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 idoform 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 idoform is understandable. Yet, Ref.2 claimed that the authors have obtained idoform 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? $\endgroup$ – Som V. Tambe Apr 6 at 9:13
  • $\begingroup$ This question is hard to answer without more details from the two papers. $\endgroup$ – Zhe Apr 6 at 13:18
  • $\begingroup$ @Zhe hey i have a mechanism which gives us great yields of iodoform. The only problem is the details about dilution. I dont get why they dilute it. $\endgroup$ – Som V. Tambe Apr 6 at 13:53
  • $\begingroup$ @Som V. Tambe: I believe dilution is to prevent some unnecessary side reactions. I'm just speculating, though. :-) $\endgroup$ – Mathew Mahindaratne Apr 6 at 21:57
  • $\begingroup$ @ Som V. Tambe: I believe the following statement in the publication may give what they meant by "the addition of iodine has to be continued": ...by means of a potassium iodide solution of iodine in conjunction with liquor ammonia, which represents nitrogen iodide in the nascent state... However, in the first paper they had prepared nitrogen iodide by adding a solution of iodine monochloride in hydrochloric acid to strong ammonia where nitrogen iodide had formed as black precipitate. $\endgroup$ – Mathew Mahindaratne Apr 8 at 18:01

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