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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.

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@shortstheory actually the question has been edited so the close votes have been retracted...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

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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 $a$

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