[Dan]

I am going to attempt to resurrect the highly enjoyable problem of the week that we used to have. I can't guarantee I'll come up with a problem every week, but I encourage others to set their own problems, so we can have some fun discussions, learn something and bring a bit of life back to this sub-forum.

So, this week I invite you to explain one of my favourite transformations: This ingenious synthesis of L-gluconolactone from its enantiomer, D-gluconolactone.

[Honclbrif]

Looks like after loss of HBr, pretty much every site is up for grabs via tautomerization.

[Dan]

Ok, can you propose a mechanism?

[azmanam]

Kudos for reviving PotW!

So the most acidic proton is prly the bottom left one to give the enone (tautomer of beta-dicarbonyl).  From there I can rationalize the switch of the 2 remaining stereocenters through base-promoted epoxide formation from the bromohydrin and gamma-enolization/protonation (though I can't rationalize why it switches stereochem).

Now, to get the other OH back (and add proton to bottom left position), I'd think the beta carbon needs to be nucleophilic to attack H+, and alpha carbon needs to be electrophilic (prly carbocation?) to be attacked by H2O... but I can't rationalize why beta carbon would be nucleophilic or why alpha carbon would be electrophilic...

[Dan]

So the most acidic proton is prly the bottom left one to give the enone (tautomer of beta-dicarbonyl).

I think the most acidic proton is the one alpha to the carbonyl. While the enone looks like a nice intermediate, it is not detected in the reaction mixture, and as you see it's very difficult to rationalise the outcome of the reaction through an enone intermediate. Presumably pot carb is not basic enough to deprotonate alpha to Br (beta to the carbonyl).

The step with pot carb is not an E2 elimination of HBr...

[azmanam]

Alright.  Let's just ignore the fact that I said a C-H proton will be more acidic than an O-H proton...

So K2CO3 deprotonates one of the OH groups, the bottom one can't geometrically form an epoxide, but the top one can.  So my new guess is the epoxide is the mystery intermediate.

I think the epoxide won't open up until the acidic part of the last step for regio/stereochem reasons.  I wouldn't have started with the alpha proton, but since you brought it up... I can do some interesting manipulations... but I'm probably missing the obvious.

Apparently all I've got is epoxides.

[Dan]

Ok, on the right track now. The diepoxide intermediate in the bottom right is a proposed intermediate in the KOH step.

The isolated product of the pot carb step is an epoxide, but not the one you drew. You are right to point out that the cis relationship between the 2-Br and 3-OH do not allow epoxide formation directly - but position 2 is not configurationally stable in base...

From the the diepoxide intermediate you drew in the bottom right, elimination is not the proposed pathway - there are no particularly acidic protons available to abstract. Given this, consider electrophilic sites that could be attacked by hydroxide.

[azmanam]

(whine, complain) epimerization (grumble, grumble) stupid epoxide (b*tch, moan) KOH gives diepoxide.

OK, so I can get 3 of the 4 stereocenters swapped... About that last one...  Is there lactone opening and/or carbocations involved?

[Dan]

Good stuff, but we're not there yet!

If you consider the diepoxide, you have chosen to attack unactivated position 3 over position 2 (which is activated to nucleophilic attack due to the alpha carbonyl), and then secondary position 5 over primary position 6. Of the epoxidic (is that a word?) positions, 2 and 6 are the most electrophilic - note that this is also why the 2,6-dibromide forms in the first step, these are the most reactive positions. So I don't think the mechanism is reasonable.

Is there lactone opening

Bullseye!

[azmanam]

But attack on the 2 & 6 positions lead to retention of stereochemistry at all positions (NGP)... how are we going to enantiomerize (while we're coining words) the cpd if we open the epoxides to keep the same stereocenters?

I knew attack on 5 was disfavored under basic conditions, which is why I've been holding off on that opening until the acidic part.  But you're saying the epoxide opens at the 6-position w/ KOH (and at the 2 position w/ KOH)?

[Dan]

But attack on the 2 & 6 positions lead to retention of stereochemistry at all positions (NGP)... how are we going to enantiomerize (while we're coining words) the cpd if we open the epoxides to keep the same stereocenters?

Sorry, should have been clearer. What I'm saying is that the epoxides do not open by intermolecular attack of hydroxide. Start with hydrolysis of the lactone - presumably that is the most electrophilic site of the diepoxide.

[Honclbrif]

If the ring opened with the a,b-epoxide intact, wouldn't it lead to decarboxylation via a Darzen-like pathway?

[Dan]

I think you usually have to acidify the carboxylate to promote that decarboxylation. In base I think it should be stable.

I know it's a painful question, but we're almost there... (that was a hint...)

[azmanam]

If this is it, then you're right, that is a clever transformation.

[Dan]

Nice one, yes. The transformation is achieved by a series of Payne rearrangements (painful... see what I did there?).

The last epimerisation was not proposed to proceed by intermolecular epoxide opening, but by a similar lactone hydrolysis pathway. As you are so close I will post the solution below.

Reference: Lundt, I.; Madsen, R. Top. Curr. Chem., 2001, 215, 177-191

From diepoxide 1 (you gave the mechanism for its formation so I won't repeat it), lactone hydrolysis followed by Payne rearrangement gives trans-epoxide 2, which is favoured over cis-epoxide 3.

Epoxide 3 then closes in a favoured 5-exo-tex fashion to give 5. In accordance with Baldwin's rules, a 6-endo-tet ring closure, which would give 4, is not observed.

Hydrolysis and Payne rearrangement of lactone 5 produces an equilibrium mixture of two trans-epoxides 6 and 7. However, intramolecular epoxide opening of 6 to give 8 is not observed as it would proceed by an unfavourable 5-endo-tet (the 4-exo-tet possibility is not shown). On the other hand, 7 could lactonise by a disfavoured 6-endo-tet (not shown) or a favoured 5-exo-tet, with the latter favourable process giving L-gluconolactone 9. 9 is hydrolysed in base - which is irreversible in base - giving carboxylate 10, in a thermodynamic sink.

Acidification of 10 finally allows acid catalysed lactonisation, giving L-gluconolactone 9, which I think is all rather astonishing.