Identify Reaction Proceeding Through Radical Mechanism Combination

To determine the correct combination that proceeds via a free radical mechanism, let us analyze the reaction given in option (D): (I) (ii) (R)

1. Starting Material (Column 1 - I): Toluene ($\text{C}_6\text{H}_5\text{CH}_3$)
2. Reaction Conditions (Column 2 - ii): $\text{Br}_2 / h\nu$ (Bromine in the presence of light)
3. Type of Reaction (Column 3 - R): Substitution

Reaction Mechanism (Free Radical Substitution):

When toluene is treated with bromine in the presence of light ($h\nu$), it undergoes free radical substitution at the benzylic position to form benzyl bromide ($\text{C}_6\text{H}_5\text{CH}_2\text{Br}$).

The step-by-step free radical mechanism is as follows:

• Initiation: Homolytic fission of the bromine molecule induced by light energy ($h\nu$) generates bromine free radicals: $\text{Br}_2 \xrightarrow{h\nu} 2\,\text{Br}^\bullet$

• Propagation:

  1. A bromine radical abstracts a hydrogen atom from the methyl group of toluene, yielding a resonance-stabilized benzyl radical: $\text{C}_6\text{H}_5\text{CH}_3 + \text{Br}^\bullet \longrightarrow \text{C}_6\text{H}_5\text{CH}_2^\bullet + \text{HBr}$
  2. The benzyl radical reacts with another bromine molecule to yield benzyl bromide and regenerate a bromine radical: $\text{C}_6\text{H}_5\text{CH}_2^\bullet + \text{Br}_2 \longrightarrow \text{C}_6\text{H}_5\text{CH}_2\text{Br} + \text{Br}^\bullet$

• Termination: Combination of any two free radicals removes the reactive species from the system, e.g.: $2\,\text{Br}^\bullet \longrightarrow \text{Br}_2$ $\text{C}_6\text{H}_5\text{CH}_2^\bullet + \text{Br}^\bullet \longrightarrow \text{C}_6\text{H}_5\text{CH}_2\text{Br}$

Thus, the combination (I) (ii) (R) correctly describes a free radical substitution reaction.

Correct Option: D