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I stumbled upon this paper when I'm trying to look at established protocol to synthesize one of the compounds (compound IIk) mentioned in the paper.

Studies on Aromatic Nitro Compounds. V. A Simple One-Pot Preparation of o-Aminoaroylnitriles from Some Aromatic Nitro Compounds Reaction with unknown mechanism

The paper described transformation of aromatic nitro compounds into o-aminoaroylnitriles by ethyl cyanoacetate in the presence of strong base. While the general scheme showed fused aromatics, this reaction also works for any electron-withdrawing (in addition to the nitro) containing single aromatics.

I have replicate the reaction for compound IIk and I was abled to purified the compound with yield consistent with what was described in the paper.

I have never seen this kind of transformation before, especially with ethyl cyanoacetate. The paper itself mentioned that the mechanism is unknown and I haven't found any subsequent papers that try to elucidate the mechanism.

Anyone has any idea how this reaction works?

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  • $\begingroup$ As a noobie, realize that you are expected to show some effort toward an answer. Then answers may ensue. $\endgroup$
    – user55119
    Jun 19, 2020 at 20:56
  • $\begingroup$ That's the problem. None of the typical organic chemistry knowledge applies here. Reduction of nitrile for example requires a reductant. Obviously ethyl cyanoacetate act as a kind of reductant here, but no known reaction (aside from this) use ethyl cyanoacetate as reductant. Ethyl cyanoacetate typically is used in the presence of base for aldol-type reaction. But I fail to see how the generated enolate interact with nitro group in the aromatic ring here, Aromatic substitution too, especially with C-H activation, typically requires metal catalyst. This reaction has none. $\endgroup$
    – hand15
    Jun 19, 2020 at 21:05
  • $\begingroup$ you are correct about ethyl cyanoacetate being a reductant. Note that the initial products contain oxalate units that are formed from the cyano ester. $\endgroup$
    – user55119
    Jun 19, 2020 at 21:11
  • $\begingroup$ This sure is interesting. I wager reaction may still be similar to aldol, with further ring closure and subsequent reopening. $\endgroup$
    – Mithoron
    Jun 19, 2020 at 21:40

1 Answer 1

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This reaction was conducted using a 3- to 5-fold excess of ethyl cyanoacetate (ECA) and KOH over the arylnitro compounds in dimethylformamide (DMF). It is also important to realize that most KOH is 15% water, which allows for a source of protons in this basic medium. Steps 1 --> 3 (Scheme 1) illustrate a possible route to nitrone 4. In so doing the ECA moiety is being oxidized as the arylnitro compound is being reduced. Addition of hydroxide to nitrone 4 forms 5, which can collapse in two ways. One pathway (another will be mentioned later) leads to reactive ethyl oxocyanoacetate 6 and an aryl hydroxylamine. Base deprotonates the hydroxylamine which adds to 6 to afford species 7 and liberates cyanide anion. Addition of cyanide to the aromatic ring of 7 affords arylamine 8 with loss of the half ester of oxalic acid. A second equivalent of 6 reacts with amine 8 to produce target compound 9.



Condensation (Scheme 2) of the anion 2 of ECA with ethyl oxocyanoacetate 6 provides ketodiester 10 which condenses with arylamine 8 to form the second targeted product 11. Scheme 3 provides another mode of collapse of intermediate 5 which leads to 9. The hydrolysis of 9 and 11 is routine. There are undoubtedly other variants of these Schemes that may apply.


Update: The reference provided by the OP was not the original paper in the series and the picture provided referred to reference 5. The original[1] reactions with 6-nitroquinoline used KCN as the base providing yields of 9 (IIa) in the 50% range and trace amounts of 11 (IIIa) in reaction times up to 5 hours. After 24 hours the yield of 9 is greatly reduced with the formation of the oxamic acid derived from ester 9 by an unspecified mechanism of formation. The appearance of 16.5% of 11 should got be construed that 11 is derived from 9. Significantly, the use of KOH as a base produces 9 in 66.5% yield without the formation of 11.

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  1. Y. Tomioka, A. Mochiike, J. Himeno and M. Yamazaki, Chem. Pharm. Bull., 1981, 29(5), 1286.
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  • $\begingroup$ Interesting and certainly possible. Given the conditions of excess cyanoacetate, why would we not see only the products of cyanide addition to the ring and not cyanoacetate anion addition? $\endgroup$
    – Waylander
    Jun 20, 2020 at 7:51
  • $\begingroup$ @Waylander: Good question! There are a lot of questions about this reaction. The yields are moderate. Why isn't structure 10 just deprotonated given its acidity? If the yields of 9 and 11 were high, then every time 6 is generated an arylnitro cmpd is required. But then again, the yields are not quantitative. I'd be interested in your thoughts. Even the authors didn't venture a mechanism. I'm still working on it. $\endgroup$
    – user55119
    Jun 20, 2020 at 11:59
  • $\begingroup$ Why is 6 not hydrolysed immediately in an environment of excess OH-? Its reactivity is similar to ethyl oxalylchloride. Wouldn't 5 be more likely to expel CN-? $\endgroup$
    – Waylander
    Jun 20, 2020 at 13:00
  • $\begingroup$ I buy that. The loss of cyanide from 5 is in Scheme 3 which explains the formation of 9 but not 11 unless 9 is saponified to 8. Does not appeal to me. As I noted earlier, there are issues here. $\endgroup$
    – user55119
    Jun 20, 2020 at 14:15
  • $\begingroup$ @Waylander: See the update. $\endgroup$
    – user55119
    Jun 20, 2020 at 19:11

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