Iodoxybenzoate as a Test Reagent for Free Phenolic Hydroxyl Groups in Organic Compounds.

While attempting to study the pharmacological action of oxidation products of morphine, I tried to induce rapid oxidation of the drug by means of ammonium iodoxybenzoate. The salts of iodoxybenzoic acid have been studied by Loevenhart and his associates 1 and their oxidizing properties are well known. Upon treating morphine salts with an aqueous solution of ammonium iodoxybenzoate the straw to garnet color characteristic of a morphine solution on eremacausis develops within a few moments. No other opium alkaloid, except apomorphine, seems to yield color with this reagent. Codeine is monomethyl morphine, with the methyl group replacing the hydrogen of the phenolic hydroxyl group of morphine. This “muzzling” of the phenolic hydroxyl group (which may explain the difference in action between the drugs), and the resulting inhibition of the oxidation color development on treatment with ammonium iodoxybenzoate, suggested that the reagent might be used for the detection of “free” phenolic hydroxyl groups in various other aromatic organic compounds.

This was found to be the case. The reactivity of the phenols in color development, especially in contact with aldehydes and inorganic acids, has been recently reviewed by Levine and Magiera. 2 The reaction system, phenol-aldehyde-acid, is so powerful, however, that “muzzled” phenolic hydroxyl groups, such as found in codeine or acetyl-salicylic acid, are opened up so that color develops. Hence such tests as Pettenkofer's 3 and Marquis' 4 for the opium alkaloids, for instance, do not differentiate morphine from its associated alkaloids most of which seem to contain “muzzled” phenolic hydroxyl groups. The advantage, then, of iodoxy benzoate as a test reagent in connection with phenolic compounds lies in the fact that it does not decompose the compounds with which it comes in contact, but merely reveals, by an oxidizing color reaction, the presence of a “free” phenolic hydroxyl group.