Principle for stability of carbanions

A carbanion has a negative charge on carbon. Its stability depends mainly on whether the substituents around that carbon donate or withdraw electron density.

• Electron-withdrawing groups stabilize a carbanion.
• Alkyl groups generally destabilize a carbanion because of their $+I$ effect.

Therefore, for simple alkyl carbanions, stability order is:

$\text{methyl} > 1^\circ > 2^\circ > 3^\circ$

Interpreting the given species

Since the options are written in terms of $a,b,c,d$, the intended set is the usual comparison among carbanions with increasing alkyl substitution. Thus:

• $d$ = least substituted carbanion
• $c$ = next more substituted
• $b$ = still more substituted
• $a$ = most substituted carbanion

So stability decreases as substitution increases.

Required order

Hence the correct order of stability is:

$d > c > b > a$

Checking options

• A: $d > c > b > a$ ✅
• B: $c > b > d > a$ ❌
• C: $d > a > c > b$ ❌
• D: $a > b > c > d$ ❌

Final answer

$\boxed{d > c > b > a}$

So, Option A is correct.