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Match each compound in List-I with its correct aromaticity description in List-II. List-I: (P) Five-membered ring with two double bonds and an NH group in the ring (pyrrole-type) (Q) Five-membered ring with two double bonds and an NH2+ group in the ring (pyrrolium-type) (R) Five-membered ring with two double bonds and a C=O group in the ring (cyclopentadienone-type) (S) Seven-membered ring with three double bonds and a positive charge (tropylium cation) List-II: (1) Aromatic (2) Non-aromatic (3) Antiaromatic (unstable at room temperature) (4) Homocyclic

1. P -> 1; Q -> 2; R -> 3; S -> 1,2,4
2. P -> 1; Q -> 2; R -> 3,4; S -> 1,4
3. P -> 1,2; Q -> 2; R -> 3; S -> 1,4
4. P -> 1; Q -> 2; R -> 3,4; S -> 1,2,3,4

Correct answer: P -> 1; Q -> 2; R -> 3,4; S -> 1,4

Solution

P (pyrrole-type): The N lone pair is delocalised into the ring, giving 6 pi electrons (4n+2, n=1) — aromatic (1). Q (pyrrolium-type, NH2+): The NH2+ group cannot donate a lone pair into the ring (positive nitrogen, lone pair used for bonding to H), leaving only 4 pi electrons from 2 double bonds — considered non-aromatic (2) due to disrupted pi system. R (cyclopentadienone): C=O exocyclic pulls electron density; ring has 4 pi electrons (4n, n=1) — antiaromatic (3); and it is homocyclic (all ring atoms are carbon) (4). S (tropylium cation, C7H7+): 6 pi electrons (4n+2, n=1) — aromatic (1); all ring atoms are carbon — homocyclic (4). Answer: P->1; Q->2; R->3,4; S->1,4.

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