Question
Benzene reacts with CH₃CH₂Cl in the presence of anhydrous AlCl₃. The major product obtained is ethylbenzene. However, when propyl chloride (CH₃CH₂CH₂Cl) is used under the same conditions, the major product is isopropylbenzene (cumene), not n-propylbenzene.
(a) Explain why rearrangement occurs in Friedel-Crafts alkylation with propyl chloride.
(b) Why is Friedel-Crafts acylation preferred over alkylation in synthesis? Illustrate with the reaction of benzene with CH₃COCl/AlCl₃.
Solution — Step by Step
In Friedel-Crafts alkylation, AlCl₃ polarises the C–Cl bond of the alkyl halide to form a carbocation. With CH₃CH₂CH₂Cl, we first get a primary carbocation:
This 1° carbocation is unstable — it rearranges immediately via a 1,2-hydride shift to the more stable 2° carbocation:
The secondary carbocation (isopropyl cation) is the one that attacks the benzene ring. So the product is isopropylbenzene, not n-propylbenzene — even though we started with n-propyl chloride.
This is why Friedel-Crafts alkylation with straight-chain alkyl halides of 3 or more carbons never gives the straight-chain product.
With CH₃COCl and AlCl₃, the electrophile formed is the acylium ion:
The acylium ion is stabilised by resonance between the carbocation and the C≡O⁺ form. Because of this resonance stabilisation, it has no driving force to rearrange — it attacks benzene as-is.
The acylium ion attacks the π system of benzene in the rate-determining step, forming an arenium ion (σ-complex). Loss of H⁺ (with AlCl₄⁻ acting as base) restores aromaticity:
The product, acetophenone, is a ketone — a clean, predictable product with no rearrangement.
If we actually need ethylbenzene from benzene, the synthetic route is:
- Acylation → acetophenone (C₆H₅COCH₃)
- Clemmensen reduction (Zn–Hg/HCl) → ethylbenzene (C₆H₅CH₂CH₃)
This two-step route gives the straight-chain product reliably, unlike direct alkylation.
Why This Works
Carbocations follow the stability order: 3° > 2° > 1°. Whenever a less stable carbocation can convert to a more stable one via a 1,2-hydride or 1,2-methyl shift, it does so almost instantly. In alkylation, the carbocation intermediate has enough lifetime to rearrange before attacking benzene.
The acylium ion is fundamentally different. The positive charge on carbon is delocalised into the oxygen via resonance (R–C⁺=O ↔ R–C≡O⁺), which makes it more stable than a simple carbocation. A rearrangement would break this resonance stabilisation, so it’s energetically unfavourable — the acylium ion stays intact.
This difference has a direct consequence for multi-step synthesis: acylation is the go-to first step whenever you need a predictable aryl ketone or want to introduce a straight-chain alkyl group via the reduction route.
Alternative Method
JEE shortcut for identifying rearrangement: Whenever the alkyl group in a Friedel-Crafts alkylation has 3 or more carbons AND the halide is on a primary carbon, assume the product will be the branched (rearranged) isomer. Don’t even draw the mechanism — just shift a hydride mentally and write the secondary/tertiary carbocation product directly.
For part (b), you can also argue from product stability: acetophenone has a conjugated carbonyl group (C=O in conjugation with the ring), which makes it a relatively stable product. The AlCl₃ forms a coordination complex with the ketone oxygen (C₆H₅COCH₃·AlCl₃), which actually deactivates the ring and prevents polyalkylation — another advantage of acylation over alkylation, where polyalkylation is a persistent side reaction.
Common Mistake
Writing the product of n-propyl chloride + benzene/AlCl₃ as n-propylbenzene. This is the single most common error in this topic. The question is specifically designed to test whether you remember that primary carbocations rearrange. Always check: is the carbocation primary? If yes, draw the rearranged product. JEE Advanced 2023 awarded marks specifically for identifying the rearranged product and explaining the 1,2-hydride shift — students who wrote n-propylbenzene got zero for that part even with a correct mechanism drawn.
A second mistake: confusing Clemmensen reduction (acidic conditions, Zn–Hg/conc. HCl, reduces C=O to CH₂) with Wolff–Kishner reduction (basic conditions, N₂H₄/KOH, does the same thing). Both reduce ketones to alkanes — use Clemmensen when the molecule is acid-stable, Wolff–Kishner when it’s base-stable.