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

The addition of HBr (1 equivalent) to 1-pentene-4-yne is ${\mathrm{H}}_3{\mathrm{CCHBrCH}}_2\mathrm{C}\equiv \mathrm{CH}$.

${\mathrm{CH}}_2={\mathrm{CHCH}}_2\mathrm{C}\equiv \mathrm{CH}\stackrel{{\mathrm{H}}^{\oplus }}{\to }{{\mathrm{CH}}_3}^{\oplus }{\mathrm{CHCH}}_2\mathrm{C}\equiv \mathrm{CH}\stackrel{{\mathrm{Br}}_{\to }}{\to }{\mathrm{CH}}_3\underset{\underset{{\mathrm{B}}_{\mathrm{r}}}{|}}{\mathrm{C}}{\mathrm{HCH}}_2\mathrm{C}\equiv \mathrm{CH}$

${\mathrm{CH}}_2={\mathrm{CHCH}}_2\mathrm{C}\equiv \mathrm{CH}\underset{\mathrm{less} \mathrm{stable} \mathrm{vinyl} \mathrm{carbocation} (\mathrm{It} \mathrm{is} \mathrm{not} \mathrm{formed})}{\overset{\mathrm{stable} 2^{°}\mathrm{alkyl} \mathrm{carbocation} H^{\oplus }}{\to }} {\mathrm{CH}}_2={\mathrm{CHCH}}_2\stackrel{\oplus }{\mathrm{C}}={\mathrm{CH}}_2$

Hence, option C is correct.