Q.Show two canonicals of benzene by drawing. Benzene is stored in a bottle. Is there existence of the two canonicals in the benzene of the bottle? Answer with reason. [2 + 1] OR Between (CH₃)₃C–Cl and CH₃–Cl, which compound undergoes heterolytic fission readily in water? Why? [1 + 2]

C6H5NH3+ (the anilinium/phenylammonium ion) shows no resonance effect because nitrogen's lone pair, which would otherwise conjugate with the ring, is used up forming a bond to the extra proton.

Check each species for a lone pair (or pi electrons) available to conjugate with the benzene ring's pi system:

1. C6H5NH2 (aniline): nitrogen has a lone pair that conjugates into the ring, donating electron density by resonance (+M effect) — this is why aniline is a stronger base... actually more relevantly here, why the ring is activated towards electrophilic substitution. Resonance effect present.
2. C6H5OH (phenol): oxygen's lone pair conjugates into the ring similarly (+M effect), activating the ring. Resonance effect present. …

Resonance structures (canonical forms) represent the same molecule/ion with the same arrangement of atoms and same connectivity, differing only in how the electrons (and formal/negative charge) are distributed — exactly what the two given structures show for the enolate anion of acetone.

The two structures given are:

(1) (-)CH2-C(=O)-CH3 — the carbanion form, with the negative charge localised on the terminal CH2 carbon and a C=O double bond to the middle carbon.

(2) CH2=C(-O^-)-CH3 — the enolate form, with a C=C double bond between the same two carbons and the negative charge now on the oxygen.

In both structures, the positions of every atom (the three carbons, the oxygen, and all hydrogens) are exactly the same — only the arrangement of the pi electrons and the location of the formal negative charge differ (delocalised between the carbanion carbon and the oxygen). This delocalisation of electron density over a fixed framework of atoms is precisely what defines resonance (canonical) structures — they are not two different, isolable compounds, but two ways of drawing the same real hybrid species.

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The anilinium ion, C6H5NH3+, does not show the substituent resonance effect because its nitrogen has no lone pair left to donate.

In aniline (C6H5NH2), the lone pair on nitrogen conjugates with the benzene ring's π system, giving resonance structures that place negative charge at the ortho/para carbons — this is why aniline is more reactive than benzene towards electrophilic substitution and less basic than typical aliphatic amines.

In the anilinium ion (C6H5NH3+), the nitrogen has already used its lone pair to bond the extra proton (forming the fourth N-H bond), so there is no lone pair left on nitrogen to conjugate with the ring. Hence this species does not exert the (lone-pair) resonance effect the way aniline, phenol, and chlorobenzene do.

By contrast: …