Question

X and Z respectively in the following reaction sequence are: \[ C_3H_6 \xrightarrow{X} Y \xrightarrow{C_6H_6/AlCl_3} Z \]

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Peroxides lead to anti-Markovnikov addition of HBr, forming 1-bromopropane instead of 2-bromopropane.

Answer 1

The Correct Option is C

Step 1: Propene (\( C_3H_6 \)) reacts with HBr in the presence of ROOR to form 1-bromopropane via anti-Markovnikov addition. Step 2: 1-bromopropane undergoes Friedel-Crafts alkylation with benzene and AlCl3, forming n-propylbenzene. Thus, the correct sequence is HBr/ROOR.