enter image description here The closest reaction I can find is Lossen rearrangement but in that hydroxyamic acid is first converted to its derivative but how will that be possible here?

enter image description here

Moreover, even if I make an amine in this step I cannot proceed further, how will the amine and alcohol react in the next step? Any hints would be appreciated.

  • 7
    $\begingroup$ First deprotonate and then lose carboxylate. The remaining neutral species reranges to an isocyanate. In methanol the isocyanate forms the methyl ester of a carbamic acid. Compare the similarity of the Lossen and Hofmann rearrangements $\endgroup$
    – user55119
    Commented May 9, 2019 at 21:16
  • 5
    $\begingroup$ My thoughts. I think you have to deprotonate the NH of the hydroxamic acid and lose OH- to give the isocyanate analogous to the mechanism of the Lossen that you have drawn. This is the only option that makes any kind of sense, and it is known that deprotonation of hydroxamic acids proceeds at NH rather than OH. $\endgroup$
    – Waylander
    Commented May 10, 2019 at 18:04
  • $\begingroup$ Seems like methanol/methoxide nucleophilic attack on the isocyanate (instead of water/hydroxide) would lead to the carbamate. Not sure if that squares with the supposedly two-step reaction scheme though. $\endgroup$
    – Curt F.
    Commented Jul 1, 2019 at 13:59
  • 1
    $\begingroup$ Now that a bounty has been placed on this post and a reputable source for the answer is requested, I seriously suspect that one will not be found. Any version of the Lossen rearrangement requires activation of the oxygen into a viable leaving group before base can effect deprotonation and alpha-elimination can occur to form the isocyanate. If the methyl carbamate is the desired product, simply treat the the activated hydroxamic acid with MeONa/MeOH rather than aqueous base. $\endgroup$
    – user55119
    Commented May 20, 2021 at 20:16
  • $\begingroup$ ![enter image description here](i.sstatic.net/be2bA.png) This is from March 6th edition page 1610. Does this help? :) $\endgroup$
    – anomeric
    Commented May 21, 2021 at 0:02

3 Answers 3


In your question, you have mentioned:

The closest reaction I can find is Lossen rearrangement but in that hydroxyamic acid is first converted to its derivative but how will that be possible here?

You have recognized the reaction correctly but the Lossen rearrangement is the conversion of a hydroxamate ester to an isocyanate (only the first step of given mechanism). Although the original Lossen rearrangement needed hydroxamate ester $\left(\ce{R(C=O)NH-O(C=O)R'}\right)$ to rearrange, recent research has shown that the rearrangement is possible with free hydroxamic acids $\left(\ce{R(C=O)NH-OH}\right)$ as well (Ref.1):

Without using activating agents, a variety of free aromatichydroxamic acids could be rearranged to aromatic amines in the presence of base alone:

Rearrangement of hydroxamate ester to amine Source of the image: Ref.1

According to $\bf{(a)}$, free aromatichydroxamic acid $\left(\bf{1}; \ce{R(C=O)NH-OH}\right)$ react with a base to give the corresponding carboxylate $\left(\ce{R(C=O)-O-}\right)$ as intermediate and hydroxyamine $\left(\ce{NH2-OH}\right)$ byproduct. The intermediate $\left(\ce{R(C=O)-O-}\right)$ reacts with another aromatichydroxamic acid $(\bf{1})$ to give necessary hydroxamate ester $\bf{A}$, which undergoes Lossen rearrangement in basic conditions to give the isocyanate $\bf{B}$. Mechanism for aromatic amine from isocyanate is depicted in $\bf{(b)}$.

For instance, the base-mediated rearrangement of free 4-methylphenylhydroxamic acid $\bf{(1a)}$ to 4-methylaniline $\bf{(2a)}$ is given as follows:

A mixture of N-hydroxy-4-methylbenzamide $\bf{(1a)}$ $(\pu{0.363 g}, \ \pu{2.4 mmol}),$ $\ce{K2CO3}$ $(\pu{0.332 g},$ $\pu{2.4 mmol})$, and DMSO $(\pu{2 mL})$ was heated to $\pu{90 ^\circ C}$ and stirred at that temperature for $\pu{2 h}$. The mixture was cooled to rt, and then treated with $\pu{2 M} \ \ce{ HCl}$ $(ca. \pu{3 mL})$. After the mixture became the clear solution, $\pu{2 M} \ \ce{NaOH}$ $(ca. \pu{3 mL})$ was added and extracted with $\ce{Et2O}$ $(\pu{15 mL} \times 3)$. The combined organic layers were dried over anhydrous $\ce{Na2SO4}$, filtered and evaporated under reduced pressure. The residue was purified by silica gel column chromatography $(\ce{hexane/Et2O}, 1:1)$ to yield the pure 4-methylaniline $(\pu{0.253 g}, \ 98\%)$ as a white crystalline solid.

The effects of the amount of base, reaction time, and presence of water are summarized in following diagram (Ref.1):

Preparation of tolylamine from tolylhydroxamic acid

Thus, as shown in above diagram, presence of water may have direct effect on the conversion of the substrate to corresponding amine (even in DMSO). Further, Ref.1 clearly shows that this reaction requires polar aprotic solvent to proceed through Lossen rearrangement to give thiocyanate intermediate, which eventually converted to corresponding amine in the presence of aqueous base. For instance, if $\bf{(1a)}$ is reacted with $\ce{K2CO3}$ in methanol instead of in DMSO or DMF at for $\pu{90 ^\circ C}$ for $\pu{2 h}$, the yield was found to be only $10\%$ while it was $98\%$ in DMSO and $92\%$ in DMF under the same conditions. The authors reported $86\%$ of the starting compound has been recovered unreacted (none of it recovered in DMSO or DMF).

Note that the first reaction given in the question is in aqueous medium. Thus, according to the results found in Ref.1, the resulting product in the given reaction would be a trace of corresponding amine $\left(\ce{R-NH2}\right)$ and most of substrate would be recovered unreacted. Consequently, there won't be a reactive isocyanate $\left(\ce{R-N=C=O}\right)$, thus the reaction would not proceed to give the anticipated product in methanol as indicated.

Suppose this first reaction has been done in $\ce{OH-/DMSO}$ in place of $\ce{OH-/H2O}$. That would give you the anticipated isocyanate intermediate $\left(\ce{R-N=C=O}\right)$. It is well known that isocyanates reacts very differently in the presence of water, alcohol, and amine. In the presence of water, it would give you corresponding amine with evolution of $\ce{CO2}$. However, in the presence of alcohol or amine, it would give you corresponding carbamate or urea derivative, respectively$:

Isocyanate reaction

Since the given second reaction is in methanol (an alcohol), the product would be a methyl carbamate as shown in the scheme above (Ref.2). For example, polyurethane is prepared from diisocyantes:



  1. Yujiro Hoshino, Moriaki Okuno, Eri Kawamura, Kiyoshi Hondab and Seiichi Inoue, "Base-mediated rearrangement of free aromatichydroxamic acids (ArCO–NHOH) to anilines," Chem. Commun. 2009, (17), 2281-2283 (DOI: https://doi.org/10.1039/B822806J).
  2. Greet Raspoet, Minh Tho Nguyen, Michelle McGarraghy, and Anthony Frank Hegarty, "The Alcoholysis Reaction of Isocyanates Giving Urethanes:  Evidence for a Multimolecular Mechanism," J. Org. Chem. 1998, 63(20), 6878–6885 (DOI: https://doi.org/10.1021/jo9806411).

This is like to proceed by a isocyanate intermediate.

The Lossen reaction has been shown to be slow but possible in free base, without the addition of a leaving group or metal catalyst, by Hoshino et. al. in 2009[1] - see Scheme 1(a)

Isocyanates, when attacked by alcohol nucleophiles, form carbamate esters

See also: Org. Biomol. Chem., 2016,14, 9046-9054

[1] Chem. Commun., 2009, 2281-2283

  • $\begingroup$ Will try to add diagram etc later. $\endgroup$
    – user213305
    Commented May 20, 2021 at 21:43
  • 1
    $\begingroup$ While this Lossen rearrangement employs base (K2CO3), the reaction is run in DMSO, not hydroxylic solvent. A careful reading of the mechanism illustrates that an effective leaving group is generated in the dimerization. There is still no evidence that the hydroxyl group in hydroxylic base is an effective leaving group. $\endgroup$
    – user55119
    Commented May 21, 2021 at 0:35
  • $\begingroup$ The paper you mention in your answer contains the answer to the question. However, I can award the bounty to only answer so I am awarding it to Mathew's answer as it is more elaborate even though both answers technically refer to the same paper. $\endgroup$
    – S R Maiti
    Commented May 24, 2021 at 19:23

Mechanistically, the N-H is deprotonated to form an amide ion. In basic medium, the hydroxyl group may be slightly deprotonated as well, but the pKa are not drastically different and this is not mechanistically useful. You can envision an isocyanate forming upon rearrangement, which is then subject to nucleophilic attack by methanol. Graphically:

enter image description here

It is also worth nothing that hydroxide ion is the base and the leaving group from the hydrodyxamic acid is also hydroxide. This is likely by design and not accident to prevent unnecessary by-products from existing.

Does this help? :)

  • 2
    $\begingroup$ Could you please add some reference for the mechanism? This has been unanswered for a long time, so I want to make sure that the answers can be backed up by a good reference. $\endgroup$
    – S R Maiti
    Commented May 20, 2021 at 21:11
  • $\begingroup$ Did you read my Comment of today? What is the evidence that no activation of the hydroxamic acid is required? If you have hydroxide in methanol, then you also have methoxide. Uncharged methanol is not as good a nucleophile as methanol. $\endgroup$
    – user55119
    Commented May 20, 2021 at 21:12
  • $\begingroup$ @user55119 Sorry, I missed the comment. The way OP posted the reaction scheme showed two different arrows so I guess it can be assumed that they are two different steps i.e. there is no hydroxide in methanol. And are you saying the reaction OP posted won't happen? $\endgroup$
    – S R Maiti
    Commented May 20, 2021 at 21:29
  • 2
    $\begingroup$ @Shoubhik R Maiti: That is what I am saying. I would like to have a reference for this reaction as described. I suspect there isn't one and that the question was manufactured ... and quite poorly. $\endgroup$
    – user55119
    Commented May 20, 2021 at 21:36
  • $\begingroup$ I don't have a reference other than my own knowledge. It follows precedent analogously of Hofmann and Curtius rearrangements to form isocyanates. The isocyanate undergoes nucleophilic addition in a second, separate step to form the carbamate (urethane). If you consider Hofmann rearrangement, OH group (here) and Br group serve to increase acidity of the N-H hydrogen. Interception of the isocyanate intermediate gives the final product. Another similar precedent is the Lossen rearrangement, where OR is used instead of OH. These proposals work mechanistically but may not in practice. $\endgroup$
    – anomeric
    Commented May 20, 2021 at 23:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.