When CH_3CH_2CHCl_2 is treated with NaNH_2,th | vedclass

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(B) The reaction of a geminal dihalide like $CH_3CH_2CHCl_2$ with a strong base like $NaNH_2$ leads to dehydrohalogenation. First,one molecule of $HCl

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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 $NaNH_2$ leads to dehydrohalogenation. First,one molecule of $HCl$ is eliminated to form a vinyl halide $(CH_3CH=CHCl)$. Then,a second molecule of $HCl$ is eliminated to form an alkyne. Thus,$CH_3CH_2CHCl_2 + 2NaNH_2 \rightarrow CH_3-C \equiv CH + 2NaCl + 2NH_3$. The final product is propyne $(CH_3-C \equiv CH)$.

When $CH_3CH_2CHCl_2$ is treated with $NaNH_2$,the product formed is

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