# How to determine the most stable carbanion?

Which of the following carbanions is the most stable?

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.

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.

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 $a$

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.

The stability of carbanion is mostly determined by:

Inductive effect:

If the electronegative atom attached to carbanion and it withdraw electrons from the negative charge specie and reduce the electronic density , hence this factor stabilize the carbanion. The carbanion which have more electronegative atoms will be more stable and vice vera for electropositive atoms.

Resonance:

The carbanions in resonance will be more stable , the electron will delocalize and hence stabilize the carbanions .

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...

• 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. Sep 9 '15 at 4:44

In (a) the carbonation is 2degree which means there will be inductive effect from two carbon atoms thereby destabilising it.

But in (c) the carbanion is at the end so it would be more stable than (a).