JEE Main 2021 · 27 July · evening · Single correct · +4 / −1
Given below are two statements :
Statement I : Hyperconjugation is a permanent effect.
Statement II : Hyperconjugation in ethyl cation (CH₃-C(+)H₂) involves the overlapping of C_(sp²)-H₁ₛ bond with empty 2p orbital of other carbon.
Choose the correct option :
- (a) Both statement I and statement II are false
- (b) Statement I is incorrect but statement II is true
- (c) Statement I is correct but statement II is false
- (d) Both statement I and statement II are true.
Answer and solution
Answer (c) Statement I is correct but statement II is false
Solution
Chemistry ma’am · 4 steps
Look at which carbon donates the C–H bond: the CH₃ carbon is sp³, not sp².
- Statement I. Hyperconjugation is present in the molecule all the time, even with no attacking reagent around. It is a delocalisation of σ electrons in the ground state, so it is a permanent effect. Statement I is correct.
- Find the hybridisation. In CH₃-C(+)H₂, the positive carbon is sp² and holds the empty 2p orbital. The C–H bonds that donate electrons belong to the neighbouring CH₃ carbon, which has four single bonds and so is sp³.
- Statement II. The overlap is between a C_(sp³)-H₁ₛ σ bond and the empty 2p orbital of the cationic carbon. Statement II says C_(sp²)-H₁ₛ, so it is false.
- So. Statement I is correct and Statement II is false, which is option (c).
The idea to keep. In hyperconjugation the donating C–H bond is always on the α-carbon next to the cation, radical or double bond. That carbon is sp³, so the bond is C_(sp³)-H.
Answer from the NTA final answer key. Solution by PrepWiser.