Which of the following compunds will show retention in configuration on nucleophilic substitution by OH – ion?

Solution

Chemistry ma’am · 3 steps

First find the stereocentre. If OH⁻ attacks a different carbon, that configuration cannot change.

1. What can change. Inversion happens only when the nucleophile attacks the chiral carbon itself. So does racemisation. If no bond to the stereocentre breaks, its configuration is kept.

   Option	Carbon bearing Br	Where is the stereocentre?	Result with OH⁻
   (a) C₆H₅CH(Br)CH₃	secondary, benzylic	the C–Br carbon	S_N1 through a stable benzylic carbocation gives racemisation
   (b) (CH₃)₂CHBr	secondary	none, since this carbon carries two CH₃ groups	achiral, so there is no configuration to keep
   (c) CH₃CH(C₂H₅)CH₂Br	primary CH₂Br	the next carbon, not the C–Br carbon	S_N2 at CH₂ leaves the stereocentre untouched, so retention
   (d) CH₃CH(Br)C₆H₁₃	secondary	the C–Br carbon	S_N2 with backside attack gives inversion

2. Look at (c). In 1-bromo-2-methylbutane, OH⁻ replaces Br on the primary CH₂ carbon. The chiral carbon keeps its H, CH₃, C₂H₅ and CH₂⁻ bonds, so its spatial arrangement stays the same.
3. So. Only (c) shows retention of configuration. The answer is (c).

The idea to keep. Retention or inversion is decided only at the carbon that is attacked. If the chiral centre is a neighbouring carbon, none of its bonds break, so its configuration is kept whatever the mechanism.

Answer from the NTA final answer key. Solution by PrepWiser.