chemistry : class11 Activating Group Influence : Friedel-Crafts acylation : Hydrocarbons MCQ Questions and Answers
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Activating Group Influence
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MCQ Questions
1. When anisole undergoes Friedel–Crafts acylation with ethanoyl chloride and anhydrous AlCl3, which product is expected to predominate and why?
A.
2-methoxyacetophenone, because the methoxy group directs acylation exclusively to the ortho position
B.
4-methoxyacetophenone, because the methoxy group activates both ortho and para positions, while para substitution has less steric crowding
C.
3-methoxyacetophenone, because the methoxy group is meta-directing in the presence of AlCl3
D.
Acetophenone, because the methoxy group is removed completely by AlCl3 before acylation
Correct Answer:
4-methoxyacetophenone, because the methoxy group activates both ortho and para positions, while para substitution has less steric crowding
Explanation:
The methoxy group is strongly activating through resonance donation and directs electrophilic substitution to the ortho and para positions. The para position is favored over the ortho position because it is less sterically hindered by both the methoxy and incoming acyl groups.
2. Which substrate is expected to undergo Friedel–Crafts acylation most readily under comparable conditions?
A.
Benzene
B.
Toluene
C.
Nitrobenzene
D.
Benzoic acid
Correct Answer:
Toluene
Explanation:
The methyl group activates the benzene ring through hyperconjugation and its electron-releasing inductive effect. Therefore, toluene reacts more readily than benzene and gives mainly ortho- and para-acylation, whereas nitrobenzene and benzoic acid are strongly deactivated.
3. Why is direct Friedel–Crafts acylation of aniline generally unsuccessful with an acyl chloride and AlCl3?
A.
The amino group strongly deactivates the ring by resonance withdrawal
B.
The amino group directs the acyl group only to the meta position
C.
The amino group coordinates strongly with AlCl3, reducing its activating effect and interfering with formation of the effective acylating system
D.
The amino group prevents formation of the acylium ion by reacting irreversibly with the acyl chloride only through the ring
Correct Answer:
The amino group coordinates strongly with AlCl3, reducing its activating effect and interfering with formation of the effective acylating system
Explanation:
Although an amino group strongly activates an aromatic ring by resonance donation, its lone pair coordinates with the Lewis acid AlCl3. This complexation withdraws electron density from the nitrogen and can inhibit the Friedel–Crafts acylation, so an amino group is commonly protected before the reaction.
4. Acetanilide undergoes Friedel–Crafts acylation more satisfactorily than aniline mainly because the acetamido group:
A.
is strongly meta-directing and therefore avoids interaction with AlCl3
B.
retains electron donation to the ring while its nitrogen lone pair is less available for strong complexation with AlCl3
C.
converts the benzene ring into a nonaromatic intermediate before acylation
D.
withdraws electron density from the ring more strongly than a nitro group
Correct Answer:
retains electron donation to the ring while its nitrogen lone pair is less available for strong complexation with AlCl3
Explanation:
In acetanilide, the nitrogen lone pair is delocalized toward the carbonyl group, so it is less available to bind AlCl3 than the lone pair in aniline. The acetamido group still donates electron density to the ring by resonance and directs acylation mainly to the para position.
5. In the Friedel–Crafts acylation of toluene, which statement best describes the influence of the methyl group?
A.
It deactivates the ring but directs acylation to the meta position
B.
It activates the ring and stabilizes the ortho- and para-substituted sigma complexes relative to the meta-substituted complex
C.
It activates only the carbon atom directly attached to the methyl group and prevents para substitution
D.
It has no electronic influence, so the product distribution is determined only by the acyl chloride
Correct Answer:
It activates the ring and stabilizes the ortho- and para-substituted sigma complexes relative to the meta-substituted complex
Explanation:
A methyl group is electron-releasing through hyperconjugation and the inductive effect. In the ortho and para pathways, additional resonance stabilization of the sigma complex is possible, so these positions are favored over the meta position.
6. Which predicted major product is most consistent with the activating influence of a methoxy group during Friedel–Crafts acylation of 1,3-dimethoxybenzene?
A.
Acylation at the carbon between the two methoxy groups, because it is strongly activated by both groups
B.
Acylation at a carbon ortho to one methoxy group and para to the other, because that position receives reinforcing activation from both groups
C.
Acylation at a carbon meta to both methoxy groups, because methoxy groups become meta-directing during acylation
D.
No reaction, because two methoxy groups completely deactivate the ring toward electrophilic substitution
Correct Answer:
Acylation at a carbon ortho to one methoxy group and para to the other, because that position receives reinforcing activation from both groups
Explanation:
Methoxy groups are strong resonance-donating activators and direct electrophiles to ortho and para positions. In 1,3-dimethoxybenzene, a position that is ortho to one methoxy group and para to the other is electronically activated by both substituents, although steric effects can influence the exact product ratio.
7. A benzene ring bears both a methyl group and a methoxy group. If a position is ortho to the methoxy group and para to the methyl group, while another available position is meta to the methoxy group and ortho to the methyl group, which position is generally favored for acylation?
A.
The position ortho to methoxy and para to methyl, because both substituents activate that position in the same directing sense
B.
The position meta to methoxy and ortho to methyl, because methyl groups are stronger activators than methoxy groups
C.
Both positions equally, because activating groups do not influence regioselectivity
D.
The position meta to methoxy and ortho to methyl, because acylation always occurs meta to the stronger donating group
Correct Answer:
The position ortho to methoxy and para to methyl, because both substituents activate that position in the same directing sense
Explanation:
Both methoxy and methyl groups are activating and are ortho/para directors. A position that is ortho to methoxy and para to methyl is favored because it is activated by both groups and is less sterically crowded than a position ortho to both substituents.
8. Which sequence correctly ranks the approximate ease of Friedel–Crafts acylation, assuming comparable conditions and no strong complexation complications?
Explanation:
The methoxy group strongly activates the ring by resonance donation, while the methyl group activates it more weakly through hyperconjugation and induction. Benzene has no activating substituent, and the nitro group strongly deactivates the ring, making it unsuitable for Friedel–Crafts acylation.
9. Why does para acylation usually dominate over ortho acylation when an activated aromatic substrate such as anisole reacts under Friedel–Crafts conditions?
A.
The para position is electronically activated, whereas the ortho position is electronically deactivated
B.
The para sigma complex is always more resonance-stabilized than the ortho sigma complex
C.
Both ortho and para positions are activated, but the para pathway generally experiences less steric repulsion between the methoxy group and the incoming acyl group
D.
The methoxy group can donate electron density only to the para position
Correct Answer:
Both ortho and para positions are activated, but the para pathway generally experiences less steric repulsion between the methoxy group and the incoming acyl group
Explanation:
A methoxy group activates both ortho and para positions through resonance donation. The para product is generally major because the incoming acyl group is farther from the existing methoxy group, reducing steric interactions during substitution.
10. Which modification most effectively preserves activation of an aniline ring while reducing the tendency of the nitrogen atom to bind AlCl3 during Friedel–Crafts acylation?
A.
Convert the amino group into an acetamido group before acylation
B.
Convert the amino group into a nitro group before acylation
C.
Oxidize the amino group to a carboxyl group before acylation
D.
Replace the amino group with a cyano group before acylation
Correct Answer:
Convert the amino group into an acetamido group before acylation
Explanation:
Acetylation of aniline forms acetanilide. The nitrogen lone pair becomes delocalized into the carbonyl group, decreasing strong coordination with AlCl3 while retaining resonance donation to the aromatic ring. The acetamido group therefore continues to activate the ring and directs acylation mainly to the para position.
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