Question

Choose the correct stability order of the given free radicals. 

• A. \(I>II>III>IV\)
• B. \(II>I>IV>III\)
• C. \(II = I>IV>III\)
• D. \(III>IV>II>I \)

Hint:

Free radicals are more stable when they have resonance stabilization and hyperconjugation.

Answer 1

The Correct Option is B

Step 1: Understanding Free Radical Stability Factors
The stability of free radicals is influenced by: 1. Resonance stabilization
2. Hyperconjugation
3. Inductive effects
Step 2: Evaluating Stability of Each Compound
Free radicals (II) and (I) have stabilizing effects from resonance and hyperconjugation.
Free radical (III) has a strong \(-I\) effect from \(-NO_2\), which destabilizes it.
Free radical (IV) has a weak electron-withdrawing effect from \(-CN\), making it more stable than (III).
Step 3: Ranking Stability Order
The stability order follows: \[ II>I>IV>III \] Thus, the correct answer is (B).

Answer 2

To determine the stability order of the given free radicals, we must consider the effects of substituents on the benzyl free radical. The key factors include resonance, inductive effect, and hyperconjugation.

Let's analyze each compound:

Free Radical I:
This is a benzyl radical with a methyl group (–CH₃) and a hydroxyl group (–OH) on the ring. The –OH group is an electron-donating group via resonance (+R effect), which stabilizes the radical. The methyl group is also weakly electron-donating via hyperconjugation.

Free Radical II:
Same as I, but the –OH group is in the para position, which allows full delocalization of the unpaired electron into the ring through resonance. The +R effect of –OH is stronger at the para position, leading to even more stabilization than in I.

Free Radical III:
Contains a nitro group (–NO₂), which is a strong electron-withdrawing group by both –R and –I effects. This destabilizes the free radical by pulling electron density away from the ring and the radical center.

Free Radical IV:
Has a cyano group (–CN), which is also an electron-withdrawing group, but its –R effect is weaker than –NO₂. Thus, it destabilizes the radical but not as severely as the nitro group in III.

Conclusion:
Based on the resonance and inductive effects, the correct stability order is:

II > I > IV > III

This means:
- II is the most stable due to strong resonance stabilization from the para –OH group.
- I is next due to meta –OH and hyperconjugation from –CH₃.
- IV is less stable due to electron-withdrawing –CN.
- III is least stable due to strong electron-withdrawing –NO₂.

Correct Answer: \(II > I > IV > III\)