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Quizzes > Science

EAS Practice Problems: Electrophilic Aromatic Substitution Quiz

Quick electrophilic aromatic substitution quiz with instant results and explanations.

Editorial: Review CompletedCreated By: Lauren ElderkinUpdated Aug 24, 2025
Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art style benzene ring with reagent arrows on sky blue background invites EAS practice quiz

This quiz helps you practice electrophilic aromatic substitution by predicting major products, ortho/para/meta outcomes, and how directing groups change rates. After you finish, brush up core ideas with the aromaticity test or challenge broader skills in the organic chemistry quiz. If you want more speed drills on reaction outcomes, try a short chemical reaction predictions quick check.

What is the role of the Lewis acid in Friedel - Crafts alkylation?
To oxidize the aromatic ring
To protonate the aromatic system
To act as a nucleophile attacking the aromatic ring
To generate the electrophile by coordinating to the alkyl halide
In Friedel - Crafts alkylation, the Lewis acid like AlCl? coordinates with the alkyl halide to form a more reactive carbocation electrophile. This activated electrophile then attacks the aromatic ring, allowing substitution. Without the Lewis acid, the alkyl halide would be too unreactive. See for details.
Which reagent is commonly used for the nitration of benzene?
Cl? and AlCl?
Concentrated HNO? and H?SO?
KMnO? in basic solution
Br? and FeBr?
Nitration of benzene requires generation of the nitronium ion (NO??) by combining concentrated nitric acid with concentrated sulfuric acid. The sulfonic acid protonates the nitric acid, facilitating water loss and NO?? formation. This ion is the active electrophile that attacks benzene. More at .
Which intermediate is formed during the electrophilic aromatic substitution mechanism?
Carbanion
Carbene
Radical cation
?-complex (arenium ion)
The ?-complex, also known as the arenium ion or Wheland intermediate, forms when the electrophile adds to the aromatic ring, disrupting aromaticity. It is stabilized by resonance over three carbon atoms. Loss of a proton then restores aromaticity. More on the mechanism at .
Which substituent strongly activates the aromatic ring and directs ortho/para?
-CHO
-NO?
-OH
-CF?
The hydroxyl group (-OH) donates electrons via resonance, increasing electron density on the ring and strongly activating it toward electrophilic attack at ortho and para positions. Nitro (-NO?) and trifluoromethyl (-CF?) are deactivators, while formyl (-CHO) is also a deactivator. Activation and directing effects are detailed at .
Which reagent combination is used for sulfonation of benzene?
HCl and ZnCl?
SO? and H?SO?
Hg(OAc)?
Br? and FeBr?
Sulfonation uses oleum or a mixture of sulfur trioxide (SO?) dissolved in concentrated sulfuric acid to generate the electrophile, sulfur trioxide or protonated SO? (SO?H?). This electrophile attacks benzene to give benzenesulfonic acid. Conditions are reversible at high temperature. Read more at .
What is the directing effect of a nitro group during EAS?
Activating and meta-directing
Activating and ortho/para-directing
Deactivating and meta-directing
Deactivating and ortho/para-directing
The nitro group (-NO?) is strongly electron withdrawing via both resonance and induction, which deactivates the aromatic ring toward electrophilic attack. It directs incoming electrophiles to the meta position because the ortho/para carbons have less positive charge in the arenium intermediate. Meta-director details at .
Which solvent is commonly used for Friedel - Crafts acylation?
Dichloromethane (DCM)
Water
Acetone
Methanol
Non-polar or mildly polar solvents like dichloromethane are used to dissolve both reactants and maintain the reactivity of the Lewis acid catalyst. Water or strong polar protic solvents can deactivate the Lewis acid or hydrolyze the acylium ion. Thus, DCM is a typical choice. More at .
Which of the following is the major product of bromination of toluene under FeBr? catalysis?
o-Bromotoluene
m-Bromotoluene
p-Bromotoluene
1,3- and 1,4-dibromotoluene mixture
The methyl group is an ortho/para director by hyperconjugation and weak electron donation. Steric hindrance favors the para isomer predominately, giving p-bromotoluene as the major product. Ortho is present in smaller amounts, and meta is negligible. Read more at .
What is the product when benzene reacts with acetyl chloride and AlCl??
Acetophenone
Benzoic acid
Chloroacetone
Benzaldehyde
In Friedel - Crafts acylation, acetyl chloride reacts with AlCl? to form the acylium ion, which then adds to benzene. This yields acetophenone (phenyl methyl ketone) after regeneration of the catalyst. No rearrangement occurs in acylations. More info at .
Which substituent has the strongest deactivating effect on the benzene ring?
-OCH?
-CHO
-NH?
-NO?
The nitro group is strongly electron withdrawing by both resonance and induction, making it the most deactivating substituent commonly encountered in EAS. Aldehyde (-CHO) also deactivates but less than -NO?. Amino (-NH?) and methoxy (-OCH?) activate the ring. More at .
During sulfonation of toluene, which positions are most activated for substitution?
Only para
Only ortho
Ortho and para
Meta
The methyl group directs electrophiles to ortho and para positions due to hyperconjugation and inductive effects. In sulfonation, the SO?H? electrophile attacks these activated sites. Ortho and para products form, with distribution influenced by temperature. Details at .
What is the effect of electron-withdrawing groups on the rate of EAS?
They increase the rate by stabilizing the transition state
They decrease the rate by reducing ring nucleophilicity
They convert EAS into radical substitution
They have no effect
Electron-withdrawing groups reduce the electron density of the aromatic ring, making it less nucleophilic and less reactive toward electrophiles. This slows the rate of EAS. They also direct meta due to unfavorable resonance at ortho/para. See .
Which reaction condition favors para substitution over ortho in nitration of nitrotoluene?
Use of a nonpolar solvent
Higher temperature
Lower temperature
Addition of radical initiator
Lower temperatures favor the more stable para product by reducing the contribution of sterically hindered ortho substitution. At higher temperatures, kinetic control can allow more ortho product. Nitration is reversible, so thermodynamic control (low T) gives para preference. More at .
Which of the following is not a step in the EAS mechanism?
Formation of ?-complex
Radical generation
Regeneration of aromaticity
Electrophile formation
Electrophilic aromatic substitution does not involve radical intermediates. It proceeds via formation of an electrophile, attack on the ring to give a ?-complex, then deprotonation to restore aromaticity. Radical mechanisms are radical substitutions, not EAS. See .
Why does nitration of phenol proceed without a strong acid catalyst?
Phenol undergoes radical substitution instead
Phenol activates the ring enough to generate NO?? with mild nitric acid
Phenol easily loses a proton to form phenoxide free radical
Phenol forms a diazonium ion under these conditions
The hydroxyl group greatly activates the aromatic ring via resonance, making it nucleophilic enough that milder nitrating mixtures can produce enough nitronium ion for substitution. No strong acid medium is needed for electrophile generation. Phenoxide is highly activated, speeding EAS. See .
Which reaction yields a mixture of ortho and para products due to lack of steric bias?
Sulfonation of nitrobenzene
Chlorination of chlorobenzene
Nitration of t-butylbenzene
Nitration of anisole
Anisole's methoxy group strongly activates the ring and directs ortho/para. Steric hindrance is minimal for small OCH?, giving a mix of ortho and para nitroanisole. In t-butyl case, sterics favor para. Nitrobenzene is meta director and deactivated. More at .
What is the Hammett ? value used to quantify?
Electronic effect of substituents on reaction rates
Concentration of electrophile
Steric hindrance in EAS
Solvent polarity
Hammett ? constants describe the substituent electronic effect (inductive and resonance) on reaction rates and equilibria compared to unsubstituted benzene. A positive ? indicates electron withdrawal, negative indicates donation. They allow correlation of structure and reactivity. More at .
How does a strong deactivator like -SO?H affect the rate constant (k) of EAS compared to benzene?
It has no effect on k
It increases k by resonance donation
It significantly lowers k by decreasing electron density
It changes the mechanism to radical substitution
Sulfonic acid group is strongly electron withdrawing, reducing electron density of the aromatic ring, which decreases the nucleophilicity and lowers the rate constant for electrophilic attack compared to benzene. EAS rate drops significantly. Reaction still follows SEAr mechanism. Learn more at .
Which of the following can reverse Friedel - Crafts alkylation rearrangements?
Adding more Lewis acid
Increasing temperature
Working in protic solvent
Using acylation instead of alkylation
Friedel - Crafts acylation avoids carbocation rearrangements that plague alkylations because the acylium ion is resonance-stabilized and cannot rearrange. This ensures the carbonyl group remains at the intended position. Alkylations often give rearranged products. More at .
In an electrophilic chlorination of nitrobenzene, which position is substituted?
Ortho
Meta
Para
No reaction
The nitro group is a meta-director and deactivates the ring toward electrophiles. Chlorination still occurs but at a much slower rate and exclusively at the meta position because ortho/para resonance structures place a positive charge next to the electron-withdrawing nitro. See .
Why is polyalkylation common in Friedel - Crafts alkylation?
Alkyl groups deactivate the ring, preventing further reaction
Alkyl substituents activate the ring further, promoting additional substitutions
Solvent reacts with the ring
Lewis acid catalyst is consumed
Alkyl groups donate electron density to the aromatic ring, increasing its reactivity for EAS. Once the first alkylation occurs, the ring is more activated, leading to further alkylation and often polyalkylation unless controlled. Use acylation to avoid this. See .
Which factor explains why bromination of benzene is slower than chlorination under similar conditions?
Lower electrophilicity of Br? compared to Cl?
Bromide ion is a better leaving group
Chlorine is nonpolar
Bromine has a higher molecular weight
The bromonium ion (Br?) is a weaker electrophile than the chloronium ion (Cl?), making electrophilic attack on benzene slower for bromination. Generation of Br? with FeBr? is also less favorable. Thus chlorination typically proceeds faster. More at .
Design a two-step synthesis of p-nitrotoluene starting from benzene. Which sequence is correct?
Nitration first, then Friedel - Crafts alkylation
Sulfonation then nitration
Friedel - Crafts alkylation with CH?Cl/AlCl?, then nitration with HNO?/H?SO?
Halogenation then alkylation
To get p-nitrotoluene, you must first install the methyl group via Friedel - Crafts alkylation, which is ortho/para-directing. Then nitration produces mainly para product. If nitration is done first, the nitro group deactivates the ring and prevents alkylation. See .
Which strategy would best prevent polyalkylation in a Friedel - Crafts reaction?
Use Friedel - Crafts acylation followed by reduction
Add excess alkyl halide
Perform reaction at high temperature
Use a stronger Lewis acid
Friedel - Crafts acylation yields a ketone which is less activating, preventing further substitution. Subsequent reduction of the carbonyl group yields the alkylated product without risk of polyalkylation. Adding excess alkyl halide promotes additional substitution. More at .
When 1,3-dimethoxybenzene undergoes nitration, which position is most reactive and why?
Position 6, because it is meta to both groups
Position 4, because it is para to one methoxy and ortho to the other
Position 5, because both methoxy groups donate electron density to ortho/para sites generating a superactivated position
Position 2, because it is between the two methoxy groups
In 1,3-dimethoxybenzene, the methoxy groups direct to ortho/para relative to each substituent. The position that is ortho to one and para to the other (position 5) receives electron donation from both, making it the most activated. This dual activation leads to selective nitration there. Further reading at .
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Study Outcomes

  1. Understand EAS Mechanisms -

    Grasp the step-by-step pathway of electrophilic aromatic substitution, including σ-complex formation and re-aromatization.

  2. Identify Activating & Deactivating Groups -

    Classify substituents by their electron-donating or electron-withdrawing effects and predict their influence on reaction rate.

  3. Predict Regioselectivity -

    Determine ortho, meta, and para product distributions based on directing effects of existing substituents.

  4. Apply Reaction Conditions -

    Select appropriate reagents, catalysts, and conditions to optimize electrophilic aromatic substitution outcomes.

  5. Solve EAS Practice Problems -

    Work through aromatic electrophilic substitution practice problems to reinforce understanding and boost confidence.

  6. Evaluate Reaction Pathways -

    Compare competing mechanisms and justify the most favorable pathway for a given aromatic substitution reaction.

Cheat Sheet

  1. Fundamental Mechanism of EAS -

    The aromatic electrophilic substitution mechanism proceeds in two main steps: formation of a sigma complex (arenium ion) via electrophile attack on the π system, followed by deprotonation to regenerate aromaticity. For example, nitration uses HNO₃/H₂SO₄ to generate NO₂❺, which attacks benzene (source: MIT OCW). A helpful mnemonic is "SEAr: Slow Electrophile Attack, rapid Rearomatization."

  2. Substituent Effects and Directing Patterns -

    Electron-donating groups (e.g., - OH, - OCH₃) activate the ring and direct ortho/para, while electron-withdrawing groups (e.g., - NO₂, - CF₃) deactivate and direct meta (IUPAC guidelines). Remember: "Activators party at 2 & 4, deactivators go at 3" to predict positions in aromatic substitution reactions quiz problems. Review EAS practice problems to reinforce directing effects derived from resonance and inductive influences.

  3. Common EAS Reaction Types and Conditions -

    Nitration, sulfonation, halogenation, and Friedel - Crafts acylation/alkylation are foundational aromatic electrophilic substitution practice problems; each uses specific catalysts (e.g., AlCl₃ for FC, H₂SO₄ for sulfonation). For instance, benzene + Cl₂/FeCl₃ yields chlorobenzene via a chloronium intermediate (ACS Org. Chem. Principles). Practice with an electrophilic aromatic substitution quiz to master reagent selection.

  4. Rate-Determining Step and Energetics -

    The slow formation of the σ-complex is the rate-determining step, as confirmed by kinetic studies and DFT calculations (Nature Chem. articles). Substituent effects alter activation energy: stronger activators lower the barrier, observed in organic chemistry practice problems datasets. Plotting Hammett reaction rates (log(kX/kH) vs σ) yields linear free-energy relationships for predictive power.

  5. Regioselectivity in Multisubstituted Arenes -

    When multiple substituents conflict, the strongest activating group often dominates directing effects, but steric hindrance can override electronic preference (university-level problem sets). For example, in m-nitrotoluene, - CH₃ directs ortho/para while - NO₂ directs meta, resulting in mixed products; use blocking/deblocking strategies to steer selectivity. Challenge yourself with aromatic substitution reactions quiz scenarios to solidify planning tactics.

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