Chemistry - How to determine the most stable carbanion?

Solution 1:

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).

Solution 2:

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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Solution 3:

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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Solution 4:

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 .