Friedel–Crafts alkylation – reactivity and limitations MCQs With Answer

Introduction: Friedel–Crafts alkylation is a fundamental electrophilic aromatic substitution reaction widely used in medicinal chemistry and B. Pharm coursework to introduce alkyl groups onto aromatic rings. This reaction involves generation of a carbocation (or equivalent) by a Lewis or Brønsted acid catalyst and its attack on an aromatic substrate. Key concepts include reactivity trends, directing effects, carbocation rearrangements, polyalkylation, catalyst choice (e.g., AlCl3), and major limitations like incompatibility with strongly deactivated rings. Understanding these aspects is crucial for designing aromatic intermediates in drug synthesis. Now let’s test your knowledge with 50 MCQs on this topic.

Q1. What is the initial electrophile in a typical Friedel–Crafts alkylation using an alkyl halide and AlCl3?

• Neutral alkyl halide
• Free radical
• Carbocation or complexed carbocation
• Carbanion

Correct Answer: Carbocation or complexed carbocation

Q2. Which catalyst is most commonly used for Friedel–Crafts alkylation?

• Sodium hydroxide
• AlCl3 (aluminum chloride)
• Pd/C (palladium on carbon)
• KMnO4

Correct Answer: AlCl3 (aluminum chloride)

Q3. Why are tertiary alkyl halides more reactive in Friedel–Crafts alkylation than primary alkyl halides?

• Tertiary alkyl halides are less sterically hindered
• Tertiary carbocations are more stable
• Tertiary halides form radicals preferentially
• Tertiary halides are stronger bases

Correct Answer: Tertiary carbocations are more stable

Q4. What major problem arises when using primary alkyl halides in Friedel–Crafts alkylation?

• Excessive oxidation of aromatic ring
• Carbocation rearrangement leading to different products
• Immediate polymerization of the aromatic substrate
• Formation of aziridines

Correct Answer: Carbocation rearrangement leading to different products

Q5. Which reagent combination can generate a carbocation from an alcohol for Friedel–Crafts alkylation?

• NaBH4 and H2O
• Pd/C and H2
• H2SO4 or HCl with a strong Lewis acid
• KMnO4 and heat

Correct Answer: H2SO4 or HCl with a strong Lewis acid

Q6. Which aromatic substituent will block Friedel–Crafts alkylation due to strong deactivation?

• –OMe (methoxy)
• –NO2 (nitro)
• –Me (methyl)
• –OH (hydroxy)

Correct Answer: –NO2 (nitro)

Q7. Friedel–Crafts alkylation proceeds via which intermediate on the aromatic ring?

• Benzyne intermediate
• Wheland (sigma) intermediate
• Carbanion complex
• Perylene intermediate

Correct Answer: Wheland (sigma) intermediate

Q8. What is a common method to avoid polyalkylation in Friedel–Crafts reactions?

• Use excess alkyl halide
• Use stoichiometric AlCl3
• Introduce a temporary deactivating group or use Friedel–Crafts acylation followed by reduction
• Increase reaction temperature

Correct Answer: Introduce a temporary deactivating group or use Friedel–Crafts acylation followed by reduction

Q9. Which of the following substrates is most reactive toward Friedel–Crafts alkylation?

• Nitrobenzene
• Benzene
• Toluene
• Fluorobenzene

Correct Answer: Toluene

Q10. How does a strongly electron-donating group on the aromatic ring affect rate of Friedel–Crafts alkylation?

• Decreases the rate drastically
• Has no effect
• Increases the rate by stabilizing the sigma complex
• Converts reaction to nucleophilic aromatic substitution

Correct Answer: Increases the rate by stabilizing the sigma complex

Q11. Which phenomenon explains the formation of unexpected isomers during Friedel–Crafts alkylation?

• Radical chain branching
• Carbocation rearrangement (hydride/alkyl shift)
• SN2 inversion
• Photochemical isomerization

Correct Answer: Carbocation rearrangement (hydride/alkyl shift)

Q12. Why is Friedel–Crafts alkylation often unsuitable for substrates bearing strong Lewis-basic groups like –OH or –NH2 without protection?

• They cause radical formation
• They complex with the Lewis acid catalyst and deactivate it
• They increase vapor pressure
• They make the ring non-aromatic

Correct Answer: They complex with the Lewis acid catalyst and deactivate it

Q13. Which alternative to alkyl halides can be used as alkylating agents in Friedel–Crafts alkylation?

• Alkenes with strong acid to generate carbocations
• Alcohols without any acid catalyst
• Epoxides under basic conditions
• Carboxylic acids with base

Correct Answer: Alkenes with strong acid to generate carbocations

Q14. Which product predominates when benzene reacts with tert-butyl chloride and AlCl3?

• n-Butylbenzene
• tert-Butylbenzene
• Isobutylbenzene
• Benzoic acid

Correct Answer: tert-Butylbenzene

Q15. Friedel–Crafts alkylation is best described as which type of reaction?

• Nucleophilic aromatic substitution
• Electrophilic aromatic substitution
• Radical substitution
• Pericyclic cycloaddition

Correct Answer: Electrophilic aromatic substitution

Q16. Which condition often leads to polymerization during Friedel–Crafts alkylation with simple olefins?

• Low catalyst loading
• Use of very dilute solutions
• Strong acid and highly reactive carbocations without trapping
• Presence of radical inhibitors

Correct Answer: Strong acid and highly reactive carbocations without trapping

Q17. How does Friedel–Crafts acylation help avoid rearrangement problems seen in alkylation?

• Acylation uses radicals which do not rearrange
• Acylation forms a resonance-stabilized acylium ion that does not rearrange
• Acylation always proceeds via SN2
• Acylation uses base catalysts only

Correct Answer: Acylation forms a resonance-stabilized acylium ion that does not rearrange

Q18. Which aromatic substitution position is favored by an electron-donating group (EDG) like –OH?

• Meta
• Para and ortho
• Only para
• No substitution occurs

Correct Answer: Para and ortho

Q19. What is the role of AlCl3 in Friedel–Crafts alkylation using alkyl chlorides?

• Hydrogen bond donor
• Oxidizing agent
• Lewis acid that polarizes the C–Cl bond forming a complex or carbocation
• Phase-transfer catalyst

Correct Answer: Lewis acid that polarizes the C–Cl bond forming a complex or carbocation

Q20. Why is nitrobenzene unreactive in Friedel–Crafts alkylation?

• Nitro group makes the ring too electron-rich
• Nitro group is strongly deactivating and withdraws electron density by resonance and inductive effects
• Nitrobenzene forms stable carbocations
• Nitrobenzene is insoluble in all solvents

Correct Answer: Nitro group is strongly deactivating and withdraws electron density by resonance and inductive effects

Q21. Which technique is often used to quench AlCl3 after a Friedel–Crafts reaction?

• Add water carefully or aqueous HCl
• Add NaOH directly as a solid
• Heat to 200 °C to decompose catalyst
• Add elemental sodium

Correct Answer: Add water carefully or aqueous HCl

Q22. Which of these statements about polyalkylation is correct?

• Polyalkylation is impossible with tertiary carbocations
• Once an alkyl group is introduced, the ring becomes more activated and susceptible to further alkylation
• Polyalkylation decreases with more EDGs
• Polyalkylation only occurs in the presence of bases

Correct Answer: Once an alkyl group is introduced, the ring becomes more activated and susceptible to further alkylation

Q23. For pharmaceutical synthesis, why might one prefer Friedel–Crafts acylation followed by reduction over direct alkylation?

• Acylation is cheaper than alkylation
• Acylation avoids carbocation rearrangement and polyalkylation, offering better regiocontrol
• Reduction converts acyl to carboxylic acid directly
• Acylation introduces radical centers useful for further reactions

Correct Answer: Acylation avoids carbocation rearrangement and polyalkylation, offering better regiocontrol

Q24. Which directing effect does a halogen (e.g., –Cl) display in electrophilic aromatic substitution?

• Strongly activating ortho/para
• Deactivating but ortho/para directing
• Meta directing and activating
• No directing effect

Correct Answer: Deactivating but ortho/para directing

Q25. Which solvent is commonly used for Friedel–Crafts alkylations with AlCl3?

• Water
• Alcohols like methanol
• Non-nucleophilic solvents such as dichloromethane or nitrobenzene
• Aqueous base

Correct Answer: Non-nucleophilic solvents such as dichloromethane or nitrobenzene

Q26. In a Friedel–Crafts reaction using benzene and ethyl chloride, what major product problem might one expect?

• Formation of benzaldehyde
• Formation of ethylbenzene accompanied by rearranged isopropylbenzene
• Exclusive formation of chlorobenzene
• Formation of phenol

Correct Answer: Formation of ethylbenzene accompanied by rearranged isopropylbenzene

Q27. Which of the following increases the likelihood of rearrangement during alkylation?

• Formation of a very stable carbocation (e.g., tertiary)
• Use of resonance-stabilized acylium ion
• Performing acylation instead of alkylation
• Using electron-poor aromatic rings

Correct Answer: Formation of a very stable carbocation (e.g., tertiary)

Q28. How does steric hindrance on the aromatic ring influence regioselectivity in Friedel–Crafts alkylation?

• Steric hindrance favors ortho substitution
• Steric hindrance favors meta substitution
• Steric hindrance often favors para substitution over ortho due to less crowding
• Steric hindrance has no effect

Correct Answer: Steric hindrance often favors para substitution over ortho due to less crowding

Q29. Can heteroaromatic compounds like pyridine undergo classical Friedel–Crafts alkylation?

• Yes, pyridine is highly reactive
• No, pyridine is deactivated and coordinates to Lewis acids, preventing reaction
• Yes, but only under basic conditions
• Yes, without catalyst at room temperature

Correct Answer: No, pyridine is deactivated and coordinates to Lewis acids, preventing reaction

Q30. Which factor does NOT typically limit Friedel–Crafts alkylation?

• Presence of strong electron-withdrawing groups on the ring
• Carbocation rearrangement
• Polyalkylation
• High aromatic stabilization energy of benzene

Correct Answer: High aromatic stabilization energy of benzene

Q31. The Wheland intermediate in Friedel–Crafts alkylation is characterized by:

• A positive charge delocalized over the ring carbons
• A free radical on the substituent
• A negatively charged aromatic system
• An sp-hybridized aromatic carbon

Correct Answer: A positive charge delocalized over the ring carbons

Q32. When using AlCl3 catalysis, why might acylated products require special work-up?

• AlCl3 cannot be quenched by water
• Acylated products complex strongly with AlCl3 and form stable salts or complexes
• Acylated products decompose instantly on contact with air
• AlCl3 converts acyl groups to alcohols

Correct Answer: Acylated products complex strongly with AlCl3 and form stable salts or complexes

Q33. Which is a safer catalytic alternative to AlCl3 that can be used in some Friedel–Crafts reactions?

• FeCl3 or BF3•OEt2 under milder conditions
• Conc. HNO3
• NaOH
• LiAlH4

Correct Answer: FeCl3 or BF3•OEt2 under milder conditions

Q34. How does an electron-withdrawing substituent at the para position affect orientation of a new substituent?

• It makes ortho/para substitution faster
• It increases meta-directing behavior if strongly withdrawing by resonance
• It always leads to nucleophilic aromatic substitution
• It transforms the ring into a saturated cyclohexane

Correct Answer: It increases meta-directing behavior if strongly withdrawing by resonance

Q35. In the context of drug synthesis, why is control of regiochemistry in Friedel–Crafts reactions important?

• Only one regioisomer is ever active or safe; impurities complicate purification and biological activity
• Regiochemistry does not matter in pharmaceuticals
• Regioisomers always have identical pharmacokinetics
• Regioselectivity only affects color of product

Correct Answer: Only one regioisomer is ever active or safe; impurities complicate purification and biological activity

Q36. What happens if a strongly nucleophilic solvent is used in a Friedel–Crafts alkylation?

• Solvent may compete with the aromatic ring, leading to side reactions or deactivation of the electrophile
• Reaction rate always increases without side products
• Solvent becomes alkylated exclusively at oxygen
• No effect; solvent choice is irrelevant

Correct Answer: Solvent may compete with the aromatic ring, leading to side reactions or deactivation of the electrophile

Q37. Which type of aromatic ring is most resistant to Friedel–Crafts alkylation?

• Rings with strong electron-donating substituents
• Highly activated heteroarenes like furan
• Rings with multiple strong electron-withdrawing groups
• Substituted benzene with alkyl groups only

Correct Answer: Rings with multiple strong electron-withdrawing groups

Q38. Which mechanistic step is usually rate-determining in Friedel–Crafts alkylation?

• Deprotonation of the sigma complex
• Formation of the carbocation/electrophile and its attack on the aromatic ring
• Final work-up with water
• Complexation of solvent molecules

Correct Answer: Formation of the carbocation/electrophile and its attack on the aromatic ring

Q39. Which of the following can be used to generate tertiary carbocations without using alkyl halides?

• Dehydration of tertiary alcohols with strong acids
• Oxidation of primary alcohols with PCC
• Hydrogenation of alkenes with Pd/C
• Treatment of alkanes with base

Correct Answer: Dehydration of tertiary alcohols with strong acids

Q40. Why might chlorobenzene undergo Friedel–Crafts alkylation more slowly than benzene?

• Chlorobenzene is more electron-rich than benzene
• Chlorine is deactivating by inductive effect despite ortho/para directing by resonance
• Chlorobenzene forms radicals preferentially
• Chlorobenzene is insoluble in organic solvents

Correct Answer: Chlorine is deactivating by inductive effect despite ortho/para directing by resonance

Q41. Which is true about using Lewis acids in catalytic vs stoichiometric amounts for Friedel–Crafts reactions?

• Stoichiometric amounts often form stable complexes with product and complicate isolation
• Catalytic amounts always give better yields
• Stoichiometric catalyst is never necessary
• Catalyst amount has no influence on outcome

Correct Answer: Stoichiometric amounts often form stable complexes with product and complicate isolation

Q42. In Friedel–Crafts alkylation, which structural feature of the alkylating agent reduces likelihood of rearrangement?

• Ability to form a highly stabilized carbocation
• Primary carbocation prone to rearrangement
• Use of resonance-stabilized cations like benzylic or allylic cations
• Use of agents that produce non-rearranging acylium ions

Correct Answer: Use of agents that produce non-rearranging acylium ions

Q43. What is a common diagnostic test to confirm electrophilic aromatic substitution (like Friedel–Crafts) occurred?

• Check for incorporation of alkyl group and loss of H at substitution site by NMR and mass spectrometry
• Look for radical signals in IR
• Measure conductivity of product
• Observe color change only

Correct Answer: Check for incorporation of alkyl group and loss of H at substitution site by NMR and mass spectrometry

Q44. Which substrate is likely to give ortho/para products predominantly in Friedel–Crafts alkylation?

• Nitrobenzene
• Benzene
• Anisole (methoxybenzene)
• Trifluoromethylbenzene

Correct Answer: Anisole (methoxybenzene)

Q45. What is the effect of adding a proton scavenger or base to a Friedel–Crafts reaction mixture?

• Accelerates formation of carbocation
• May neutralize necessary acids or catalysts and inhibit the reaction
• Converts aromatic product to a carboxylic acid
• Has no effect on catalytic Lewis acids

Correct Answer: May neutralize necessary acids or catalysts and inhibit the reaction

Q46. During scale-up of Friedel–Crafts reactions for API intermediates, which issue is most concerning?

• Exothermic reactions and heat transfer leading to runaway reactions
• Decreased reagent costs
• Excessive solubility of all reagents
• Improved selectivity at large scale

Correct Answer: Exothermic reactions and heat transfer leading to runaway reactions

Q47. Which statement about Friedel–Crafts alkylation on poly-substituted aromatic rings is correct?

• Regioselectivity is solely determined by steric factors
• Electronic and steric effects of existing substituents together determine product distribution
• All positions react equally
• Poly-substituted rings cannot undergo Friedel–Crafts reactions

Correct Answer: Electronic and steric effects of existing substituents together determine product distribution

Q48. Why might Friedel–Crafts alkylation fail with substrates containing basic nitrogen atoms unless protected?

• Basic nitrogens donate electrons and over-activate the ring only
• Basic nitrogens coordinate to Lewis acids, deactivating the catalyst and preventing electrophile formation
• Basic nitrogens are converted to nitriles
• Basic nitrogens form insoluble salts with aromatic rings

Correct Answer: Basic nitrogens coordinate to Lewis acids, deactivating the catalyst and preventing electrophile formation

Q49. Which experimental strategy can improve selectivity in Friedel–Crafts alkylation of an activated aromatic?

• Use an excess of Lewis acid and alkylating agent indiscriminately
• Lower temperature, limit equivalents of alkylating agent, or use acylation then reduction
• Perform reaction under strongly basic conditions
• Replace aromatic substrate with an aliphatic one

Correct Answer: Lower temperature, limit equivalents of alkylating agent, or use acylation then reduction

Q50. When designing a Friedel–Crafts alkylation for a pharmaceutical intermediate, which of the following is a prudent consideration?

• Ignore functional group compatibility to save steps
• Plan protection/deprotection, avoid rearrangement-prone reagents, and consider alternative methods if the ring is deactivated
• Always use AlCl3 in stoichiometric amounts for best yields
• Ensure all reactions are run at the highest possible temperature

Correct Answer: Plan protection/deprotection, avoid rearrangement-prone reagents, and consider alternative methods if the ring is deactivated

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