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It's given in my book that other than the usual compounds that answer iodoform test, resorcinol (benzene-1,3-diol) also gives positive iodoform test.

Is this because of tautomerization which gives a diketone with an active methylene group? But why should a ring lose its aromaticity by doing so?

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There is an equilibrium between the keto enol forms, however small the keto form may be as a contributor. The rest is as you say. The precipitation of iodoform drives the equilibrium further to the products, as per Le Chatelier's principle.Of course, it is not in all cases where the equilibrium is controlled by such factors. The net gibbs energy change for the reaction must be negative.

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  • $\begingroup$ Speculation aside, neither resorcinol nor phloroglucinol give positive iodoform tests. [Fuson, R.C. and Tullock, C.W., J. Am. Chem. Soc., 1934, 56, 1638]( pubs.acs.org/doi/abs/10.1021/…). $\endgroup$ – user55119 Mar 28 at 0:17
  • $\begingroup$ @user55119 Note that the conditions used by the experimenters in the study are in fact quite mild, i.e. 2 min at 60 deg C. The following study which was also not done under vigorous conditons, at temperature of 20 deg C. However, the experimenters waited long enough for iodoform reaction to occur for resorcinol to give a positive test. Hence, they concluded that resorcinol does give a "slow test". Seelye, R. N., & Turney, T. A. (1959). The iodoform reaction. Journal of Chemical Education, 36(11), 572. doi:10.1021/ed036p572 $\endgroup$ – Tan Yong Boon Apr 8 at 12:19
  • $\begingroup$ Tan Yong Boon: Yes, if you pound on the substrate, I'm not surprised. Marrian (1932) treated estrone with KOH/I2 to obtain the dicarboxylic acid from the D-ring. Heer and Miescher (Helv. Chim. Acta, 1945, 28, 156) treated estrone benzyl ether with methanolic KOH/I2 at 80 oC to obtain O-benzyl marrianolic acid. Although the reaction presumably goes through the diiodoketone, I would be reluctant to call this reaction or that of resorcinol "haloform tests". $\endgroup$ – user55119 Apr 8 at 15:10

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