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We know from SIR effect (Steric Inhibition of Resonance) that in o-nitroaniline, the nitrogen's p-orbitals will change their plane, causing the conjugation/resonance of its lone pairs to stop. So, the only effect acting on it now is −I effect from the $\ce{NO2}$ group.

Comparing with just aniline, the $\ce{NH2}$ lone pairs are in constant resonance with the benzene ring making them unavailable for donation. Thus, aniline should be less basic than o-nitroaniline. But it isn't as per experimental values. So what's going on? What's the logic behind this?

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  • $\begingroup$ Oh, do we know that? No, it's your misunderstanding - in such cases it's ortho-effect not SIR. $\endgroup$
    – Mithoron
    Mar 31 '20 at 15:16
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Yes it is true that SIR is observed in o-nitro aniline so it will not show -M effect.

But I think you're confused in the resonance of the lone pair of NH2 group with the benzene ring which in fact occurs in both cases. The thing that makes o-nitro aniline less basic is the -I effect which makes the lone pair on - NH2 group less available for donation .

In fact the complete order of basic strength of nitro substituted anilines is: [Aniline>m-nitro aniline>p-nitro aniline>o-nitro aniline ]

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  • $\begingroup$ How can resonance of Lone Pairs of Nitrogen of$NH2$ occur in ortho case?? Resonance completely stops due to SIR effect. $\endgroup$ Mar 31 '20 at 8:08
  • $\begingroup$ I think it the nitro group goes out of plane $\endgroup$ Mar 31 '20 at 11:04

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