Free-radical monochlorination of 3-methylpentane with Cl2/hnu is carried out. Counting every constitutional isomer and every stereoisomer…

3-Methylpentane, $\mathrm{CH_3CH_2CH(CH_3)CH_2CH_3}$, has four types of hydrogen: (i) the two equivalent terminal $\mathrm{CH_3}$ of the ethyl groups, giving $\mathrm{ClCH_2CH_2CH(CH_3)C_2H_5}$ with one stereocentre, so 2 stereoisomers; (ii) the two equivalent $\mathrm{CH_2}$ groups, giving $\mathrm{CH_3CHClCH(CH_3)CH_2CH_3}$ with two stereocentres and no symmetry, so 4 stereoisomers; (iii) the tertiary C-H, giving 3-chloro-3-methylpentane, which is achiral, so 1; (iv) the methyl on C-3, giving $\mathrm{ClCH_2CH(C_2H_5)_2}$, which is achiral, so 1. Total $=2+4+1+1=8$. Counting only structural isomers gives 4, and forgetting the second stereocentre gives 6.