Q.Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?

Concept and Intuition

Delocalization of $\pi$ electrons happens through resonance across a conjugated $\pi$ system. Delocalization of $\sigma$ electrons happens only through hyperconjugation — a $\sigma$(C–H) bond adjacent to a $\pi$ system (or a carbocation/radical) overlapping with it. So the key discriminator is: does the species have a C–H bond on an sp$^3$ carbon directly attached to the conjugated system?

Step-by-Step Solution

1. But-2-ene ($CH_3$–CH=CH–$CH_3$): the C=C is a $\pi$ system, and each terminal $CH_3$ has C–H $\sigma$-bonds adjacent to it → allylic hyperconjugation. Both $\sigma$ and $\pi$ delocalized.
2. Tropylium cation ($C_7H_7^+$): a fully conjugated 7-membered ring, all carbons sp$^2$, 6 $\pi$ electrons delocalized over all 7 p-orbitals (aromatic). No adjacent sp$^3$ C–H exists for hyperconjugation — only $\pi$ delocalized.
3. Nitrobenzene: the $-NO_2$ group conjugates with the ring via resonance (its lone pair/π system merges with the ring π cloud). No sp$^3$ C–H is involved anywhere — only $\pi$ delocalized. …

Concept and Intuition

Aromaticity is not restricted to benzene-derived (benzenoid) rings. Certain non-six-membered ring systems can also satisfy Huckel's rule (4n+2 pi electrons in a cyclic, planar, conjugated system) and display aromatic stability. Tropolone (2-hydroxy-2,4,6-cycloheptatrien-1-one) is a classic example: it is a seven-membered ring related to the tropylium cation system, exhibiting aromatic character without containing a benzene ring -- making it a non-benzenoid aromatic compound.

Step-by-Step Solution

1. Recall the structure of tropolone: a seven-membered ring with alternating double bonds, a ketone, and a hydroxyl group, related to cycloheptatriene.
2. Recognize that its pi-electron system can be described in terms of an aromatic tropylium-like cation stabilized by the adjacent oxygen functionalities, giving it aromatic character. …

Concept and Intuition

Aromaticity requires a cyclic, planar (or near-planar over local rings), conjugated system with delocalized $\pi$ electrons obeying Hückel-type stabilization. Diamond is purely $sp^3$-hybridized with no $\pi$ system at all. Graphite has delocalized electrons but is an extended 2D sheet, not typically described as 'aromatic' in the classic molecular sense, though it does have conjugation. Coke is impure, largely amorphous carbon with no ordered conjugated framework. Buckminsterfullerene, $C_{60}$, consists of 12 pentagons and 20 hexagons of $sp^2$ carbon atoms with a continuous conjugated $\pi$ system over the whole cage — it is explicitly described in NCERT/standard texts as having aromatic character.

Step-by-Step Solution

1. Diamond: all carbons $sp^3$, no $\pi$ electrons → not aromatic.
2. Graphite: $sp^2$ layers with delocalized electrons, but it's an infinite planar sheet, not a discrete aromatic molecule. …

Concept and Intuition

Cyclic organic compounds split into aromatic (planar, conjugated, obeying Hückel's rule, e.g. benzene derivatives and pyridine) and alicyclic (aliphatic-type rings — saturated or unsaturated but without aromatic character). Recognizing which rings are aromatic versus simply cyclic aliphatic structures is the key skill tested here.

Step-by-Step Solution

1. Cyclohexene: a six-membered carbocyclic ring with one C=C double bond, not aromatic — alicyclic.
2. Anisole: methoxybenzene — an aromatic benzene ring with an –OCH$_3$ substituent — not alicyclic.
3. Pyridine: an aromatic six-membered N-heterocycle (isoelectronic with benzene) — not alicyclic.
4. Tetrahydrofuran: a fully saturated five-membered cyclic ether, non-aromatic — alicyclic. …

Why Aromaticity? The Core Idea

Aromatic compounds are unusually stable because their π electrons are delocalised around a closed loop. For a monocyclic, planar molecule to be aromatic, it must have a continuous ring of overlapping p orbitals (full conjugation) and exactly $4n+2$ π electrons (where $n = 0,1,2,\dots$). If the ring is not fully conjugated, or if the π count is $4n$, the molecule is antiaromatic or non‑aromatic. Here we test each structure against these criteria.

Step‑by‑Step Analysis

1. Option (A) – Cyclopentadienyl anion

• Structure: A five‑membered ring with two double bonds and a negative charge on one carbon.
• Conjugation: The negative charge means that carbon has a lone pair in a p orbital. All five ring carbons are sp² hybridised, so the p orbitals form a continuous loop.
• π electron count: Each double bond contributes 2 π electrons → $2 \times 2 = 4$. The lone pair on the anionic carbon contributes 2 more π electrons. Total = $6$ π electrons.
• Hückel check: $4n+2 = 6$ gives $n=1$. Planar, cyclic, fully conjugated → aromatic.
• Result: Aromatic.

2. Option (B) – Tropylium cation (cycloheptatrienyl cation)

• Structure: A seven‑membered ring with three double bonds and a positive charge on one carbon.
• Conjugation: All seven carbons are sp² hybridised; the positive charge means the charged carbon has an empty p orbital. The p orbitals overlap all around the ring.
• π electron count: Three double bonds → $3 \times 2 = 6$ π electrons. The empty p orbital contributes 0. Total = $6$ π electrons.
• Hückel check: $4n+2 = 6$ → $n=1$. Planar, cyclic, fully conjugated → aromatic.
• Result: Aromatic.

3. Option (C) – Cyclopropenyl cation

• Structure: A three‑membered ring with one double bond and a positive charge on one carbon.
• Conjugation: All three carbons are sp² hybridised; the charged carbon has an empty p orbital. The three p orbitals overlap in a triangle.
• π electron count: One double bond → 2 π electrons. The empty p orbital contributes 0. Total = $2$ π electrons.
• Hückel check: $4n+2 = 2$ gives $n=0$. Planar, cyclic, fully conjugated → aromatic (the smallest aromatic system).
• Result: Aromatic.

4. Option (D) – Cycloheptatriene (neutral, with only two double bonds) …

Concept and Intuition

A ring system is aromatic if it is (i) cyclic, (ii) planar, (iii) fully conjugated (every ring atom has a p-orbital), and (iv) has $4n+2$ pi electrons delocalized in that ring (Huckel's rule, $n = 0,1,2,\dots$). Getting exactly $4n$ pi electrons in such a system instead makes it antiaromatic (destabilized), not merely "non-aromatic."

Step-by-Step Solution

1. Pyridine: six-membered ring, each carbon and the nitrogen contribute one p-electron each to the ring pi-system = 6pi electrons ($n=1$); nitrogen's lone pair sits in an $sp^2$ orbital in the plane of the ring (not part of the pi system). Aromatic.
2. Cyclopentadienyl cation ($\mathrm{C_5H_5^+}$): five-membered ring with only 4 pi electrons (one fewer than the aromatic cyclopentadienyl anion, which has 6pi and is famously aromatic). $4\pi$ electrons = Huckel's $4n$ pattern ($n=1$) leads to antiaromatic, i.e. NOT aromatic (in fact destabilized relative to a non-cyclic reference).
3. Anthracene: three linearly fused benzene rings, fully conjugated planar polycyclic aromatic hydrocarbon with $14$ pi electrons overall -- aromatic (each individual ring, and the whole system, is stabilized). …

Concept and Intuition

Counting hydrogens/degrees of unsaturation is a quick way to distinguish an aromatic ring from a partially saturated one bearing the same substituents. Aromaticity additionally requires a cyclic, planar, fully conjugated π system (Hückel's rule) — merely having double bonds in a ring is not sufficient if they are not conjugated around the whole ring.

Step-by-Step Solution

1. Benzene is $C_6H_6$; a 1,4-dibromobenzene (aromatic) would be $C_6H_4Br_2$.
2. The given compound is $C_6H_6Br_2$ — TWO more hydrogens than the aromatic dibromobenzene, meaning the ring is not fully unsaturated/aromatic; it corresponds to a dibromo-substituted cyclohexadiene (only 2 ring double bonds, not benzene's 3 alternating ones).
3. The problem explicitly states "two non-conjugate double bonds," confirming a 1,4-cyclohexadiene-type skeleton, not benzene's continuous conjugated system. …

Concept and Intuition: Hückel's Rule

Aromaticity is a special stability found in certain cyclic, planar molecules with a fully conjugated π-electron system. The key test is Hückel's rule: a planar, monocyclic, fully conjugated ring is aromatic if it contains 4n+2 π electrons (n = 0, 1, 2, …). Rings with 4n π electrons are antiaromatic. For each compound, check: (1) Is it cyclic and planar? (2) Is every ring atom sp²-hybridized so the p-orbitals overlap continuously? (3) Does it have 4n+2 π electrons?

Step-by-Step Analysis

1. Furan (i): Cyclic, planar, all atoms sp². Oxygen contributes 2 π electrons from one lone pair (its other lone pair lies in the ring plane and is not part of the π system); the two double bonds contribute 2 π electrons each. Total = 2+2+2 = 6 π electrons. 6 = 4(1)+2 → aromatic. ✓

2. Cyclobutadiene (ii): Cyclic, planar, two double bonds → 4 π electrons. 4 = 4(1) → antiaromatic. ✗

3. Cyclopropenyl anion (iii): Cyclic, planar. The double bond contributes 2 π electrons; the negatively charged sp²-carbon contributes 2 more from its lone pair. Total = 4 π electrons → antiaromatic (note: the cyclopropenyl cation, with 2 π electrons, is aromatic — but the anion is not). ✗ …