An alkane ${\mathrm{C}}_{8}{\mathrm{H}}_{18}$ is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide.

From Wurtz reaction of an alkyl halide gives an alkane with double the number of carbon atoms present in the alkyl halide. Here, Wurtz reaction  of a primary alkyl halide gives and alkane $\left({\mathrm{C}}_{8}{\mathrm{H}}_{18}\right)$, therefore, the alkyl halide must contain four carbon atoms. Now the two possible primary alkyl halides having four carbon atoms each are I and II.
Since, alkane ${\mathrm{C}}_{8}{\mathrm{H}}_{18}$ on monobromination yields a single isomer of tertiary alkyl halide, therefore, the alkane must contain tertiary hydrogen. This is possible, only if primary alkyl halide (which undergoes Wurtz reaction) has a tertiary hydrogen.