Test: Conjugation and Aromaticity - JEE MCQ

Compounds b, c, d, e all have (4n + 2) pie electrons hence all compounds listed are aromatic.

The most stable resonating structure is the one with no charge separation. This corresponds to P.
This is followed by Q as it has an octet of all atoms complete. None of them is electron deficient.
The resonating structure with a positive charge on a relatively electropositive atom and negative charge on the electronegative atom is more stable than the reverse case.
Order of stability is:-
P > Q > R > S

N has no d orbital, hence it can't accommodate more than 8 electrons in its valence shell.

The order of stability of resonating structures carrying no charge > carrying minimum charge and each atom having octet complete.

1. Option (a): Benzene (C₆H₆)

   • Benzene is a stable aromatic compound with no acidic hydrogen that can be easily removed. Benzene itself is very weakly acidic and does not have a significant tendency to donate a proton.

2. Option (b): 1,2-Dihydronaphthalene

   • This structure contains a naphthalene ring with one ring partially saturated (a six-membered ring with only one double bond). It’s slightly more reactive than benzene but does not have a particularly acidic proton.

3. Option (c): Indene

   • Indene has a fused benzene ring and a five-membered ring. It has a slightly acidic hydrogen on the carbon adjacent to the benzene ring (on the methylene bridge between the rings). Removal of this proton would lead to an aromatic stabilized anion due to resonance with the benzene ring, making it relatively more acidic.

4. Option (d): Fluorene

   • Fluorene has a structure with a benzene ring fused to another benzene ring through a five-membered ring. The hydrogen on the carbon between the two benzene rings (the bridgehead position in the five-membered ring) is relatively acidic. Deprotonation of this hydrogen would lead to a stable aromatic anion with resonance stabilization across the two benzene rings.

Conclusion: The most acidic compound here is fluorene (option d), because the conjugate base formed after deprotonation is stabilized by resonance across both benzene rings, making it more stable than the conjugate bases of the other compounds.

An aromatic compound has (4n + 2)π-electons, where n is an integer. Cyclopentadiene has only 4π electrons and thus will be nonaromatic.
no conjugation present in cyclopentadiene

Electron withdrawing −NO₂ substituent has an acid-strengthening effect while the electron releasing −CH₃ substituent has an acid-weakening effect on phenol. Thus, p-cresol is the weakest acid amongst the given compounds. The m - and p-nitrophenols are stronger acid than phenol (p-isomer is more strong than m-isomers). The given choice (b) satisfies this requirement.

Electron-releasing group decreases the acidity while electron-attracting group increases the acidity. Thus, the molecule III will have maximum acidity and the molecule IV will have minimum acidity. Thus, the choice (c) is correct.

(i) NO₂ group increases the acidic nature due to - I effect.
(ii) OCH₃ and CH₃ are electron donating groups (i.e., +I effect), thus decreases the acidic nature, but OCH₃ will decrease acidic nature more than of CH₃ group Hence, correct order is III < IV < I < II

A molecule is said to be aromatic if the following conditions are satisfied.

• An aromatic molecule must be cyclic.
• An aromatic molecule must be planar.
• An aromatic ring must contain only sp2-hybridized atoms that can form a delocalised system of π molecular orbitals.
• The number of π electrons in the delocalised π system must equal 4n + 2, where n is an integer.
• The “4n + 2 rule” was proposed by E. Huckel, and is known as the Hückel rule

Aromatic compounds are more stable compounds among the conjugated compounds.