(no ring substitution) Product (A) is | vedclass

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(B) The reaction involves a free radical mechanism initiated by light $(hv)$.$CBrCl_3$ undergoes homolytic fission to produce radicals. Since the $C-B

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$Ph-CH_2-Br$

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(B) The reaction involves a free radical mechanism initiated by light $(hv)$.$CBrCl_3$ undergoes homolytic fission to produce radicals. Since the $C-Br$ bond is weaker than the $C-Cl$ bond,the $C-Br$ bond breaks preferentially:$CBrCl_3 \xrightarrow{hv} Br^{\bullet} + {}^{\bullet}CCl_3$The ${}^{\bullet}CCl_3$ radical then abstracts a hydrogen atom from the methyl group of toluene $(PhCH_3)$ to form a benzyl radical $(PhCH_2^{\bullet})$:$PhCH_3 + {}^{\bullet}CCl_3 \rightarrow PhCH_2^{\bullet} + CHCl_3$The benzyl radical then reacts with $CBrCl_3$ to form the product $(A)$ and regenerate the ${}^{\bullet}CCl_3$ radical:$PhCH_2^{\bullet} + CBrCl_3 \rightarrow PhCH_2Br + {}^{\bullet}CCl_3$Thus,the major product $(A)$ is benzyl bromide $(Ph-CH_2-Br)$.

(no ring substitution) Product $(A)$ is

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