The products of the reduction of esters with $\ce {LiAlH4}$ and the products of the reduction of amides with $\ce {LiAlH4}$ are vastly different. The former reduction cleaves the ester and produces two alcohols while the latter reduction produces an amine with the carbonyl group of the original amide replaced with $\ce {CH2}$. A carbamate seems to display both chemical behaviour of esters and amides. I am curious to know what would be the mechanism by which reduction of carbamate with $\ce {LiAlH4}$ takes place and what would be the products of such a reduction.


1 Answer 1


Carbamates are usually reduced to N-methyl groups. There are numerous examples:

Reduction of methyl carbamate

J. Am. Chem. Soc. 2012, 134 (16), 6936–6939

Reduction of t-butyl carbamate / Boc

Org. Lett. 2012, 14 (18), 4834–4837

But it is not always a given. In this next example, the nitrogen is part of a three-membered ring (aziridine). These nitrogens are better leaving groups than usual, cf. Ketone/aldehyde synthesis from N-acylazetidines or aziridines where the same kind of reactivity is observed:

Reduction of aziridine carbamate

Angew. Chem. Int. Ed. 2002, 41 (24), 4683–4685

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    $\begingroup$ What if the "ester part" of the carbamate is also linked to another part of the molecule such that it doesn't simply go away after reduction? What would happen to the "ester part" of the carbamate upon reduction? $\endgroup$ Commented Oct 1, 2019 at 11:29
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    $\begingroup$ You get the alcohol. e.g. N-CO2Et -> EtOH There aren't really other ways to link a carbamate to some other part of the molecule; the central carbon (which becomes the NMe carbon) has to have four bonds to either N or O. The N bit becomes NMe and the O bit becomes the alcohol. $\endgroup$ Commented Oct 1, 2019 at 11:38
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    $\begingroup$ @orthocresol: H. C. Brown, et al., argue that it is the inability of the aziridine to form the iminium structure that represses reductive alkylation in favor of aziridine formation. J.A.C.S, 1961, 83, 4549. $\endgroup$
    – user55119
    Commented Oct 1, 2019 at 22:42
  • $\begingroup$ @user55119 thank you for the reference (I just noticed this). That is consistent with the rationale provided in the question I linked, which is good! $\endgroup$ Commented Oct 27, 2019 at 23:38

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