JEE Advanced 2017 · Paper 2 · Chemistry Q6 · Dronacharya Exams

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ChemistrySingle correct+3 / -1Question 24 of 54

The major product of the following reaction is

2'-Aminobiphenyl-3-ol (NH₂ on one ring ortho to the biaryl bond; OH meta to the biaryl bond on the other ring) treated with i) NaNO₂, HCl, 0 °C; ii) aq. NaOH.

A

2'-Chlorobiphenyl-3-ol (Cl in place of the amino group)

B

Sodium 2'-(diazonio)biphenyl-3-olate chloride (the diazonium salt with the phenoxide)

C

Benzo[c]cinnoline with OH on the ring para to the new N-C bond (intramolecular azo coupling para to OH)

D

Benzo[c]cinnoline with OH ortho to the new N-C bond (azo coupling ortho to OH)

Answer and worked solution

Official answer: C

$\ce{NaNO2/HCl}$ at 0 °C diazotises the $\ce{NH2}$ .
In aqueous NaOH the OH becomes the strongly activating phenoxide.
The diazonium group, held next to the other ring by the biaryl bond, couples intramolecularly. It attacks the carbon ortho to the biaryl bond and para to $\ce{O-}$ .

This closes a six-membered ring containing N=N: a hydroxy-benzo[c]cinnoline with the OH para to the new C-N bond. That is structure (C).

(C)

Topics:Diazonium salts Phenols

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