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.

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.