Solution:
\(n\)-butane, \(n\)-hexane, 2,3 -dimethylbutane are symmetrical alkanes. Such alkanes are prepared in good yield by Wurtz reaction as these require single alkyl halide for their preparation.
\(n\)-heptane is an unsymmetrical alkane that requires two different alkyl halides for its preparation. This will form three different alkanes, thus, lowering the yield of required alkane.
$$
\begin{aligned}
&\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl} \stackrel{\text { dry ether }}{\longrightarrow} \\
&\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \\
&+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \\
&+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}
\end{aligned}
$$
