Question

Observe the orbital diagram showing hyperconjugation in ethyl cation. Here $A, B$ and $C$ refer to

• A. $A \rightarrow$ hyperconjugation, $B \rightarrow$ empty $2 p$-orbital of carbon, $C \rightarrow C _{s p^{3}}- H _{1 s}$ bond
• B. $A \rightarrow$ inductive effect, $B \rightarrow i 2 p$-orbital of carbon, $C \rightarrow C _{s p^{2}}- H _{1 s}$ bond
• C. $A \rightarrow$ electromeric effect, $B \rightarrow$ empty $2 p$-orbital of carbon, $C \rightarrow C _{s p^{3}}- H _{1 s}$ bond
• D. $A \rightarrow$ resonance effect, $B \rightarrow i 2 p$-orbital of carbon, $C \rightarrow C _{s p^{3}}- H _{1 s}$ bond

Answer 1

Answer: (A)

In $CH _{3} \overset{+}{C}H _{2}$ (ethyl cation), the positively charged carbon atom has an empty $p$-orbital. One of the $C - H$ bonds of the methyl group can align in the plane of this empty $p$-orbital and the electrons constituting the $C - H$ bond in plane with this $p$-orbital can then be delocalised into the empty $p$-orbital as depicted below :