[RaZ]

Hi, well I know there are a million applications to this reaction, but I was wondering if someone can label some of the more important ones.  Also, I was wondering if someone could help me with this mechanism (attached). I'm wondering about the selectivity of the ring opening, how is the PhS attached (does the OH just leave?), and the last step (why does K2CO3 give syn and DIBAL-H give anti?). 

Thanks a lot, I might have more questions on sharpless expoidation later.

[RaZ]

Also, just wondering why sharpless epoxidation has to happen at ~ -20 degree C??

[Custos]

Where did you get this synthesis from (author, journal)? There are a few curious reactions in the scheme. As you point out, the second step is hard to follow (unless I'm missing something) - I can't believe PhSH in NaOH would displace -OH ... there must be something else going on. Also, in the next step... the trans ketalisation with 2,2-dimethoxypropane is fine, but treating the product with m-CPBA I would have expected the sulfur to oxidise to the sulfoxide or even the sulfone long before an alpha hydroxylation. In fact I've done very similar reactions and the sulfur of a PhS- substituent oxidises very rapidly with m-CPBA, or even much weaker oxidants like bleach.

As for your final question, it's possible that potassium carbonate could catalyse an epimerization of the carbon alpha to the aldehyde, but 100% syn? I don't think so.

[RaZ]

Hi, the website with the picture of the reaction is: http://www.chem.harvard.edu/groups/myers/handouts/Sharpless-AE.pdf

The original journal is:
Ko, S. Y.; Lee, A. W. M.; Masamune, S; Reed, L. A., III; Sharpless, K. B.; Walker, F. J.
Tetrahedron 1990, 46, 245-264.

[movies]

Okay, first things first: The SAE is done at low temp because that tends to increase the level of enantioselectivity.  If there is less energy available to the system you are less likely to get side reactions that have poor selectivity.  By lowering the T you force the lowest energy pathway to predominate.

Second, step two is tricky.  It's not direct displacement of OH.  The first thing that happens is deprotonation of the hydroxyl group followed by a Payne rearrangement to the terminal epoxide.  This epoxide is less sterically hindered and is therefore more susceptible to attack by the thiolate.  So basically you migrate the epoxide and then open the epoxide with the nucleophile.  Clever, no?

Third, the m-CPBA step does indeed oxidize the sulfur to the sulfoxide.  The next step (Ac₂O, NaOAc), however, converts it back to a thioether by means of a Pummerer rearrangement.

Finally, I agree that K₂CO₃ epimerizes the stereocenter alpha to the carbonyl after the mixed acetal is cleaved.  In the DIBAL version I think the acetate is reduced of giving an alkoxide that then kicks out the thiolate to make the aldehyde.  Since DIBAL isn't really that basic and is very hindered to boot, it doesn't epimerize the same way.  It's also possible that all the DIBAL is consumed in reducing the acetate and the resulting alkoxide doesn't collapse until you warm the reaction up (in fact I'm almost certain that is the case because the aldehyde doesn't get reduced).

[RaZ]

Thanks a lot Movies and Custos...really helps (esp. the Payne rearrangement).  I might have a few questions later:)

[RaZ]

Hi, just wondering if this looks alright...

[movies]

The DIBAL route looks okay, but the K₂CO₃ route is a bit off.  You probably don't make ketene.  Think about your solvent and what it would do in the presence of a base and an acetate.

[RaZ]

Ah right, you deprotonate the methanol which would attack the carbonyl on the acetate...making the aldehyde and an ester.  thanks.

[movies]

Yep.

[RaZ]

One more simple question about protection/deprotection of the diol..I'm not sure if I'm getting the mechanism right.  The protection is done with 2,2-dimethoxypropane + POCl₃.  Deprotection is done with trifluoroacetic acid-water.  Thanks.

[movies]

The protection is a little odd.  Usually you would use acid, not POCl₃.  At any rate, I agree with your mechanism except for your last step which you have drawn as an S_N2 displacement, but it should go through an oxocarbenium ion (kicking out MeOH) like when you added the first hydroxyl group to the PG.

Cleavage mechanism looks perfect.