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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?

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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.

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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.

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

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    $\begingroup$ Sources, please. $\endgroup$ – Jan Oct 16 '15 at 17:05
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When you compare the ortho and para positions, both have a higher electron density vs the rest of the positions. So for now, we can attack both the positions. Now, look at the position of the alcohol group. The alcohol group pulls electrons by the inductive effect so it will decrease the electron density at the ortho position in comparison to the ortho position, making the ortho position a much more suitable position for the weak electrophile NO+ to attack, which needs high electron density to attack as it is a weak nucleophile.

TL;DR : The ortho position has lower electron density in comparison to the para position due to the inductive effect of the alcohol group.

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  • $\begingroup$ This is almost entirely wrong. $\endgroup$ – Jan Jul 2 at 16:02
  • $\begingroup$ May I please know what is incorrect. I know better explanations have been given but what is wrong with mine. Just want to know so that I don't repeat the mistake. Sorry again. $\endgroup$ – Chaitanya Gambali Jul 3 at 1:50
  • $\begingroup$ The slight effect of oxygen’s higher electronegativity is not enough to explain no attack at all on the ortho position. (Also, my brain autocorrected but you only talk about ortho even when you probably mean para. That doesn’t really help your case though.) $\endgroup$ – Jan Jul 3 at 5:55
  • $\begingroup$ Also note that older sources such as this (unfortunately in German) mention simple nitrations of phenol giving $\pu{40\%}$ ortho and $\pu{18\%}$ para product. $\endgroup$ – Jan Jul 3 at 5:57
  • $\begingroup$ Hmmm. I posted what I was taught in my chemistry class, but you obviously have a much greater expertise in this matter. I'll share your explanation with my teachers. Thank you for the info and I apologize for the incorrect answer. Shall I take it down? $\endgroup$ – Chaitanya Gambali Jul 3 at 6:23

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