CHEM 332 Exam 4 Version A Questions
And Answers With Rationales/ Graded
A+/2026 Update
Section 1: Aromaticity & Benzene Derivatives (Questions 1-15)
1. Which of the following compounds is aromatic?
a) Cyclooctatetraene
b) Cyclopentadienyl cation
c) Cyclopropenyl anion
d) Cyclobutadiene
Answer: c) Cyclopropenyl anion
Rationale: Aromaticity requires a cyclic, planar, fully conjugated system with 4n+24n+2 π electrons
(Hückel's rule). Cyclopropenyl anion has 2 π electrons (n=0n=0), is cyclic, planar, and conjugated, making
it aromatic. Cyclooctatetraene is tub-shaped (non-planar) and non-aromatic. Cyclopentadienyl cation
has 4 π electrons (anti-aromatic). Cyclobutadiene is anti-aromatic.
2. What is the major product of the nitration of benzoic acid?
a) Nitrobenzene
b) m-Nitrobenzoic acid
c) o-Nitrobenzoic acid
d) p-Nitrobenzoic acid
Answer: b) m-Nitrobenzoic acid
Rationale: The carboxylic acid group is a strong deactivating, meta-directing group. The electrophile
(NO2+2+) will attack the meta position relative to the -COOH group.
,3. Which of the following groups is ortho/para-directing and deactivating?
a) -NO22
b) -CF33
c) -Cl
d) -COOH
Answer: c) -Cl
Rationale: Halogens are unique: they are deactivating due to their strong -I effect, but they
are ortho/para-directing due to resonance donation of lone pairs into the ring.
4. Which of the following best describes the product of Friedel-Crafts acylation of anisole?
a) p-Methoxyacetophenone (major) and o-Methoxyacetophenone (minor)
b) m-Methoxyacetophenone
c) p-Methylanisole
d) No reaction occurs.
Answer: a) p-Methoxyacetophenone (major) and o-Methoxyacetophenone (minor)
Rationale: The methoxy group is a strong activating, ortho/para-directing group. Steric hindrance makes
the para product the major one.
5. Which compound has the highest boiling point?
a) Toluene
b) Ethylbenzene
c) Phenol
d) Benzene
Answer: c) Phenol
Rationale: Phenol can form strong intermolecular hydrogen bonds, significantly raising its boiling point
compared to alkylbenzenes which only exhibit London dispersion forces.
6. Which of the following is anti-aromatic?
a) Benzene
b) Cyclooctatetraene dianion
c) Cyclobutadiene
d) Pyridine
Answer: c) Cyclobutadiene
Rationale: Cyclobutadiene is cyclic, planar, and conjugated, but has 4 π electrons, fitting the 4n4n rule
(n=1n=1) for anti-aromaticity. The cyclooctatetraene dianion has 10 π electrons (n=2n=2) and is
aromatic.
7. What is the IUPAC name for the compound commonly known as TNT?
a) 2,4,6-Trinitrotoluene
, b) 2,4,6-Trinitrophenol
c) 1,3,5-Trinitrobenzene
d) 2,4-Dinitrotoluene
Answer: a) 2,4,6-Trinitrotoluene
Rationale: TNT is methylbenzene (toluene) with three nitro groups at the 2, 4, and 6 positions.
8. In electrophilic aromatic substitution, the rate-determining step is:
a) Formation of the σ-complex (Wheland intermediate)
b) Deprotonation to restore aromaticity
c) Formation of the electrophile
d) Loss of the leaving group
Answer: a) Formation of the σ-complex (Wheland intermediate)
Rationale: The attack of the electrophile on the aromatic ring to form the non-aromatic, high-energy σ-
complex is the slowest, highest-energy step.
9. Which reagent would be used to reduce a nitro group to an amino group?
a) LiAlH44
b) NaBH44
c) H22, Pd/C
d) KMnO44
Answer: c) H22, Pd/C
Rationale: Catalytic hydrogenation (H22 with a metal catalyst like Pd, Pt, or Ni) or Sn/HCl effectively
reduces -NO22 to -NH22.
10. Which of the following is a heterocyclic aromatic compound?
a) Cyclohexane
b) Aniline
c) Pyridine
d) Styrene
Answer: c) Pyridine
Rationale: Pyridine is a six-membered aromatic ring containing one nitrogen atom. The nitrogen’s lone
pair is in an sp22 orbital and does not participate in the aromatic sextet.
11. In the sulfonation of benzene, the electrophile is:
a) SO33
b) H22SO44
c) SO33H++
d) SO22
And Answers With Rationales/ Graded
A+/2026 Update
Section 1: Aromaticity & Benzene Derivatives (Questions 1-15)
1. Which of the following compounds is aromatic?
a) Cyclooctatetraene
b) Cyclopentadienyl cation
c) Cyclopropenyl anion
d) Cyclobutadiene
Answer: c) Cyclopropenyl anion
Rationale: Aromaticity requires a cyclic, planar, fully conjugated system with 4n+24n+2 π electrons
(Hückel's rule). Cyclopropenyl anion has 2 π electrons (n=0n=0), is cyclic, planar, and conjugated, making
it aromatic. Cyclooctatetraene is tub-shaped (non-planar) and non-aromatic. Cyclopentadienyl cation
has 4 π electrons (anti-aromatic). Cyclobutadiene is anti-aromatic.
2. What is the major product of the nitration of benzoic acid?
a) Nitrobenzene
b) m-Nitrobenzoic acid
c) o-Nitrobenzoic acid
d) p-Nitrobenzoic acid
Answer: b) m-Nitrobenzoic acid
Rationale: The carboxylic acid group is a strong deactivating, meta-directing group. The electrophile
(NO2+2+) will attack the meta position relative to the -COOH group.
,3. Which of the following groups is ortho/para-directing and deactivating?
a) -NO22
b) -CF33
c) -Cl
d) -COOH
Answer: c) -Cl
Rationale: Halogens are unique: they are deactivating due to their strong -I effect, but they
are ortho/para-directing due to resonance donation of lone pairs into the ring.
4. Which of the following best describes the product of Friedel-Crafts acylation of anisole?
a) p-Methoxyacetophenone (major) and o-Methoxyacetophenone (minor)
b) m-Methoxyacetophenone
c) p-Methylanisole
d) No reaction occurs.
Answer: a) p-Methoxyacetophenone (major) and o-Methoxyacetophenone (minor)
Rationale: The methoxy group is a strong activating, ortho/para-directing group. Steric hindrance makes
the para product the major one.
5. Which compound has the highest boiling point?
a) Toluene
b) Ethylbenzene
c) Phenol
d) Benzene
Answer: c) Phenol
Rationale: Phenol can form strong intermolecular hydrogen bonds, significantly raising its boiling point
compared to alkylbenzenes which only exhibit London dispersion forces.
6. Which of the following is anti-aromatic?
a) Benzene
b) Cyclooctatetraene dianion
c) Cyclobutadiene
d) Pyridine
Answer: c) Cyclobutadiene
Rationale: Cyclobutadiene is cyclic, planar, and conjugated, but has 4 π electrons, fitting the 4n4n rule
(n=1n=1) for anti-aromaticity. The cyclooctatetraene dianion has 10 π electrons (n=2n=2) and is
aromatic.
7. What is the IUPAC name for the compound commonly known as TNT?
a) 2,4,6-Trinitrotoluene
, b) 2,4,6-Trinitrophenol
c) 1,3,5-Trinitrobenzene
d) 2,4-Dinitrotoluene
Answer: a) 2,4,6-Trinitrotoluene
Rationale: TNT is methylbenzene (toluene) with three nitro groups at the 2, 4, and 6 positions.
8. In electrophilic aromatic substitution, the rate-determining step is:
a) Formation of the σ-complex (Wheland intermediate)
b) Deprotonation to restore aromaticity
c) Formation of the electrophile
d) Loss of the leaving group
Answer: a) Formation of the σ-complex (Wheland intermediate)
Rationale: The attack of the electrophile on the aromatic ring to form the non-aromatic, high-energy σ-
complex is the slowest, highest-energy step.
9. Which reagent would be used to reduce a nitro group to an amino group?
a) LiAlH44
b) NaBH44
c) H22, Pd/C
d) KMnO44
Answer: c) H22, Pd/C
Rationale: Catalytic hydrogenation (H22 with a metal catalyst like Pd, Pt, or Ni) or Sn/HCl effectively
reduces -NO22 to -NH22.
10. Which of the following is a heterocyclic aromatic compound?
a) Cyclohexane
b) Aniline
c) Pyridine
d) Styrene
Answer: c) Pyridine
Rationale: Pyridine is a six-membered aromatic ring containing one nitrogen atom. The nitrogen’s lone
pair is in an sp22 orbital and does not participate in the aromatic sextet.
11. In the sulfonation of benzene, the electrophile is:
a) SO33
b) H22SO44
c) SO33H++
d) SO22
