The molecule is reacting with sodamide, therefore, it should be terminal alkyne. Now, this alkyne is subjected ozonolysis followed by oxidation giving two carboxylic acids in which one carboxylic acid is optically active among them. Hence, the alkyne should be:$3-\mathrm{Methylpent}-1-\mathrm{yne}$.

The reaction is shown below: ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}\left({\mathrm{CH}}_3\right)\mathrm{C}\equiv \mathrm{CH}\underset{\mathrm{Zn}/{\mathrm{H}}_2{\mathrm{O}}_2}{\overset{{\mathrm{O}}_3}{\to }}\underset{\mathrm{Optically} \mathrm{active}}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}\left({\mathrm{CH}}_3\right)\mathrm{COOH}}+\mathrm{HCOOH}$