Question

What is \(Z\) in the given sequence of reactions?

• A. \(\mathrm{C_6H_5COCl}\)
• B. \(\mathrm{C_6H_5COCH_3}\)
• C. \(\mathrm{C_6H_5OCH_3}\)
• D. \(\mathrm{C_6H_5COCH_2CH_3}\)

Hint:

Remember the sequence: \[ \boxed{ \text{Anti-Markovnikov addition} \rightarrow \text{Wurtz reaction} \rightarrow \text{Aromatization} \rightarrow \text{Friedel--Crafts acylation}. } \]

Answer 1

The Correct Option is B

Step 1: Identify compound \(X\). Propene undergoes anti-Markovnikov addition of HBr in the presence of benzoyl peroxide to give \[ \mathrm{CH_3CH_2CH_2Br} \] (1-bromopropane). On treatment with sodium in dry ether (Wurtz reaction), \[ 2\mathrm{CH_3CH_2CH_2Br} \rightarrow \mathrm{CH_3(CH_2)_4CH_3}. \] Thus, \[ X=\mathrm{n\text{-}hexane}. \]

Step 2: Identify compound \(Y\). On passing \(n\)-hexane over \[ \mathrm{Mo_2O_3} \] at \[ 773\,\mathrm{K}, \] aromatization occurs to form \[ \boxed{Y=\mathrm{C_6H_6}\ (\text{benzene}).} \]

Step 3: Identify compound \(Z\). Benzene undergoes Friedel--Crafts acylation with acetyl chloride in the presence of anhydrous \(\mathrm{AlCl_3}\): \[ \mathrm{C_6H_6+CH_3COCl \rightarrow C_6H_5COCH_3+HCl.} \] Hence, \[ \boxed{Z=\mathrm{C_6H_5COCH_3}.} \] Therefore, the correct option is \(\boxed{(B)}\).