NEETAceToday's NEET Daily

Electrophilic aromatic substitution and directing effects: NEET MCQs with answers

50 questions on this topic of Hydrocarbons (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Hydrocarbons freeAll Hydrocarbons questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. In the electrophilic substitution of acetanilide (C₆H₅NHCOCH₃) versus aniline (C₆H₅NH₂), acetanilide is significantly less reactive because:

  1. The carbonyl oxygen withdraws electron density away from nitrogen through resonance, reducing nitrogen lone pair donation into the ring
  2. The acetyl group increases steric hindrance preventing electrophilic attack completely
  3. The nitrogen atom becomes sp³ hybridized in acetanilide
  4. Acetophenone forms an insoluble complex with the incoming electrophile
Show answer and explanation

Answer: (A) The carbonyl oxygen withdraws electron density away from nitrogen through resonance, reducing nitrogen lone pair donation into the ring

In acetanilide, the lone pair of electrons on nitrogen is delocalized into the adjacent carbonyl group via resonance (–N–C=O ↔ –N⁺=C–O⁻). Consequently, the availability of the lone pair to conjugate with the aromatic ring is reduced compared to aniline, making acetanilide less activating and moderating its reactivity.

medium

2. What is the major product obtained when benzene is treated with 1-chloropropane in the presence of anhydrous AlCl₃?

  1. Isopropylbenzene (Cumene)
  2. n-Propylbenzene
  3. 1,2-Diphenylpropane
  4. 1-Chloro-2-propylbenzene
Show answer and explanation

Answer: (A) Isopropylbenzene (Cumene)

The reaction is a Friedel-Crafts alkylation. The primary carbocation (CH₃CH₂CH₂⁺) initially generated undergoes a 1,2-hydride shift to form the more stable secondary carbocation (CH₃-CH⁺-CH₃), which then acts as the electrophile to yield isopropylbenzene (cumene) as the major product.

hard

3. How many structurally isomeric monobromo derivatives are formed when 1,4-dimethylbenzene (p-xylene) undergoes nuclear electrophilic bromination using Br₂ and FeBr₃?

  1. 2
  2. 1
  3. 3
  4. 4
Show answer and explanation

Answer: (B) 1

In 1,4-dimethylbenzene, all four unsubstituted ring carbon positions (C-2, C-3, C-5, C-6) are chemically and symmetrically equivalent due to D₂h molecular symmetry. Substitution of a hydrogen atom by bromine at any of these positions produces the exact same compound, 2-bromo-1,4-dimethylbenzene.

medium

4. What is the major organic product formed when benzene is treated with propanoyl chloride in the presence of anhydrous AlCl₃?

  1. 1-Phenylpropane
  2. Cumene
  3. 1-Phenylpropan-2-one
  4. Propiophenone (1-phenylpropan-1-one)
Show answer and explanation

Answer: (D) Propiophenone (1-phenylpropan-1-one)

The reaction is Friedel-Crafts acylation. Benzene reacts with propanoyl chloride (CH₃CH₂COCl) in the presence of anhydrous AlCl₃ via an acylium ion intermediate (CH₃CH₂CO⁺) to give propiophenone (C₆H₅COCH₂CH₃). Acylium ions do not undergo carbocation rearrangements like alkylations often do.

easy

5. Which one of the following substituted benzene derivatives undergoes nitration at the fastest rate under identical conditions?

  1. Benzoic acid
  2. Nitrobenzene
  3. Chlorobenzene
  4. Toluene
Show answer and explanation

Answer: (D) Toluene

Toluene contains a methyl group (-CH₃) which activates the benzene ring toward electrophilic aromatic substitution via hyperconjugation and inductive effect (+I). In contrast, -NO₂, -Cl, and -COOH are deactivating groups, which decrease the electron density of the ring. Therefore, toluene undergoes nitration at the highest rate.

hard

6. Benzene is subjected to the following sequence of reactions: Benzene + CH₃COCl (in presence of anhydrous AlCl₃) → Compound X Compound X + HNO₃ / H₂SO₄ (Δ) → Compound Y Compound Y is:

  1. m-Nitroacetophenone
  2. o-Nitroacetophenone
  3. p-Nitroacetophenone
  4. m-Dinitrobenzene
Show answer and explanation

Answer: (A) m-Nitroacetophenone

The first step is Friedel-Crafts acylation of benzene to produce acetophenone (Compound X). The acetyl group (-COCH₃) is an electron-withdrawing carbonyl group that deactivates the aromatic ring and directs incoming electrophiles to the meta position. Nitration of acetophenone therefore gives m-nitroacetophenone as the major product (Compound Y).

medium

7. When 4-chlorotoluene is subjected to monobromination in the presence of FeBr₃, the major product obtained is:

  1. 2-Bromo-4-chlorotoluene (methyl directs)
  2. 1-Bromo-4-chlorotoluene
  3. 3-Bromo-4-chlorotoluene (chloro directs)
  4. α-Bromotoluene
Show answer and explanation

Answer: (A) 2-Bromo-4-chlorotoluene (methyl directs)

Both -CH₃ and -Cl are ortho/para directors. When two groups oppose each other, the strongly activating or activating group (-CH₃ via hyperconjugation/+I) dominates over the deactivating halogen (-Cl via -I). Thus, substitution occurs ortho to the methyl group, yielding 2-bromo-4-chlorotoluene.

medium

8. Given below are two statements: Statement I: Direct fluorination of benzene with F₂ gas is violently exothermic and usually not carried out directly. Statement II: Direct iodination of benzene with I₂ is reversible and requires the presence of an oxidizing agent like HNO₃ or HIO₃ to oxidize HI. In light of the above statements, choose the correct answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Both statements are correct. Direct fluorination is extremely vigorous and explosive due to the high reactivity of F₂, so it is not used directly. Iodination is reversible because HI is a strong reducing agent; an oxidizing agent (HNO₃ or HIO₃) is added to oxidize HI to I₂ and drive the reaction forward.

hard

9. During the nitration of benzene, when benzene is replaced by hexadeuterobenzene (C₆D₆), the observed rate of nitration is virtually identical to that of C₆H₆ (k_H / k_D ≈ 1.0). What does this kinetic isotope effect indicate regarding the reaction mechanism?

  1. Carbon-hydrogen bond cleavage occurs in the rate-determining step
  2. Proton loss from the sigma complex is slower than electrophilic attack
  3. The reaction proceeds via a free-radical intermediate
  4. Formation of the arenium ion (sigma complex) is the rate-determining step
Show answer and explanation

Answer: (D) Formation of the arenium ion (sigma complex) is the rate-determining step

The absence of a primary kinetic isotope effect (k_H / k_D ≈ 1) indicates that C-H (or C-D) bond breaking is not involved in the rate-determining step. Instead, the first step—attack of the nitronium ion (NO₂⁺) on the benzene ring to form the arenium ion (sigma complex)—is the slow, rate-determining step, followed by a rapid loss of a proton/deuteron.

medium

10. In Friedel-Crafts alkylation of benzene with CH₃Cl, the catalyst used is:

  1. Anhydrous AlCl₃
  2. Concentrated H₂SO₄
  3. Ni
  4. Zn/HCl
Show answer and explanation

Answer: (A) Anhydrous AlCl₃

Anhydrous AlCl₃ (a Lewis acid) helps generate the electrophile CH₃⁺ from CH₃Cl, which attacks the benzene ring to form toluene.

NCERT Class 11

hard

11. Given below are two statements: Statement I: Sulphonation of benzene is a reversible electrophilic aromatic substitution reaction. Statement II: Unlike nitration, sulphonation shows a significant kinetic primary isotope effect (k_H / k_D > 1). In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Sulphonation is reversible because the benzenesulphonic acid easily desulphonates in dilute acid with steam. Since the second step (loss of proton) has a significant activation barrier and is partially rate-determining, it exhibits a primary kinetic isotope effect (k_H / k_D > 1).

hard

12. What is the correct order of decreasing directing power and activation towards electrophilic aromatic substitution among the following meta-directing groups? (I) -NO₂ (II) -CHO (III) -COOH (IV) -CN

  1. IV > I > II > III
  2. III > II > IV > I
  3. I > IV > III > II
  4. II > III > IV > I
Show answer and explanation

Answer: (D) II > III > IV > I

All these are deactivating and meta-directing. Reactivity towards electrophiles decreases as electron deficiency on the attached atom increases: -CHO > -COOH > -CN > -NO₂. Therefore, the least deactivated (most reactive among these meta directors) is -CHO, followed by -COOH, -CN, and the most strongly deactivating is -NO₂.

Keep going with Hydrocarbons. All 379 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Hydrocarbons