# Weerman degradation on alpha-hydroxy amides

I came across this reaction scheme in the solutions of a test paper:

I understand the formation of isocyanate intermediate as the first step. In the second step, I realized that hydrolysis will be more favourable at the benzyl carbon, hence forming benzaldehyde (given as the major product).

However, I cannot understand how aniline is being formed. There must be another migration/rearrangement occurring, but I cannot figure out how. It might be some reaction I am unaware of.

• The proton attached to the benzyl position appears to be acidic. It may be removed by OH-. Something similar to the first hoffmann rearrangement (where Br was the leaving group) may occur with the C=O, with phenyl shifting to N atom. Aug 19 at 8:36
• @AdityaRoychowdhury more acidic than the hydroxy proton? Aug 19 at 9:16
• @TRC: The reaction is formally known as Weerman degradation so that I changed the title accordingly. If you don't like what I did, please feel free to roll back. Aug 19 at 21:26
• @TRC Yeah the leaving group given in the accepted answer is $\ce{H2NCO2H}$ so maybe that is what is happening, but it's a good leaving group nevertheless and aldehyde will form. Aug 20 at 3:46
• @MathewMahindaratne - I'm always glad to learn new reaction names. Saves a lot of trouble and frustration in case I see it in the exam!
– TRC
Aug 20 at 6:00

Since OP did not give any reference to the rearrangement in the question, I doubted the formation of aniline is true even if as a minor product. The conversion of $$\alpha$$-hydroxy amides to corresponding aldehydes is a well-known degradation reaction, commonly known as Weerman degradation: $$\ce{R-CH(OH)-C=O-NH2 ->[NaOBr][\Delta] RCHO + H2NCO2H}$$ Although this degradation is independently discovered (Ref.1), Some authors have regarded it as an extension of the Hofmann rearrangement (Ref.2) because of the similarity in proposed mechanism:

Accordingly, to my knowledge, there is no place to have an rearrangement other than proposed. However, even if it is most unlikely, there could be a possibility to have 1,3-phenyl shift at the intermediate $$\bf{I}$$ stage:

The resultant phenyl $$\alpha$$-ketoamide could undergoes base hydrolysis under the reaction condition to give aniline.

Note: The 1,3-phenyl shift has been known for other reactions (Ref.3).

References:

1. R. A. Weerman, "Sur une synthèse d'aldéhydes et de l'indol," Recueil des Travaux Chimiques des Pays-Bas et de la Belgique 1910, 29(1-2), 18-21 (DOI: https://doi.org/10.1002/recl.19100290104).
2. C. L. Arcus and D. B. Greenwood, "398. The Hofmann reaction with α- and β-hydroxy-amides: reactions of the intermediate isocyanates," J. Chem. Soc. 1953, 1937-1940 (DOI: https://doi.org/10.1039/JR9530001937).
3. Colin Eaborn, Karen L. Jones, Paul D. Lickiss, and Włodzimierz A. Stańczyk, "Proportion of 1,3-phenyl migration in trifluoroethanolysis, methanolysis, and hydrolysis of $$\ce{(Me3Si)2C(SiMe2Ph)(SiEt2I)}$$," J. Chem. Soc., Perkin Trans. 2 1993, (3), 395-397 (DOI: https://doi.org/10.1039/P29930000395).