JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Haloalkanes and Haloarenes

What happens to the stereocentre depends on *where* the substitution occurs and by *which* mechanism.

(P) (−)-1-Bromo-2-ethylpentane, ${\mathrm{S}}_{\mathrm{N}}2$: the bromine is on C-1 ($\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$), which is not the stereocentre — the stereocentre is C-2. Substitution at C-1 leaves the bonds at C-2 untouched, so the product keeps its configuration → (2) retention of configuration.

(Q) (−)-2-Bromopentane, ${\mathrm{S}}_{\mathrm{N}}2$: here the bromine sits on the stereocentre (C-2). Back-side attack by $\mathrm{O}{\mathrm{H}}^{-}$ inverts it → (1) inversion of configuration.

(R) (−)-3-Bromo-3-methylhexane, ${\mathrm{S}}_{\mathrm{N}}1$: the tertiary bromide ionises to a planar carbocation at the only stereocentre; ${\mathrm{H}}_{2}O/O{\mathrm{H}}^{-}$ attacks either face equally, giving both configurations → (3) mixture of enantiomers (racemisation).

(S) The substrate has two stereocentres: $\mathrm{CH}(\mathrm{Me})$ and the tertiary $\mathrm{C}(\mathrm{Me})(\mathrm{Br})$. ${\mathrm{S}}_{\mathrm{N}}1$ makes a planar carbocation at the C–Br centre only; attack from both faces gives both configurations there while the other centre is unchanged. Two products differing at one of two stereocentres are → (5) mixture of diastereomers.

P → 2, Q → 1, R → 3, S → 5.

Answer: (B)