JEE Main 2025 Jan 22 Shift 2, Chemistry Q8: Preparation of Alkyl Halides (Halo Alkanes) & Aryl Halides (Halo Arenes)

The reaction is the standard Grignard protonolysis: $RBr\stackrel{Mg}{\to }RMgBr\stackrel{H_{2}O}{\to }RH$. So the alkyl group of $RBr$ becomes the alkane skeleton of 2-methylbutane: $(C{H}_3{)}_{2}CH-C{H}_2-C{H}_3$.

Counting structural isomers of $C_5{H}_{11}Br$ whose carbon skeleton is exactly that of 2-methylbutane: Br can be placed on any of the four distinct carbon environments of 2-methylbutane:

(i) On C-1 (primary $C{H}_3$ adjacent to the branch) $\to$ 1-bromo-2-methylbutane.

(ii) On C-2 (tertiary CH carbon) $\to$ 2-bromo-2-methylbutane.

(iii) On C-3 (secondary $C{H}_2$) $\to$ 2-bromo-3-methylbutane (i.e. 3-bromo-2-methylbutane).

(iv) On C-4 (terminal primary CH3 of the ethyl side) $\to$ 1-bromo-3-methylbutane.

The two methyl groups on C-2 are equivalent, so substituting them gives the same compound as (i). Hence there are exactly 4 distinct $RBr$ structures.