Question

Arrange the following reaction in decreasing order of reactivity with NBS/heat:

• A. $1>2>3>4$
• B. $2>1>3>4$
• C. $1>2>4>3$
• D. $4>3>2>1$

Answer 1

The correct option is C $1>2>4>3$

NBS/heat reaction with alkenes follows free radical mechanism. If free radical is more stable more is the reactivity.

(1) It forms a tertiary allylic free radical which is very stable due to resonance and electron donating effect of $-C{H}_3$

(2) It forms 2 same type of free radical but at secondary carbon, the resonance from both side

(3) resonance only from one side

(4) radical more stable than (3) due to $-C{H}_3$ group.

1 is most reactive as it produces 3∘3∘, allylic free radical and 3 is least reactive as it produces least stable free radical.