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The major product $(P)$ formed in the below reaction is
$$
\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\mathrm{CCl}_4, 253 \mathrm{~K}]{\mathrm{Br}_2(1 \mathrm{~mol})} P
$$
Options:
$$
\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\mathrm{CCl}_4, 253 \mathrm{~K}]{\mathrm{Br}_2(1 \mathrm{~mol})} P
$$
Solution:
1289 Upvotes
Verified Answer
The correct answer is:
$\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}(\mathrm{Br})-\mathrm{CH}_2(\mathrm{Br})$
Treatment of alkene with $\mathrm{Br}_2$ in $\mathrm{CCl}_4$ solvent gives vicinal dibromides. Bromine added to opposite faces of the double bond (anti addition).
$$
\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2
$$
$$
\mathrm{CH} \equiv \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2
$$
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