An alkyl bromide A ($\mathrm{C}{}_4\mathrm{H}{}_9\mathrm{Br}$) is treated with sodium in dry ether to give an alkane B. Free radical monochlorination of B gives exactly three constitutional isomers (ignore stereoisomers). A is

1-Bromo-2-methylpropane couples to 2,5-dimethylhexane, which has three kinds of hydrogen ($\mathrm{CH}{}_3$, $\mathrm{CH}$, $\mathrm{CH}{}_2$), so three monochlorides. 1-Bromobutane gives octane (4 types of H), 2-bromobutane gives 3,4-dimethylhexane (4 types) and 2-bromo-2-methylpropane would give 2,2,3,3-tetramethylbutane (1 type).