CH_3-C(Br)_2-C(Br)_2-CH_3 xrightarrow{x NaNH_ | vedclass

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(D) The reaction proceeds as follows:$1$. Dehydrohalogenation: $CH_3-C(Br)_2-C(Br)_2-CH_3$ reacts with $4 \ moles$ of $NaNH_2$ to form the conjugated

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$8$

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(D) The reaction proceeds as follows:$1$. Dehydrohalogenation: $CH_3-C(Br)_2-C(Br)_2-CH_3$ reacts with $4 \ moles$ of $NaNH_2$ to form the conjugated diyne $CH_3-C \equiv C-C \equiv C-H$ (or the corresponding terminal acetylide salt).$2$. Further deprotonation: The terminal protons of the diyne are acidic and react with $2 \ moles$ of $NaNH_2$ to form the disodium salt $Na^+ C^- \equiv C-C \equiv C^- Na^+$.$3$. Total moles of $NaNH_2$ consumed: $x = 4 + 2 = 6$.$4$. Alkylation: The disodium salt reacts with $2 \ moles$ of $CH_3I$ via an $S_N2$ mechanism to yield $CH_3-C \equiv C-C \equiv C-CH_3$.$5$. Total moles of $CH_3I$ consumed: $y = 2$.$6$. Therefore,$x + y = 6 + 2 = 8$.

$CH_3-C(Br)_2-C(Br)_2-CH_3$ $\xrightarrow{x NaNH_2}$ $\xrightarrow{y CH_3I} CH_3-C \equiv C-C \equiv C-CH_3$
$x$ and $y$ moles are consumed. The value of $x + y$ is:

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$CH_3-C(Br)_2-C(Br)_2-CH_3$ $\xrightarrow{x NaNH_2}$ $\xrightarrow{y CH_3I} CH_3-C \equiv C-C \equiv C-CH_3$$x$ and $y$ moles are consumed. The value of $x + y$ is: