CH_{3}-C equiv CH xrightarrow{2 HBr} xrightar | vedclass

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(B) The reaction proceeds as follows:$1$. Addition of $2 	ext{ moles}$ of $HBr$ to propyne $(CH_{3}-C \equiv CH)$ follows Markovnikov's rule,resultin

Vedclass · Accepted Answer

$CH_{3}-C(=O)-CH_{3}$

Vedclass · Answer

(B) The reaction proceeds as follows:$1$. Addition of $2 	ext{ moles}$ of $HBr$ to propyne $(CH_{3}-C \equiv CH)$ follows Markovnikov's rule,resulting in the formation of a geminal dibromide: $CH_{3}-C(Br)_{2}-CH_{3}$.$2$. Hydrolysis of the geminal dibromide with $H_{2}O$ replaces the bromine atoms with hydroxyl groups,forming a geminal diol (gem-diol): $CH_{3}-C(OH)_{2}-CH_{3}$.$3$. Gem-diols are unstable and lose a molecule of water to form a ketone: $CH_{3}-C(OH)_{2}-CH_{3} ightarrow CH_{3}-C(=O)-CH_{3} + H_{2}O$.Therefore,the final product is acetone $(CH_{3}-C(=O)-CH_{3})$.

$CH_{3}-C \equiv CH$ $\xrightarrow{2 HBr}$ $\xrightarrow{H_{2}O} \text{Product}$. The product is:

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