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In this video it is mentioned that urea and sodium hydroxide will react to produce ammonia. However, I can't seem to find a clear explanation of the reaction and the other products generated.

One explanation I found says that the NaOH causes the urea to hydrolyze, yielding sodium carbamate, which will then further hydrolyze to sodium carbonate.

Another says that the urea tautomerizes to ammonium cyanate, which preforms a substitution with the NaOH to yield ammonium hydroxide and sodium cyanate.

Which of the two reactions occurs? Do they both occur? Neither?

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Urea can be attacked by nucleophiles like any carbonyl. The immediate intermediate is a tetrahedral orthoacid which will break down to liberate ammonia once. The same process can be repeated a second time to liberate the second molecule of ammonia.

hydrolysis of urea

Technically, this is a hydrolysis catalysed by hydroxide. Hydroxide may be the reactive species but it is regenerated in the course of the reaction.

For urea to tautomerise to ammonium cyanate, a significant rearrangement and breaking a $\ce{C-N}$ bond would be required. Neither is likely in any way.

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  • $\begingroup$ Does the carbamic acid react further and we end up with carbonate ion + more ammonia? $\endgroup$ – Oscar Lanzi Sep 24 '17 at 15:53
  • $\begingroup$ Carbamic acid is inherently unstable and will decompose to ammonia and carbon dioxide rapidly @OscarLanzi $\endgroup$ – Jan Sep 25 '17 at 8:19
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Well, in practice adding sodium hydroxide granules to a concentrated urea solution at room temperature results in the rapid growth of needle crystals, most likely urea being pushed out of solution. Further heating does not result in $\ce{NH4+}$ or $\ce{NH3}$. However the melting of the dry reactants might produce a different result.

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  • $\begingroup$ There are a lot of possible experiments using urea and sodium hydroxide; you are presenting just one variety. Of course, given concentrated solutions of both, chances are that one of the two becomes over-concentrated and crystallises out but there are a lot more variables. $\endgroup$ – Jan Feb 7 '17 at 16:19

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