One of the drugs I work with is a beta-lactam (4-membered ring with an amide bond) fused to a sulfone ring, tazobactam.

skeletal diagram of tazobactam

It's relatively stable in water; the lactam is not significantly hydrolyzed without being catalyzed. When the lactam is hydrolyzed, however, typically by an enzyme, and the lactam ring opened, the sulfone ring also opens cleaving the bond between the sulfur and bridgehead carbon. A paper by Kuzin gives the following reaction scheme (this is part of figure 2, click for the whole thing)

Kuzin's reaction scheme

and says of it (emphasis mine):

The Ser70-bound moiety (2) is thought to be the initial intermediate [...] Ring opening after departure of the sulfinate from C5 produces a reactive imine (3) and the more inert tautomeric enamine forms (4, 5).

But why is the sulfuryl (sulfone, sulfonyl(?)...) leaving only when the lactam is opened and not spontaneously beforehand?

Other fused bicyclic beta-lactam drugs don't open the other ring, where in lieu of the sulfone there is commonly a thioether (as in penicillin). The sulfone group seems more electron hungry, but I don't see how the broken lactam bond affects this.

I've heard some people suggest that the lone pair on the nitrogen is able to make this happen, but why doesn't it then happen in the intact drug?

  • $\begingroup$ By "intact drug", you mean its crystallized form? If so, I can imagine that the proton transfer doesn't occur so well in solid form, whereas in solution it is facilitated, which could explain the compounds crystalline stability. Also, your sulphur is $sp^3d^2$ hybdridized, which leaves 3 vacant $d$ orbitals. This might explain its reactivity. I'd like to see what would happen if you swapped SO2- for PO2- (different geometry). $\endgroup$
    – CHM
    Commented May 5, 2012 at 18:36
  • $\begingroup$ @CHM I mean the with the lactam bond not yet hydrolyzed. AFAIK the drug isn't terribly reactive (from the point of view of the sulfone ring opening) in water. $\endgroup$
    – Nick T
    Commented May 5, 2012 at 21:27
  • $\begingroup$ The lactam will be hydrolyzed in water, under acidic/alkaline conditions. What's the optimum pH of your enzyme? $\endgroup$
    – CHM
    Commented May 5, 2012 at 22:39
  • $\begingroup$ @CHM biological, so about 7. Spontaneous (non-enzyme catalyzed) hydrolysis is negligible. $\endgroup$
    – Nick T
    Commented May 5, 2012 at 22:41
  • $\begingroup$ This is interesting. But I'm a bit worried about his mechanism proposal, in step 3. He shows two electrons moving to the sp2 carbon adjacent to the N, but the carbon stays sp2. There's something missing. Have you got an idea of the other AA residues near Ser70? $\endgroup$
    – CHM
    Commented May 5, 2012 at 23:21

1 Answer 1


Here's a draft:

First, read this paper. Considering we're under alkaline conditions (blood pH between 7.35-7.45) we'll take their alkaline mechanism proposal.


enter image description here

This is obviously incomplete, with good reason. Without knowing what the active site of the enzyme looks like (except for Ser70), it's difficult to devise of a mechanism. I could imagine another alkoxy attacking the sp3 carbon between S and N, and shifting two electrons to the sulfur thus generating your SO2- in one step, but that would require a proper active site.

The geometry of the molecule inside the active site of the enzyme is obviously optimal for this reaction to proceed. I'd like to know more. For some reason my school doesn't give me access to this paper.

  • $\begingroup$ The intramolecular proton transfer you show in your last step is unlikely. At biological pH, that carboxylic acid is already deprotonated. $\endgroup$
    – Ben Norris
    Commented May 25, 2012 at 17:09

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