Product (A) is:

Options

A. `Ph-CH_(2)-Cl`
B. `Ph-CH_(2)-Br`
C. `Ph-CH_(2)-C Cl_(3)`
D. `Ph-CH_(2)-CBrCl_(2)`

Answer

b

Explanation

The use of light suggest a free radical mehanism. This means that the methane derivative will undergo homolytic. Since the C- Br bond is weaker than the C-Cl bond, it is resonable to suppose that the former will be broken. Hence :
`CBrcl_(3)overset(hv)to Br^(**)+**C Cl_(3)`
`PhCH_(3)overset(**C Cl_(3))toCHCl_(3)+PhCH_(2)**overset(CBrCl_(3))toPhCH_(2)Br+**C Cl_(3)`,etc.
Attack by the free radical on toluene occurs at the methyl side -chain and not in the ring beacasue the `C-H` bond in Me is weaker than that of a ring -hydrogen atom and the benzyl free radical is far more stable than an aryl free radical.
The other point that requires explanation is why toluence is attacked by the `C Cl_(3)` free radical and not by the bromine free radical. Activation energies involving free radicals are usually very low and so the controlling factors is the heat of reaction (or, more correctly, the free energy of reaction). The more exothermic the reaction (greater is `DeltaG`),the more favoured is that reaction. If the bromine atom attacks, is HBr, the bond of which is much weaker than the `C-H` bond formed when `**C Cl_(3)` attacks.Hence, reaction proceeds by the later more.