[kelaklub]

The answer to the following question is choice D. I understand that choices B, C, and E are more acidic due to the electron withdrawing effects of the halogens. But I thought alkyl groups were electron donating so why is choice A, the one with the longer alkyl chain, the answer? Is it that the inductive effect stabilizes the conjugate acid?

[Winga]

If you want to explain this question in terms of the stability of the conjugate acid, both A and D can be stabilized by inductive effect.

About PhNH3+ and CH3NH3+, it is difficult to tell that which inductive effect is greater(in fact, sp2 hybridized C of Ph ring is more electronegative than that of methyl C, but I think no one will use it for explanation). By compare their acidity, PhNH3+ is more acidic because its conjugate base can be stabilized by delocalization of N's lone pair into Ph ring.

Actually, I will use the availability of the lone pair of N to explain. As I mentioned above, the lone pair of N of PhNH2 can delocalize on Ph ring (resonance effect), its lone pair now is not quite available to abstract a proton, while CH3NH2 has no such effect.

[kelaklub]

Oh man. That's right, choice A represents a cyclic system, and it's conjugate acid is more resonance stabilized.

Thanks.

[Dan]

Oh man. That's right, choice A represents a cyclic system, and it's conjugate acid is more resonance stabilized.

No, D is the most basic. You've got the wrong end of the stick, the lone pair on N in choice A has it's bascisity suppressed by delocalisation. The conjugate acid of A, PhNH3+ is NOT resonance stabilised - read Winga's post again.