Author Topic: Urea amination of tert-alcohols  (Read 31009 times)

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  • Guest
Urea amination of tert-alcohols
« on: October 10, 2004, 07:04:00 PM »


[url=" class="bbc_link" target="_blank" rel="noopener noreferrer">Post 513173

(synthon: "Oh, what a great idea, Ning...", Methods Discourse)

"to one molecular quantity of urea, [...] two molecular quantities of concentrated sulphuric acid (above 94%) are added gradually [...]"

what do you think about using another acid than concentrated H2SO4 ?

- 70% H2SO4 (boiled 30%H2SO4)
- deshydrated toluenesulfonic acid
- tosyl chloride + h2o
- tosyl chloride directly
- H3PO4

the product obtained is HSO4-NH3-CO-NH3-HSO4, then wich is reacted with tert-alcohol
(2 H2SO4 + NH2-CO-NH2 --> 2 HSO4- + NH3-CO-NH3 2+)

I guess the H2O is not really free in the case of use of 30% h2so4, but can it be separated from the "sulfuric urea" ?

and finally  will some MgSO4 work to obtain a product HSO4-NH3-CO-NH3-HSO4 like ?

merci d'avance...


  • Guest
« Reply #1 on: October 12, 2004, 06:53:00 AM »

I think concentrated acid is a good thing, because it needs to protonate the tertiary alcohol.

What's really happening, as I'm sure Nicodem will tell you  ;)  is a variant of the Ritter reaction. Urea is a sufficiently good nucleophile (??) to attack the tertiary alcohol ion.

Which, by the way, is why it should be tertiary.

Although, it's possible a secondary alcohol might work too.

Someone should try it with a 4:1 molar excess of isopropanol to urea, then hydrolyze it to see if they get a decent yield of diisopropylamine, a useful item indeed.

I think tosic acid would be a good bet, being that it's less oxidizing than sulfuric acid.

As far as I remember, the reaction should be done cold, to prevent that annoying oxidation from occuring, and the production of ethers, too.

Feel free to try. It's not like urea is hard to find.


  • Guest
How the reaction works?
« Reply #2 on: October 12, 2004, 06:30:00 PM »
I’m not in the mood for answering so many questions so I will just tell how I see it (that is, I see it similar to the Ritter reaction*) so you can give answers yourself.

The first requirement for the reaction is that the alcohol forms a carbocation with the help of the sulphuric acid:
1.)    R3C-OH + H+  <==>  R3C+ + H2O

This carbocation has then at least three ways to react further.

The first is the reverse reaction of the above, the nucleophylic addition of water formed above (alcohols can also add but we will ignore that as the reaction is analogous). This reaction is kept at its minimum with the water being removed by sulphuric acid:
2.)   R3C+ + H2O  <==>  R3C-OH + H+

The second is the beta proton elimination reaction:
3.)   R2C+-CH2-R  <==>  R2C=CH-R + H+

The third option is the addition of a nucleophyle. You want the product to bee an amine but ammonia or amines cannot be used due to the highly acidic environment needed for the reaction 1 to occur. Amines are fully protonated in sulphuric acid and as such are not nucleophyles. So you need a nitrogen source with a lone electron pair that will not get completely protonated at the reaction conditions, thus retaining some nucleophylicity. Urea (or R-CN in the Ritter reaction) is a really bad nucleophyle and as such does not posses much basicity (pKa=0.1) and is not completely protonated even by the strongest acids (a proton is also a very strong electrophyle). Thus, as a compromise, even though the nucleophylicity of urea is so low there is at least always some of it free (unprotonated) and this can add to the electrophylic carbocation (similarly like water in the reaction 2):
4.)   R3C+ + H2N-CO-NH2  ==>  R3C-NH-CO-NH2 + H+

But given that both reactions 2 and 3 are fully reversible (alkenes also add a proton to the double bond forming back the tertiary carbocation) the urea product becomes the major product of the altogether reaction. This is because the urea group is a much less effective leaving group than the protonated alcohol group, making the reaction 4 the least reversible of all the other possible.

I hope this also makes it clear why this reaction works so well only with the tertiary alcohols and why concentrated sulphuric acid is used. The secondary carbocations are considerably harder to produce or put in other words their concentration in the reaction medium is possibly to low for a good reaction rate and satisfying yields.

* see


  • Guest
thanks... (A) R3C-OH + H+ <==> R3C+ +...
« Reply #3 on: October 12, 2004, 11:15:00 PM »

(A)   R3C-OH + H+  <==>  R3C+ + H2O
(B)   R3C+  <==>  R2C=CH-R + H+
(C)   R2C=CH-R + H+ <==> R3C+
tertiaries alcohols react easily with "not diluted" acids
whereas secondaries react only with concentrated acids...

so the reaction (A) will work with H2SO4(30%) without problems...
the reaction (B) doesn't depend of the concentration of the H2SO4
but with H2SO4 <80% or something the reaction (C) will not exist!!!!

i wonder if the yield of this urea amination will go from 50percent when using 70%-H2SO4 to 2percent when using 69%-H2SO4....