# Formamide and Hoffmann Bromamide Reaction

I encountered a problem to identify the compound which does not give Hoffmann bromamide degradation reaction. One of the options was formamide, and it was given as the answer.

Why can't formamide give Hoffmann bromamide reaction? I feel that the answer is wrong (it should produce ammonia).

• I guess hydride isn't sufficiently nucleophilic to do the 1,2-migration?
– PCK
Mar 26 '19 at 15:08
• If hydride were to migrate, isocyanic acid from hydride migration would be deprotonated under the basic conditions of the Hofmann rearrangement. Mar 26 '19 at 23:11

I have spent a lot of hours to find the evidence to support OP's idea of formamide may undergoes Hoffmann degradation reaction, but I can't find any. However, I find at least one publication, which may suggest formamide may decompose to formic acid and ammonia (ironically) in hot alkaline medium, before exchanging amide $$\ce{N-H}$$ to $$\ce{N-Br}$$ (please keep in mind that $$\mathrm{p}K_\mathrm{a}$$ of $$\ce{HC(O)NH2}$$ is about 24 at $$\pu{25 ^{\circ}C}$$ in DMSO, Evan's pKa Table):

Ref.1 has described the rates of decomposition rates of acetamide and formamide, which have been studied by individually decomposing them in pressurized hot water. According to the studies, while apparent decomposition rates for both amides were more represented by the second order rather than the first order reaction kinetics, the rate constant for formamide was more than 10 times higher than that for acetamide (Ref.1).

Even though, the work has been done in high temperature and high pressure, the authors observed the decomposition without using acids or bases, and throughout wide range of concentrations. They also found that rate increases with increasing concentrations. Thus, one can expect that in highly alkaline media used in Hoffmann degradation, some amide may hydrolyzed before rearrangement. According to a mechanistic study of the rearrangement, hydrolysis of even aromatic amides under alkaline conditions has been observed:

It has been found that in the presence of excess alkali at $$\pu{30 ^{\circ}C}$$, p-nitrobromobenzamide gives approximately as much p-nitrobenzoic acid by hydrolysis as p-nitroaniline by rearrangement; at $$\pu{90\!-\! 100 ^{\circ}C}$$, however, a $$90\%$$ yield of p-nitroaniline is obtained.

All of these evidence directed to the theory of hydrolysis of formamide (which is aliphatic and most susceptible to hydrolyze under the conditions due to its size) before it converted to N-bromformamide. Further more, the Ref.3 described a theoretical study of the Curtius rearrangement, which also believed to be going through same mechanism as Hoffmann degradation. Although authors have discussed various pathways of rearrangement including going via oxazirene ion, and identified the most stable intermediate is isocyanic acid during the rearrangement, their conclusion wasn't clear on the hydride ion shift.

References:

1. M. Okazaki, T. Funazukuri, “Decomposition of acetamide and formamide in pressurized hot water,” Journal of Materials Science 2006, 41(5), 1517–1521 (DOI: 10.1007/s10853-006-4616-1).
2. C. R. Hauser, W. B. Renfrow, Jr., “The Removal of $$\ce{HX}$$ from Organic Compounds by Means of Bases. III. The Rates of Removal of Hydrogen Bromide from Substituted N-Bromobenzamides and their Relative Ease of Rearrangement in the Presence of Alkali. The Hofmann Rearrangement,” J. Am. Chem. Soc. 1937, 59(1), 121–125 (DOI: 10.1021/ja01280a029).
3. A. Rauk, P. F. Alewood, “A theoretical study of the Curtius rearrangement. The electronic structures and interconversions of the $$\ce{CHNO}$$ species,” Canadian Journal of Chemistry 1977, 55(9), 1498–1510 (https://doi.org/10.1139/v77-209).
• So can we conclude that if in place of an alkyl group we have H or some electron withdrawing group then, hydrolysis would take place before rearrangement because of increase in electrophilicity of the carbonyl carbon? Mar 27 '19 at 14:05
• Can't make it universal rule on everything. My argument is pure speculating, but have some evidence in literature. Also, if you look at reactivity of aldehyde vs ketone, my argument make more sense. Keep in mind also that, $|ce{H}$ is not electron withdrawing (compare electronegativity of $|ce{H}$ vs $|ce{C}$). I believe it's more of its size. Mar 27 '19 at 15:36