Question

Propene on reaction with reagent (A) gave \(X\) as major product. \(X\) undergoes Wurtz reaction to give \(Y\). In the presence of (B) at \(773\,\mathrm{K}\) and \(10{-}20\,\mathrm{atm}\) pressure, \(Y\) undergoes aromatization. It also undergoes isomerisation in the presence of (C). What are \(A\), \(B\), \(C\) respectively?

• A. \(\mathrm{HBr;\ anhyd.\ AlCl_3;\ V_2O_5}\)
• B. \(\mathrm{HCl;\ MoO_3;\ AlCl_3/HCl}\)
• C. \(\mathrm{HBr/(C_6H_5CO)_2O_2;\ Cr_2O_3;\ anhyd.\ AlCl_3/HCl}\)
• D. \(\mathrm{HBr/(C_6H_5CO)_2O_2;\ AlCl_3;\ CrO_3}\)

Hint:

Remember: \[ \boxed{ \mathrm{HBr + peroxide} \Rightarrow \text{Anti-Markovnikov addition} } \] \[ \boxed{ \mathrm{Cr_2O_3} \Rightarrow \text{Aromatization catalyst} } \] \[ \boxed{ \mathrm{AlCl_3/HCl} \Rightarrow \text{Isomerisation catalyst} } \]

Answer 1

The Correct Option is C

Step 1: Identify reagent \(A\). Propene reacts with HBr in the presence of peroxide by the anti-Markovnikov addition. \[ \mathrm{CH_3CH=CH_2} \xrightarrow[\mathrm{(C_6H_5CO)_2O_2}]{\mathrm{HBr}} \mathrm{CH_3CH_2CH_2Br} \] Thus, \[ X=\mathrm{1\!-\!bromopropane}. \]

Step 2: Formation of \(Y\). Wurtz reaction gives \[ 2\mathrm{CH_3CH_2CH_2Br} \xrightarrow{\mathrm{Na/dry\ ether}} \mathrm{n\!-\!hexane}. \] Hence, \[ Y=\mathrm{n\!-\!hexane}. \]

Step 3: Identify reagents \(B\) and \(C\). Aromatization of \(n\)-hexane at \[ 773\,\mathrm{K},\ 10{-}20\,\mathrm{atm} \] is carried out using \[ \boxed{\mathrm{Cr_2O_3}}. \] Isomerisation of alkanes occurs in the presence of \[ \boxed{\mathrm{anhyd.\ AlCl_3/HCl}}. \] Therefore, \[ \boxed{ A=\mathrm{HBr/(C_6H_5CO)_2O_2},\; B=\mathrm{Cr_2O_3},\; C=\mathrm{anhyd.\ AlCl_3/HCl} } \] Hence, \[ \boxed{(C)} \] is the correct answer.