Take the 2-minute tour ×
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

Can anyone please tell me what the correct answer will be and why?
I think the answer is either a or c as in d there are no chances for carbanion to have its charge become less .in b i think there is + inductive effect due to methyl group which gives it more -ve charge . The ones left are a and c . N is an electronegative compound so it might lessen the charge on carbanion but in both a and c carbanion is same close to it. I am unable to reason out it from here.

share|improve this question
This is actually a decent question, it deserves to be on this site IMO. –  shortstheory May 13 '14 at 16:05
@shortstheory actually the question has been edited so the close votes have been retracted... –  G M May 13 '14 at 17:08

2 Answers 2

up vote 4 down vote accepted

Wikipedia says:

The stability and reactivity of a carbanion is determined by several factors. These include

1)The inductive effect. Electronegative atoms adjacent to the charge will stabilize the charge;

2)Hybridization of the charge-bearing atom. The greater the s-character of the charge-bearing atom, the more stable the anion;

3)The extent of conjugation of the anion. Resonance effects can stabilize the anion. This is especially true when the anion is stabilized as a result of aromaticity.

You can see that point 3) is not relevant here. We do not have a conjugated system, nor is there any chance for resonance.

Keeping in mind point 1), the $\ce{N^+}$ has a $\ce{-I}$ effect, but inductive effect reduces over distance. The negative charges in options a), b) and d) are either in the meta or para position.

Also, remember that $\ce{+I}$ effect destabilises the carbanion.

In c), the carbanion is primary, but in a), it is secondary. Primary carbanions are more stable due to lesser $\ce{+I}$ effect.

So your answer is c).

share|improve this answer

In $a$ or $c$ $\ce N$ atom is equidistant from $-$ charge. Hence, $c$ is most stable as it has no $+I$ effect from other ends like $a$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.