Question

Bromination of toluene in the presence of anhydrous \(\mathrm{AlCl_3}\) gave \(X\) and bromination of cyclohexene in the presence of light yielded \(Y\). \(X\) and \(Y\) respectively are

• A. Aryl bromide, Allyl bromide
• B. Benzyl bromide, Allyl bromide
• C. Aryl bromide, Vinyl bromide
• D. Benzyl bromide, Vinyl bromide

Hint:

Remember: \[ \boxed{ \mathrm{Br_2/AlCl_3} \rightarrow \text{Electrophilic aromatic substitution} } \] \[ \boxed{ \mathrm{Br_2/h\nu} \rightarrow \text{Free radical allylic substitution} } \]

Answer 1

The Correct Option is A

Step 1: Identify \(X\). Bromination of toluene in the presence of \[ \mathrm{Br_2/AlCl_3} \] proceeds by electrophilic aromatic substitution on the benzene ring. Hence, the product is \[ \boxed{\text{Aryl bromide}.} \]

Step 2: Identify \(Y\). Bromination of cyclohexene in the presence of light proceeds through free radical substitution at the allylic position. Hence, the product is \[ \boxed{\text{Allyl bromide}.} \]

Step 3: Write the answer. Therefore, \[ X=\text{Aryl bromide}, \] and \[ Y=\text{Allyl bromide}. \] Thus, \[ \boxed{(A)} \] is the correct answer.