I am facing a similar synthetic problem as Boekman Jr. et al noted in their synthesis of Tetronolide.[1] I will quote their’s rather than disclosing my actual synthetic problem, but note that the key element — an α-hydroxy-γ-lactone is identical.

The authors’ intended convergent synthesis involves coupling the two compounds 1 and 2 shown in scheme 1 in a ‘tandem ketene-trapping/[4+2] cycloaddition’, which requires the free hydroxy group in 1 to nucleophilicly attack the ketene.

Reaction scheme
Scheme 1: Tandem reaction attempted by Boekman Jr. et al; would include a nucleophilic attack of 1’s hydroxide onto 2’s ketene motif.

However, as indicated this reaction did not produce 3. The authors note:

However, despite considerable experimentation, all attempts to couple lactone [1] with [2] to afford β-ketoester lactone [3] failed. We surmised that the failure stemmed from impaired nucleophilicity of the tertiary OH group owing to a stereoelectronic effect resulting from the enforced alignment of the lactone carbonyl π-orbitals and the adjacent CO bond. (Compound numbers adapted to fit this scheme.)

Instead, the authors used alcohol 4, which readily reacted with the masked ketene 2 to give the ester 5 as shown in scheme 2.

Reaction scheme
Scheme 2: Successful tandem reaction attempted by Boekman Jr. et al; included a nucleophilic attack of 4’s hydroxide onto 2’s ketene motif.

My experimental findings have confirmed a similar effect for my compound. However, I have a bit of trouble discerning which electronic effects exactly cause this unreactivity. My first assumption is that the conformation of the five-membered ring forces the hydroxy group into an angle of approximately $90^\circ$ with respect to the carbonyl-π system. This should allow mixing of σ and π orbitals of whichever kind.

My second assumption was tracing the problem back to the anomeric effect. In that effect, a nonbonding electron pair interacts with an adjacent σ* orbital, lowering its own energy but reducing the σ bond strength. The required electrons could be supplied by the π system, resulting in the effect being π and σ*. However, I do not quite see yet, how this kind of mixing would reduce the nucleophilicity of the oxygen atom.

Thus, the questions are:

  • Which electronic interactions exactly are responsible for this effect?
  • Has this been the topic of a research paper?


R. K. Boekman Jr., P. Shao, S. T. Wrobleski, D. J. Boehmler, G. R. Heintzelman, A. J. Barbosa, J. Am. Chem. Soc. 2006, 128, 10572. DOI: 10.1021/ja0581346.

  • 1
    $\begingroup$ Do you have a tertiary alcohol as well? My first thought is that sterics are the dominant factor. What is the track record of alpha-hydroxy esters (hindered and unhindered) adding to ketenes in general? I'd want to know that before speculating on a stereoelectronic effect. $\endgroup$
    – jerepierre
    Sep 21, 2016 at 15:10
  • $\begingroup$ @jerepierre The paper goes on to state that they decided to couple a different alcohol with the ketene. The carboxyl was isopropyl-protected, the alcohol that would form the lactone vinyl-protected. In my case, the alcohol is secondary, but there are very bulky groups in beta- and gamma-position probably fixing the conformation of the gamma-lactone. $\endgroup$
    – Jan
    Sep 21, 2016 at 15:14
  • $\begingroup$ I am extremely dubious of the stereoelectronic effect. I can't recall ever seeing an example where an antibonding orbital aligns with a CO sigma, and that reduces nucleophilicity. I would be much more inclined to think that this is steric in nature. Have you modeled what the two substrates looks like (or your reactant)? You'd have to pretty convincingly rule out sterics for this kind of heavily substituted system before you want to make a claim of lone pair stabilization... $\endgroup$
    – Zhe
    Oct 5, 2016 at 21:45
  • $\begingroup$ One more thought. When we open the lactone in compound 4, is the hydroxyl group even pseudo-axial? It might be pseudo-equatorial and less sterically encumbered. $\endgroup$
    – Zhe
    Oct 6, 2016 at 14:40
  • $\begingroup$ I'd like to second the concerns of Zhe and jerepierre, an electronic explanation would certainly explain a slow reaction. If that reaction doesn't go at 140°C (certainly enough to disrupt a hydrogen bond) I think the steric explanation fits better. $\endgroup$
    – BiggChemT
    Nov 21, 2016 at 21:43

1 Answer 1


If we focus on the 2-oxo-3s-hydroxy-tetrahydrofuran the electronic effects present are the hyperconjugation and the dipole minimization. "Both effects contribute to the preferred (Z)-conformation of esters over the (E)-conformation. In the (Z) conformation, the lone pair of electrons in the alpha oxygen can donate into the neighboring σ* C-O orbital. In addition, the dipole is minimized in the (Z)-conformation and maximized in the (E)-conformation. But this electronic effects occur between the carbonyl carbon and the oxygen of the ring but the hydroxyl group." [a]

enter image description here

In contrast, I calculated the PES of hydroxy group in the following two systems:

  • 2-oxo-3s-hydroxy-tetrahydrofuran [S1]
  • (1S,2S)-isopropyl 1-hydroxy-2-((vinyloxy)methyl)cyclohexanecarboxylate

The latter one is:

enter image description here

The oxygen of iprO-(C=O)- group was pointed away from [S2] and towards to [S3] cyclohexane. The PES is: (It was calculated with am1 due to computational resources, it was done with my laptop.)

enter image description here

It can be observed that the intramolecular hydrogen bond (IHB) in [S3] is more stable than in the other systems. The order of intramolecular hydrogen bond preference is [S3]>[S2]>[S1]. In other words, the IHB formed with the oxygen SP$^3$ of iprO-(C=O)- group is preferred over the carbonyl oxygen. [b] This can be seen on the rotational barrier between both minima. This conformational preference allows the lone pair of hydroxyl oxygen to be more nucleophilic than in the other systems. As it can be seen in the following figures:


enter image description here


enter image description here


enter image description here

Related to the electronic effects present in this IHB, it can be complemented that "this is a consequence of interacting the lone pair on the accepting oxygen with an empty hydride antibond of the donating O-H atoms, $n_O$ → $\sigma^*OH$. Occupancy of the antibonding orbital raises the energy of the O-H species while stabilizing the oxygen." As Albrecht et. al. concluded in her paper.[c]


  • [a] Cuevas, Eusebio Juaristi, Gabriel (1995). The anomeric effect. Boca Raton: CRC Press. ISBN 0-8493-8941-0.
  • [b] Rincon et al. J Solution Chem (2011) 40: 656.
  • [c] Albrecht et al. J. Phys. Chem. A. (2012) 116: 3946
  • $\begingroup$ By the way; how did you define the angle in the graph? I.e. what corresponds to $0^\circ$, what corresponds to $90^\circ$? $\endgroup$
    – Jan
    Nov 29, 2016 at 16:09
  • $\begingroup$ Additionally, scanning through Albrecht’s paper it appears your quotation is not a literal one. You should use quotation markes only for fully literal quotes. $\endgroup$
    – Jan
    Nov 29, 2016 at 16:26

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