Question

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

• A. $A\to$ hyperconjugation, $B\to$ empty $2p$-orbital of carbon, $C\to C_{s{p}^3}-H_{1s}$ bond
• B. $A\to$ inductive effect, $B\to i2p$-orbital of carbon, $C\to C_{s{p}^2}-H_{1s}$ bond
• C. $A\to$ electromeric effect, $B\to$ empty $2p$-orbital of carbon, $C\to C_{s{p}^3}-H_{1s}$ bond
• D. $A\to$ resonance effect, $B\to i2p$-orbital of carbon, $C\to C_{s{p}^3}-H_{1s}$ bond

Answer 1

Answer: (A)

In $C{H}_3\stackrel{+}{C}{H}_2$ (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 :