Practice important
Substituent Effects
MCQ questions and answers.
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MCQ Questions
1. Under standard anhydrous AlCl3 conditions, which substrate is most likely to undergo Friedel-Crafts acylation readily and predominantly give a para-acylated product?
A.
Nitrobenzene
B.
Anisole
C.
Benzoic acid
D.
Fluorobenzene
Correct Answer:
Anisole
Explanation:
The methoxy group strongly activates the ring through resonance and directs electrophilic substitution to the ortho and para positions. The para product predominates because it is less sterically hindered. Nitrobenzene and benzoic acid are strongly deactivated, while halobenzenes are deactivated and commonly fail to undergo Friedel-Crafts acylation.
2. Why does nitrobenzene generally fail to undergo Friedel-Crafts acylation with an acyl chloride and AlCl3?
A.
The nitro group strongly activates the ring but directs substitution exclusively to the meta position
B.
The nitro group prevents formation of the acylium ion
C.
The nitro group strongly withdraws electron density and makes formation of the arenium ion unfavorable
D.
The nitro group is replaced by the acyl group before electrophilic substitution occurs
Correct Answer:
The nitro group strongly withdraws electron density and makes formation of the arenium ion unfavorable
Explanation:
The nitro group exerts strong inductive and resonance electron withdrawal, greatly lowering the nucleophilicity of the aromatic ring. Consequently, the ring cannot readily attack the acylium ion, so Friedel-Crafts acylation is strongly suppressed.
3. In Friedel-Crafts acylation of anisole using ethanoyl chloride and AlCl3, which product is expected to be the major regioisomer?
A.
2-methoxyacetophenone
B.
3-methoxyacetophenone
C.
4-methoxyacetophenone
D.
1,4-dimethoxybenzene
Correct Answer:
4-methoxyacetophenone
Explanation:
The methoxy substituent is an ortho-para director and strongly activates both positions. Para substitution is favored over ortho substitution because the approaching acylium ion experiences less steric repulsion from the methoxy group.
4. A student predicts that chlorobenzene should undergo Friedel-Crafts acylation rapidly because chlorine is an ortho-para director. What is the best correction?
A.
Chlorine activates the ring strongly by resonance, so only the ortho product forms
B.
Chlorine deactivates the ring overall, and coordination with AlCl3 can further suppress the reaction
C.
Chlorine directs substitution to the meta position and prevents acylium-ion formation
D.
Chlorobenzene undergoes acylation only after chlorine is oxidized to chlorate
Correct Answer:
Chlorine deactivates the ring overall, and coordination with AlCl3 can further suppress the reaction
Explanation:
Halogens are ortho-para directors because of resonance donation, but their strong inductive withdrawal makes the ring less reactive overall. In addition, interaction of the halogen with the Lewis acid can further reduce the ring's ability to undergo Friedel-Crafts acylation.
5. When m-xylene undergoes Friedel-Crafts acylation, which ring position is generally favored for introduction of the acyl group?
A.
The position para to both methyl groups
B.
A position meta to both methyl groups
C.
A position ortho to both methyl groups
D.
A position para to one methyl group and ortho to the other
Correct Answer:
The position para to both methyl groups
Explanation:
Each methyl group activates the ring by hyperconjugation and directs substitution toward its ortho and para positions. In m-xylene, the position para to both methyl groups receives strong reinforcement from both substituents and is less sterically congested than positions ortho to a methyl group.
6. In the Friedel-Crafts acylation of tert-butylbenzene, why is the para-acylated product expected to predominate over the ortho-acylated product?
A.
The tert-butyl group is meta-directing because of its large size
B.
The tert-butyl group deactivates only the ortho position by resonance
C.
Both positions are activated, but para substitution minimizes steric interaction with the bulky tert-butyl group
D.
The tert-butyl group is removed by AlCl3 before acylation occurs
Correct Answer:
Both positions are activated, but para substitution minimizes steric interaction with the bulky tert-butyl group
Explanation:
An alkyl group activates the aromatic ring and directs electrophilic substitution to ortho and para positions. Because tert-butyl is especially bulky, approach at the para position is much less hindered, making the para product the major product.
7. Which substituent most strongly prevents normal Friedel-Crafts acylation of its benzene derivative under standard AlCl3 conditions?
A.
Methyl
B.
Ethyl
C.
Methoxy
D.
Carboxylic acid
Correct Answer:
Carboxylic acid
Explanation:
The carboxylic acid group is strongly electron withdrawing and greatly deactivates the aromatic ring toward electrophilic substitution. It can also interact with AlCl3, further reducing the ring's reactivity. Alkyl and methoxy groups are activating substituents.
8. What is the principal substituent effect that makes further Friedel-Crafts acylation of acetophenone much less favorable than acylation of benzene?
A.
The acyl group donates electron density strongly by resonance
B.
The acyl group withdraws electron density through both inductive and resonance effects
C.
The acyl group changes the ring into a saturated cyclohexane derivative
D.
The acyl group directs all subsequent substitution to the ortho position while activating the ring
Correct Answer:
The acyl group withdraws electron density through both inductive and resonance effects
Explanation:
An acyl group attached to the ring is strongly electron withdrawing because of the carbonyl group. It deactivates the aromatic ring, so the product of the first acylation is less reactive toward a second Friedel-Crafts acylation; this is why polyacylation is generally limited.
9. Under standard Friedel-Crafts conditions, which outcome is most likely when N,N-dimethylaniline is treated with an acyl chloride and AlCl3?
A.
Rapid para ring acylation because the dimethylamino group is a powerful activator unaffected by AlCl3
B.
Predominant meta ring acylation because the dimethylamino group becomes a meta director
C.
Suppressed normal ring acylation because the amino group coordinates strongly with AlCl3 and loses its activating effect
D.
Exclusive formation of a para-diacylated product because the first acyl group activates the ring
Correct Answer:
Suppressed normal ring acylation because the amino group coordinates strongly with AlCl3 and loses its activating effect
Explanation:
Although a free dimethylamino group strongly activates an aromatic ring, its lone pair coordinates with the Lewis acid AlCl3. This complex removes the group's effective electron donation and can strongly deactivate the ring, preventing the expected facile Friedel-Crafts acylation.
10. Which comparison correctly describes the relative suitability of ethylbenzene and nitrobenzene for Friedel-Crafts acylation?
A.
Nitrobenzene reacts more readily because the nitro group stabilizes the arenium ion
B.
Ethylbenzene reacts more readily because the ethyl group activates the ring, whereas the nitro group strongly deactivates nitrobenzene
C.
Both react at comparable rates because substituents affect only product orientation, not reaction rate
D.
Ethylbenzene is unreactive because alkyl groups prevent formation of the acylium ion
Correct Answer:
Ethylbenzene reacts more readily because the ethyl group activates the ring, whereas the nitro group strongly deactivates nitrobenzene
Explanation:
An ethyl group increases aromatic electron density through inductive donation and hyperconjugation, facilitating attack on the acylium ion. The nitro group has the opposite effect and makes formation of the arenium ion highly unfavorable, so nitrobenzene is unsuitable for the reaction.
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