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

Weerman degradation

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.

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    $\begingroup$ 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. $\endgroup$ Aug 19, 2021 at 8:36
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    $\begingroup$ @AdityaRoychowdhury more acidic than the hydroxy proton? $\endgroup$
    – Waylander
    Aug 19, 2021 at 9:16
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    $\begingroup$ @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. $\endgroup$ Aug 19, 2021 at 21:26
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    $\begingroup$ @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. $\endgroup$
    – Ashish
    Aug 20, 2021 at 3:46
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    $\begingroup$ @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! $\endgroup$
    – TRC
    Aug 20, 2021 at 6:00

1 Answer 1


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:

Weerman degradation

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:

Phenyl a-ketoamide formation

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).


  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).

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