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CHEM 219 Module 5 Exam: Aromatic Compounds 2026/2027 UPDATE

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CHEM 219 Module 5 Exam: Aromatic Compounds 2026/2027 UPDATE

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CHEM 219 Module 5 Exam: Aromatic Compounds 2026/2027 UPDATE


1. What is the hybridization of all carbon atoms in the benzene ring?

A. sp3

B. sp

C. sp2

D. dsp2

Answer: C
Rationale: In benzene, each carbon is bonded to two other carbons and one hydrogen,
forming a planar hexagonal structure with sp2 hybridization.

2. According to Hückel’s Rule, a compound is aromatic if it is cyclic, planar, fully
conjugated, and contains how many pi electrons?

A. 4n

B. 4n + 2

C. 2n + 2

D. 3n + 1

Answer: B
Rationale: Hückel’s rule states that an aromatic system must have (4n + 2) pi electrons,
where n is a non-negative integer.

3. Which of the following describes the bond lengths in benzene?

A. Alternating long single and short double bonds

B. All bonds are of equal length, between a single and double bond

C. All bonds are equal to standard C=C double bonds

D. Three long bonds and three short bonds

Answer: B

,Rationale: Due to resonance delocalization, all C-C bonds in benzene are identical in length
(1.39 Å), which is intermediate between a single and double bond.

4. What is the common name for hydroxybenzene?

A. Aniline

B. Phenol

C. Toluene

D. Anisole

Answer: B
Rationale: Phenol is the IUPAC-accepted common name for a benzene ring with a hydroxyl
(-OH) group.

5. Which reagent is typically used for the nitration of benzene?

A. NaNO2 and HCl

B. HNO3 and H2SO4

C. NH3 and KMnO4

D. HNO2 and H2O

Answer: B
Rationale: Nitration requires a mixture of concentrated nitric acid and sulfuric acid to
generate the nitronium ion (NO2+) electrophile.

6. In the chlorination of benzene, what is the purpose of adding FeCl3?

A. It acts as a solvent

B. It stabilizes the benzene ring

C. It acts as a base to remove a proton

D. It acts as a Lewis acid catalyst to generate the Cl+ electrophile

Answer: D
Rationale: FeCl3 is a Lewis acid that reacts with Cl2 to create a more powerful
electrophilic species capable of attacking the stable aromatic ring.

, 7. Which of the following substituents is a strong activator and an ortho/para
director?

A. -NO2

B. -Cl

C. -NH2

D. -COOH

Answer: C
Rationale: The amino group (-NH2) has a lone pair that can be donated into the ring by
resonance, making it a strong activator and directing substitution to the ortho and para
positions.

8. What is the electrophile in the Friedel-Crafts alkylation reaction?

A. A carbocation

B. An acylium ion

C. A halogen atom

D. A nitronium ion

Answer: A
Rationale: In Friedel-Crafts alkylation, an alkyl halide reacts with a Lewis acid (like AlCl3)
to form a carbocation, which serves as the electrophile.

9. Which molecule is considered ‘anti-aromatic’?

A. Cyclobutadiene

B. Pyridine

C. Benzene

D. Naphthalene

Answer: A
Rationale: Cyclobutadiene is cyclic, planar, and conjugated but has 4 pi electrons (4n
where n=1), making it highly unstable and anti-aromatic.

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