Friedel Crafts

8 MCQs9-step worked example

Source: NCERT Aldehydes, Ketones and Carboxylic AcidsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

Friedel-Crafts reactions are electrophilic aromatic substitutions that install carbon groups directly onto a benzene ring. Two variants exist: alkylation (attaches an alkyl group) and acylation (attaches an acyl group, R–C=O).

The core mechanism: A Lewis acid catalyst (anhydrous AlCl₃ or FeCl₃) generates the electrophile. In alkylation, AlCl₃ polarises an alkyl halide (R–Cl) to produce a carbocation-like species R⁺. In acylation, it activates an acyl halide (RCOCl) to form the acylium ion RC≡O⁺. This electrophile attacks the electron-rich benzene π-system, forming an arenium ion intermediate, followed by loss of H⁺ to restore aromaticity.

Critical differences between the two variants:

Polyalkylation problem — Alkylation makes the ring MORE electron-rich (alkyl is activating), so the product reacts faster than the starting material. Multiple alkyl groups attach. Acylation does NOT suffer this: the acyl group (–COR) is deactivating, so the mono-acylated product is less reactive — reaction stops cleanly at mono-substitution.
Carbocation rearrangement — In alkylation, the carbocation intermediate can rearrange (hydride/methyl shift) to a more stable form, giving an unexpected product. Acylation produces a resonance-stabilised acylium ion that does NOT rearrange.
Substrate limitations — Friedel-Crafts reactions FAIL on strongly deactivated rings (those bearing –NO₂, –CN, –SO₃H, or multiple halogens). They also fail on amines (–NH₂, –NHR, –NR₂) because the lone pair on nitrogen coordinates with AlCl₃, deactivating the catalyst.

NCERT reference: NCERT Class 11 Chemistry, Chapter 10 (Hydrocarbons), Part 2, page 26 documents Friedel-Crafts as a named reaction of benzene.

Watch-out for NEET: Questions often test whether you recognise that acylation avoids the two problems (polysubstitution and rearrangement) that plague alkylation. If the question asks "which gives a single, well-defined product" — the answer is acylation.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Which catalyst is required for Friedel-Crafts alkylation of benzene?

MCQ 2Easy RecallPractice

Friedel-Crafts reactions fail on benzene rings bearing which of the following groups?

MCQ 3Easy RecallPractice

Why does Friedel-Crafts acylation stop at mono-substitution while alkylation gives polysubstituted products?

MCQ 4Direct ApplicationPractice

When n-propyl chloride is treated with benzene in the presence of anhydrous AlCl₃, the major product is:

MCQ 5Direct ApplicationPractice

A chemist wants to attach a straight-chain propyl group to benzene without rearrangement. Which approach succeeds?

MCQ 6Direct ApplicationPractice

Friedel-Crafts alkylation of benzene with CH₃Cl/AlCl₃ often gives a mixture of toluene, xylenes, and higher alkylated products. What is the reason?

MCQ 7Concept TrapPractice

An aromatic compound bearing an –NH₂ group does not undergo Friedel-Crafts alkylation. The most accurate explanation is:

MCQ 8CalculationPractice

Benzene is first treated with CH₃COCl/AlCl₃ (Friedel-Crafts acylation), and the product is then reduced with Zn-Hg/conc. HCl (Clemmensen reduction). If the final product is subsequently treated with CH₃COCl/AlCl₃ again, the major product is:

Worked Example

1
Given
Benzene is treated with 2-chloro-2-methylpropane (tert-butyl chloride) in the presence of anhydrous AlCl₃.
2
Required
Identify the major product and explain whether rearrangement occurs.
3
Concept
Friedel-Crafts alkylation generates a carbocation from the alkyl halide via Lewis acid activation. Carbocations may rearrange to a more stable form. However, a tertiary carbocation is already the most stable alkyl carbocation — no rearrangement is needed.
4
Formula/Principle
- Lewis acid activation: R–Cl + AlCl₃ → R⁺ + AlCl₄⁻ - Carbocation stability: 3° > 2° > 1° > methyl - If the carbocation is already tertiary, no rearrangement occurs.
5
Substitution
(CH₃)₃C–Cl + AlCl₃ → (CH₃)₃C⁺ + AlCl₄⁻ The tert-butyl cation is tertiary — maximum stability. No hydride or methyl shift occurs.
6
Calculation/Reasoning
The stable (CH₃)₃C⁺ electrophile attacks benzene's π-system at any position (all equivalent). The arenium ion intermediate loses H⁺ to regenerate aromaticity. Product: tert-butylbenzene. Note: No rearrangement here because the starting carbocation is already tertiary. This contrasts with n-propyl or n-butyl halides where primary → secondary (or secondary → tertiary) shifts occur.
7
Final answer
**Major product: tert-butylbenzene (C₆H₅C(CH₃)₃)** No rearrangement because the tert-butyl carbocation is already at maximum stability (tertiary).
8
Common trap
The trap here is expecting rearrangement when none occurs. Students who memorise "Friedel-Crafts alkylation always rearranges" get confused by tertiary halides. The rule is: rearrangement occurs only when a MORE STABLE carbocation is accessible. Tertiary is already the ceiling for simple alkyl systems. A second trap: if the question instead gave neopentyl chloride ((CH₃)₃CCH₂Cl), the primary carbocation WOULD rearrange (methyl shift) to give the same tert-butyl cation — same product, different starting material.
9
Similar NEET-style question
"When neopentyl chloride is treated with benzene/AlCl₃, the product obtained is the same as with tert-butyl chloride. Explain." (Answer: 1,2-methyl shift converts the neopentyl primary carbocation to the tert-butyl tertiary carbocation.) ---

Before solving, remember these

Key Fact

Electrophilic aromatic substitution

(1) Halogenation (Cl₂/FeCl₃, Br₂/FeBr₃). (2) Nitration (HNO₃/H₂SO₄). (3) Sulphonation (oleum, SO₃). (4) Friedel-Crafts alkylation (RX/AlCl₃) and acylation (RCOX/AlCl₃).

-- NCERT, p. 26

Formulas

Markovnikov's rule (and anti-Markovnikov)

H^{+} to C with more H

Without peroxide: ionic mechanism — H goes to carbon with MORE hydrogens (carbocation stability rule). With peroxide (HBr only, Kharasch): radical mechanism — anti-Markovnikov.

Symbol	Quantity	SI Unit
H,X	added atoms	-

Valid when

Asymmetric alkene
H-X with X = Cl, Br, I
Without peroxide for Markovnikov

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Markovnikov reverses with peroxide

Category: Organic Reaction Conditions

HBr addition to alkene: WITHOUT peroxide → Markovnikov (H to C with more H). WITH peroxide (Kharasch effect) → anti-Markovnikov (radical mechanism). Specific to HBr only — not HCl, HI.

When it triggers

Question gives HX addition to alkene with explicit peroxide condition or hints (e.g. ROOR, light).

How to avoid

Without peroxide: ionic mechanism, carbocation stability → Markovnikov. With peroxide: radical mechanism, radical stability → anti-Markovnikov. Effect ONLY for HBr (HCl too strong, HI too weak).

Trap

Solvent changes product distribution

Category: Organic Reaction Conditions

Same starting materials give different products depending on solvent. Polar protic (water, alcohols): SN1/E1 favoured. Polar aprotic (DMSO, DMF): SN2 favoured. Affects substitution vs elimination.

When it triggers

Question contrasts product when solvent is changed; or specifies solvent type.

How to avoid

Polar protic stabilises carbocation → SN1/E1 (3° preferred). Polar aprotic doesn't solvate nucleophile → strong SN2 nucleophile (1°/2° preferred). Bulky base (t-BuOK) favours E2 over SN2.

Mistake

Confuses ortho/para directors with meta directors.

Root cause: concept gap

Correction

o,p-directors (activators except halogens): -OH, -OR, -NH₂, -NHR, alkyl. m-directors (deactivators): -NO₂, -CN, -COOH, -CHO. Halogens: o,p-directors but DEACTIVATORS.

Mistake