# Why is the para product major in the nitrosation of phenol?

When phenol reacts with nitrous acid ($\ce{NaNO2 + \mathrm{conc}\ H2SO4}$), 4-nitrosophenol and not 2-nitrosophenol is formed. I cannot understand why the para-isomer should be preferred over the ortho-isomer.

I understand the reaction mechanism proceeds by electrophilic substitution by the $\ce{NO+}$ cation, and that the $\ce{OH}$ group activates the ring, which leads to ortho and para substitution.

I don't understand why the para-isomer should be preferred over ortho?

Ortho and para positions are activated due to the +R effect of the $\ce{-OH}$ but the preference of the para product can be explained by the mechanism.

The mechanism proceeds through a dienone. The ortho dienone intermediate will experience strain as $\ce{=O}$ annd$\ce{=NOH}$ lie in the same plane. Para nitrosated intermediate on the other hand is free from strain.

Reference: González-Mancebo S., Lacadena J., García-Alonso Y., Hernández-Benito J., Calle E., Casado J. Monatsh. Chem. 2002, 133 (2), 157-166. DOI: 10.1007/s706-002-8245-0.

• I feel this answer would be improved a lot by a drawing of the strained ortho-intermediate. – Jan Jul 2 '20 at 16:03

Due to steric hinderance and also there is some electronegativity of alcohol group due to induction it takes electron from sigma bond though not dominating therefore

These are the 2 reasons for more p- substitution

• Sources, please. – Jan Oct 16 '15 at 17:05