To carry out the above conversion,(A) and (B) | vedclass

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(C) The reaction involves the alkylation of a terminal alkyne. Step $A$: The terminal alkyne is treated with a strong base like $NaNH_2$ to form an ac

Vedclass · Accepted Answer

$NaNH_2$,$I-CH_2-CH_2-CH_2-Br$

Vedclass · Answer

(C) The reaction involves the alkylation of a terminal alkyne. Step $A$: The terminal alkyne is treated with a strong base like $NaNH_2$ to form an acetylide ion. Step $B$: The acetylide ion acts as a nucleophile and undergoes an $S_N2$ reaction with an alkyl halide. To ensure the reaction occurs selectively at the $Br$ end,we use an alkyl halide with a better leaving group on the other side,such as $I-CH_2-CH_2-CH_2-Br$. Since $I^-$ is a better leaving group than $Br^-$,the nucleophilic attack occurs at the carbon attached to the iodine atom. Therefore,$(A) = NaNH_2$ and $(B) = I-CH_2-CH_2-CH_2-Br$.

To carry out the above conversion,$(A)$ and $(B)$ respectively are:

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