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Does the lone pair of nitrogen in nitrosobenzene take part in resonance? If not, then why? If it takes part in resonance then the hybridisation of nitrogen is sp. Can a sp hybridised atom be attached to the ring?

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2 Answers 2

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Let's look at nitrosobenzene.

enter image description here

In order to be delocalized (participate in resonance), the lone pair on the nitrogen needs to be in a $\ce{p}$ orbital. As I have draw it above, the lone pair is in a $\ce{sp^2}$ orbital, and the $\ce{N=O}$ pi bond is conjugated with the ring:

enter image description here

If the lone pair on the nitrogen atom is delocalized, then yes the nitrogen atom is $\ce{sp}$ hybridized. As a consequence, the $\ce{N=O}$ pi bond is perpendicular to the pi system. It is no longer conjugated:

enter image description here

Either the lone pair or the $\ce{N=O}$ pi bond is conjugated, not both.

If the lone pair is conjugated, the nitroso group can be electron donating by resonance:

enter image description here

If the $\ce{N=O}$ bond is conjugated, the nitroso group is electron withdrawing by resonance:

enter image description here

Finally, because the nitrogen is more electronegative than carbon, the nitroso group is electron withdrawing by induction:

enter image description here

It is more likely that the $\ce{N=O}$ bond is conjugated because that gives a higher degree of conjugation: 8 atoms versus 7 atoms.

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    $\begingroup$ does this mean that there are two types of nitrosobenzene? Or somehow the one pair conjugation and pi bond conjugation interconvert? $\endgroup$ Commented Sep 28, 2020 at 3:14
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The nitroso group can assume either state of conjugation with the ring. In electrophilic substitution, where the ring is made electron deficient, the nitroso group will engage its lone pair in the conjugation providing an additional $\pi$ electrons. This resonance directs the donated electrons to the ortho/para positions, making the nitroso group ortho/para directing even as the electronegativity of nitrogen and oxygen make it deactivating. It's the same situation as we see with halogen atoms.

In nucleophilic substitution the ring has an excess of $\pi$ electron density. Now the nitroso group will turn to conjugate its $\pi$ bond with the ring and withdraw some of the electron density. This mode of resonance combines with electronegativity to make the nitroso group strongly activating and ortho/para directing.

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