I was told that urea on reaction with nitrous acid gives nitrogen gas, carbon dioxide and water. While on the other hand, thiourea on reaction with nitrous acid gives $\ce{H+}$, thiocyanate ion and water.

I tried looking for the mechanism and thought that this would be related:

enter image description here

I carried out the above procedure on both the nitrogens of urea and was able to obtain the products of the first reaction enlisted above. Now, my questions are:

• Isn't the nucleophilicity of the lone pairs of urea's nitrogens negligible, and any reaction should proceed via attack by oxygen's lone pairs? (I thought that this was similar to amidines and amides)

• Why are the products of the reactions with urea and thiourea different, as the carbonyl oxygen, or sulphur doesn't seem involved in the reaction according to me

• More importantly, is this even the correct mechanism?

  • $\begingroup$ It is similar as in amidines and amides and it doesn't matter. Any attacks on O are unproductive, so effect of conjugation is rather kinetical. $\endgroup$ – Mithoron Jan 12 at 19:17

As implied in your link, the reaction of urea and nitrous acid will follow the mechanism you posted, repeating a second time with the nitrogen on the remaining amide. As for why the nitrogen is more nucleophilic, I can speculate, but I don't want to be wrong.

The difference for thiourea is that sulfur can also act as a nucleophile instead of the nitrogen in the thioamide. The product is dependent on the pH. You can read more about it here. Here is the mechanism:

enter image description here

That's a thiourea that I squeezed in on the far right as a product from the strong acid. Sorry for the mess.


What I've so far is that Thiourea is a stronger nucleophile (Sulfur is less electronegative than oxygen, thus more nucleophilic). Thiourea reacts at the site of the sulfur whereas urea reacts the the amine sites.

  • $\begingroup$ Thanks for the idea, but it would be great help if you could provide a mechanism scheme $\endgroup$ – YUSUF HASAN Jan 10 at 13:06

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.