For the following Friedel Craft's alkylation reaction, which of the statements are correct? [figure] Choose the correct answer from the options given below :
- A and D only
- B and C only
- A and C only
- B and D only
Answer
(2) B and C only
Solution
CH3CH2CH2Cl + AlCl3 CH3CH2 CH2 + AlCl4
The 1° carbocation rearranges by a 1,2-hydride shift to the more stable 2° isopropyl carbocation:
CH3CH2 CH2 CH3 CHCH3
⇒ Major product is isopropylbenzene (cumene), not n-propylbenzene – A incorrect, B correct.
The alkyl group introduced activates the ring (+I, hyperconjugation), so the product reacts faster than benzene ⇒ polyalkylation always accompanies the reaction – C correct.
Electron-donating groups activate the ring, so alkylbenzenes are formed even more easily – D incorrect.
∴ Correct statements: B and C
Answer: (2)