In the synthesis of Panacene by Canesi et al. [1], a step involves the installation of an $\ce{-OH}$ group on a double bond, using the oxymercuration-demercuration procedure. However, the product happens to be a hemiacetal, as shown in the picture. With a 78% yield, it seems that the hemiacetal is not reactive towards the added sodium borohydride. Is this true in general, that hemiacetals will not be reduced to diols by $\ce{NaBH_4}$? What about $\ce{LiAlH_4}$?

Reduction of a hemiacetal


  1. Cyrille Sabot, Didier Bérard, Sylvain Canesi, "Expeditious Total Syntheses of Natural Allenic Products via Aromatic Ring Umpolung," Organic Letters 2008, 10(20), 4629-4632 (DOI: 10.1021/ol801921d).
  • 3
    $\begingroup$ Hemiacetals generally reduce with NaBH4 because of the equilibrium with the open hydroxyaldehyde in base. Given some time, I think 15 will reduce to the diol. LAH should work but perhaps not for the demercuration. Notice in the paper the NaBH4 rxn was conducted at 0oC for 1 minute. Apparently, over eduction is a problem. $\endgroup$ – user55119 Apr 9 at 21:36
  • $\begingroup$ I agree. It makes one wonder what the other reaction products are. I strongly suspect that there will be a significant amount of the diol reduction product. $\endgroup$ – Waylander Apr 9 at 21:47
  • $\begingroup$ The $\pu{600 MHz}$ proton NMR of 15 is not clean, and a lot of unrelated peaks in it. I couldn't imagine how they predicted the product by just looking at it. $\endgroup$ – Mathew Mahindaratne Apr 9 at 23:46

The major purpose of $\ce{NaBH4}$ in the second step of given reaction sequence is the demercuration, which stops at the hemiacetal as the applied condition is the limit (at $\pu{0 ^\circ C}$ for $\pu{1 min}$).

In general under usual conditions (at $\pu{25 ^\circ C}$ in ethanol for for an hour or two), $\ce{NaBH4}$ reacts with hemiacetals and reduce them to corresponding diols. For instance, $\ce{NaBH4}$ reacts with glucose, which is a hemiacetal in aqueous solutions, to give sorbitol, the corresponding diol (Ref.1). The reaction has been done in water and at room temperature ($\pu{20-25 ^\circ C}$) for $\pu{1-2 h}$. Thus, running the given reduction in $\pu{0 ^\circ C}$ for shorter time make sense.

Also, although lithium aluminum hydride ($\ce{LiAlH4}$) is too vigorous a reagent for reduction, under carefully controlled conditions, it can be used to convert lactone to lactol (cyclic hemiacetal such as monosaccaride) without reacting further to give the corresponding diol. However, excess $\ce{LiAlH4}$ would give the corresponding diol after reacting further with lactol (Ref.2,3):

reduction of hemiacetal with LiAlH4


  1. M. Abdek-Akher, J. K. Hamilton, F. Smith, "The Reduction of Sugars with Sodium Borohydride," J. Am. Chem. Soc. 1951, 73(10), 4691–4692 (DOI: https://doi.org/10.1021/ja01154a061). 2.Glen E. Arth, "Reduction of Lactones to Hydroxyaldehydes with Lithium Aluminum Hydride," J. Am. Chem. Soc. 1953, 75(10), 2413–2415 (DOI: https://doi.org/10.1021/ja01106a040).
  2. Robert A. W. Johnstone, "Chapter 1.11: Reduction of Carboxylic Acids to Aldehydes by Metal Hydrides," In Comprehensive Organic Synthesis - Selectivity, Strategy and Efficiency in Modern Organic Chemistry, Volumes 8: Reduction; Ian Fleming, Volume Editor; Barry M. Trost, Ian Fleming, Editors-in-Chiefs; Pergamon Press: Oxford, United Kingdom, 1991, pp. 259-282 (ISBN: 978-0-08-040599-1).

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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