Question

Consider the following compound (X):
 
The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding C - H bond are:

• A. I, IV
• B. III, II
• C. II, IV
• D. II, and I

Hint:

Radical stability increases with the number of alkyl groups attached to the carbon carrying the radical. The more alkyl groups, the more stable the radical.

Answer 1

The Correct Option is D

Step 1: Analyze Radical from C-H bond I

Cleavage of the C-H bond at position I results in a radical on an sp-hybridized carbon atom.

\[ \text{H}-\overset{\text{I}}{\text{C}}\equiv\text{C}-\text{CH}_2-\text{CH(CH}_3)_2 \xrightarrow{\text{cleavage}} \cdot\text{C}\equiv\text{C}-\text{CH}_2-\text{CH(CH}_3)_2 \]

This is an alkynyl radical. The unpaired electron resides in an sp orbital, which has 50% s-character. Radicals on sp-hybridized carbons are extremely unstable.

Step 2: Analyze Radical from C-H bond II

Cleavage of a C-H bond at position II results in a radical on the carbon atom adjacent to the triple bond.

\[ \text{HC}\equiv\text{C}-\overset{\text{II}}{\text{CH}_2}-\text{CH(CH}_3)_2 \xrightarrow{\text{cleavage}} \text{HC}\equiv\text{C}-\dot{\text{C}}\text{H}-\text{CH(CH}_3)_2 \]

This is a secondary (2°) radical. More importantly, it is a propargylic radical. The unpaired electron is adjacent to the C≡C triple bond and can be delocalized through resonance, which provides significant stabilization.

\[ \text{HC}\equiv\text{C}-\dot{\text{C}}\text{H}-\text{R} \longleftrightarrow \text{H}\dot{\text{C}}=\text{C}=\text{CH}-\text{R} \]

Step 3: Analyze Radical from C-H bond III

Cleavage of the C-H bond at position III results in a radical on a tertiary carbon atom.

\[ \text{HC}\equiv\text{C}-\text{CH}_2-\overset{\text{III}}{\text{CH}}(\text{CH}_3)_2 \xrightarrow{\text{cleavage}} \text{HC}\equiv\text{C}-\text{CH}_2-\dot{\text{C}}(\text{CH}_3)_2 \]

This is a tertiary (3°) radical. It is stabilized by hyperconjugation with the hydrogen atoms of the adjacent methyl and methylene groups. However, it is not directly adjacent to the triple bond, so it is not stabilized by resonance.

Step 4: Analyze Radical from C-H bond IV

Cleavage of a C-H bond at position IV (on one of the methyl groups) results in a radical on a primary carbon atom.

\[ \text{HC}\equiv\text{C}-\text{CH}_2-\text{CH(CH}_3)\overset{\text{IV}}{\text{CH}_3} \xrightarrow{\text{cleavage}} \text{HC}\equiv\text{C}-\text{CH}_2-\text{CH(CH}_3)\dot{\text{C}}\text{H}_2 \]

This is a primary (1°) radical, stabilized only by hyperconjugation.

Conclusion:

Most Stable Radical: Comparing the radicals, the propargylic radical (from II) is stabilized by resonance, while the tertiary radical (from III) is stabilized only by hyperconjugation. Resonance is a much stronger stabilizing effect than hyperconjugation. Therefore, the radical formed at position II is the most stable.

Least Stable Radical: The alkynyl radical (from I) has its unpaired electron in an sp orbital, making it far less stable than any of the radicals on sp³-hybridized carbons (II, III, and IV). Therefore, the radical formed at position I is the least stable.

The most stable and least stable carbon radicals are produced from the cleavage of C-H bonds at positions II and I, respectively.