Questions: Friedel-Crafts Alkylation Mechanism

Question 1

You attempt to make n-propylbenzene by reacting benzene with 1-chloropropane and AlCl₃. What is the predominant product?

• A. n-propylbenzene — the primary carbocation attacks directly before rearrangement
• B. isopropylbenzene — the primary carbocation rearranges to a secondary carbocation via a 1,2-hydride shift
• C. no reaction — AlCl₃ cannot ionize primary alkyl halides
• D. a dialkylated benzene — polyalkylation occurs before any monosubstituted product forms

Answer: (B)

A primary carbocation is highly unstable and undergoes rapid 1,2-hydride shift to the more stable secondary carbocation before the benzene ring can attack. The secondary carbocation then reacts with benzene, giving isopropylbenzene. This rearrangement problem means Friedel-Crafts alkylation cannot reliably install straight-chain primary alkyl groups longer than ethyl.

Question 2

Why does Friedel-Crafts alkylation tend to produce polyalkylated products even when a 1:1 ratio of alkyl halide to benzene is used?

• A. AlCl₃ is regenerated after each reaction, so it can catalyze unlimited substitutions
• B. The alkyl group installed on the ring is electron-donating, activating the ring toward further electrophilic attack
• C. The carbocation intermediate attacks the product faster than benzene because of ring strain
• D. Polyalkylation is a rearrangement artifact and only occurs with primary alkyl halides

Answer: (B)

Alkyl groups are electron-donating (via hyperconjugation and induction), which activates the ring and makes the monoalkylated product more reactive toward EAS than the starting benzene. The product therefore reacts faster than the starting material, making it hard to stop cleanly at monosubstitution. A large excess of benzene is used to dilute this effect.

Question 3

Friedel-Crafts alkylation of benzene with 1-chloropropane and AlCl₃ gives predominantly n-propylbenzene.

• T. True
• F. False

Answer: False

False. The primary carbocation formed from 1-chloropropane undergoes a 1,2-hydride shift to the more stable secondary carbocation before attacking benzene. The predominant product is isopropylbenzene (cumene), not n-propylbenzene. This rearrangement is unavoidable with primary alkyl halides longer than ethyl.

Question 4

Friedel-Crafts alkylation fails on nitrobenzene because AlCl₃ is not a strong enough Lewis acid to ionize the alkyl halide in the presence of the nitro group.

• T. True
• F. False

Answer: False

False. The reaction fails because the nitro group is a strong electron-withdrawing group that deactivates the benzene ring. The ring is no longer nucleophilic enough to attack the carbocation electrophile — not because the electrophile fails to form. Any strongly electron-withdrawing substituent (–NO₂, –CF₃, –COR) deactivates the ring to the point where EAS cannot proceed.

Question 5

Why is Friedel-Crafts acylation often preferred over alkylation when the synthetic goal is to install a straight-chain alkyl group on a benzene ring?

Model answer: Acylation generates a resonance-stabilized acylium cation (RC≡O⁺) that does not rearrange, giving a predictable product. The resulting ketone also deactivates the ring (the carbonyl is electron-withdrawing), preventing polyacylation. The ketone can then be reduced to the desired alkyl group. Alkylation suffers from both carbocation rearrangement and polyalkylation.

The acylium cation is stabilized by resonance with the oxygen lone pair, which distributes the positive charge and prevents 1,2-shifts. The carbonyl product deactivates the ring so cleanly that a second acylation rarely occurs. After Clemmensen or Wolff-Kishner reduction, you obtain the straight-chain alkyl group that alkylation could never deliver reliably.