I recently encountered two old publications on action of nitrogen iodide on various compounds (Ref.1 and Ref.2). Between them, Ref.1 described the iodoform formation by methyl ketones upon treatment of the conditions:

Ketones containing a methyl group react very readily with nitrogen iodide, iodoform, ammonia, an acid, and an amide being produced. In these reactions the methyl group appears first to be completely substituted by iodine, a tri-iodomethyl ketone being formed, which in presence of the ammonia simultaneously set free is hydrolysed to iodoform and an acid, a similar reaction between the substituted ketone and ammonia leading to the formation of iodoform and an amide.

The reactions were suggested as follows: \begin{align} \ce{RCOCH3 + NH3.NI3 &-> RCOCI3 + 2NH3}\\ \ce{RCOCI3 + H2O &-> RCO2H + CHI3}\\ \ce{RCOCI3 + NH3 &-> RCONH2 + CHI3} \end{align}

This formation of iodoform is understandable. Yet, Ref.2 claimed that the authors have obtained iodoform from diethylketone (3-pentanone):

Like the other ketones, diethylketone when treated with nitrogen iodide, yields the iodoketones, having very irritable odor. On effecting the interaction in a dilute solution, iodoform is formed in good quantity. It has been observed that as we rise higher and higher in the homologous series of the ketones, the tendency of the formation of iodoform gradually diminishes and the yields also gradually become less.

The reported method the authors had been used: $\pu{2 mL}$ of diethylketone are taken in a beaker and diluted with $\pu{150 mL }$ of weak ammonia. The mixture is warmed on the water bath, and to it, iodide solution is added when at first nitrogen iodide is precipitated, which is immediately taken up. The addition of iodine has to be continued, and finally a copious yellow crystalline precipitate deposits. This is iodoform. On recrystallization from alcohol, this has been obtained quite pure, melting at $\pu{119 ^{\circ}C}$. The yield amounts to $\pu{0.8 g}$.

I’m very curious about their claim and like to hear from your theory of mechanism.

Cited References:

  1. F. D. Chattaway, R. R. Baxter, “CCX.—The Action of Nitrogen Iodide on Methyl Ketones,” J. Chem. Soc., Trans. 1913, 103, 1986–1988 (DOI: 10.1039/CT9130301986).
  2. R.L. Datta, N. Prosad, “Halogenation. XVI. Iodination by Means of Nitrogen Iodide or by Means of Iodine in the Presence of Ammonia,” J. Am. Chem. Soc. 1917, 39(3), 441–456 (DOI: 10.1021/ja02248a013).
  • $\begingroup$ In the second reaction i dont understand the significance of adding iodine continuously. If NI3 is already formed it has a property of dissociating again to I2. Then what is the reason for adding I2? $\endgroup$ Apr 6, 2019 at 9:13
  • $\begingroup$ This question is hard to answer without more details from the two papers. $\endgroup$
    – Zhe
    Apr 6, 2019 at 13:18
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    $\begingroup$ @ KG3: I cannot pinpoint how they distinguish those chemicals. Yet, chemistry is in prictice way before 1917. For example, how would you think Claisen recognized the rearrangement without NMR or IR or MS? It's not easy, but they have done that and results are very fruitful. $\endgroup$ Apr 11, 2019 at 21:57
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    $\begingroup$ @MathewMahindaratne you are completely missing the point. The proposed reaction in reference #2 contradicts the entirety of the literature on the haloform reaction since 1870. Either we should (A) discard all subsequent literature examples that utilize the haloform test to identify methyl ketones as incorrect (B) presume Datta & Prosad made incorrect but completely understandable assumption. If you're going to make extraordinary claims you must supply extraordinary evidence. Claisen did. Datta and Prosad did not. $\endgroup$
    – KG3
    Apr 11, 2019 at 22:59
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    $\begingroup$ @KG3 and Mathew: My first thought about diethylketone giving CHI3 was that they had 2-pentanone. I am incredulous with the statement, "It has been observed that as we rise higher and higher in the homologous series of the ketones, the tendency of the formation of iodoform gradually diminishes and the yields also gradually become less." It is conceivable that 3-pentanone is converted to the 2-iodide, hydrolyzed to the acyloin and oxidized to 2,3-pentanedione that gives iodoform. This procedure is improbable for higher symmetrical ketones. Of course, 2-pentanone is still a option. $\endgroup$
    – user55119
    Sep 18, 2019 at 20:30


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