- ✓$4$ methyl$-2 $ pentyne
- B$4, 4-$dimethyl$-2-$butyne
- CMethyl isopropyl acetylene
- D$2-$methyl$-4-$pentyne
Now, we start numbering from the leftmost carbon as by doing that, the triple bond will be on $2^{nd}$ carbon.
Hence, the name of the compound will be $4$methyl-$2$-pentyne as there are $5$ carbons on the parent chain.
Hence, (a) is correct answer.
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$ - C{H_2} - \mathop C\limits_{\mathop |\limits_{OH} } H - C{H_3}$
$\begin{array}{*{20}{c}}
{^{C{H_3}}} \\
{_H}
\end{array}\begin{array}{*{20}{c}}
{{\text{ }}\backslash {\text{ }}} \\
/
\end{array}\mathop C\limits^{} {\mkern 1mu} = \mathop C\limits^{} {\mkern 1mu} \begin{array}{*{20}{c}}
/ \\
{{\text{ }}\backslash {\text{ }}}
\end{array}_{\mathop C\limits^{} {\kern 1pt} \equiv \mathop C\limits^{} {\kern 1pt} - \mathop C\limits^{} {\kern 1pt} {H_2}\mathop C\limits^{} {\kern 1pt} {H_3}}^H{\mkern 1mu} $