
because of its electron-donating ability.) Choice A is thus the most electron-donating
and has the most destabilizing effect on the conjugate base (the anion), thereby
disfavoring dissociation of the proton. The compound para-aminobenzoic
acid is therefore the least acidic, and has the smallest Ka.
(B)
Distortion. Aldehydes are electron withdrawing
groups. They stabilize the anion, making it more acidic and subsequently
have a high Ka.
(C)
Opposite. Nitro groups are the most electron
withdrawing group. Therefore it would be the most acidic and subsequently
have the highest Ka.
(D)
Distortion. Chlorine is weakly electron withdrawing
and will raise Ka because it stabilizes the anion somewhat.
(E)
Distortion. Methyl groups are electron donating
and will lower Ka but not as much as amine groups.
