We know,generally, keto tautomeric is more stable corresponding enol.(please explain it apart from thermodynamic explanation)

Nitroso is analogous to keto form and oxime is analogous to enol form.

Then why oxime is more stable than nitroso?


In keto-enol tautomerism generally the keto tautomer is more stable than its enol counterpart because in the former a significant ionic character exists which stabilises it. However, there are certain special cases. For instance if the enol form is aromatic(phenol), or very bulky mesityl groups are attached, etc.

In oxime-nitroso tautomerism however, the electronegativity difference comes into play. enter image description here

In oxime, nitrogen is attached to carbon by a double bond;enter image description here

and in nitroso, nitrogen is attached to oxygen by a double bond.

Now it is known that greater the difference in the electronegativity values, stronger is the bond. This property makes the nitrogen carbon double bond stronger hence making oximes more stable than nitroso.

(electronegativity values: N=3.04, O=3.44, C=2.55)

  • $\begingroup$ sorry but I can not at all agree with the above answer! The greater the difference in electronegativity between 2 atoms, the weaker is the bond and not the reverse, when we compare atoms of the same period (C, N, O, F). For example H-F is the weakest (most acidic) linkage i.e., F-H <O-H <N-H <C-H. The C-H bond is the strongest and therefore the most difficult to break. $\endgroup$ – lemos Nov 29 '16 at 14:38
  • 1
    $\begingroup$ @lemos The farther away the electronegativity values of 2 atoms, the stronger the bond generally is. Cesium has the lowest, and Fluorine has the highest and the make the strongest ionic bond. Be nice to people. $\endgroup$ – Piyush Mishra Dec 18 '16 at 12:00
  • $\begingroup$ For your acidic strength thingy: chemistry.stackexchange.com/questions/39711/… $\endgroup$ – Piyush Mishra Dec 18 '16 at 12:05

I wonder if the $\alpha$ effect is in play here. Two hetero atoms with lone pairs bonded to each other have a weaker bond because of lone pair repulsion. This probably destabilizes the corresponding $\pi$ bond as well. That would favor the oxime.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.