Friedel-Crafts Alkylation and Acylation: Mechanism and Limits

ChemistryClass 11-12, Hydrocarbons; Aldehydes and KetonesFor JEE and NEET aspirants

The Friedel-Crafts reaction puts an alkyl group (alkylation) or an acyl group (acylation) on a benzene ring. An alkyl halide or an acyl chloride is activated by anhydrous $AlC l_{3}$ , the electrophile it releases attacks the ring, and the ring then loses $H^{+}$ : an electrophilic aromatic substitution. Benzene with chloromethane gives toluene; with ethanoyl chloride it gives acetophenone. Friedel-Crafts reactions appear in the Class 11 Hydrocarbons chapter and the Class 12 chapters on haloarenes and aromatic ketones, and in JEE Main, JEE Advanced and NEET.

At a glance

Alkylation:
$C_{6} H_{6} + C H_{3} Cl anhyd. AlC l_{3} C_{6} H_{5} C H_{3} + HCl$
Acylation:
$C_{6} H_{6} + C H_{3} COCl anhyd. AlC l_{3} C_{6} H_{5} COC H_{3} + HCl$
Acid anhydrides work too:
$C_{6} H_{6} + (C H_{3} CO)_{2} O anhyd. AlC l_{3} C_{6} H_{5} COC H_{3} + C H_{3} COOH$
Electrophile: a carbocation $R^{+}$ (or the polarised $R - Cl \dots AlC l_{3}$ complex) in alkylation; the acylium ion $RC \equiv O^{+}$ in acylation.
How much $AlC l_{3}$ : a little (catalytic) for alkylation; at least one mole per mole for acylation, because the ketone product holds on to $AlC l_{3}$ until water is added (more with an anhydride, whose acid by-product binds it too).
Type: electrophilic substitution through a $σ$ -complex (arenium ion). $FeC l_{3}$ and $B F_{3}$ also work; alkenes or alcohols with acid also give the carbocation.
Fails on: rings carrying a meta-directing (deactivating) group: $N O_{2}$ (strongly), CN, CHO, COR, COOH, $S O_{3} H$ (moderately); and on aniline, whose $N H_{2}$ binds $AlC l_{3}$ .

1. Mechanism

Attack on benzene costs it its aromatic stability, so only a strong electrophile can do it. The job of $AlC l_{3}$ , a Lewis acid with an empty orbital, is to make that electrophile by pulling a chloride off the halide.

Figure 1: Friedel-Crafts acylation of benzene. Step 2 is slow because the ring loses its aromaticity; step 3 restores it. Alkylation follows the same three steps with a carbocation (or the polarised

R - Cl \dots AlC l_{3}

complex) as the electrophile.

Making the electrophile. The Cl of the acyl chloride gives a lone pair to $AlC l_{3}$ , and the C–Cl bond breaks: the acylium ion forms, with $AlC l_{4}^{-}$ . It is stabilised by resonance, with the + shared by C and O.
Attack (slow). Two $π$ electrons of the ring bond to the acylium carbon. The ring is no longer aromatic: the $σ$ -complex has one $s p^{3}$ carbon, and its + charge is shared by the carbons ortho and para to it.
Loss of $H^{+}$ (fast). A Cl of $AlC l_{4}^{-}$ removes the H from the $s p^{3}$ carbon. The C–H electrons rebuild the aromatic ring, and $HCl$ and $AlC l_{3}$ are released (the $AlC l_{3}$ then binds the ketone's C=O until water is added).

Alkylation follows the same three steps. With tertiary and secondary halides the electrophile is a carbocation; with primary halides it is the polarised $R - Cl \dots AlC l_{3}$ complex, which can still rearrange whenever a more stable cation is in reach.

Exam Trick

Need a straight chain on the ring? Acylate, then reduce. Alkylation with 1-chloropropane gives isopropylbenzene, because the n-propyl electrophile rearranges to the isopropyl cation before it attacks. Propanoyl chloride puts on $COC H_{2} C H_{3}$ without rearrangement, and Clemmensen reduction (Zn-Hg, HCl) turns the C=O into $C H_{2}$ .

2. Conditions and scope

2.1 Which rings react

Ring	Reacts?	Why
Benzene	Yes	The reference case
Toluene, anisole	Yes, faster	Activating groups; ortho and para products, para usually major
Chlorobenzene	Yes, slower	Cl deactivates but directs ortho and para; with $C H_{3} Cl$ : 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene (para major)
Nitrobenzene	No	$N O_{2}$ deactivates strongly; nitrobenzene is even used as the solvent
Benzoic acid, benzaldehyde, acetophenone	No	Deactivating C=O groups; COOH also binds $AlC l_{3}$
Aniline	No	The $N H_{2}$ gives its lone pair to $AlC l_{3}$ ; the positive N then deactivates the ring

The halide matters too. Aryl and vinyl halides cannot be used: their cations do not form, so chlorobenzene and benzene do not give biphenyl.

2.2 The three problems of alkylation

Rearrangement. The cation rearranges to a more stable one before it attacks. 1-Chloropropane gives isopropylbenzene (cumene), not n-propylbenzene, by a 1,2-hydride shift.

C_{6} H_{6} + C H_{3} C H_{2} C H_{2} Cl anhyd. AlC l_{3} C_{6} H_{5} CH (C H_{3})_{2} + HCl

Polyalkylation. The alkyl group put on activates the ring, so the product reacts faster than benzene. A large excess of benzene keeps the product mostly monoalkylated.
Deactivated rings fail (table above).

Acylation avoids the first two: the acylium ion is resonance-stabilised and does not rearrange, and the ketone formed deactivates the ring, so only one acyl group goes on.

3. Variants

3.1 Intramolecular acylation

A carboxylic acid or acid chloride on a side chain can acylate its own ring when the new ring is five- or six-membered. Polyphosphoric acid (PPA), or concentrated $H_{2} S O_{4}$ , makes the acylium ion from the acid itself. 3-Phenylpropanoic acid gives 1-indanone (2,3-dihydro-1H-inden-1-one); 4-phenylbutanoic acid gives 1-tetralone.

C_{6} H_{5} C H_{2} C H_{2} COOH PPA C_{9} H_{8} O + H_{2} O

3.2 Cyclic anhydrides

A cyclic anhydride acylates the ring and leaves a free COOH on the chain. Phthalic anhydride and benzene give 2-benzoylbenzoic acid, which closes to anthraquinone with concentrated $H_{2} S O_{4}$ . Succinic anhydride gives 4-oxo-4-phenylbutanoic acid, the first step of the route to 1-tetralone in Solved Example 3.

3.3 Alkenes and alcohols as the alkyl source

Any reaction that makes a carbocation next to benzene can alkylate it. Propene with phosphoric or sulphuric acid gives cumene, the industrial start of the cumene route to phenol and propanone:

C_{6} H_{6} + C H_{3} CH = C H_{2} H^{+} C_{6} H_{5} CH (C H_{3})_{2}

3.4 Gattermann-Koch formylation

Methanoyl chloride ( $HCOCl$ ) is too unstable to use, so a CHO group is put on with CO and $HCl$ over anhydrous $AlC l_{3}$ and CuCl: benzene gives benzaldehyde. It is a Friedel-Crafts-type acylation, but exams name it separately.

JEE Advanced

Alkylation is reversible; acylation is not. With excess $AlC l_{3}$ and heat, alkyl groups can come off and go back on, so the more stable isomer builds up: dialkylbenzenes drift towards the meta isomer. Acyl groups stay where they are put.

Rearrangements to look for. Isobutyl chloride gives tert-butylbenzene (hydride shift); neopentyl chloride, $(C H_{3})_{3} CC H_{2} Cl$ , gives 2-methyl-2-phenylbutane (methyl shift to a tertiary cation).

The one acylium exception. A tertiary acylium ion can lose CO: pivaloyl chloride, $(C H_{3})_{3} CCOCl$ , with benzene and $AlC l_{3}$ gives mainly tert-butylbenzene, not the ketone.

4. Worked examples

Solved Example 1

Benzene is treated with 1-chloropropane and anhydrous

AlC l_{3}

. What is the major product, and how would you make n-propylbenzene instead?

Solution:

The primary propyl cation (or complex) rearranges by a 1,2-hydride shift to the secondary isopropyl cation, so the major product is isopropylbenzene (cumene), $C_{6} H_{5} CH (C H_{3})_{2}$ .

For n-propylbenzene, acylate first with propanoyl chloride and $AlC l_{3}$ to get propiophenone, $C_{6} H_{5} COC H_{2} C H_{3}$ (no rearrangement), then reduce the C=O with Zn-Hg and concentrated HCl (Clemmensen) to $C_{6} H_{5} C H_{2} C H_{2} C H_{3}$ .

Solved Example 2

Which two of these do not give a Friedel-Crafts product?
(A) toluene with

C H_{3} COCl

AlC l_{3}

(B) nitrobenzene with

C H_{3} Cl

AlC l_{3}

C H_{3} COCl

AlC l_{3}

(D) aniline with

C H_{3} Cl

AlC l_{3}

Solution:

Answer: (B) and (D). Nitrobenzene is too strongly deactivated. Aniline's $N H_{2}$ forms a complex with $AlC l_{3}$ , and the positive nitrogen deactivates the ring. Toluene and anisole are activated and react mainly at the para position.

Solved Example 3

Benzene reacts with succinic anhydride and

AlC l_{3}

to give A. A is heated with Zn-Hg and concentrated HCl to give B. B is warmed with concentrated

H_{2} S O_{4}

to give C. Identify A, B and C.

Solution:

A: the anhydride acylates the ring and opens, leaving a COOH: 4-oxo-4-phenylbutanoic acid, $C_{6} H_{5} COC H_{2} C H_{2} COOH$ .

B: Clemmensen reduction turns the ketone C=O into $C H_{2}$ and leaves the COOH alone: 4-phenylbutanoic acid, $C_{6} H_{5} C H_{2} C H_{2} C H_{2} COOH$ .

C: the acid acylates its own ring (intramolecular Friedel-Crafts), closing a six-membered ring: 1-tetralone (3,4-dihydronaphthalen-1(2H)-one), $C_{10} H_{10} O$ .

5. Exam traps

Watch out

When the electrophile can rearrange, the rearranged product is the major one: 1-chloropropane gives mainly isopropylbenzene, isobutyl chloride mainly tert-butylbenzene.
Polyalkylation happens; polyacylation does not.
Nitrobenzene, benzoic acid and aniline do not undergo Friedel-Crafts reactions. Nitrobenzene is a common solvent for them.
Aryl and vinyl halides cannot alkylate a ring: chlorobenzene with benzene and $AlC l_{3}$ does not give biphenyl.
Acylation needs at least one mole of $AlC l_{3}$ per mole of acyl chloride. When $AlC l_{3}$ is the catalyst it must be anhydrous: water destroys it.
Not every "benzene with $AlC l_{3}$ " is Friedel-Crafts: $C l_{2}$ with $AlC l_{3}$ or $FeC l_{3}$ is ring halogenation, and CO with $HCl$ is Gattermann-Koch formylation.
The mechanism is electrophilic substitution, not addition and not nucleophilic substitution.
Activated rings react at ortho and para positions; toluene with ethanoyl chloride gives mainly 4-methylacetophenone.

Past-paper questions on Friedel-Crafts reaction

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 16 questions in them that test this:

JEE Main 8 (2026)
JEE Advanced 5 (2022-2023, 2026)
NEET 3 (2024, 2026)

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