Free-radical monochlorination of 2-methylbutane is carried out and every possible constitutional isomer of C5H11Cl is formed. What is the total…

$2$-Methylbutane, $\mathrm{(CH_3)_2CH{-}CH_2{-}CH_3}$, has four types of hydrogens, so four constitutional monochlorides arise. Substitution at the methyl groups on $\mathrm{C2}$ gives $1$-chloro-$2$-methylbutane, whose $\mathrm{C2}$ carries $\mathrm{H}$, $\mathrm{CH_3}$, $\mathrm{CH_2CH_3}$ and $\mathrm{CH_2Cl}$, so it is chiral: $2$ stereoisomers. Substitution at $\mathrm{C2}$ gives $2$-chloro-$2$-methylbutane (two methyls on the carbon, achiral): $1$. Substitution at $\mathrm{C3}$ gives $2$-chloro-$3$-methylbutane, with the new stereocentre at the carbon bearing $\mathrm{Cl}$, $\mathrm{H}$, $\mathrm{CH_3}$ and isopropyl: chiral, $2$ stereoisomers. Substitution at the terminal $\mathrm{C4}$ gives $1$-chloro-$3$-methylbutane, which is achiral: $1$. The total is $2+1+2+1=6$.