Organic Chemistry 2 Quiz: Activating Deactivating Groups And Directing Effects
20 questions · exam conditions
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Activating Deactivating Groups And Directing EffectsQuestion 1 of 20

In EAS, how does an -NH2 substituent affect benzene reactivity compared with benzene?

Strongly increases rate
Strongly decreases rate
No significant change
Stops reaction completely
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Activating Deactivating Groups And Directing Effects

Practice Activating Deactivating Groups And Directing Effects in Organic Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Activating Deactivating Groups And Directing Effects, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

In EAS, how does an -NH2 substituent affect benzene reactivity compared with benzene?

  1. Strongly increases rate (correct answer)
  2. Strongly decreases rate
  3. No significant change
  4. Stops reaction completely
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Activating groups like -NH2 significantly increase the reactivity of benzene rings by donating electrons through resonance, making EAS faster than in unsubstituted benzene. The -NH2 group is one of the strongest activators due to its powerful +R effect from nitrogen lone pairs. The correct answer choice highlights how -NH2 strongly increases the rate, consistent with its role in stabilizing the sigma complex at ortho/para positions. A common distractor could be assuming deactivation, stemming from confusing amines with electron-withdrawing groups. To help students: Emphasize the rate enhancement by activators in EAS kinetics. Practice drawing resonance structures to see how electron donation boosts reactivity.

Question 2

Which substituent is an activating ortho/para director but weaker than -OH and -NH2?

  1. Alkoxy (-OR) (correct answer)
  2. Nitro (-NO2)
  3. Carboxyl (-CO2H)
  4. Trifluoromethyl (-CF3)
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Alkoxy groups like -OR activate via +R but are weaker than -OH or -NH2 due to alkyl insulation. Still, they direct ortho/para effectively. The correct answer choice selects -OR, unlike meta directors. A common distractor could be nitro, confusing types. To help students: Rank activators by strength. Use examples to show relative reactivities.

Question 3

Which substituent is most likely to prevent Friedel-Crafts alkylation due to strong deactivation?

  1. Nitro (-NO2) (correct answer)
  2. Methyl (-CH3)
  3. Methoxy (-OCH3)
  4. Ethyl (-CH2CH3)
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Strong deactivators like -NO2 make the ring too unreactive for Friedel-Crafts alkylation. This prevents carbocation electrophile attack. The correct answer choice identifies -NO2, contrasting with activators like -CH3. A common distractor might be methyl, overlooking deactivation. To help students: Note limitations of FC on deactivated rings. Relate to overall EAS reactivity.

Question 4

Which substituent withdraws by resonance (-R) and therefore deactivates the ring toward EAS?

  1. Methoxy (-OCH3)
  2. Amino (-NH2)
  3. Formyl (-CHO) (correct answer)
  4. Methyl (-CH3)
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Resonance withdrawers like -CHO deactivate via -R, directing meta. This depletes electron density at ortho/para. The correct answer choice picks -CHO, differing from +R like -OCH3. A common distractor might be methoxy, mixing effects. To help students: Visualize -R in carbonyls. Compare activation/deactivation mechanisms.

Question 5

For chlorobenzene, what is the directing effect of -Cl in EAS?

  1. Meta-directing and activating
  2. Ortho/para-directing and deactivating (correct answer)
  3. Meta-directing and deactivating
  4. Ortho/para-directing and strongly activating
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Halogens like -Cl are deactivating via -I but ortho/para-directing by +R. This leads to slower but regioselective EAS. The correct answer choice describes -Cl accurately, as deactivating ortho/para. A common distractor could be meta-deactivating, treating like -NO2. To help students: Remember halogens' unique profile. Balance inductive and resonance effects.

Question 6

Which substituent is meta-directing mainly due to a strong -I (inductive withdrawal) effect?

  1. Trifluoromethyl (-CF3) (correct answer)
  2. Methoxy (-OCH3)
  3. Methyl (-CH3)
  4. Amino (-NH2)
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. The -CF3 group is meta-directing primarily through strong -I effect from electronegative fluorines, lacking resonance. This deactivates the ring and favors meta positions. The correct answer choice picks -CF3, differing from +R donors like -OCH3. A common distractor might be methoxy, confusing effects. To help students: Emphasize inductive withdrawal in halogenated alkyls. Compare to resonance-based meta directors.

Question 7

A benzene ring with -CO2H is nitrated; what is the major directing outcome?

  1. Ortho substitution dominates
  2. Meta substitution dominates (correct answer)
  3. Para substitution dominates
  4. Side-chain substitution dominates
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. The -CO2H group is deactivating and meta-directing due to its electron-withdrawing carbonyl, leading to meta nitration. This occurs because meta sigma complex is least destabilized. The correct answer choice indicates meta dominance, unlike ortho/para for activators. A common distractor might be para, confusing with donors. To help students: Examine resonance in carboxylic acids for withdrawal. Practice predicting products based on directors.

Question 8

Which substituent is activating and ortho/para-directing mainly by hyperconjugation?

  1. Methyl (-CH3) (correct answer)
  2. Nitro (-NO2)
  3. Acetyl (-COCH3)
  4. Cyano (-C≡N)
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Groups like -CH3 activate via hyperconjugation, donating electron density to ortho/para positions without strong resonance. This makes toluene more reactive than benzene, with substitution mainly at ortho/para. The correct answer choice highlights -CH3's hyperconjugative activation, unlike resonance withdrawers like -NO2. A common distractor might be nitro, confusing activation mechanisms. To help students: Differentiate hyperconjugation from resonance in alkyl groups. Draw hyperconjugative structures to visualize directing effects.

Question 9

Which substituent is strongly deactivating and meta-directing in EAS?

  1. Methoxy (-OCH3)
  2. Amino (-NH2)
  3. Sulfonic acid (-SO3H) (correct answer)
  4. Methyl (-CH3)
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Strongly deactivating meta directors like -SO3H withdraw electrons via resonance and induction, favoring meta substitution. This group significantly reduces ring reactivity compared to benzene. The correct answer choice selects -SO3H, contrasting with activators like -OCH3. A common distractor could be methoxy, from misclassifying effects. To help students: Classify substituents by strength and directing type. Analyze sigma complex energies for meta preference.

Question 10

In EAS, which substituent on benzene is an activating group that directs ortho/para?

  1. Nitro (-NO2)
  2. Trifluoromethyl (-CF3)
  3. Methoxy (-OCH3) (correct answer)
  4. Carboxyl (-CO2H)
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Activating groups like -OCH3 increase the reactivity of benzene rings by donating electrons through resonance, directing electrophiles to ortho/para positions. In this case, the methoxy group (-OCH3) is a classic example of a strong activator due to its +R effect from the oxygen lone pairs. The correct answer choice reflects the established activating and ortho/para-directing effects of -OCH3, distinguishing it from deactivating groups like -NO2 or -CF3. A common distractor might be nitro, a misconception from confusing electron-withdrawing with donating substituents. To help students: Focus on classifying substituents based on their resonance and inductive effects. Encourage memorizing key examples of activators versus deactivators to predict EAS outcomes accurately.

Question 11

Which group is deactivating yet still directs ortho/para in EAS?

  1. Fluoro (-F) (correct answer)
  2. Nitro (-NO2)
  3. Cyano (-C≡N)
  4. Trifluoromethyl (-CF3)
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Halogens like -F are unique as they are deactivating due to -I effect but ortho/para-directing via +R from lone pairs. This duality arises because resonance donation stabilizes ortho/para sigma complexes despite overall deactivation. The correct answer choice identifies -F as deactivating yet ortho/para-directing, unlike pure meta directors like -NO2. A common distractor could be nitro, confusing strong deactivators with halogen behavior. To help students: Highlight halogens as exceptions in directing effects. Practice resonance structures to visualize +R outweighing -I in directing but not in activation.

Question 12

Which substituent is meta-directing primarily due to strong resonance withdrawal?

  1. Methyl (-CH3)
  2. Methoxy (-OCH3)
  3. Nitro (-NO2) (correct answer)
  4. Amino (-NH2)
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Meta-directing groups like -NO2 withdraw electrons through resonance, deactivating the ring and favoring meta substitution. The -NO2 group's strong -R effect destabilizes ortho/para sigma complexes more than meta. The correct answer choice identifies -NO2 as meta-directing via resonance withdrawal, unlike activators like -OCH3. A common distractor could be methoxy, confusing donors with withdrawers. To help students: Focus on resonance structures showing electron density depletion. Practice predicting directing effects by substituent classification.

Question 13

Which EAS reaction is reversible and used to install a temporary blocking group on benzene?

  1. Sulfonation (correct answer)
  2. Nitration
  3. Bromination
  4. Friedel–Crafts acylation
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Sulfonation is reversible, allowing -SO3H as a temporary meta-directing blocker. This aids in synthesis control. The correct answer choice identifies sulfonation, distinct from irreversible like nitration. A common distractor might be nitration, overlooking reversibility. To help students: Explore synthetic applications of reversible EAS. Tie to directing strategies.

Question 14

Which substituent is deactivating and meta-directing because it contains a carbonyl group?

  1. Acetyl (-COCH3) (correct answer)
  2. Ethyl (-CH2CH3)
  3. Methoxy (-OCH3)
  4. Amino (-NH2)
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Carbonyl-containing groups like -COCH3 deactivate and direct meta due to -R withdrawal. This stabilizes meta sigma complex relatively. The correct answer choice selects -COCH3, unlike activators like -OCH3. A common distractor could be methoxy, from effect confusion. To help students: Analyze carbonyl resonance for deactivation. Predict directing based on functional groups.

Question 15

Which EAS reaction introduces an acyl group using RCOCl/AlCl3\mathrm{RCOCl/AlCl_3}?

  1. Friedel–Crafts acylation (correct answer)
  2. Friedel–Crafts alkylation
  3. Sulfonation
  4. Nitration
Explanation: This question tests intermediate-level organic chemistry skills: understanding activating/deactivating groups and directing effects in electrophilic aromatic substitution. Friedel-Crafts acylation uses RCOCl/AlCl3 to introduce acyl groups via acylium ion. This reaction is influenced by ring activation state. The correct answer choice specifies acylation, separate from alkylation. A common distractor might be alkylation, confusing Lewis acids. To help students: Recall electrophile formation in FC reactions. Consider how deactivators prevent them.

Question 16

Predict the major monosubstitution product when 4-methylphenol (p-cresol) is treated with one equivalent of Br₂ in CCl₄.

  1. 2-Bromo-4-methylphenol (correct answer)
  2. 3-Bromo-4-methylphenol
  3. Bromomethyl-phenol
  4. 2,6-Dibromo-4-methylphenol
Explanation: The ring has two activating, ortho,para-directing groups: a hydroxyl group (-OH) and a methyl group (-CH₃). The hydroxyl group is a much stronger activating group than the methyl group, so it will control the position of substitution. The -OH group directs incoming electrophiles to its ortho positions (C2 and C6) and its para position (which is blocked by the methyl group). Since positions 2 and 6 are equivalent, bromination will occur at one of these positions to yield 2-bromo-4-methylphenol. Distractor C suggests benzylic bromination, which requires NBS and light/heat, not Br₂/CCl₄. Distractor D suggests disubstitution, which would require excess bromine.

Question 17

The relative rates of electrophilic bromination for three compounds were measured against benzene (relative rate = 1). Compound X had a relative rate of 3 × 10⁻⁵, and Compound Y had a relative rate of 40. Which pair of structures most plausibly corresponds to X and Y?

  1. X = Chlorobenzene; Y = Phenol
  2. X = Benzoic acid; Y = Toluene (correct answer)
  3. X = Nitrobenzene; Y = Anisole
  4. X = Toluene; Y = Benzoic acid
Explanation: A relative rate much less than 1 indicates a deactivated ring, while a rate greater than 1 indicates an activated ring. Benzoic acid (-COOH) is a strongly deactivating group, consistent with a very low relative rate (3 × 10⁻⁵). Toluene (-CH₃) is a moderately activating group, consistent with a relative rate of 40. While nitrobenzene and anisole (C) are deactivating and activating respectively, the values fit better with the moderate effects of -COOH and -CH₃ compared to the very strong deactivating -NO₂ group and very strong activating -OH group (A) or -OCH₃ group (C). The values provided are typical for benzoic acid and toluene.

Question 18

The nitration of tert-butylbenzene produces a mixture of ortho, meta, and para isomers. Which statement best describes the expected product distribution?

  1. The ortho and para isomers are formed in roughly a 2:1 ratio, reflecting the number of available positions.
  2. The meta isomer is the major product because the tert-butyl group is deactivating.
  3. The para isomer is the major product, with very little ortho isomer formed due to significant steric hindrance. (correct answer)
  4. The ortho isomer is the major product because it is electronically favored over the para position.
Explanation: The tert-butyl group is an alkyl group, which is an activating, ortho,para-director. However, it is extremely bulky. This steric hindrance makes it very difficult for the electrophile (and the subsequent reaction complex) to approach the ortho positions, which are adjacent to the tert-butyl group. As a result, substitution occurs almost exclusively at the sterically accessible para position. The meta product is formed in very small amounts because the group is an o,p-director.

Question 19

A student performs a reaction on thioanisole (C₆H₅SCH₃) and finds the product is a strong meta-director in subsequent electrophilic nitration. What reaction was most likely performed on the thioanisole?

  1. Deprotonation of the methyl group with a strong base.
  2. Alkylation of the sulfur atom to form a sulfonium salt.
  3. Reduction of the sulfide to a thiol with a reducing agent.
  4. Oxidation of the sulfide group with an agent like H₂O₂ or KMnO₄. (correct answer)
Explanation: Thioanisole (-SCH₃) is a sulfide, which is an activating ortho,para-director due to resonance donation from sulfur's lone pairs. To convert it into a meta-director, the sulfur atom must be made electron-withdrawing. This is achieved by oxidation. Strong oxidation converts the sulfide (-S-) to a sulfone (-SO₂-). The methylsulfonyl group (-SO₂CH₃) is strongly electron-withdrawing due to the electronegative oxygen atoms and is a meta-directing deactivator. Alkylation (B) would form a sulfonium salt, which is also a meta-director, but oxidation is a more common transformation in this context. Reduction (C) or deprotonation (A) would not result in a meta-director.

Question 20

For the electrophilic bromination of 3-methoxybenzoic acid, which position is the most likely site of substitution?

  1. Position 2
  2. Position 4
  3. Position 5
  4. Position 6 (correct answer)
Explanation: The molecule has two competing directing groups: a methoxy group (-OCH₃) at C1 and a carboxylic acid group (-COOH) at C3. The methoxy group is a strong activating, o,p-director. The carboxylic acid group is a strong deactivating, m-director. The powerful activating effect of the -OCH₃ group will control the position of substitution. It directs to its ortho (C2, C6) and para (C4) positions. The -COOH group directs to its meta position (C5). The positions most activated for substitution are C2, C4, and C6. However, positions C2 and C4 are ortho to the deactivating -COOH group, which disfavors substitution there. Position C6 is ortho to the activator and meta to the deactivator, making it the most favorable site for electrophilic attack.