Question

For the following Friedel Craft's alkylation reaction, which of the statements are correct?}

A. Major product is n-propyl benzene.
B. iso-propyl carbocation intermediate is also generated.
C. Multiple substitution is inevitable.
D. Introducing electron-donating substituent on benzene will not produce any alkyl benzene.

• A. A and D only
• B. B and C only
• C. A and C only
• D. B and D only

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The reaction shown is benzene reacting with n-propyl chloride in the presence of anhydrous \(AlCl_3\). This is a standard electrophilic aromatic substitution (Friedel-Crafts alkylation).
Step 2: Detailed Explanation:
Statement A: When n-propyl chloride reacts with \(AlCl_3\), the initial n-propyl carbocation (\(1^\circ\)) undergoes a 1,2-hydride shift to form a more stable isopropyl carbocation (\(2^\circ\)). Thus, the major product is isopropylbenzene (cumene), not n-propylbenzene. Incorrect.
Statement B: As explained above, the rearrangement produces the isopropyl carbocation intermediate. Correct.
Statement C: Alkyl groups are activating groups. Once the first alkyl group is attached, the benzene ring becomes more reactive than the starting benzene. Therefore, further alkylation (polyalkylation) is a common and often inevitable side reaction. Correct.
Statement D: Electron-donating groups (EDG) like \(-OH, -OCH_3, -CH_3\) activate the benzene ring towards electrophilic attack. They would make the production of alkyl benzene easier, not prevent it. (Deactivating groups like \(-NO_2\) prevent Friedel-Crafts). Incorrect.
Step 3: Final Answer:
Statements B and C are correct.