Question

Arrange the following free radicals in the increasing order of their stability:

• A. \( A<B<D<C \)
• B. \( C<D<B<A \)
• C. \( A<B<D<A \)
• D. \( A<B<C<D \)

Hint:

Free radical stability increases with hyperconjugation and resonance. General order: benzylic \(>\) allylic \(>\) tertiary \(>\) secondary \(>\) primary.

Answer 1

The Correct Option is D

Step 1: Identify radical A.
Radical A is an isopropyl type radical:
\[ (CH_3)_2\dot{C}H \]
It is a secondary alkyl radical.

Step 2: Identify radical B.
Radical B is a tertiary butyl radical:
\[ (CH_3)_3\dot{C} \]
It is a tertiary alkyl radical and is more stable than a secondary radical due to greater hyperconjugation.

Step 3: Identify radical C.
Radical C is an allylic radical:
\[ CH_2=CH-\dot{CH}_2 \]
It is stabilized by resonance, so it is more stable than ordinary alkyl radicals.

Step 4: Identify radical D.
Radical D is a benzyl radical:
\[ C_6H_5-\dot{CH}_2 \]
It is highly stabilized due to resonance with the benzene ring.

Step 5: Compare alkyl radical stability.
For alkyl radicals:
\[ 3^\circ>2^\circ>1^\circ \]
Thus, radical B is more stable than radical A.

Step 6: Compare resonance-stabilized radicals.
Allylic and benzylic radicals are resonance stabilized.
Benzylic radical is more stable than allylic radical because the unpaired electron is delocalized over the benzene ring.
\[ \text{Benzylic radical}>\text{Allylic radical} \]

Step 7: Write the increasing order.
Therefore, the increasing order of stability is:
\[ A<B<C<D \]
Final Answer:
\[ \boxed{A<B<C<D} \]