3-Methylpentane, $CH X_{3} CH X_{2} CH (CH X_{3}) CH X_{2} CH X_{3}$ , undergoes free radical monochlorination. Counting every stereoisomer (each enantiomer and each diastereomer) as a separate compound, how many different monochloro products are possible? Give an integer.

Numerical answer type, tolerance 0.5.

Answer: 8

Cl at

C X_{1}

gives 1-chloro-3-methylpentane with one stereocentre (2 isomers). Cl at

C X_{2}

gives 2-chloro-3-methylpentane with two stereocentres (4 isomers). Cl at

C X_{3}

gives 3-chloro-3-methylpentane (1) and Cl on the methyl gives 1-chloro-2-ethylbutane (1, achiral). Total

2 + 4 + 1 + 1 = 8