A pure sample of the branched alkane 2-methylbutane (isopentane) undergoes free-radical monochlorina — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Let's systematically replace hydrogens on 2-methylbutane ($C^1H_3 - C^2H(CH_3) - C^3H_2 - C^4H_3$). 1. Replacing $H$ on C1 or the identical methyl group on C2 gives 1-chloro-2-methylbutane. C2 becomes a chiral center. This exists as 2 enantiomers ($R$ and $S$). 2. Replacing $H$ on C2 gives 2-chloro-2-methylbutane. This molecule has a plane of symmetry and lacks a chiral center. It exists as 1 achiral product. 3. Replacing $H$ on C3 gives 2-chloro-3-methylbutane. C2 is a chiral center. This exists as 2 enantiomers ($R$ and $S$). 4. Replacing $H$ on C4 gives 1-chloro-3-methylbutane. This lacks a chiral center. It exists as 1 achiral product. Total stereoisomers = $2 + 1 + 2 + 1 = 6$.