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(A) The reaction of vinyl bromide $(CH_2=CHBr)$ with a strong base like $NaNH_2$ is a dehydrohalogenation reaction. $NaNH_2$ is a very strong base tha

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$H-C \equiv C-H$

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(A) The reaction of vinyl bromide $(CH_2=CHBr)$ with a strong base like $NaNH_2$ is a dehydrohalogenation reaction. $NaNH_2$ is a very strong base that abstracts a proton from the vinyl bromide,leading to the elimination of $HBr$ and the formation of the triple bond in acetylene $(HC \equiv CH)$. The reaction is: $CH_2=CHBr + NaNH_2 \rightarrow HC \equiv CH + NaBr + NH_3$ Thus,the major product is acetylene $(HC \equiv CH)$.

The major product in the following reaction is

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