(S)-1-Bromo-2-methylbutane reacts with NaCN in DMSO by S_N2 at the CH2Br carbon. The product 3-methylpentanenitrile is

(S)-3-methylpentanenitrile, because the stereocentre is not the reacting carbon

(R)-3-methylpentanenitrile, because S_N2 always inverts the label

JEE Main levelSingle correct optionDifficulty 2 of 3hx045

(S)-1-Bromo-2-methylbutane reacts with $NaCN$ in DMSO by $S_{N} 2$ at the $CH X_{2} Br$ carbon. The product 3-methylpentanenitrile is

A(S)-3-methylpentanenitrile, because the stereocentre is not the reacting carbon
Bracemic 3-methylpentanenitrile
C(R)-3-methylpentanenitrile, because $S_{N} 2$ always inverts the label
Dan achiral product

The reacting carbon (

CH X_{2} Br

) is not a stereocentre, so the bonds to the stereocentre are untouched and its spatial arrangement is retained. Priorities at the stereocentre:

CH X_{2} Br > C X_{2} H X_{5} > CH X_{3} > H

in the substrate and

CH X_{2} CN > C X_{2} H X_{5} > CH X_{3} > H

in the product (at the second sphere, (N,N,N) beats (H,H,H)). The order is unchanged, so the label remains (S).