What is the product formed when CH_3CH_2CHCl_ | vedclass

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(B) The reaction of a geminal dihalide like $CH_3CH_2CHCl_2$ with a strong base like $NaNH_2$ (sodamide) proceeds via dehydrohalogenation. First,one m

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$CH_3-C\equiv CH$

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(B) The reaction of a geminal dihalide like $CH_3CH_2CHCl_2$ with a strong base like $NaNH_2$ (sodamide) proceeds via dehydrohalogenation. First,one molecule of $HCl$ is eliminated to form a vinyl halide $(CH_3CH=CHCl)$. Then,the second molecule of $HCl$ is eliminated by the excess $NaNH_2$ to form an alkyne. Thus,$CH_3CH_2CHCl_2 \xrightarrow{NaNH_2} CH_3-C\equiv CH$.

What is the product formed when $CH_3CH_2CHCl_2$ is treated with $NaNH_2$?

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