Organic Chemistry II: Synthesis & Spectroscopy Exam 2026:
Comprehensive University-Level Practice Examination — 200
Multiple-Choice Questions with Answers & Detailed Rationales
SECTION 1: AROMATICITY, ELECTROPHILIC AROMATIC SUBSTITUTION, AND DIRECTING
EFFECTS (Q1–Q30)
Question 1
Which of the following species is aromatic according to Hückel's rule?
A. Cyclohexane
B. Cyclopentadienyl anion
C. Cyclobutadiene
D. Cyclohexene
Correct Answer: B
Explanation: The cyclopentadienyl anion has 6 π-electrons in a planar, cyclic, fully conjugated
system (4n+2, n=1), satisfying Hückel's rule; the others are not fully conjugated or lack the
correct count.
Question 2
The pyridine nitrogen lone pair resides in which orbital, and how does this affect basicity?
A. An sp³ orbital in the ring plane; strongly basic
B. An unhybridized p orbital in the π system; non-basic
C. An sp² orbital in the ring plane; weakly basic
D. A d orbital; variable basicity
Correct Answer: C
,Explanation: Pyridine's nitrogen is sp²-hybridized with its lone pair in the σ-plane, orthogonal
to the aromatic π system, so it is available for protonation but is weakly basic (pKaH ≈ 5.2).
Question 3
A student nitrates anisole and isolates predominantly para and ortho products. This
regiochemistry is best explained by:
A. Steric hindrance of the meta position only
B. Resonance donation by the methoxy group stabilizing the arenium ion at o/p positions
C. Inductive electron withdrawal by oxygen
D. Formation of a benzyne intermediate
Correct Answer: B
Explanation: The –OCH₃ group donates electron density by resonance, stabilizing the
intermediate cation at ortho and para positions, directing substitution there (with para
favored sterically).
Question 4
Which substituent is meta-directing and deactivating?
A. –NH₂
B. –CH₃
C. –NO₂
D. –OH
Correct Answer: C
,Explanation: The nitro group is strongly electron-withdrawing by both induction and
resonance, deactivating the ring and directing meta by destabilizing o/p arenium ions.
Question 5
In the Friedel–Crafts acylation of benzene with acetyl chloride, what is the essential role of
AlCl₃?
A. It donates a hydride to the ring
B. It abstracts chloride to generate the acylium electrophile
C. It acts as a radical initiator
D. It protonates the aromatic ring
Correct Answer: B
Explanation: AlCl₃ is a Lewis acid that coordinates chloride, generating the resonance-
stabilized acylium ion (CH₃C≡O⁺), the true electrophile.
Question 6
A researcher attempts Friedel–Crafts alkylation on nitrobenzene. The reaction fails primarily
because:
A. Nitrobenzene is too electron-rich
B. The nitro group strongly deactivates the ring toward electrophiles
C. AlCl₃ destroys the nitro group
D. The reaction requires UV light
Correct Answer: B
, Explanation: Strongly deactivating groups like –NO₂ render the ring insufficiently
nucleophilic; Friedel–Crafts reactions fail on strongly deactivated arenes.
Question 7
Which of the following is the correct order of reactivity toward electrophilic aromatic
substitution (most to least reactive)?
A. Toluene > benzene > chlorobenzene > nitrobenzene
B. Nitrobenzene > chlorobenzene > benzene > toluene
C. Benzene > toluene > nitrobenzene > chlorobenzene
D. Chlorobenzene > toluene > benzene > nitrobenzene
Correct Answer: A
Explanation: Activating groups (CH₃) increase reactivity; halogens weakly deactivate; nitro
strongly deactivates—giving toluene > benzene > chlorobenzene > nitrobenzene.
Question 8
In the sulfonation of benzene, the reaction is reversible. Which condition favors
desulfonation?
A. Cold concentrated H₂SO₄
B. Hot dilute aqueous acid
C. Excess SO₃
D. Anhydrous AlCl₃
Correct Answer: B
Comprehensive University-Level Practice Examination — 200
Multiple-Choice Questions with Answers & Detailed Rationales
SECTION 1: AROMATICITY, ELECTROPHILIC AROMATIC SUBSTITUTION, AND DIRECTING
EFFECTS (Q1–Q30)
Question 1
Which of the following species is aromatic according to Hückel's rule?
A. Cyclohexane
B. Cyclopentadienyl anion
C. Cyclobutadiene
D. Cyclohexene
Correct Answer: B
Explanation: The cyclopentadienyl anion has 6 π-electrons in a planar, cyclic, fully conjugated
system (4n+2, n=1), satisfying Hückel's rule; the others are not fully conjugated or lack the
correct count.
Question 2
The pyridine nitrogen lone pair resides in which orbital, and how does this affect basicity?
A. An sp³ orbital in the ring plane; strongly basic
B. An unhybridized p orbital in the π system; non-basic
C. An sp² orbital in the ring plane; weakly basic
D. A d orbital; variable basicity
Correct Answer: C
,Explanation: Pyridine's nitrogen is sp²-hybridized with its lone pair in the σ-plane, orthogonal
to the aromatic π system, so it is available for protonation but is weakly basic (pKaH ≈ 5.2).
Question 3
A student nitrates anisole and isolates predominantly para and ortho products. This
regiochemistry is best explained by:
A. Steric hindrance of the meta position only
B. Resonance donation by the methoxy group stabilizing the arenium ion at o/p positions
C. Inductive electron withdrawal by oxygen
D. Formation of a benzyne intermediate
Correct Answer: B
Explanation: The –OCH₃ group donates electron density by resonance, stabilizing the
intermediate cation at ortho and para positions, directing substitution there (with para
favored sterically).
Question 4
Which substituent is meta-directing and deactivating?
A. –NH₂
B. –CH₃
C. –NO₂
D. –OH
Correct Answer: C
,Explanation: The nitro group is strongly electron-withdrawing by both induction and
resonance, deactivating the ring and directing meta by destabilizing o/p arenium ions.
Question 5
In the Friedel–Crafts acylation of benzene with acetyl chloride, what is the essential role of
AlCl₃?
A. It donates a hydride to the ring
B. It abstracts chloride to generate the acylium electrophile
C. It acts as a radical initiator
D. It protonates the aromatic ring
Correct Answer: B
Explanation: AlCl₃ is a Lewis acid that coordinates chloride, generating the resonance-
stabilized acylium ion (CH₃C≡O⁺), the true electrophile.
Question 6
A researcher attempts Friedel–Crafts alkylation on nitrobenzene. The reaction fails primarily
because:
A. Nitrobenzene is too electron-rich
B. The nitro group strongly deactivates the ring toward electrophiles
C. AlCl₃ destroys the nitro group
D. The reaction requires UV light
Correct Answer: B
, Explanation: Strongly deactivating groups like –NO₂ render the ring insufficiently
nucleophilic; Friedel–Crafts reactions fail on strongly deactivated arenes.
Question 7
Which of the following is the correct order of reactivity toward electrophilic aromatic
substitution (most to least reactive)?
A. Toluene > benzene > chlorobenzene > nitrobenzene
B. Nitrobenzene > chlorobenzene > benzene > toluene
C. Benzene > toluene > nitrobenzene > chlorobenzene
D. Chlorobenzene > toluene > benzene > nitrobenzene
Correct Answer: A
Explanation: Activating groups (CH₃) increase reactivity; halogens weakly deactivate; nitro
strongly deactivates—giving toluene > benzene > chlorobenzene > nitrobenzene.
Question 8
In the sulfonation of benzene, the reaction is reversible. Which condition favors
desulfonation?
A. Cold concentrated H₂SO₄
B. Hot dilute aqueous acid
C. Excess SO₃
D. Anhydrous AlCl₃
Correct Answer: B
