Friedel-Crafts Acylation: How to Prepare Acetophenone and Aromatic Ketones
Chemistry · Aldehydes, Ketones And Carboxylic Acid · NEET
Friedel-Crafts acylation adds a C=O group (an acyl group) to a benzene ring. When benzene reacts with an acid chloride like CH3COCl (or an acid anhydride) using anhydrous AlCl3 as catalyst, you get an aromatic ketone. With CH3COCl you get acetophenone (C6H5COCH3); with C6H5COCl you get benzophenone (C6H5COC6H5). Memory hook: "Acyl chloride + benzene + AlCl3 = aryl Ketone" (acylation makes Ketones, alkylation makes alkyl arenes).
Benzene reacts with acetyl chloride (CH3COCl) using anhydrous AlCl3 as a Lewis-acid catalyst. The acetyl group (CH3CO-) replaces one ring hydrogen, giving the aromatic ketone acetophenone plus HCl. Swap CH3COCl for benzoyl chloride (C6H5COCl) to get benzophenone instead.
Your doubts, answered
What exactly do you get when benzene reacts with CH3COCl and anhydrous AlCl3?
You get acetophenone (C6H5COCH3), an aromatic ketone. The acetyl group (CH3CO-) replaces one H atom on the benzene ring. Reaction: C6H6 + CH3COCl --AlCl3--> C6H5COCH3 + HCl. This is the classic way to attach a -COCH3 group to a benzene ring, and it is a very common NEET question.
What is the difference between Friedel-Crafts alkylation and acylation?
Alkylation uses an alkyl halide (like CH3Cl) and gives an alkylbenzene (like toluene). Acylation uses an acyl halide (like CH3COCl) or an acid anhydride and gives an aromatic ketone (like acetophenone) that has a C=O group. Key point for NEET: acylation makes a KETONE (C=O present); alkylation makes an alkyl side chain (no C=O). Both need anhydrous AlCl3.
Why must AlCl3 be anhydrous (dry)?
AlCl3 is a Lewis acid. It pulls Cl from the acyl chloride to make the acylium electrophile (CH3CO+), which is the real attacking species. If water is present, water reacts with AlCl3 and destroys its Lewis-acid power, so the reaction fails. That is why the AlCl3 must be completely dry (anhydrous).
How do I make benzophenone instead of acetophenone?
Use benzoyl chloride (C6H5COCl) in place of acetyl chloride. Benzene + C6H5COCl with anhydrous AlCl3 gives benzophenone (C6H5COC6H5), a ketone with a benzene ring on each side of the C=O. This exact conversion (benzene to benzophenone) was matched in NEET 2024.
Can I use an acid anhydride instead of an acid chloride?
Yes. NCERT says acylation works with an acyl halide OR an acid anhydride, using AlCl3 (a Lewis acid). For example, benzene + (CH3CO)2O / AlCl3 also gives acetophenone. So both acetyl chloride and acetic anhydride are valid acylating agents.
Why does acylation not happen on benzene rings that carry -COOH, -NO2 or -NH2 groups?
Deactivated rings resist Friedel-Crafts reactions. Strong deactivating groups like -COOH and -NO2 pull electron density out of the ring, so the ring is too electron-poor to attack the electrophile. For aniline (-NH2), the basic N lone pair bonds to AlCl3 forming a salt, which makes N positive and strongly deactivates the ring. So aniline and benzoic acid do not undergo Friedel-Crafts reactions.
⚠️ The NEET trap ✗ Benzene + CH3COCl / AlCl3 gives toluene or an alkylbenzene (an alkyl side chain, no C=O). ✓ Benzene + CH3COCl / AlCl3 gives acetophenone C6H5COCH3, a KETONE with a C=O group. Acylation always makes a ketone; alkylation makes an alkyl arene. 🧠 See 'acyl chloride' (RCOCl) -> the product has C=O -> it is a KETONE, not an alkane side chain. Acetophenone, benzophenone = acylation products.
Solution: The key match is (a)-(i): benzene + benzoyl chloride (C6H5COCl) with anhydrous AlCl3 undergoes Friedel-Crafts acylation to give benzophenone (C6H5COC6H5). This is the pure acylation match. Alkene ozonolysis is (b)-(iv), phenol oxidation by CrO3 is (c)-(ii), and side-chain oxidation of ethylbenzene by KMnO4/KOH gives benzoic acid, (d)-(iii). So (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii) = option C.
ReNEET 2026
For the sequence: Benzene --(i) CH3COCl, AlCl3--> --(ii) NaOCl--> P + Q, choose the correct option.
A · If P is the sodium salt of a carboxylic acid, Q is a primary alcohol
B · P and Q are aromatic compounds
C · If P gives a carboxylic acid on acidification, Q gives a poisonous gas on exposure to air and light ✓
D · Both P and Q are carbonyl compounds
Solution: Step (i) is Friedel-Crafts acylation: benzene + CH3COCl / AlCl3 gives acetophenone (C6H5COCH3). Acetophenone is a methyl ketone (CH3CO- group), so with NaOCl it undergoes the haloform reaction, giving sodium benzoate (P) and chloroform CHCl3 (Q). Acidifying P gives benzoic acid. Chloroform CHCl3, when exposed to air and light, is oxidised to the poisonous gas phosgene COCl2. So option C is correct.
NEET 2018
Compound A (C8H10O) reacts with NaOI (from Y + NaOH) and gives a yellow precipitate with a characteristic smell. A and Y are respectively:
A · C6H5CH(OH)CH3 and I2 ✓
B · C6H5CH2CH2OH and I2
C · H3C-C6H4-CH2OH and I2
D · (CH3)2C6H3OH and I2
Solution: NaOI forms from I2 + 2NaOH, so Y = I2. A yellow precipitate with a characteristic smell is iodoform CHI3 (positive iodoform test). 1-Phenylethanol C6H5CH(OH)CH3 (a secondary alcohol) is first oxidised by NaOI to acetophenone C6H5COCH3 (the same aromatic ketone made by Friedel-Crafts acylation). The CH3CO- group of acetophenone then gives iodoform. So A = C6H5CH(OH)CH3, Y = I2, option A.
Solved Aldehydes, Ketones And Carboxylic Acid NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the product of Friedel-Crafts acylation of benzene with acetyl chloride?
Acetophenone, C6H5COCH3. The reaction is C6H6 + CH3COCl --anhydrous AlCl3--> C6H5COCH3 + HCl.
Which catalyst is used in Friedel-Crafts acylation?
Anhydrous aluminium chloride (AlCl3), a Lewis acid. It generates the acylium electrophile from the acid chloride or anhydride. It must be dry.
Is acetophenone an aldehyde or a ketone?
Acetophenone (C6H5COCH3) is a ketone. Its carbonyl carbon is bonded to two carbon groups (a benzene ring and a CH3 group), so it is a ketone, not an aldehyde.
Why can Friedel-Crafts acylation not be done on nitrobenzene or benzoic acid?
Because -NO2 and -COOH are strong deactivating groups. They make the ring electron-poor, so it cannot attack the electrophile. Aniline also fails because its N bonds to AlCl3 and deactivates the ring.
Does acylation give only mono-substituted product?
Yes, usually. The C=O group introduced is itself deactivating, so it discourages a second acylation. This is an advantage over alkylation, which can over-substitute.
