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Question

(D) The reaction of a vicinal dihalide with alcoholic $KOH$ followed by $NaNH_2$ results in double dehydrohalogenation to form an alkyne.Step $1$: $CH

Vedclass · Accepted Answer

$CH_3-CH_2-C \equiv CH$

Vedclass · Answer

(D) The reaction of a vicinal dihalide with alcoholic $KOH$ followed by $NaNH_2$ results in double dehydrohalogenation to form an alkyne.Step $1$: $CH_3-CH_2-CH(Br)-CH_2-Br \xrightarrow{KOH(alc)} CH_3-CH_2-CH=CHBr$ (Vinyl halide).Step $2$: $CH_3-CH_2-CH=CHBr \xrightarrow{NaNH_2/liq. NH_3} CH_3-CH_2-C \equiv CH$ (But$-1-$yne).The major product is but$-1-$yne.

The major product of the following reaction is: $CH_3-CH_2-CH(Br)-CH_2-Br \xrightarrow[(ii) NaNH_2 \text{ in liq. } NH_3]{(i) KOH \text{ alc.}}$

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