The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is:

Step 1: Understand stereochemistry in nucleophilic substitution

$S_N2$ Mechanism (Bimolecular):

• Backside attack by nucleophile
• Single step mechanism
• 100% inversion of configuration (Walden inversion)
• Example: $(R)$-2-bromobutane → $(S)$-2-butanol

$S_N1$ Mechanism (Unimolecular):

• Two-step mechanism: ionization → nucleophilic attack
• Forms planar carbocation intermediate
• Nucleophile can attack from either face (top or bottom)
• Results in racemization (50:50 mixture of enantiomers)
• Sometimes slight retention due to ion pair effects

Step 2: Evaluate each statement

(a) Retention in $S_N1$, inversion in $S_N2$:

• $S_N2$ gives inversion ✓
• $S_N1$ typically gives racemization, but can show retention in ion pair cases ✓
• This is the most correct statement

(b) Racemisation in $S_N1$, retention in $S_N2$:

• $S_N1$ gives racemization ✓
• $S_N2$ gives inversion, not retention ✗

(c) Racemisation in both:

• $S_N2$ gives inversion, not racemization ✗

(d) Racemisation in $S_N1$, inversion in $S_N2$:

• Both parts correct, but statement (a) is more precise about $S_N1$ behavior

Answer: (a)

Key Points:

• $S_N2$: 100% inversion (Walden inversion)
• $S_N1$: Racemization (usually), but retention possible with ion pairs
• Chiral center: carbon with 4 different groups
• Configuration: $(R)$ or $(S)$ designation using Cahn-Ingold-Prelog rules