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JEE Main levelSingle correct optionDifficulty 2 of 3o4134

Phenyl 2-phenoxyethyl sulfone, $PhSO X_{2} CH X_{2} CH X_{2} OPh$ , reacts with $NaOEt$ in ethanol. Which description is correct?

AThe $PhO X^{-}$ group ionises first to give a primary carbocation, then a proton is lost.
BThe $H$ on the carbon next to $OPh$ is removed, giving a vinyl ether.
C $EtO X^{-}$ attacks the carbon bearing $OPh$ in a single-step SN2 reaction giving $PhSO X_{2} CH X_{2} CH X_{2} OEt$ only.
DThe $H$ on the carbon next to $SO X_{2} Ph$ is removed first, then $PhO X^{-}$ leaves, giving phenyl vinyl sulfone.

Answer: D

The sulfonyl group stabilises an adjacent carbanion strongly, so that alpha-H is far more acidic than the one next to

OPh

. The carbanion

PhSO X_{2} CH X^{-} CH X_{2} OPh

then expels the poor leaving group

PhO X^{-}

, an E1cB path giving

PhSO X_{2} CH = CH X_{2}

. A primary carbocation would not form.

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