JEE Main 2026 · 24 January · morning · Single correct · +4 / −1
Given below are two statements :
Statement I : ' C-Cl ' bond is stronger in CH₂=CH-Cl than CH₃-CH₂-Cl
Statement II : The given optically active molecule,
on
hydrolysis gives a solution that can rotate the plane polarized light.
In the light of the above statements, choose the correct answer from the options given below
- (a) Statement I is false but Statement II is true
- (b) Statement I is true but Statement II is false
- (c) Both Statement I and Statement II are true
- (d) Both Statement I and Statement II are false
Answer and solution
Answer (b) Statement I is true but Statement II is false
Solution
Chemistry ma’am · 5 steps
Vinyl C–Cl gets partial double bond character, but this tertiary benzylic chloride racemises by S_N1.
- Statement I. In CH₂=CH-Cl, a lone pair on Cl conjugates with the π bond (⁻CH₂-CH=Cl⁺), so the C–Cl bond gets partial double bond character. The carbon is also sp², which holds electrons closer than the sp³ carbon of CH₃CH₂Cl. So the bond is shorter and stronger, and Statement I is true.
- Look at the substrate. The chiral carbon carries Ph, Me, Et and Cl, so it is a tertiary, benzylic chloride. Crowding blocks backside attack. The phenyl ring and the alkyl groups stabilise the carbocation, so hydrolysis goes by S_N1.
- Planar carbocation. After Cl⁻ leaves, the carbocation is sp² and flat. Water attacks the top and bottom faces equally, giving equal amounts of the two enantiomers of the alcohol, a racemic mixture.
- Statement II. In a racemic mixture the rotations of the two enantiomers cancel, so the solution is optically inactive. Statement II is false.
- So. Statement I is true and Statement II is false, which is option (b).
The idea to keep. A chiral tertiary or benzylic halide reacts by S_N1 through a flat carbocation, so it racemises. Halogen on an sp² carbon has partial double bond character, so its C–X bond is stronger.
Answer from the NTA final answer key. Solution by PrepWiser.