Free-radical monochlorination of 3-methylpentane with Cl2 under ultraviolet light is allowed to go to completion at every hydrogen position. The…

3-Methylpentane, $\mathrm{(CH_3CH_2)_2CHCH_3}$, has four different types of hydrogen, so there are four constitutional monochloro isomers. (i) Chlorination of the methyl on C3 gives 1-chloro-2-ethylbutane, $\mathrm{(CH_3CH_2)_2CHCH_2Cl}$, which has a plane of symmetry: 1 product. (ii) Chlorination at C3 gives 3-chloro-3-methylpentane, with two identical ethyl groups, so it is achiral: 1 product. (iii) Chlorination at C2 gives 2-chloro-3-methylpentane, which has two stereocentres (C2 and C3) and no symmetry, so $2^2=4$ stereoisomers. (iv) Chlorination at C1 gives 1-chloro-3-methylpentane, with one stereocentre (C3 now bears H, methyl, ethyl and $\mathrm{CH_2CH_2Cl}$): 2 enantiomers. The total is $1+1+4+2=8$.