The major product of the following reaction, if it occurs by S N 2 mechanism is :

Solution

Chemistry ma’am · 4 steps

Phenoxide hits the CH₂Br carbon directly, and in S_N2 nothing else in the chain moves.

1. Make the nucleophile. Phenol is acidic enough for K₂CO₃ to remove its O–H proton, giving phenoxide, C₆H₅O⁻. Acetone is a polar aprotic solvent, so it suits an S_N2 reaction. This is a Williamson ether synthesis.
2. Find the attack site. In S_N2 the nucleophile attacks the carbon that carries the leaving group, from the back side. Here that is the CH₂ next to Br. It is a primary allylic carbon, so it is open to attack and reacts fast.

   C₆H₅O⁻ + (CH₃)₂C=CH-CH₂Br → (CH₃)₂C=CH-CH₂-O-C₆H₅ + Br⁻

3. Rest stays put. Br leaves as Br⁻ in the same step that the new C–O bond forms. The double bond and the two methyl groups are not touched, so the C=C stays between C-2 and C-3.
4. So. The product is phenyl 3-methylbut-2-enyl ether, C₆H₅-O-CH₂-CH=C(CH₃)₂, which is option (d).

The idea to keep. In a plain S_N2 on an allylic halide, the nucleophile takes exactly the carbon that held the halogen, and the double bond stays where it was. Look for that product before anything rearranged.

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