JEE Main · 2024 · 30 Jan 2024 · Shift 2 · General Organic Chemistry

Free-radical chlorination of 2-chlorobutane, ${\text{CH}}_3{\text{–CHCl–CH}}_2{\text{–CH}}_3$, can substitute at any of its four carbons. Substitution at C-1 gives 1,2-dichlorobutane, with a single stereocentre at C-2 (bonded to Cl, ${\text{CH}}_2\text{Cl}$, ${\text{C}}_2{\text{H}}_5$, H) — a chiral pair of enantiomers, both optically active. Substitution at C-2 (the carbon already bearing Cl) gives 2,2-dichlorobutane, with two identical chlorines on that carbon, so it has no stereocentre and is achiral. Substitution at C-3 gives 2,3-dichlorobutane, which now carries two stereocentres with an identical set of substituents on each — this gives a chiral (R,R)/(S,S) enantiomeric pair (both optically active) and a separate achiral meso (R,S) form. Substitution at C-4 gives 1,3-dichlorobutane, ${\text{CH}}_3{\text{–CHCl–CH}}_2{\text{–CH}}_2\text{Cl}$, whose remaining stereocentre (Cl, ${\text{CH}}_3$, H, ${\text{CH}}_2{\text{CH}}_2\text{Cl}$, all different) again gives a chiral pair, both optically active. Counting the optically active members: 2 (from C-1) + 2 (the d,l pair from C-3) + 2 (from C-4) = 6.