Question

Consider the following statements and select the correct option: Statement-I : \(CH_3^-\) lacks hyperconjugative stability. Statement-II : There is no vacant \(p\)-orbital in \(CH_3^-\) and as such it cannot participate in hyperconjugation.

• A. Statement-I is correct and Statement-II is the correct explanation of Statement-I.
• B. Statement-I is correct but Statement-II is incorrect.
• C. Statement-I is incorrect but Statement-II is correct.
• D. Both Statement-I and Statement-II are incorrect.

Hint:

Hyperconjugation requires an adjacent empty \(p\)-orbital or a \(\pi\)-bond. \[ CH_3^+ \;\text{shows hyperconjugation} \] \[ CH_3^- \;\text{does not show hyperconjugation} \] because it has a lone pair and no vacant \(p\)-orbital.

Answer 1

The Correct Option is A

Concept: Hyperconjugation requires an adjacent empty \(p\)-orbital or a \(\pi\)-system to overlap with a \(\sigma\)-bond. Examples showing hyperconjugation: \[ CH_3^+,\qquad CH_2=CH_2,\qquad \text{free radicals} \] A methyl carbanion possesses a lone pair and does not contain a vacant \(p\)-orbital.

Step 1: Analyze Statement-I. The methyl carbanion is \[ CH_3^- \] Carbon is \(sp^3\)-hybridized and contains a lone pair. Since hyperconjugation requires an empty orbital, \(CH_3^-\) does not exhibit hyperconjugative stabilization. Therefore, Statement-I is Correct

Step 2: Analyze Statement-II. In \(CH_3^-\),

• Carbon is \(sp^3\)-hybridized.

• No vacant \(p\)-orbital is present.

• Hence \(\sigma\)-electrons cannot delocalize through hyperconjugation.
Therefore, Statement-II is Correct

Step 3: Check whether Statement-II explains Statement-I. Absence of a vacant \(p\)-orbital is precisely the reason why \(CH_3^-\) cannot undergo hyperconjugation. Hence Statement-II correctly explains Statement-I. \[ \boxed{\text{Statement-I and Statement-II are correct, and Statement-II explains Statement-I}} \] Therefore, option \(\mathbf{(A)}\) is correct.