For the given reaction: What is A?

Step 1: Identify Reaction Conditions

$\ce{Br2}$ with UV light (hv) = free radical bromination (not electrophilic aromatic substitution which requires Lewis acid).

Free radical conditions: bromination occurs at the most stable free radical position.

Step 2: Identify Most Stable Radical Position

Starting material: ethylbenzene derivative with $\ce{-CN}$ group.

• Benzylic position ($\ce{-CHBr-}$ on the $\ce{CH2}$ adjacent to ring): benzylic radical is stabilized by resonance with the aromatic ring. Very stable.
• Ring C-H: requires homolytic cleavage of strong $\ce{sp^2}$ C-H (less favored under free radical conditions).

Bromine radical preferentially abstracts H from the benzylic position.

Step 3: Product

$\ce{C6H4(CH2CH3)(CN) + Br. -> C6H4(CHBr-CH3)(CN)}$

$A =$ 1-(1-bromoethyl)-cyanobenzene = benzylic monobromination product.

Key Points:

• UV light + $\ce{Br2}$: free radical mechanism (not EAS)
• Free radical bromination: highly selective, attacks most stable radical position
• Benzylic > tertiary > secondary > primary for radical stability
• Ring bromination requires $\ce{FeBr3}$/Lewis acid (ionic mechanism)