Arrange the ortho (I), meta (II) and para (III) methoxybenzyl carbanions in decreasing order of stability.
(I)
(II)
(III)

Question

Arrange the ortho (I), meta (II) and para (III) methoxybenzyl carbanions in decreasing order of stability. (I) (II) (III)

TestHub · Accepted Answer

For a carbanion the effects reverse. OCH₃ pushes electrons by +R from ortho and para, which destabilises the negative charge. At meta only the -I effect works, and that stabilises the anion, so meta is the most stable.

The -I effect is stronger at ortho than at para, so ortho $>$ para.

Order : II $>$ I $>$ III.

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