Organic Chemistry 2 Quiz: Electrophilic Aromatic Substitution Eas General Mechanism

What this quiz covers

This quiz focuses on Electrophilic Aromatic Substitution Eas General Mechanism, 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.

Question 1

1. C-H bond electrons re-form $\pi$ bond (correct answer)
2. $\mathrm{NO_2^+}$ donates electrons to ring
3. Ring $\pi$ electrons attack $\mathrm{HSO_4^-}$
4. Nitrate adds to ring then eliminates water

Question 2

1. Ring $\pi$ electrons attack NO$_2^+$ (correct answer)
2. NO$_2^+$ attacks ring by donating electrons
3. Nitrate donates electrons to form C-N bond
4. Benzene attacks HSO$_4^-$ to form C-S bond

Question 3

1. A C-N bond between ring and NO$_2$ (correct answer)
2. An O-H bond between ring and sulfuric acid
3. A C-C bond from benzene dimerization
4. A C-Br bond from bromination conditions

Question 4

1. A new C-N bond forms to $\mathrm{NO_2}$ (correct answer)
2. A new N-N bond forms to $\mathrm{NO_2^+}$
3. A new C-O bond forms to $\mathrm{HSO_4^-}$
4. A new C-H bond forms at substitution site

Question 5

1. Loss of H$^+$ from the sigma complex (correct answer)
2. Addition of water across the ring
3. Homolytic cleavage of the N-O bond
4. Rearrangement of NO$_2^+$ to NO$^+$

Question 6

1. It consumes the nitronium ion irreversibly
2. It generates NO$_2^+$ from nitric acid (correct answer)
3. It adds directly across benzene's double bond
4. It forms a free-radical initiator for nitration

Question 7

1. It acts as a nucleophile via its $\pi$ electrons (correct answer)
2. It acts as an electrophile via its $\pi$ electrons
3. It acts as a radical trap forming cyclohexane
4. It acts as a leaving group to form NO$_2^+$

Question 8

1. Re-aromatization before electrophile formation
2. Generation of NO$_2^+$ in acidic mixture (correct answer)
3. Radical substitution at the benzene ring
4. Direct nucleophilic attack of nitrate on benzene

Question 9

1. A resonance-stabilized cyclohexadienyl cation (correct answer)
2. A resonance-stabilized phenyl anion intermediate
3. A bridged bromonium ion on aromatic ring
4. A neutral cyclohexadiene formed by addition

Question 10

1. Aromaticity is lost temporarily in sigma complex (correct answer)
2. Aromaticity is never disrupted during EAS
3. Aromaticity is lost permanently after substitution
4. Aromaticity increases during electrophile attack

Question 11

1. It remains sp² throughout the mechanism, as the π system is never fully broken.
2. It changes from sp² to sp³, and then returns to sp² in the final product. (correct answer)
3. It changes from sp² to sp, forming a linear transition state before reverting to sp².
4. It changes from sp³ to sp², as the ring must be activated before electrophilic attack.

Question 12

1. Drawing an arrow from the center of the aromatic ring to the electrophile to show the attack by the π system.
2. Showing the formation of a resonance-stabilized carbocation where one carbon atom of the ring is sp³-hybridized.
3. Drawing an arrow that originates from the positive charge of the electrophile and points towards the benzene ring. (correct answer)
4. Using a weak base, such as the conjugate base of the acid catalyst, to deprotonate the carbocation intermediate.

Question 13

1. Removal of the electrophile E⁺ would simply be the reverse of the rate-determining step, leading back to the starting materials. (correct answer)
2. The C-H bond is always weaker than the newly formed C-E bond, making its cleavage kinetically more favorable.
3. The bases present in EAS reactions are specifically designed to be strong enough to remove H⁺ but too weak to remove E⁺.
4. The electrophile E⁺ is typically too sterically hindered for a base to access it for removal from the ring.

Question 14

1. The C-H (or C-D) bond is broken in the rate-determining step of the reaction, but the bond strengths are nearly identical.
2. The formation of the sigma complex is the slow, rate-determining step, and the subsequent C-H (or C-D) bond cleavage is fast. (correct answer)
3. The reaction proceeds through a concerted mechanism where C-E bond formation and C-H bond breaking occur simultaneously.
4. The nitronium ion (NO₂⁺) is not a sufficiently strong electrophile to differentiate between the vibrational frequencies of C-H and C-D bonds.

Question 15

1. The deprotonation of the sigma complex derived from phenol is much faster because the hydroxyl group lowers the pKa of the ring proton.
2. The hydroxyl group of phenol actively participates in generating the nitronium ion (NO₂⁺) electrophile, increasing its concentration.
3. The transition state leading to the formation of the sigma complex is stabilized by electron donation from the hydroxyl group of phenol. (correct answer)
4. The aromatic stabilization energy of benzene is substantially greater than that of phenol, making its π system harder to disrupt.

Question 16

1. The intermediate from cyclohexene is a stable bromonium ion, while the intermediate from benzene is an unstable carbocation.
2. The arenium ion intermediate from benzene can eliminate a proton to regenerate a highly stable aromatic ring, a pathway unavailable to the carbocation from cyclohexene. (correct answer)
3. The FeBr₃ catalyst specifically inhibits the addition pathway for benzene by forming a bulky complex that blocks the second attack.
4. Benzene is a weaker nucleophile than cyclohexene, so it can only react via substitution, which has a lower activation energy than addition.

Question 17

1. Step 1 is unlikely, as aromatic rings are generally poor nucleophiles and require activation.
2. Step 2 is mechanistically flawed; the arenium ion is an electrophile, not a nucleophile, but the step is otherwise reasonable.
3. Step 2 violates the fundamental driving force of EAS; the arenium ion eliminates a proton to restore aromaticity, rather than undergo further attack. (correct answer)
4. Step 3 is incorrect; the elimination of the original electrophile E⁺ would be more favorable than eliminating a proton.

Question 18

1. The sigma complex intermediate is achiral due to a plane of symmetry passing through the sp³ carbon.
2. The final deprotonation step, which restores aromaticity, eliminates the very carbon atom that was the potential stereocenter.
3. The reaction is stereospecific and proceeds with inversion of configuration at the sp³ carbon.
4. The product, brominated ethylbenzene, is achiral, so it cannot exist as a racemic mixture. (correct answer)

Question 19

1. The ketone product formed is a Lewis base that strongly coordinates to the AlCl₃, deactivating it and preventing it from acting catalytically. (correct answer)
2. The acylium ion electrophile is much less reactive than an alkyl carbocation, requiring a higher catalyst concentration to drive the reaction.
3. The AlCl₃ catalyst is consumed during the deprotonation step of the acylation mechanism, forming an inactive byproduct.
4. The C-Cl bond of the acyl chloride is much stronger than in an alkyl chloride, requiring more catalyst to facilitate its cleavage.

Question 20

1. The reaction would proceed to give 1-propylbenzene, but at a significantly reduced rate.
2. The reaction would yield exclusively 2-propylbenzene due to the spontaneous rearrangement of the alkyl halide.
3. Essentially no reaction would occur because the C-Cl bond in 1-chloropropane is not sufficiently polarized to create an effective electrophile. (correct answer)
4. A nucleophilic aromatic substitution would occur, where benzene attacks the chlorine atom of 1-chloropropane.