In the reaction CH_3-CH_2-C equiv CH + HCl to | vedclass

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(C) Step $1$: Reaction of $1$-butyne with $HCl$. According to Markovnikov's rule,the electrophile $H^+$ adds to the terminal carbon,and the nucleophil

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$CH_3-CH_2-CCl(I)-CH_3$

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(C) Step $1$: Reaction of $1$-butyne with $HCl$. According to Markovnikov's rule,the electrophile $H^+$ adds to the terminal carbon,and the nucleophile $Cl^-$ adds to the more substituted carbon. The product $B$ is $CH_3-CH_2-CCl=CH_2$ ($2$-chloro$-1-$butene).Step $2$: Reaction of $B$ with $HI$. Again,following Markovnikov's rule,the $H^+$ adds to the $CH_2$ group and the $I^-$ adds to the $CCl$ carbon. The final product $C$ is $CH_3-CH_2-CCl(I)-CH_3$ ($2$-chloro$-2-$iodobutane).

In the reaction $CH_3-CH_2-C \equiv CH + HCl \to B \xrightarrow{HI} C$,what is the product $C$?

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