Abstract
In order to clarify the photodecomposition mechanism of riboflavin, detection and determination of photolytic products were made.
(1) In neutral solution, the photolytic products, calculated on the basis of 1M of initial riboflavin, were as follows: 0.82M of lumichrome, 0.74M of formaldehyde, 0.09M of formic acid and about 1M of sugar, probably erythrose.
Lumiflavin, acetaldehyde, acetic acid and volatile alcohols could not be detected.
(2) From the solution irradiated in an alkaline medium, 0.65M of lumiflavin, 0.09M of lumichrome, 0.1M of formaldehyde and 0.06M formic acid were detected. Besides, four carbon compounds, probably mixtures of erythrose and erythronic acid, were produced. Acetaldehyde, acetic acid and volatile alcohols were absent.
From these results, it may be considered that formaldehyde is produced accompanying lumichrome. Accordingly, the following photodecomposition mechanism of riboflavin is conceived:
At first, hydroxylated carbon at position 2′ is oxidized to carbonyl group. Then, hydrolytic rupture of carbon-carbon bond occurs between 1′ and 2′. At that time, hydroxyl group is bound to flavin-methylene radical, forming 9-hydroxymethyl flavin which converts instantly to lumichrome by liberating formaldehyde in lumichromic cleavage. On the other hand, in lumiflavinic cleavage, flavin-methylene group combines with proton, forming flavin-methyl compound known as lumiflavin. Existence of formic acid may be resulted by a further oxidation of a part of formaldehyde once produced.
