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The addition of HBr (1 equivalent) to 1-pentene-4-yne is

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a

H_{2} C = {CHCH}_{2} C (Br) = {CH}_{2}

b

H_{2} C = {CHCH}_{2} CH = CHBr

c

H_{3} {CCHBrCH}_{2} C \equiv CH

d

{BrH}_{2} {CCH}_{2} {CH}_{2} C \equiv CH

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detailed solution

Correct option is C

The addition of HBr (1 equivalent) to 1-pentene-4-yne is $H_{3} {CCHBrCH}_{2} C \equiv CH$ .

${CH}_{2} = {CHCH}_{2} C \equiv CH \overset{H^{\oplus}}{\to} {CH}_{3}^{\oplus} {CHCH}_{2} C \equiv CH \overset{{Br}_{\to}}{\to} {CH}_{3} \underset{\underset{B_{r}}{|}}{C} {HCH}_{2} C \equiv CH$

${CH}_{2} = {CHCH}_{2} C \equiv CH \to_{less stable vinyl carbocation (It is not formed)}^{stable 2^{°} alkyl carbocation H^{\oplus}} {CH}_{2} = {CHCH}_{2} \overset{\oplus}{C} = {CH}_{2}$

Hence, option C is correct.

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