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I don't understand the mechanism of the first reaction; I tried different ways but I can usually find how to make the ring without the double bonds, but it's obviously not the right thing; I think HNO2 reacts as the nitrosonium ion, bonding with one of the oxygens (since it has free electrons) but I can't never make it work.

EDIT: Probably the acid is formed by nitric acid (like in one of the articles orthocresol linked), that I think is not written because it can be formed by the decomposition of nitrous acid; I still don't understand the mechanism of the formation of the ring.

The problem is the first reaction

Image: https://commons.wikimedia.org/wiki/File%3AIsocarboxazid_synthesis.png by Nuklear (Own work) [CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0)], via Wikimedia Commons

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    $\begingroup$ Where did you get this scheme from? Based on the papers I'm looking at, I believe it's wrong. The initial substituent should be COMe instead of CO2H, and the oxidation to CO2H is effected using nitric acid. See J. Heterocycl. Chem. 2009, 46, 909. Either that, or you use HNO3 instead of HNO2 to get to the carboxyl isoxazole in one step. See J. Med. Chem. 2004, 47, 3642. $\endgroup$
    – orthocresol
    Commented Nov 19, 2017 at 15:33
  • $\begingroup$ What is "Bn" In that diagonal arrow? $\endgroup$
    – Oscar Lanzi
    Commented Nov 19, 2017 at 15:36
  • $\begingroup$ I found it in a book I used : Vardanyan, R.S.; Hruby, V.J.; Synthesis of Essential Drugs; Elsevier (2006); I also found other sources online (like this image from Wikipedia) that uses that synthesis. $\endgroup$
    – Jake Q
    Commented Nov 19, 2017 at 15:40
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    $\begingroup$ I think "Bn" is short for benzyl-, so it's benzylhidrazine $\endgroup$
    – Jake Q
    Commented Nov 19, 2017 at 15:43
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    $\begingroup$ You have misdrawn your final product, it should be the benzyl hydrazine not the phenylhydrazine $\endgroup$
    – Waylander
    Commented Nov 19, 2017 at 16:08

2 Answers 2

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As you may know from diazotisation reactions, $\ce{HNO2}$ is a generator of the nitrosonium cation, $\ce{NO+}$. Electrophiles like these can react with the enol tautomer of hexanedione:

Reaction of hexanedione with nitrosonium cation

Following this, you tautomerise the nitroso compound to the oxime. The mechanism is exactly analogous to keto-enol tautomerism:

Nitroso-oxime tautomerism

Note that I've elected to draw the oxime with a (E)-configuration. Both (E) and (Z) oximes can form, but the (Z)-configured oxime cannot cyclise in the following step to form an isoxazole:

Isoxazole formation

Following this, to get to the compound in your question, you supposedly need to oxidise the acetyl substituent ($\ce{COCH3}$) to a carboxyl substituent ($\ce{COOH}$). Nitrous acid isn't an oxidant, which casts doubts on whether your source is veritable (perhaps it's just a typo).

In the literature, I find that this oxidation is accomplished using nitric acid, $\ce{HNO3}$, which is a perfectly fine oxidant. In one case (J. Heterocycl. Chem. 2009, 46, 909), the diketone is first treated with nitrous acid to get the 3-acetylisoxazole, and is then treated with nitric acid to form the 3-carboxylisoxazole. In a separate case (J. Med. Chem. 2004, 47, 3642), the diketone is simply refluxed with nitric acid to form the 3-carboxylisoxazole. I suspect that this works because nitrous acid is present in the nitric acid, so the mechanistic pathway is exactly the same (a similar effect has been observed with the nitration of phenols using dilute nitric acid).

Exactly how nitric acid oxidises an acetyl group to a carboxyl group, though, I don't know (and I'm not sure if it is known).

Edit: There is, in fact, a mechanism by which nitrous acid may oxidise the acetyl group to a carboxyl; see user55119's answer.

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  • $\begingroup$ Can you use iodoform reaction here to convert acetyl to carboxylate? Just curious since there is a misstep in the question. $\endgroup$
    – Mathew Mahindaratne
    Commented May 16, 2020 at 13:43
  • $\begingroup$ @MathewMahindaratne, I suppose you could, although my heterocyclic chemistry is nowhere near good enough to comment on whether the isoxazole will be stable under the basic conditions. $\endgroup$
    – orthocresol
    Commented May 16, 2020 at 13:53
  • $\begingroup$ With my experience, they are stable. I'm worrying about its methyl substitution.:-) $\endgroup$
    – Mathew Mahindaratne
    Commented May 16, 2020 at 15:27
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Orthocresol did a great job describing how to get to the isoxazole methyl ketone. However, nitrous acid can be an oxidant. The methyl ketone in the presence of mineral acid, which you need to get the nitrosyl cation, can tautomerize the methyl ketone to the enol which reacts with the cation to get an α-nitroso ketone. The nitroso group tautomerizes to an oxime as orthocresol has shown. Acid-catalyzed dehydration to a nitrile leads to an α-ketonitrile. This species is similar in reactivity to an acyl halide. Aqueous hydrolysis gives the desired carboxylic acid with the loss of HCN. In nitrous acid (HONO), nitrogen is +3. In HCN, nitrogen is -3. This is a 6-electron reduction for nitrogen and a 6-electron oxidation of the substrate. The nitrosation is a 4-electron oxidation and the oxime dehydration is a 2-electron oxidation. Nitric acid is not required.

Brief reaction mechanism

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    $\begingroup$ The truth depends on how the nitrous acid is made. You don't have concentrated stock solutions of $\ce{HNO2}$. You combine a nitrite salt and a strong acid in situ. Use $\ce{HCl}$ for the strong acid and you have the mechanism here. With $\ce{HNO3}$ you can use the alternate approach. $\endgroup$
    – Oscar Lanzi
    Commented Nov 20, 2017 at 1:50

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