JEE Main 2023 · 13 April · morning · Single correct · +4 / −1
2-Methyl propyl bromide reacts with C₂ H₅ O⁻ and gives 'A' whereas on reaction with C₂ H₅ OH it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are :
- (a) S_N 1, A= tert-butyl ethyl ether; S_N 2, B= iso-butyl ethyl ether
- (b) S_N 1, A= tert-butyl ethyl ether; S_N 1, B= 2-butyl ethyl ether
- (c) S_N 2, A= iso-butyl ethyl ether; S_N 1, B= tert-butyl ethyl ether
- (d) S_N 2, A= 2-butyl ethyl ether; S_N 2, B= iso-butyl ethyl ether
Answer and solution
Answer (c) S_N 2, A= iso-butyl ethyl ether; S_N 1, B= tert-butyl ethyl ether
Solution
Chemistry ma’am · 4 steps
Same primary bromide, two nucleophiles: ethoxide attacks directly, ethanol waits for a rearranged carbocation.
- Read the substrate. 2-Methylpropyl (isobutyl) bromide is (CH₃)₂CH-CH₂Br. The bromine sits on a primary carbon, so there is little crowding around it for a backside attack.
- With ethoxide. C₂H₅O⁻ is a strong, charged nucleophile, and it attacks the primary carbon directly in one step. That is S_N2, and the carbon skeleton stays the same. So A is iso-butyl ethyl ether, (CH₃)₂CH-CH₂-O-C₂H₅.
- With ethanol. C₂H₅OH is a weak, neutral nucleophile and a polar protic solvent, so the reaction goes by S_N1 instead. The Br⁻ leaves first, and the primary carbocation (CH₃)₂CH-CH₂⁺ at once undergoes a 1,2-hydride shift to the much more stable tertiary cation.
(CH₃)₂CH-CH₂⁺ → (CH₃)₃C⁺ →(C₂H₅OH, -H⁺) (CH₃)₃C-O-C₂H₅
- So. A forms by S_N2 as iso-butyl ethyl ether, and B forms by S_N1 as tert-butyl ethyl ether. That is option (c).
The idea to keep. Strong nucleophile on a primary halide means S_N2 with no change in skeleton. Weak nucleophile in a protic solvent means S_N1, so always check whether the carbocation can rearrange first.
Answer from the NTA final answer key. Solution by PrepWiser.