Question

Benzene reacts with propyl chloride in presence of \(AlCl_3\) to give product \(X\). Mark correct statement(s) for the given reaction. 


(a) One of the intermediate is formed due to rearrangement. 
(b) Major product is n-propylbenzene. 
(c) Polysubstitution of substrate is also possible. 
(d) Electron releasing group decreases rate of reaction.

• A. b and d are correct
• B. a and c are correct
• C. b and c are correct
• D. a and d are correct.

Answer 1

The Correct Option is B

Concept: 
This reaction is an example of Friedel–Crafts alkylation. In this reaction, an alkyl halide reacts with benzene in presence of a Lewis acid catalyst such as \(AlCl_3\) to generate a carbocation intermediate which then attacks the benzene ring. 

Step 1: Formation of carbocation intermediate. \[ CH_3CH_2CH_2Cl + AlCl_3 \rightarrow CH_3CH_2CH_2^+ + AlCl_4^- \] The primary carbocation formed is unstable and rearranges to a more stable secondary carbocation. \[ CH_3CH_2CH_2^+ \rightarrow (CH_3)_2CH^+ \] Thus, rearrangement occurs. Step 2: Electrophilic substitution on benzene. The rearranged secondary carbocation attacks benzene to form isopropylbenzene (cumene) as the major product. \[ \text{Benzene} + (CH_3)_2CH^+ \rightarrow \text{Isopropylbenzene} \] Therefore, the major product is not n-propylbenzene. Step 3: Polysubstitution possibility. The alkyl group introduced into benzene is electron-donating and activates the ring toward further substitution. Hence polysubstitution can occur. Step 4: Effect of substituents. Electron-releasing groups increase the rate of electrophilic substitution rather than decrease it. Step 5: Final evaluation of statements. (a) True — rearrangement occurs. 
(b) False — major product is isopropylbenzene. 
(c) True — polysubstitution is possible. 
(d) False — electron releasing groups increase reaction rate. \[ \boxed{\text{Correct statements: a and c}} \]