Which among the following compounds will undergo fastest S N 2 reaction.

Solution

Chemistry ma’am · 3 steps

For SN2, look only at the carbon holding Br: the fewer groups crowding it, the faster the backside attack.

1. What SN2 needs. In SN2 the nucleophile attacks the carbon bearing Br from the side opposite to Br, in a single step. Every alkyl group on that carbon blocks this backside approach, so the rate order is 1° > 2° > 3°.
2. Classify each bromide. Count the carbon atoms attached directly to the carbon that carries Br. The ring counts as one such group wherever it is joined to that carbon.

   Option	Groups on the C–Br carbon	Type	SN2 rate
   (a) 1-bromoethylcyclobutane	cyclobutyl, CH₃, H	2°	moderate
   (b) 2-bromo-2-cyclobutylpropane	cyclobutyl, two CH₃	3°	very slow
   (c) bromomethylcyclobutane	cyclobutyl, two H	1°	fastest
   (d) 1-bromo-1-methylcyclobutane	two ring carbons, CH₃	3°	very slow

3. So. Only (c) has Br on a CH₂ carbon, a primary centre with two small H atoms. The nucleophile reaches it most easily, so (c) undergoes the fastest SN2 reaction.

The idea to keep. SN2 rate is decided by crowding at the carbon holding the leaving group: CH₃X > 1° > 2° > 3°. Classify that carbon first; the rest of the molecule matters much less.

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