Ph-CH_2-CH=CH-CH_3 xrightarrow[(ii) text{ Alc | vedclass

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(C) Step $1$: The reaction of an alkene with $Br_2$ (bromination) results in the formation of a vicinal dibromide.$Ph-CH_2-CH=CH-CH_3 + Br_2 \rightarr

Vedclass · Accepted Answer

$Ph-CH_2-C \equiv C-CH_3$

Vedclass · Answer

(C) Step $1$: The reaction of an alkene with $Br_2$ (bromination) results in the formation of a vicinal dibromide.$Ph-CH_2-CH=CH-CH_3 + Br_2 ightarrow Ph-CH_2-CH(Br)-CH(Br)-CH_3$Step $2$: Treatment of the vicinal dibromide with alcoholic $KOH$ (a strong base) leads to dehydrohalogenation via an $E2$ mechanism,resulting in the formation of an alkyne.$Ph-CH_2-CH(Br)-CH(Br)-CH_3 \xrightarrow{	ext{Alc. KOH}} Ph-CH_2-C \equiv C-CH_3 + 2HBr$Thus,the final product is $Ph-CH_2-C \equiv C-CH_3$.

$Ph-CH_2-CH=CH-CH_3$ $\xrightarrow[(ii) \text{ Alc. KOH}]{(i) \text{ } Br_2}$ ?

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