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Why is the para product major in the nitrosation of phenol?
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Marc Brevoort
Why is the para product major in the nitrosation of phenol?
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.
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 More
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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
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
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.
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.
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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
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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