Question

Identify the end product \(Z\) in the given sequence of reactions \[ C_3H_6 \xrightarrow[\left(C_6H_5COO\right)_2]{HBr} X \] \[ X \xrightarrow[\text{Anhy. }AlCl_3]{C_6H_6} Y \] \[ Y \xrightarrow[(iii)\ NaOH/CaO,\Delta]{(i)\ KMnO_4/OH^- \;\; (ii)\ H_3O^+} Z \]

• A. Toluene
• B. Cumene
• C. Benzene
• D. Xylene

Hint:

Key reactions used: \[ \text{Propene + HBr (Peroxide)} \rightarrow \text{Anti-Markovnikov product} \] \[ \text{Alkyl benzene} \xrightarrow{KMnO_4} \text{Benzoic acid} \] Soda-lime decarboxylation \[ ArCOONa \rightarrow ArH \] Thus, \[ \text{Cumene} \rightarrow \text{Benzoic acid} \rightarrow \text{Benzene}. \]

Answer 1

The Correct Option is C

Step 1: Formation of \(X\). Propene reacts with HBr in the presence of peroxide. This follows the anti-Markovnikov addition (Kharasch effect). \[ CH_3CH=CH_2 \xrightarrow[\text{Peroxide}]{HBr} CH_3CH_2CH_2Br \] Thus, \[ \boxed{X=n\text{-propyl bromide}} \]

Step 2: Formation of \(Y\). \(n\)-Propyl bromide undergoes Friedel--Crafts alkylation with benzene. The initially formed primary carbocation rearranges to the more stable secondary carbocation. \[ C_6H_6 + CH_3CH_2CH_2Br \xrightarrow{AlCl_3} C_6H_5CH(CH_3)_2 \] Hence, \[ \boxed{Y=\text{Cumene (isopropylbenzene)}} \]

Step 3: Oxidation of cumene. Any alkyl benzene having at least one benzylic hydrogen is oxidized by alkaline \(KMnO_4\) to benzoic acid. \[ C_6H_5CH(CH_3)_2 \xrightarrow{KMnO_4} C_6H_5COOH \] After acidification, \[ \boxed{\text{Benzoic acid is formed}} \]

Step 4: Soda-lime decarboxylation. Benzoic acid forms sodium benzoate, which on heating with soda lime undergoes decarboxylation. \[ C_6H_5COONa \xrightarrow[\Delta]{NaOH/CaO} C_6H_6 \] Therefore, \[ \boxed{Z=\text{Benzene}} \]

Final Answer: \[ \boxed{\text{Benzene}} \] \[ \boxed{\text{Answer = (C)}} \]