Questions: Aromaticity and Benzene

Question 1

The cyclopentadienyl anion (C₅H₅⁻) has five sp2 carbons in a ring. How many pi electrons does it have, and is it aromatic?

• A. 4 pi electrons; antiaromatic
• B. 6 pi electrons; aromatic
• C. 5 pi electrons; non-aromatic
• D. 6 pi electrons; non-aromatic because it carries a charge

Answer: (B)

Each sp2 carbon contributes one p orbital electron, giving 5 from the ring carbons, plus the lone pair on the carbanion adds one more for a total of 6 pi electrons. 6 = 4(1)+2, satisfying Hückel's rule with n=1. The ring is planar and fully conjugated, so it is aromatic. The charge is irrelevant to aromaticity — what matters is the pi electron count and the geometry.

Question 2

When benzene reacts with a halogen in the presence of a Lewis acid catalyst, it undergoes addition rather than substitution, because addition reactions are generally more thermodynamically favorable.

• T. True
• F. False

Answer: False

Benzene overwhelmingly undergoes electrophilic aromatic substitution, not addition. Addition would require converting two of the six sp2 carbons to sp3, which destroys the aromatic pi system and forfeits approximately 36 kcal/mol of resonance stabilization energy. Substitution is favored precisely because it restores aromaticity in the product. The thermodynamic driver is preservation of the aromatic system, not the inherent favorability of addition.

Question 3

Cyclobutadiene (C₄H₄) is a cyclic, fully conjugated ring, yet it is extremely unstable. Why does Hückel's rule predict this?

Model answer: Cyclobutadiene has 4 pi electrons (4n with n=1), satisfying the antiaromatic criterion. Antiaromatic systems experience destabilization from unfavorable electron filling of degenerate orbitals, making them highly reactive and essentially unobservable under normal conditions.

Hückel's rule states that (4n+2) pi electrons confer aromatic stability, while 4n pi electrons confer antiaromatic destabilization. Cyclobutadiene's 4 pi electrons place it in the 4n category (n=1), so instead of extra stabilization it experiences strong destabilization. Molecular orbital theory shows that two electrons must occupy degenerate (equal-energy) orbitals, leading by Hund's rule to a diradical ground state — extremely reactive and unstable.