Organic Chemistry 2 Quiz: Eas Reactions
20 questions · exam conditions
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Eas ReactionsQuestion 1 of 20

Which statement correctly describes halogens as substituents in EAS regarding rate and directing effects?

Activating and meta-directing
Deactivating yet ortho/para-directing
Activating and ortho/para-directing
Deactivating and meta-directing
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Organic Chemistry 2 Quiz

Organic Chemistry 2 Quiz: Eas Reactions

Practice Eas Reactions 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 Eas Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for Organic Chemistry 2.

How to use this quiz

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

Which statement correctly describes halogens as substituents in EAS regarding rate and directing effects?

  1. Activating and meta-directing
  2. Deactivating yet ortho/para-directing (correct answer)
  3. Activating and ortho/para-directing
  4. Deactivating and meta-directing
Explanation: This question tests the understanding of electrophilic aromatic substitution reactions, focusing on nitration, sulfonation, halogenation, and Friedel-Crafts (CCSS.OC.2.EAS). These reactions involve the substitution of an electrophile onto an aromatic ring, where the substituents affect both rate and regioselectivity due to their activating or deactivating nature. In this specific case, the context describes the rate and directing effects of halogens in EAS. The correct answer, B, reflects the accurate understanding that halogens are deactivating but ortho/para-directing due to inductive withdrawal and resonance donation. A common distractor, D, might mislead students by assigning meta direction, typical for other deactivators. To assist students, instructors can emphasize the importance of understanding the role of substituents in determining regioselectivity and how conditions influence the reversibility of reactions like sulfonation. Practice identifying the role of catalysts and recognizing common reaction pitfalls.

Question 2

In Friedel-Crafts alkylation, which substrate typically fails because it strongly deactivates the ring?

  1. Nitrobenzene (correct answer)
  2. Toluene
  3. Anisole
  4. Ethylbenzene
Explanation: This question tests the understanding of electrophilic aromatic substitution reactions, focusing on nitration, sulfonation, halogenation, and Friedel-Crafts (CCSS.OC.2.EAS). These reactions involve the substitution of an electrophile onto an aromatic ring, where the substituents affect both rate and regioselectivity due to their activating or deactivating nature. In this specific case, the context describes which substrate fails in Friedel-Crafts alkylation due to deactivation. The correct answer, A, reflects the accurate understanding that nitrobenzene is strongly deactivated by the nitro group. A common distractor, C, might mislead students by selecting an activator like anisole. To assist students, instructors can emphasize the importance of understanding the role of substituents in determining regioselectivity and how conditions influence the reversibility of reactions like sulfonation. Practice identifying the role of catalysts and recognizing common reaction pitfalls.

Question 3

Which group is strongly activating and ortho/para-directing due to resonance donation of a lone pair?

  1. CF3_3
  2. NO2_2
  3. OMe (correct answer)
  4. CHO
Explanation: This question tests the understanding of electrophilic aromatic substitution reactions, focusing on nitration, sulfonation, halogenation, and Friedel-Crafts (CCSS.OC.2.EAS). These reactions involve the substitution of an electrophile onto an aromatic ring, where the substituents affect both rate and regioselectivity due to their activating or deactivating nature. In this specific case, the context describes which group is strongly activating and ortho/para-directing via lone pair resonance. The correct answer, C, reflects the accurate understanding that OMe donates via resonance. A common distractor, B, might mislead students by selecting NO₂, a deactivator. To assist students, instructors can emphasize the importance of understanding the role of substituents in determining regioselectivity and how conditions influence the reversibility of reactions like sulfonation. Practice identifying the role of catalysts and recognizing common reaction pitfalls.

Question 4

In chlorination of benzene, why is FeCl3_3 used with Cl2_2 under standard EAS conditions?

  1. Generates Cl+^+ (or polarized Cl2_2) by Lewis acid activation (correct answer)
  2. Provides radical initiator for chain chlorination
  3. Acts as Brønsted base to deprotonate benzene first
  4. Is consumed stoichiometrically to form Fe metal
Explanation: This question tests the understanding of electrophilic aromatic substitution reactions, focusing on nitration, sulfonation, halogenation, and Friedel–Crafts (CCSS.OC.2.EAS). These reactions involve the substitution of an electrophile onto an aromatic ring, where the substituents affect both rate and regioselectivity due to their activating or deactivating nature. In this specific case, the context describes the role of FeCl₃ in benzene chlorination with Cl₂. The correct answer, A, reflects the accurate understanding that FeCl₃ generates Cl⁺ or polarized Cl₂ as the electrophile. A common distractor, B, might mislead students by suggesting a radical mechanism, which requires light or peroxides. To assist students, instructors can emphasize the importance of understanding the role of substituents in determining regioselectivity and how conditions influence the reversibility of reactions like sulfonation. Practice identifying the role of catalysts and recognizing common reaction pitfalls.

Question 5

In Friedel–Crafts alkylation using tert-butyl chloride and AlCl3_3, what electrophile attacks benzene?

  1. tert-Butyl carbocation (or equivalent complexed cation) (correct answer)
  2. tert-Butyl anion generated by AlCl3_3
  3. Chloride radical from homolysis
  4. AlCl4−_4^- acting as electrophile
Explanation: This question tests the understanding of electrophilic aromatic substitution reactions, focusing on nitration, sulfonation, halogenation, and Friedel–Crafts (CCSS.OC.2.EAS). These reactions involve the substitution of an electrophile onto an aromatic ring, where the substituents affect both rate and regioselectivity due to their activating or deactivating nature. In this specific case, the context describes the electrophile in Friedel–Crafts alkylation with tert-butyl chloride and AlCl₃. The correct answer, A, reflects the accurate understanding that the tert-butyl carbocation (or complex) attacks the ring. A common distractor, B, might mislead students by suggesting an anion, which is not electrophilic. To assist students, instructors can emphasize the importance of understanding the role of substituents in determining regioselectivity and how conditions influence the reversibility of reactions like sulfonation. Practice identifying the role of catalysts and recognizing common reaction pitfalls.

Question 6

Which step is rate-determining in nitration of toluene: NO2+_2^+ formation or -complex formation?

  1. Deprotonation to restore aromaticity
  2. Electrophile generation from mixed acid
  3. Attack of ring on NO2+_2^+ to form -complex (correct answer)
  4. Diffusion-controlled association of reagents
Explanation: This question tests the understanding of electrophilic aromatic substitution reactions, focusing on nitration, sulfonation, halogenation, and Friedel–Crafts (CCSS.OC.2.EAS). These reactions involve the substitution of an electrophile onto an aromatic ring, where the substituents affect both rate and regioselectivity due to their activating or deactivating nature. In this specific case, the context describes the rate-determining step in the nitration of toluene. The correct answer, C, reflects the accurate understanding that the attack of the aromatic ring on NO₂⁺ to form the σ-complex is typically rate-determining for activated rings like toluene. A common distractor, B, might mislead students by confusing it with electrophile generation, which is fast in mixed acid nitration. To assist students, instructors can emphasize the importance of understanding the role of substituents in determining regioselectivity and how conditions influence the reversibility of reactions like sulfonation. Practice identifying the role of catalysts and recognizing common reaction pitfalls.

Question 7

Why can polyalkylation occur in Friedel–Crafts alkylation of benzene under typical conditions?

  1. First alkyl group activates ring, accelerating further EAS (correct answer)
  2. AlCl3_3 permanently deactivates ring after first substitution
  3. Arenium ion rearranges to add multiple alkyls in one step
  4. Reaction proceeds by SN2 on aromatic carbon repeatedly
Explanation: This question tests the understanding of electrophilic aromatic substitution reactions, focusing on nitration, sulfonation, halogenation, and Friedel–Crafts (CCSS.OC.2.EAS). These reactions involve the substitution of an electrophile onto an aromatic ring, where the substituents affect both rate and regioselectivity due to their activating or deactivating nature. In this specific case, the context describes why polyalkylation can occur in Friedel–Crafts alkylation. The correct answer, A, reflects the accurate understanding that the first alkyl group activates the ring, promoting further substitution. A common distractor, B, might mislead students by claiming deactivation from AlCl₃. To assist students, instructors can emphasize the importance of understanding the role of substituents in determining regioselectivity and how conditions influence the reversibility of reactions like sulfonation. Practice identifying the role of catalysts and recognizing common reaction pitfalls.

Question 8

Which statement best describes sulfonation of benzene with fuming H2_2SO4_4 and heat?

  1. Irreversible; SO3_3 addition cannot be undone
  2. Reversible; higher temperature favors desulfonation (correct answer)
  3. Requires peroxides to initiate radical chain
  4. Proceeds by SN1 at sp2^2 carbon
Explanation: This question tests the understanding of electrophilic aromatic substitution reactions, focusing on nitration, sulfonation, halogenation, and Friedel–Crafts (CCSS.OC.2.EAS). These reactions involve the substitution of an electrophile onto an aromatic ring, where the substituents affect both rate and regioselectivity due to their activating or deactivating nature. In this specific case, the context describes the sulfonation of benzene with fuming H₂SO₄ and heat. The correct answer, B, reflects the accurate understanding that sulfonation is reversible, with higher temperatures favoring desulfonation. A common distractor, A, might mislead students by claiming irreversibility, ignoring the equilibrium nature. To assist students, instructors can emphasize the importance of understanding the role of substituents in determining regioselectivity and how conditions influence the reversibility of reactions like sulfonation. Practice identifying the role of catalysts and recognizing common reaction pitfalls.

Question 9

What is the primary role of sulfuric acid (H₂SO₄) in the electrophilic nitration of an aromatic ring using a mixture of nitric acid (HNO₃) and sulfuric acid?

  1. To act as a solvent for the aromatic substrate and nitric acid.
  2. To protonate the aromatic ring, making it more susceptible to attack by the nitrate ion.
  3. To act as a Brønsted-Lowry acid, protonating nitric acid to facilitate the formation of the nitronium ion (NO₂⁺). (correct answer)
  4. To form a bulky complex with the aromatic ring, directing the nitro group to the para position.
Explanation: Sulfuric acid is a stronger acid than nitric acid. Its role is to protonate the hydroxyl group of nitric acid. This creates a good leaving group (water), which departs to form the highly electrophilic nitronium ion (NO₂⁺). The nitronium ion is the active electrophile that is attacked by the electron-rich aromatic ring. Sulfuric acid does not protonate the benzene ring directly, nor does it act primarily as a solvent.

Question 10

Polyalkylation is a significant problem in Friedel-Crafts alkylation but not in Friedel-Crafts acylation. What is the fundamental reason for this difference?

  1. The acyl group sterically hinders further substitution, while the smaller alkyl group does not.
  2. The acylium ion is a much weaker electrophile than a carbocation, leading to only one substitution.
  3. The alkyl group added during alkylation activates the ring, while the acyl group added during acylation deactivates the ring. (correct answer)
  4. The carbocation intermediates in alkylation can rearrange to more reactive forms, promoting further reaction.
Explanation: The key difference lies in the electronic effect of the newly introduced substituent. An alkyl group is electron-donating and activates the aromatic ring, making the product (e.g., toluene) more reactive towards further electrophilic attack than the starting material (benzene). This leads to polyalkylation. In contrast, an acyl group (R-C=O) is strongly electron-withdrawing and deactivates the ring, making the product (an acylbenzene) much less reactive than the starting material. This effectively prevents further acylation.

Question 11

The sulfonation of naphthalene at 80°C yields naphthalene-1-sulfonic acid, but heating the reaction mixture to 160°C changes the major product to naphthalene-2-sulfonic acid. This observation is best explained by:

  1. A shift from an EAS mechanism at low temperature to a nucleophilic substitution mechanism at high temperature.
  2. The desulfonation of the 1-isomer being impossible at 80°C but rapid at 160°C, while the 2-isomer is stable at all temperatures.
  3. Increased steric hindrance at the 1-position at higher temperatures, which forces the incoming electrophile to the 2-position.
  4. The reaction being under kinetic control at 80°C to form the faster, less stable product, and under thermodynamic control at 160°C to form the slower, more stable product. (correct answer)
Explanation: When you encounter a temperature-dependent product distribution in organic reactions, think about kinetic versus thermodynamic control. This fundamental concept explains how reaction conditions determine which product predominates. At 80°C, the sulfonation favors naphthalene-1-sulfonic acid because the 1-position is more reactive toward electrophilic aromatic substitution. The 1-position has greater electron density and forms the electrophile-naphthalene intermediate faster. However, this kinetic product is less thermodynamically stable due to steric crowding and electronic factors. At 160°C, the higher energy allows the reaction to reach equilibrium. Sulfonation becomes reversible - both products can form and decompose. Under these thermodynamic conditions, naphthalene-2-sulfonic acid predominates because it's more stable. The 2-position has less steric hindrance and better orbital overlap, making this isomer thermodynamically favored despite forming more slowly initially. Choice A is wrong because both temperatures involve the same EAS mechanism - only the control regime changes. Choice B incorrectly suggests the 2-isomer is always stable; actually, both isomers can undergo desulfonation at high temperatures, but equilibrium favors the 2-isomer. Choice C misunderstands the cause - steric hindrance doesn't increase with temperature; rather, higher temperature enables equilibration to the more stable product. Remember this pattern: when you see temperature affecting product ratios in reversible reactions, consider kinetic control (lower temperature, faster-forming product) versus thermodynamic control (higher temperature, more stable product). This principle applies broadly across organic chemistry.

Question 12

Direct iodination of benzene with I₂ is an endothermic and reversible process that gives a poor yield. Which reagent is commonly added to I₂ to make the electrophilic iodination of benzene proceed efficiently?

  1. FeI₃, a Lewis acid catalyst
  2. H₂O₂, an oxidizing agent (correct answer)
  3. KI, a source of iodide ions
  4. NaOH, a strong base
Explanation: The reaction of benzene with I₂ produces HI as a byproduct. The reaction is reversible, and HI is a strong reducing agent that can reduce the iodinated product back to benzene. To drive the reaction forward, an oxidizing agent such as hydrogen peroxide (H₂O₂), nitric acid (HNO₃), or copper(II) chloride (CuCl₂) is added. The oxidizing agent reacts with the HI byproduct, removing it from the reaction mixture and shifting the equilibrium towards the products according to Le Châtelier's principle. A Lewis acid like FeI₃ (A) is not sufficient to overcome the unfavorable thermodynamics. KI (C) and NaOH (D) would not facilitate the formation of the required electrophile.

Question 13

When benzene is treated with one equivalent of chloromethane and AlCl₃, a mixture of toluene, xylenes, and other polymethylated benzenes is obtained. What is the primary reason for this lack of selectivity for the monosubstituted product?

  1. The methyl group of the toluene product activates the ring, making it more reactive than benzene. (correct answer)
  2. Carbocation rearrangements lead to the formation of multiple, differently substituted products.
  3. The reaction is highly exothermic, and the lack of temperature control leads to side reactions.
  4. The AlCl₃ catalyst becomes more active after the first substitution has occurred.
Explanation: This phenomenon is known as polyalkylation. The methyl group introduced onto the benzene ring is an electron-donating group, which activates the ring towards further electrophilic aromatic substitution. As a result, the product, toluene, is more nucleophilic and more reactive than the starting material, benzene. Consequently, as soon as some toluene is formed, it competes with benzene for the electrophile, leading to the formation of disubstituted (xylenes) and higher substituted products. Carbocation rearrangements (B) are not an issue with a methyl group.

Question 14

Which of the following halides would be LEAST suitable for a Friedel-Crafts alkylation reaction with benzene and AlCl₃ due to the inability to form the required electrophile?

  1. tert-Butyl chloride
  2. Isopropyl chloride
  3. Chlorobenzene (correct answer)
  4. Ethyl chloride
Explanation: Friedel-Crafts alkylations require the formation of a carbocation or a strongly polarized complex that can act as an electrophile. This process does not work with aryl halides (like chlorobenzene) or vinyl halides. The C-Cl bond in chlorobenzene has partial double-bond character due to resonance, making it stronger and preventing the formation of the highly unstable phenyl cation. Alkyl halides, including primary (D), secondary (B), and tertiary (A), are all suitable substrates, although primary and secondary halides may be prone to rearrangement.

Question 15

A student attempts to synthesize benzophenone by reacting benzene with benzoyl chloride and AlCl₃. After the reaction is complete, they add water to quench it. What is the role of the water in this final step?

  1. To hydrolyze any unreacted benzoyl chloride and to decompose the aluminum chloride complex of the product. (correct answer)
  2. To protonate the aromatic ring and facilitate the release of the catalyst.
  3. To act as a catalyst for the elimination of HCl from the sigma complex.
  4. To dissolve the benzophenone product, which is insoluble in the organic solvent.
Explanation: In Friedel-Crafts acylation, the ketone product forms a stable Lewis acid-base complex with the AlCl₃ catalyst. This complex must be broken to isolate the final product. Adding water (or dilute acid) accomplishes this by hydrolyzing the Al-O bond, releasing the free ketone (benzophenone). The water also serves to quench any unreacted AlCl₃ and benzoyl chloride. This workup step is essential for isolating the product from the reaction mixture.

Question 16

Consider the nitration (HNO₃, H₂SO₄) of the following compounds: benzene, chlorobenzene, and nitrobenzene. Which statement accurately compares their relative reaction rates and the major product for chlorobenzene?

  1. Rate: Benzene > Chlorobenzene > Nitrobenzene. Product: m-chloronitrobenzene.
  2. Rate: Chlorobenzene > Benzene > Nitrobenzene. Product: o,p-chloronitrobenzene.
  3. Rate: Benzene > Chlorobenzene > Nitrobenzene. Product: o,p-chloronitrobenzene. (correct answer)
  4. Rate: Benzene > Nitrobenzene > Chlorobenzene. Product: m-chloronitrobenzene.
Explanation: The chlorine atom in chlorobenzene is an electron-withdrawing group via induction, which deactivates the ring towards electrophilic aromatic substitution compared to benzene. The nitro group in nitrobenzene is a very strong electron-withdrawing group, deactivating the ring even more significantly. Thus, the order of reactivity is Benzene > Chlorobenzene > Nitrobenzene. However, the chlorine atom has lone pairs that can donate electron density through resonance, making it an ortho,para-director. Therefore, the major products of nitration are o-chloronitrobenzene and p-chloronitrobenzene.

Question 17

In a Friedel-Crafts acylation reaction, a stoichiometric amount (or slight excess) of the AlCl₃ catalyst is required. Why is a catalytic amount, which is sufficient for Friedel-Crafts alkylation, not effective for acylation?

  1. The acylium ion intermediate is less stable than a carbocation and requires more catalyst to form at an appreciable rate.
  2. The ketone product of the reaction is a Lewis base that forms a stable complex with AlCl₃, deactivating the catalyst. (correct answer)
  3. The acyl chloride starting material coordinates strongly with AlCl₃, preventing it from activating other molecules.
  4. Polyacylation is a rapid side reaction that consumes the catalyst, requiring a stoichiometric amount to drive the main reaction.
Explanation: The carbonyl oxygen of the ketone product has lone pairs and acts as a Lewis base. It coordinates strongly with the Lewis acid catalyst, AlCl₃. This complex formation effectively removes the AlCl₃ from the reaction mixture, preventing it from activating more acyl chloride molecules. Therefore, at least one equivalent of AlCl₃ is needed for each equivalent of the carbonyl compound formed. In contrast, the alkylbenzene product of alkylation is a much weaker Lewis base and does not deactivate the catalyst, allowing it to be used in catalytic amounts.

Question 18

A chemist wants to synthesize 4-bromo-2-nitrotoluene from toluene. Direct bromination of toluene produces a mixture of ortho and para isomers, making separation difficult. Which of the following sequences best utilizes a blocking group to favor the desired product?

    1. Toluene + fuming H₂SO₄; 2) Br₂, FeBr₃; 3) dilute H₂SO₄, heat; 4) HNO₃, H₂SO₄
     (correct answer)
    1. Toluene + HNO₃, H₂SO₄; 2) Br₂, FeBr₃; 3) Protect methyl group; 4) Deprotect
    1. Toluene + Br₂, FeBr₃; 2) fuming H₂SO₄; 3) HNO₃, H₂SO₄; 4) dilute H₂SO₄, heat
    1. Toluene + fuming H₂SO₄; 2) HNO₃, H₂SO₄; 3) Br₂, FeBr₃; 4) dilute H₂SO₄, heat
Explanation: This synthesis uses the reversible nature of sulfonation. 1) Toluene is sulfonated, and the bulky -SO₃H group preferentially blocks the less hindered para position, forming 4-methylbenzenesulfonic acid. 2) The remaining ortho position is then brominated. 3) The sulfonyl group is removed by treatment with dilute acid and heat. 4) The resulting o-bromotoluene is nitrated. The methyl group directs para (to the nitro group's final position), and the bromo group also directs para (to the same position), yielding the target 4-bromo-2-nitrotoluene. The other sequences either produce the wrong isomer or involve unrealistic steps.

Question 19

During sulfonation, which step restores aromaticity after the arenium ion forms?

  1. Loss of SO3_3 to regenerate benzene directly
  2. Deprotonation by base (e.g., HSO4−_4^-) to re-aromatize (correct answer)
  3. Hydride shift to stabilize carbocation
  4. Nucleophilic substitution by water at ipso carbon
Explanation: This question tests the understanding of electrophilic aromatic substitution reactions, focusing on nitration, sulfonation, halogenation, and Friedel–Crafts (CCSS.OC.2.EAS). These reactions involve the substitution of an electrophile onto an aromatic ring, where the substituents affect both rate and regioselectivity due to their activating or deactivating nature. In this specific case, the context describes the step restoring aromaticity in sulfonation after arenium ion formation. The correct answer, B, reflects the accurate understanding that deprotonation by a base like HSO₄⁻ restores aromaticity. A common distractor, A, might mislead students by suggesting direct SO₃ loss without deprotonation. To assist students, instructors can emphasize the importance of understanding the role of substituents in determining regioselectivity and how conditions influence the reversibility of reactions like sulfonation. Practice identifying the role of catalysts and recognizing common reaction pitfalls.

Question 20

In nitration of toluene (HNO3$/H_3$/H_2SOSO_4$), how does Me affect rate and o/p orientation?

  1. Deactivates; directs meta via inductive withdrawal
  2. Activates; directs ortho/para via hyperconjugation (correct answer)
  3. Deactivates; directs ortho/para via resonance withdrawal
  4. Activates; directs meta by stabilizing meta -complex
Explanation: This question tests the understanding of electrophilic aromatic substitution reactions, focusing on nitration, sulfonation, halogenation, and Friedel–Crafts (CCSS.OC.2.EAS). These reactions involve the substitution of an electrophile onto an aromatic ring, where the substituents affect both rate and regioselectivity due to their activating or deactivating nature. In this specific case, the context describes the effect of the methyl group in toluene on the nitration reaction using HNO₃/H₂SO₄. The correct answer, B, reflects the accurate understanding of the methyl group as an activator that directs ortho/para through hyperconjugation and inductive donation. A common distractor, A, might mislead students by suggesting that alkyl groups deactivate and direct meta, which is actually characteristic of electron-withdrawing groups like nitro. To assist students, instructors can emphasize the importance of understanding the role of substituents in determining regioselectivity and how conditions influence the reversibility of reactions like sulfonation. Practice identifying the role of catalysts and recognizing common reaction pitfalls.