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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 the carbanion to have its charge become less. In b I think there is a + inductive effect due to methyl group which gives it more negative charge.
The ones left are a and c. Nitrogen is an electronegative element so it might lessen the charge on the carbanion, but in both a and c the carbanion is same distance from 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 '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@shortstheory actually the question has been edited so the close votes have been retracted...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

3 Answers 3

up vote 5 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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The stability of a carbanion also depends on the nature of hybridized orbitals used by the negatively charge carbon atom.A more electronegative carbon atom accomodates a negative charge in a betterways and electronegativity of the C atom depends on the nature of hybridization its uses.the primary carbanions more stable than secondary and tertiary carbanions...

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Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. Currently your post does not answer the question, which of the carbanions is the most stable one, it reads more like a general comment on the matter of carbanions itself. Maybe you can edit your post to include some more details about the question, otherwise it might get deleted. – Martin - マーチン Sep 9 at 4:44

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