Why is more than a catalytic amount of AlCl3 needed in Friedel-Crafts acylation?

The ketone formed has a basic carbonyl oxygen that binds AlCl3 in a stable complex. Each molecule of product therefore removes one AlCl3 from the reaction, so at least one full equivalent is required. Adding water at the end breaks the complex and releases the free ketone.

Preparation of Aromatic Aldehydes & Ketones

ChemistryAldehydes And KetonesFor JEE aspirants

The preparation of aromatic aldehydes and ketones follows two routes: attach a carbonyl group directly to a benzene ring, or oxidise a methyl side chain that is already on the ring. The ring route uses the Gattermann-Koch reaction, the Gattermann reaction and Friedel-Crafts acylation. The side-chain route uses the Etard reaction, chromium trioxide in acetic anhydride, or side-chain chlorination followed by hydrolysis. These methods for the preparation of aromatic aldehydes and ketones are asked regularly in JEE Main, JEE Advanced and NEET.

Key reactions: quick reference

Gattermann-Koch
$C_{6} H_{6} CO, HCl anhyd. AlCl_{3}, CuCl C_{6} H_{5} CHO$
Gattermann
$C_{6} H_{6} (i) HCN, HCl, AlCl_{3} (ii) H_{2} O C_{6} H_{5} CHO$
Friedel-Crafts acylation
$C_{6} H_{6} + RCOCl anhyd. AlCl_{3} C_{6} H_{5} COR + HCl$
Etard reaction
$C_{6} H_{5} C H_{3} (i) CrO_{2} Cl_{2}, CS_{2} (ii) H_{3} O^{+} C_{6} H_{5} CHO$
Chromic oxide route
$C_{6} H_{5} C H_{3} CrO_{3}, (CH_{3} CO)_{2} O 273-283 K C_{6} H_{5} CH (OCOC H_{3})_{2} H_{3} O^{+} C_{6} H_{5} CHO$
Side-chain chlorination
$C_{6} H_{5} C H_{3} Cl_{2}, h ν C_{6} H_{5} CHC l_{2} H_{2} O 373 K C_{6} H_{5} CHO$
Benzophenone
$2 C_{6} H_{6} + COC l_{2} anhyd. AlCl_{3} C_{6} H_{5} CO C_{6} H_{5} + 2 HCl$

1. Two Strategies for Aromatic Carbonyl Compounds

An aromatic aldehyde or ketone has its carbonyl carbon joined directly to a benzene ring, as in benzaldehyde (C₆H₅CHO), acetophenone (C₆H₅COCH₃) and benzophenone (C₆H₅COC₆H₅). There are only two ways to build that bond.

Put the carbonyl group on the ring. Benzene acts as a nucleophile and attacks a carbon electrophile such as the formyl cation or an acylium ion. These are electrophilic aromatic substitution reactions.
Oxidise a side chain. Start from toluene, whose methyl carbon is already attached to the ring, and oxidise it to the aldehyde level without going all the way to benzoic acid.

Figure 1: Two routes for the preparation of aromatic aldehydes and ketones: attach a carbonyl group to benzene (Gattermann-Koch reaction, Friedel-Crafts acylation) or oxidise the methyl group of toluene (Etard reaction, CrO₃ in acetic anhydride, side-chain chlorination and hydrolysis).

2. Gattermann-Koch Reaction

Gattermann-Koch reaction: benzene or a substituted benzene is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride and a small amount of copper(I) chloride. A formyl group (-CHO) replaces one ring hydrogen, giving an aromatic aldehyde.

C_{6} H_{6} CO, HCl anhyd. AlCl_{3}, CuCl C_{6} H_{5} CHO

Formyl chloride (HCOCl), the obvious acylating agent, is too unstable to store. A mixture of CO and HCl acts as its equivalent. Aluminium chloride is used in about equimolecular quantity, and the trace of CuCl helps carry carbon monoxide into the reaction so that it works at ordinary pressure.

Mechanism

Electrophile forms. CO, HCl and AlCl₃ give the formyl cation, $HC \equiv O^{+}$ :
$CuCl \cdot CO + HCl + AlC l_{3} ⇌ H C + O + CuCl + AlC l_{4}^{-}$
Ring attacks (slow step). The $π$ electrons of benzene attack the formyl carbon. Aromaticity is lost and a $σ$ -complex (arenium ion) forms.
Proton is lost (fast step). $AlC l_{4}^{-}$ removes the ring hydrogen, aromaticity returns, and benzaldehyde, HCl and AlCl₃ are formed.

Figure 2: Gattermann-Koch reaction mechanism: the formyl cation forms from CO, HCl and AlCl₃, benzene attacks it to give a

σ

-complex, and loss of a proton gives benzaldehyde.

Scope: the Gattermann-Koch method is not applicable to phenols and phenol ethers, and it fails on strongly deactivated rings such as nitrobenzene. For phenols and phenol ethers, use the Gattermann reaction.

Solved Example 1

Toluene is subjected to the Gattermann-Koch reaction. Name the electrophile and give the major product.

Solution:

The electrophile is the formyl cation, $HC \equiv O^{+}$ , generated from CO, HCl and anhydrous AlCl₃ with CuCl.

The methyl group of toluene is activating and ortho/para directing. The two ortho positions are crowded by the methyl group, so attack at the para position dominates.

Major product: 4-methylbenzaldehyde (p-tolualdehyde), CH₃C₆H₄CHO.

3. Gattermann Reaction

Gattermann reaction: an aromatic compound is treated with hydrogen cyanide and hydrogen chloride in the presence of a Lewis acid such as AlCl₃. The electrophile is

[HC = NH]^{+}

instead of

HC \equiv O^{+}

. Substitution gives an aldimine, which is hydrolysed to the aldehyde.

HCN + HCl + AlC l_{3} ⇌ [HC = NH]^{+} + AlC l_{4}^{-}

C_{6} H_{6} [HC = NH]^{+} C_{6} H_{5} CH = NH H_{2} O C_{6} H_{5} CHO + N H_{3}

In acid the ammonia ends up as NH₄Cl. Because HCN is extremely poisonous, a common variation generates it inside the flask from zinc cyanide and HCl.

Figure 3: Gattermann reaction:

HCl

adds to

HCN

and

AlC l_{3}

pulls off chloride to give the formimidoyl cation; the ring attacks it (slow), loses

H^{+}

(fast) to give an aldimine, and acid hydrolysis turns the imine into benzaldehyde.

Solved Example 2

4-Methoxybenzaldehyde is to be prepared from anisole (methoxybenzene). Which reaction should be used, the Gattermann or the Gattermann-Koch, and why?

Solution:

Anisole is a phenol ether, and the Gattermann-Koch reaction does not work for phenols and phenol ethers. The Gattermann reaction (HCN, HCl, AlCl₃, then water) is used instead.

The -OCH₃ group is strongly activating and ortho/para directing, so the formyl group enters mainly at the para position. Hydrolysis of the aldimine gives 4-methoxybenzaldehyde (anisaldehyde).

4. Friedel-Crafts Acylation: Aromatic Ketones

Friedel-Crafts acylation: benzene or a substituted benzene reacts with an acid chloride (RCOCl) or an acid anhydride in the presence of anhydrous aluminium chloride. An acyl group (RCO-) replaces a ring hydrogen, giving an aromatic ketone.

C_{6} H_{6} + C H_{3} COCl anhyd. AlCl_{3} C_{6} H_{5} COC H_{3} + HCl

The product C₆H₅COCH₃ is acetophenone, IUPAC name 1-phenylethanone. Acetic anhydride works in the same way:

C_{6} H_{6} + (C H_{3} CO)_{2} O anhyd. AlCl_{3} C_{6} H_{5} COC H_{3} + C H_{3} COOH

Mechanism

Acylium ion forms. AlCl₃ pulls chlorine off the acid chloride, giving $C H_{3} C \equiv O^{+}$ and $AlC l_{4}^{-}$ . The acylium ion is resonance stabilised, so it does not rearrange. This is a big advantage over Friedel-Crafts alkylation, where carbocations often rearrange.
Ring attacks. Benzene attacks the acylium carbon to form a $σ$ -complex.
Proton is lost. $AlC l_{4}^{-}$ removes the ring hydrogen, giving the ketone and HCl.
Catalyst is trapped. The carbonyl oxygen of the ketone binds AlCl₃ strongly. At least one full equivalent of AlCl₃ is therefore needed, and water is added at the end to free the ketone.

Figure 4: Friedel-Crafts acylation of benzene:

AlC l_{3}

pulls chloride off acetyl chloride to give the resonance-stabilised acylium ion, the ring attacks it (slow) to form the

σ

complex, and loss of

H^{+}

(fast) gives acetophenone. The ketone then binds

AlC l_{3}

, so a full equivalent is used and water releases it.

Solved Example 3

Benzene is treated with an excess of acetyl chloride and anhydrous AlCl₃. Why is acetophenone the only product, with no diacetylbenzene formed?

Solution:

The acetyl group withdraws electrons from the ring by resonance, so the ring of acetophenone is much less reactive than benzene. Once the ketone is formed, its oxygen also binds AlCl₃, which makes the ring even more electron-poor.

A second acylation therefore does not occur, and Friedel-Crafts acylation stops cleanly at the monoacylated product. Friedel-Crafts alkylation is different: an alkyl group activates the ring, so polyalkylation is a common problem.

Preparation of benzophenone

Benzophenone (diphenyl ketone) is made by Friedel-Crafts acylation in two ways. Benzene can be acylated with benzoyl chloride:

C_{6} H_{6} + C_{6} H_{5} COCl anhyd. AlCl_{3} C_{6} H_{5} CO C_{6} H_{5} + HCl

Alternatively, phosgene (COCl₂) reacts with excess benzene. The first acylation gives benzoyl chloride, which then acylates a second benzene molecule:

2 C_{6} H_{6} + COC l_{2} anhyd. AlCl_{3} C_{6} H_{5} CO C_{6} H_{5} + 2 HCl

Figure 5: Preparation of benzophenone by Friedel-Crafts acylation of benzene with benzoyl chloride, or with phosgene and excess benzene through benzoyl chloride.

Which rings can be acylated?

Friedel-Crafts acylation needs a ring that is at least as electron-rich as benzene.

Activating groups (-CH₃, -OCH₃) speed the reaction up. The acyl group enters mainly para to the substituent.
Strongly deactivating groups (-NO₂, -CN, -COR) stop the reaction. Nitrobenzene is so unreactive that it is sometimes used as a solvent for Friedel-Crafts reactions.
Basic groups such as -NH₂ bind AlCl₃ and deactivate the ring, so aniline does not undergo Friedel-Crafts acylation.

Figure 6: Friedel-Crafts acylation depends on the ring: anisole gives 4-methoxyacetophenone, while nitrobenzene does not react.

Solved Example 4

Predict the major product, if any: (a) anisole + CH₃COCl / anhydrous AlCl₃; (b) nitrobenzene + CH₃COCl / anhydrous AlCl₃; (c) toluene + (CH₃CO)₂O / anhydrous AlCl₃.

Solution:

(a) -OCH₃ is activating and ortho/para directing, and para substitution is less hindered. Product: 4-methoxyacetophenone.

(b) -NO₂ strongly deactivates the ring. No Friedel-Crafts acylation takes place.

(c) Acetic anhydride is also an acylating agent, and -CH₃ directs to the para position. Product: 4-methylacetophenone, with acetic acid as the by-product.

Solved Example 5

Benzophenone can be made from benzene using either benzoyl chloride or phosgene. How many moles of HCl are released per mole of benzophenone in each case?

Solution:

With benzoyl chloride there is one acylation, and one ring hydrogen combines with one chlorine. 1 mol HCl.

C_{6} H_{6} + C_{6} H_{5} COCl \to C_{6} H_{5} CO C_{6} H_{5} + HCl

Phosgene has two C-Cl bonds and acylates two benzene rings in turn. 2 mol HCl.

2 C_{6} H_{6} + COC l_{2} \to C_{6} H_{5} CO C_{6} H_{5} + 2 HCl

5. Oxidation of Toluene: Stopping at the Aldehyde

Toluene (methylbenzene) already has a carbon attached to the ring, so oxidising the methyl group looks like the easiest route to benzaldehyde. The difficulty is that aldehydes are oxidised faster than the methyl group itself. Strong oxidants such as alkaline KMnO₄ or acidified K₂Cr₂O₇ carry toluene all the way to benzoic acid.

The successful methods lock the aldehyde carbon in a form that the oxidant cannot attack, then release it by hydrolysis.

Figure 7: Why toluene oxidation must be stopped at benzaldehyde: the aldehyde carbon is protected as benzylidene diacetate or as the Etard complex, then released by acid hydrolysis.

Chromium trioxide in acetic anhydride

Toluene is oxidised with chromic oxide (CrO₃) in acetic anhydride at 273-283 K. As soon as benzaldehyde forms, acetic anhydride traps it as benzylidene diacetate (also called benzal diacetate), which is not oxidised further. Hydrolysis with aqueous acid regenerates benzaldehyde.

C_{6} H_{5} C H_{3} CrO_{3}, (CH_{3} CO)_{2} O 273-283 K C_{6} H_{5} CH (OCOC H_{3})_{2}

C_{6} H_{5} CH (OCOC H_{3})_{2} + H_{2} O H_{3} O^{+} C_{6} H_{5} CHO + 2 C H_{3} COOH

Figure 8: Benzaldehyde from toluene using chromium trioxide in acetic anhydride at 273-283 K, through benzylidene diacetate.

Etard reaction (chromyl chloride)

Etard reaction: toluene is oxidised by chromyl chloride (CrO₂Cl₂) in carbon disulphide or carbon tetrachloride. The methyl group becomes a brown chromium complex, the Etard complex, which gives benzaldehyde on hydrolysis.

C_{6} H_{5} C H_{3} CrO_{2} Cl_{2} CS_{2} or CCl_{4} C_{6} H_{5} CH (OCrOHC l_{2})_{2} H_{3} O^{+} C_{6} H_{5} CHO

When the side chain is longer than a methyl group, chromyl chloride oxidises the end carbon of the chain to -CHO. Ethylbenzene, for example, gives phenylacetaldehyde:

C_{6} H_{5} C H_{2} C H_{3} (i) CrO_{2} Cl_{2} (ii) H_{2} O C_{6} H_{5} C H_{2} CHO

Figure 9: Etard reaction: toluene and chromyl chloride give the brown Etard complex, which hydrolyses to benzaldehyde; ethylbenzene gives phenylacetaldehyde.

Solved Example 6

Why can benzaldehyde not be obtained by heating toluene with alkaline KMnO₄, while chromyl chloride gives it in good yield?

Solution:

KMnO₄ is a strong oxidant. Any benzaldehyde formed is oxidised even faster than toluene, so the product is benzoate, which gives benzoic acid on acidification.

With chromyl chloride, the methyl group is converted into the Etard complex, C₆H₅CH(OCrOHCl₂)₂. The carbon is held at the aldehyde oxidation level and is not attacked further. Hydrolysis with H₃O⁺ then releases benzaldehyde.

Solved Example 7

p-Xylene (1,4-dimethylbenzene) is treated with chromyl chloride in CS₂, and the product is hydrolysed. Identify the aldehyde formed.

Solution:

Chromyl chloride converts one methyl group into the Etard complex. The complex separates out of the solution as a solid, so the second methyl group is not attacked.

Hydrolysis gives 4-methylbenzaldehyde (p-tolualdehyde), CH₃C₆H₄CHO.

6. Side-Chain Chlorination Followed by Hydrolysis

Chlorine attacks the methyl group of toluene (not the ring) when the reaction is carried out in sunlight or at the boiling point without a Lewis acid. This is a free-radical substitution, and the hydrogens are replaced one at a time: benzyl chloride, then benzal chloride, then benzotrichloride.

Stopping at benzal chloride (C₆H₅CHCl₂) and hydrolysing it with water at 373 K gives benzaldehyde. The two chlorines are first replaced by two -OH groups on the same carbon. This gem-diol is unstable and loses water at once. Benzaldehyde is manufactured commercially by this method.

C_{6} H_{5} C H_{3} Cl_{2}, h ν C_{6} H_{5} CHC l_{2} H_{2} O 373 K C_{6} H_{5} CHO + 2 HCl

Figure 10: Side-chain chlorination of toluene in sunlight, and hydrolysis of benzal chloride at 373 K to benzaldehyde; benzotrichloride gives benzoic acid instead.

Solved Example 8

Identify A and B. What would be obtained if 3 mol of Cl₂ were used?

C_{6} H_{5} C H_{3} Cl_{2} (2 mol), h ν A H_{2} O, 373 K B

Solution:

Two methyl hydrogens are replaced, so A = benzal chloride, C₆H₅CHCl₂. Hydrolysis gives the gem-diol C₆H₅CH(OH)₂, which loses water, so B = benzaldehyde.

With 3 mol Cl₂ the product is benzotrichloride, C₆H₅CCl₃. Its hydrolysis gives benzoic acid, not an aldehyde.

C_{6} H_{5} CC l_{3} + 2 H_{2} O \to C_{6} H_{5} COOH + 3 HCl

Do not mix up the conditions. Cl₂ with sunlight or heat substitutes on the side chain. Cl₂ with FeCl₃ or AlCl₃ in the dark substitutes on the ring and gives o- and p-chlorotoluene, which cannot give benzaldehyde.

7. General Methods That Also Work for Aromatic Compounds

Several methods used for aliphatic aldehydes and ketones work just as well when the group is attached to a benzene ring. The aromatic examples most often asked are summarised below.

Method	Starting material	Reagent	Product
Rosenmund reduction	Benzoyl chloride, C₆H₅COCl	H₂, Pd-BaSO₄ (partially poisoned)	Benzaldehyde
Stephen reduction	Benzonitrile, C₆H₅CN	(i) SnCl₂, HCl (ii) H₃O⁺	Benzaldehyde
DIBAL-H reduction	Benzonitrile or a benzoate ester	(i) DIBAL-H, low temperature (ii) H₂O	Benzaldehyde
Nitrile + Grignard reagent	Benzonitrile	(i) CH₃MgBr (ii) H₃O⁺	Acetophenone
Oxidation of alcohols	Benzyl alcohol, C₆H₅CH₂OH	PCC in CH₂Cl₂	Benzaldehyde

Solved Example 9

Identify A, B and C in the following sequence:

PhC O_{2} H SOC l_{2} A M e_{2} NH B C PhCHO

Solution:

Thionyl chloride converts the acid into its acid chloride: A = benzoyl chloride, C₆H₅COCl.

Dimethylamine converts the acid chloride into a tertiary amide: B = N,N-dimethylbenzamide, C₆H₅CON(CH₃)₂.

A mild, bulky hydride reduces a tertiary amide only as far as the aldehyde: C = lithium diethoxyaluminium hydride, LiAlH₂(OEt)₂. DIBAL-H at low temperature also works. LiAlH₄ would not do, because it reduces the amide to an amine.

8. Choosing the Right Method

You want	Start from	Best method	Watch for
Benzaldehyde or an alkylbenzaldehyde	Benzene, toluene	Gattermann-Koch (CO, HCl, AlCl₃, CuCl)	Not for phenols or phenol ethers
Hydroxy- or methoxybenzaldehyde	Phenol ethers, phenols	Gattermann (HCN, HCl, Lewis acid; then H₂O)	HCN is highly toxic
Aryl alkyl or diaryl ketone	Benzene or activated ring	Friedel-Crafts acylation	Fails on deactivated rings; needs at least 1 equiv AlCl₃
Benzaldehyde from toluene (lab)	Toluene	Etard reaction or CrO₃/(CH₃CO)₂O	Hydrolysis step is essential
Benzaldehyde (industry)	Toluene	Side-chain chlorination, then hydrolysis	Over-chlorination gives benzoic acid

Solved Example 10

Match each reagent set with its named method.
(P) CO, HCl, anhyd. AlCl₃, CuCl
(Q) (i) CrO₂Cl₂, CS₂ (ii) H₃O⁺
(R) (i) HCN, HCl, AlCl₃ (ii) H₂O
(S) (i) Cl₂,

h ν

(ii) H₂O, 373 K
Methods: (1) Etard reaction (2) Gattermann reaction (3) Gattermann-Koch reaction (4) side-chain chlorination and hydrolysis

Solution:

CO with HCl is the formyl-cation source of the Gattermann-Koch reaction, so P-3. Chromyl chloride identifies the Etard reaction, so Q-1. HCN replaces CO in the Gattermann reaction, so R-2. Chlorine in light acts on the side chain, so S-4.

Answer: P-3, Q-1, R-2, S-4.

Common Mistakes to Avoid

Watch out

Mixing up the two Gattermann reactions. Gattermann-Koch uses CO + HCl. Gattermann uses HCN + HCl. Neither is the Gattermann reaction of diazonium salts (Cu powder + HX), which makes aryl halides.
Using Gattermann-Koch on phenols or anisole. It is not applicable to phenols and phenol ethers; choose the Gattermann reaction.
Writing a catalytic amount of AlCl₃ for acylation. The ketone product binds AlCl₃, so at least one full equivalent is consumed.
Expecting acylation of nitrobenzene or aniline. Strongly deactivated rings do not react, and -NH₂ ties up the Lewis acid.
Writing benzoic acid as the Etard product, or forgetting the hydrolysis step. The reagent is chromyl chloride, CrO₂Cl₂, not CrO₃.
Using Cl₂/FeCl₃ for side-chain chlorination. A Lewis acid sends chlorine onto the ring. Side-chain substitution needs light or heat.
Over-chlorinating toluene. Benzotrichloride hydrolyses to benzoic acid, not benzaldehyde.

Frequently Asked Questions

What is the Gattermann-Koch reaction?

The Gattermann-Koch reaction converts benzene or an alkylbenzene into an aromatic aldehyde using carbon monoxide and hydrogen chloride with anhydrous AlCl₃ and a little CuCl. The electrophile is the formyl cation, HCO⁺, which substitutes a ring hydrogen. Benzene gives benzaldehyde and toluene gives mainly 4-methylbenzaldehyde.

What is the difference between the Gattermann and Gattermann-Koch reactions?

Both put a -CHO group on an aromatic ring. Gattermann-Koch uses CO and HCl, with the formyl cation as electrophile, and does not work on phenols or phenol ethers. The Gattermann reaction uses HCN and HCl, forms an aldimine that must be hydrolysed, and does work on phenols and phenol ethers.

Why is more than a catalytic amount of AlCl₃ needed in Friedel-Crafts acylation?

The ketone formed has a basic carbonyl oxygen that binds AlCl₃ in a stable complex. Each molecule of product therefore removes one AlCl₃ from the reaction, so at least one full equivalent is required. Adding water at the end breaks the complex and releases the free ketone.

How is benzaldehyde prepared from toluene?

There are three standard routes. The Etard reaction uses chromyl chloride in CS₂, then hydrolysis. Chromium trioxide in acetic anhydride at 273-283 K gives benzylidene diacetate, which is then hydrolysed. Side-chain chlorination to benzal chloride is followed by hydrolysis at 373 K, and this is the commercial method.

What is the Etard reaction?

The Etard reaction oxidises a methyl group on a benzene ring to an aldehyde group using chromyl chloride, CrO₂Cl₂, in CS₂ or CCl₄. A brown Etard complex separates out and gives the aldehyde on acid hydrolysis. Toluene gives benzaldehyde, and p-xylene gives 4-methylbenzaldehyde.

Why does Friedel-Crafts acylation not give polyacylated products?

The acyl group withdraws electrons from the ring by resonance, and its complex with AlCl₃ withdraws even more. The ketone ring is much less reactive than the starting arene, so a second acyl group does not enter. This is why acylation is cleaner than Friedel-Crafts alkylation.

Which methods of preparing aromatic aldehydes and ketones are most useful for JEE Main and JEE Advanced?

JEE questions usually match reagents with named reactions (Gattermann-Koch, Etard, Rosenmund, Stephen) or ask for the product of a sequence. Know each reagent set, the electrophile in Gattermann-Koch and Friedel-Crafts acylation, and the limits: no phenols in Gattermann-Koch and no deactivated rings in acylation.

What should NEET students remember from this topic?

NEET follows NCERT closely. Learn the NCERT methods with conditions: Etard reaction with CrO₂Cl₂, CrO₃ in acetic anhydride at 273-283 K, side-chain chlorination with hydrolysis at 373 K, Gattermann-Koch with CO and HCl, and Friedel-Crafts acylation for acetophenone and benzophenone.

Previous year questions on Preparation of Aromatic Aldehydes & Ketones

2 questions from past papers, each with a step-by-step solution.

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