Methylcyclobutane undergoes free-radical monochlorination at every distinct type of hydrogen. The total number of different monochloro products…

Methylcyclobutane has four types of hydrogen, so consider chlorination at each position. (i) The methyl group gives (chloromethyl)cyclobutane, which is achiral: 1 product. (ii) Ring carbon C1 (bearing the methyl) gives 1-chloro-1-methylcyclobutane, also achiral: 1 product. (iii) Ring carbon C2 gives 1-chloro-2-methylcyclobutane, in which C1 and C2 are both stereocentres and the two substituents differ, so there are $2^2=4$ stereoisomers (cis and trans, each as a pair of enantiomers). (iv) Ring carbon C3 gives 1-chloro-3-methylcyclobutane, where the plane of symmetry through C1 and C3 makes both carbons non-stereogenic centres and only cis and trans geometric isomers exist: 2 products. The total is $1+1+4+2=8$.