Exercises — Aromaticity — Hückel's rule (4n + 2 π electrons); examples (benzene, naphthalene, pyridine, furan, cyclopentadienyl anio

Benzene has 3 C=C double bonds inside the ring. Each contributes $2$ π electrons: $3\times 2=6$. Solve $4n+2=6\Rightarrow 4n=4\Rightarrow n=1$. 6 π electrons, $n=1$. Aromatic.

Test each: $4n+2$ gives $2(n=0),6(n=1),10(n=2),14(n=3)$.

• $2$ → yes ($n=0$)
• $4$ → no ($4=4n$, antiaromatic count)
• $6$ → yes ($n=1$)
• $8$ → no ($4n$)
• $10$ → yes ($n=2$)
• $12$ → no ($12=4\cdot 3$, this is $4n$; the next aromatic number is $14$) Aromatic: 2, 6, 10.

$5$ double bonds $\times 2=10$ π electrons. $4n+2=10\Rightarrow n=2$. Aromatic.

Five-membered ring. Four of the carbons form 2 ring double bonds → $2\times 2=4$ π electrons. The fifth carbon carries the negative charge as a lone pair; to join the flat, conjugated loop it becomes $s{p}^2$ and places that lone pair into a ring-aligned p-orbital → +2. Total $=4+2=6$. $4n+2=6\Rightarrow n=1$. Aromatic.

Four carbons → 2 ring double bonds → $4$ π electrons. Oxygen has two lone pairs. One sits in a p-orbital aligned with the ring (+2); the other stays in-plane (counts 0). Total $=4+2=6$. $n=1$. Aromatic.

Pyridine is benzene with one CH swapped for N. There are 3 ring double bonds → $6$ π electrons → $n=1$, aromatic. The nitrogen lone pair sits in an in-plane $s{p}^2$ orbital, perpendicular to the p-system, so it contributes 0 to the π count and is free to act as a base. 6 π electrons; the N lone pair is NOT in the sextet.

Flat count: $4$ double bonds $\times 2=8$ π electrons. $8=4n$ with $n=2$ — a $4n$ count, so a planar ring would be antiaromatic (destabilised, open-shell). To escape this penalty the molecule puckers into a non-planar "tub", breaking the planarity condition. With conjugation broken it is non-aromatic and reacts like an ordinary polyene. Predicted flat: antiaromatic (8 = 4n). Actual: non-aromatic tub.

• Pyrrole ($C_4{H}_{5}N$): the four carbons give $2$ double bonds ($4$ π e⁻); the ring needs $2$ more to reach the aromatic sextet, so the N lone pair is forced into a p-orbital and becomes part of the 6 π electrons. Donating it to a proton would destroy aromaticity → very weak base.
• Pyridine: reaches 6 π electrons from ring double bonds alone; the N lone pair stays in an in-plane $s{p}^2$ orbital, available for bonding → much stronger base. Pyridine is the stronger base; its lone pair is "spare," pyrrole's is "spent."

Cyclic ✓, planar (3 atoms are always coplanar) ✓, and every carbon offers a p-orbital (two in the double bond, one empty on C⁺) → fully conjugated ✓. π electrons: one double bond $=2$; the empty p-orbital contributes 0. Total $=2$. $4n+2=2\Rightarrow n=0$. Aromatic — the smallest aromatic system.

Removing one proton from the $C{H}_2$ carbon leaves a carbanion whose lone pair drops into a ring p-orbital. That gives the cyclopentadienyl anion with $4+2=6$ π electrons → aromatic (Problem 2.1). Deprotonation is unusually favourable because the conjugate base is stabilised by aromaticity, so the acid is unusually strong for a C–H bond. Aromatic stabilisation of the anion drives the low $\mathrm{p}{K}_a$. (See Acidity and basicity.)

Seven-membered ring, all $s{p}^2$, planar, fully conjugated. Three double bonds $\times 2=6$ π electrons; the C⁺ contributes an empty p-orbital (0 electrons). Total $=6$. $4n+2=6\Rightarrow n=1$. Aromatic — this is why tropylium salts are surprisingly stable.

Two ring double bonds $=4$ π electrons; the C⁺ carries an empty p-orbital (0). Total $=4$. $4=4n$ with $n=1$ → a $4n$ count. A planar, conjugated ring with $4n$ π electrons is antiaromatic (destabilised). Antiaromatic — the opposite of the anion. This is exactly why removing electrons (cation) and adding electrons (anion) from cyclopentadienyl land on opposite classifications.

Fourteen carbons, each $s{p}^2$ with a p-orbital, planar and fully conjugated. Seven double bonds $\times 2=14$ π electrons. $4n+2=14\Rightarrow 4n=12\Rightarrow n=3$. Aromatic (a polycyclic aromatic hydrocarbon). 14 π electrons, $n=3$.

Protonation happens on the in-plane $s{p}^2$ lone pair of pyridine (the one that was not in the π system). Doing so touches nothing in the p-orbital loop. The ring still has its 3 double bonds → 6 π electrons, still planar and conjugated. $n=1$. Still aromatic. Yes — aromaticity is untouched because the consumed lone pair was never part of the sextet.

Count: a fully conjugated $C_{10}{H}_{10}$ ring has 5 double bonds $\to 10$ π electrons $=4n+2$ with $n=2$ — so the count alone predicts aromatic. Failure: to be planar, the inner hydrogens of the all-cis or trans ring clash sterically (severe angle strain / H–H repulsion), so the ring cannot stay flat. Breaking planarity breaks condition 2 → conjugation is disrupted → effectively non-aromatic. The magic number is necessary but not sufficient — geometry (planarity) must also hold.

$4n+2=18\Rightarrow 4n=16\Rightarrow n=4$. Since $n=4$ is a whole number, 18 is a valid aromatic count (this is the real molecule [18]annulene). $n=4$.