The following transformation is involved in a step in the synthesis of diazepam:


What is the mechanism of the transformation from the second to the third intermediate (N-methylisatoic anhydride to 1 via glycine in triethylamine)?

There are a few possible electrophiles and nucleophiles that could be involved here. Either the aniline could attack the glycine carboxyllic acid, or the glycine could attack either of the anhydride carbonyls. The paper, however, says that the glycinate is formed, so I don't think the former option is viable. Based on the pKa's involved, what might be the most reasonable proposal?

(†) J. Org. Chem., Vol. 45, No. 9, 1980, 1679

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    $\begingroup$ I wonder if they mischaracterized the intermediate 1. The addition of acetic anhydride would still lead to 2 if we had the aniline-acid before right closure. I am a bit suspicious that the aniline would be able to add to a carboxylic acid with just triethylamine. Of course, I could be missing some kind of activation pathway, but I don't see it at the moment. $\endgroup$ – Zhe Jan 21 '17 at 18:32
  • $\begingroup$ well in the paper they say that they use triethylamine to generate the primary amine. I was thinking that the primary amine attacked the anhydride carbonyl, not the aniline attacked the glycinate carbonyl. Would you mind drawing the aniline acid you had in mind? $\endgroup$ – gannex Jan 21 '17 at 18:48
  • $\begingroup$ Welcome to chem.SE. Please see our policy on homework, and amend your question to include your thoughts on the answer. Otherwise, it will likely be closed by the community. $\endgroup$ – Todd Minehardt Jan 21 '17 at 19:00
  • $\begingroup$ @ToddMinehardt This doesn't apply to self-questions so strictly right? $\endgroup$ – DHMO Jan 21 '17 at 19:01
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    $\begingroup$ Generally, homework assignments don't constitute references to papers from JOC... $\endgroup$ – Zhe Jan 21 '17 at 20:40

A mixture of finely ground 5-chloro-N-methylisatoic anhydride (5.19 g), 2.25g of glycine, 4.15 ml of triethylamine. and 30 ml of water was stirred at room temp for 5 hours. All solid material had disappeared after 3.5-4 h. Voilatile material was removed as completely as possible on a rotavap and the residue treated with 60 ml glacial acetic acid and heated to reflux for 4.5 hours. After the mixture cooled, as much acetic acid as possible was removed on the rotary evaporator, and the tan oily residue was treated with 30 ml of ether. On brief swirling of the mixture, crystallization set in, and the colorless crystalline materiual was collecting after standing overnight and was washed with ether (4.60g, mp 176.5-178°C). The etheral filtrate (two phases) was diluted with enough ethyl acetate to render it homogenous, washed twice with dilute sodium carbonate, then with water, filtered through anhydrous sodium sulfate, and concentrated. Recrystallization of the crystalline residue (0.53g) gave 0.43 g of product, mp 177-179°C. Total yield 5.03g (91.8%).

This implies that we have an initial basic aqueous environment, and we have an acidic aqueous environment afterwards.


The glyine is deprotonated by the triethylamine to form $(1)$.

The nitrogen in the primary amine carboxylate attacks the carbon of one of the carboxylates in the 5-chloro-N-methylisatoic anhydride to form the complex $(2)$.

Some rearrangements of the electrons give $(3)$.

Decarboxylation gives $(4)$.

Then, our environment becomes acidic. All the carboxylate anions are protonated to become carboxylic acids. The ketone is not protonated. We have a neutral complex $(5)$.

It proceeds through addition/elimination to give us our desired product.


  1. In $(4)$, we have a negative charge on the nitrogen, in a basic environment, which shouldn't be favoured.
  2. The nitrogen in $(5)$ isn't exactly nucleophilic enough.
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    $\begingroup$ An amide nitrogen (N-phenylated too) acting as a nucleophile in a 7-membered transition state - unusual. $\endgroup$ – ron Jan 21 '17 at 19:13
  • $\begingroup$ The nitrogen in compound (4) is not very nucleophilic. $\endgroup$ – Zhe Jan 21 '17 at 20:42
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    $\begingroup$ Indeed, on reading the procedure more carefully, refluxing glacial acetic acid (!) is what's required to close the ring. In this case, you might well form the mixed anhydride as an intermediate to promote the ring closure. $\endgroup$ – Zhe Jan 21 '17 at 21:02
  • $\begingroup$ The nucleophilic nitrogen is part of a carbamate and has an electron-withdrawing aryl group on it; the electrophile is an unactivated carboxylic acid (!). I'd go one step further than ron and say it's not legitimate. I agree with Zhe, there probably has to be some sort of activation of the carboxylic acids. Hmm, under acidic conditions, the carbamate should spontaneously lose CO2 anyway. That may pave the way to a cyclisation of aniline N onto an anhydride. I'm not sure, but it's definitely not what's presented above. The alternative is cyclisation of an enol onto the carbamate. $\endgroup$ – orthocresol Jan 21 '17 at 21:59
  • $\begingroup$ Here's a possibility 1) cyclize [3] to the expected uracil analogue, 2) then cyclize to generate a 5-membered lactone appended to the uracil, 3) eliminate $\ce{CO2}$. I don't know if the resultant nitrogen ylide is a TS or intermediate, but it would readily ring expand to the desired product. $\endgroup$ – ron Jan 21 '17 at 22:05

I propose the following mechanism. Contrary to the answer provided by DHMO, I do not believe that the hemicarbamate intermediate is stable. Instead, I assume that it breaks down into carbon dioxide and an amido anion immediately upon formation or very shortly afterwards. This gives us a nucleophilic amido residue which can attack glycine’s carboxylate function. If need be, we can supply catalytic protons to the carboxylic group before the amido nitrogen’s attack but I have omitted that in the scheme.

proposed reaction scheme for the diazepam intermediate
Scheme 1: Proposed reaction pathway.

I consider the amino group of glycine to be the first attacker since it is most nucleophilic, and I consider it attacking the ester group since that should be more susceptible than a substituted hemicarbamate. Upon breakdown of the tetrahedric intermediate to give the first amide group, the free hemicarbamate anion is liberated which should immediately liberate carbon dioxide as stated above. The formation of the final amide linkage is thermodynamically favourable, even in spite of the unfavourable ring size.

  • $\begingroup$ That would necessitate the nucleophilic attack before protonation, though, since there's no other way the aniline gets deprotonated with mild base. $\endgroup$ – orthocresol Jan 22 '17 at 12:59
  • $\begingroup$ @orthocresol But even if the methylaniline is protonated, its lone pair is still sufficiently nucleophilic to attack a carboxy group. It’s just less nucleophilic than a free amine. $\endgroup$ – Jan Jan 22 '17 at 14:07
  • $\begingroup$ Yeah I agree with your mechanism in general. I think this (or some slight variation of it) makes the most sense. $\endgroup$ – orthocresol Jan 22 '17 at 14:10
  • $\begingroup$ I think this is like what I did in "combining some steps...", I think. Good point about the anionic hemicarbamate intermediate. You are saying that DHMO's intemediate (2) would not exist right? It makes sense that CO2 would be liberated immediately, but for me this is a finer point. The essential points were that the ester is more electrophilic than the substituted hemicarbamate and that the glycine amino group is the nucleophile. Thanks, @Jan ! $\endgroup$ – gannex Jan 23 '17 at 2:57

I came up with this, but DHMO proposed the same mechanism first.

enter image description here


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