$C X_{6} H X_{5} SO X_{2} CH X_{2} CH X_{2} OCH X_{3}$ is treated with a strong base such as $NaOEt$ in ethanol. The organic product is

A $C X_{6} H X_{5} SO X_{2} C \equiv CH$
B $C X_{6} H X_{5} SO X_{2} CH = CH X_{2}$
C $C X_{6} H X_{5} SO X_{2} CH X_{2} CH X_{2} OC X_{2} H X_{5}$
D $C X_{6} H X_{5} SO X_{2} X^{-} + CH X_{2} = CH - OCH X_{3}$

Answer: B

The sulfonyl group makes the adjacent CH2 hydrogens acidic (pKa about 29), so the base forms an alpha-sulfonyl carbanion. This expels methoxide from the next carbon (E1cB) to give phenyl vinyl sulfone,

C X_{6} H X_{5} SO X_{2} CH = CH X_{2}

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Answer: B

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$C X_{6} H X_{5} SO X_{2} CH X_{2} CH X_{2} OCH X_{3}$ is treated with a strong base such as $NaOEt$ in ethanol. The organic product is