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(A) The reaction proceeds in three steps: $1$. Treatment of $1,2-dibromo-1-phenylethane$ with excess alcoholic $KOH$ leads to dehydrohalogenation,form

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$Ph-C \equiv CH$

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(A) The reaction proceeds in three steps: $1$. Treatment of $1,2-dibromo-1-phenylethane$ with excess alcoholic $KOH$ leads to dehydrohalogenation,forming $bromostyrene$ $(Ph-CH=CHBr)$. $2$. Further treatment with $NaNH_2$ (a strong base) causes another dehydrohalogenation to form the acetylide intermediate $(Ph-C \equiv C^- Na^+)$. $3$. Finally,protonation with $H_3O^+$ yields the terminal alkyne,$phenylacetylene$ $(Ph-C \equiv CH)$.

The major product of the following reaction is:

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