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

Step 1: Recall what hyperconjugation needs.
Hyperconjugation is the overlap of a sigma bond with an adjacent empty $p$ orbital or a pi system. So an empty or partly empty orbital next door is essential.

Step 2: Look at the methyl carbanion.
In $CH_3^-$ the carbon is $sp^3$ hybridised and holds a lone pair. There is no empty $p$ orbital available.

Step 3: Judge Statement-I.
Because there is no empty orbital to accept electron density, $CH_3^-$ cannot gain hyperconjugative stability. So Statement-I is correct.

Step 4: Judge Statement-II.
Statement-II says there is no vacant $p$ orbital in $CH_3^-$ so it cannot take part in hyperconjugation. This is the true reason. So Statement-II is correct.

Step 5: Check the link.
The lack of a vacant $p$ orbital is exactly why hyperconjugation fails, so Statement-II correctly explains Statement-I.

Step 6: Choose the option.
Both statements are correct and II explains I.
\[ \boxed{\text{I correct, II correct and explains I}} \]