[Rutherford]

How to achieve the attached transformation. I thought about protecting O(1) with TIPSCl then O(3) and O(4) with acetone and finally O(2) with TBSCl. Deprotection of O(5) with PBr₃ shouldn't destroy the other groups (or maybe it will???). O(3) and O(4) could be deprotected under mildly acidic conditions, but I don't see how to methylate only O(4) and not O(3). Any hints?

[phth]

look more closely and draw the chair.  I think acetonation of C2, C3 will happen.  up-up is an axial-equitorial conformation.  You will get the down-up ring

[Rutherford]

How do you mean that an up-down ring will be obtained from an up-up relationship of OH groups?

[phth]

I don't think this is a trick question, and sorry if I was not clear b4.  the acetonation is favored to the hydroxyls that sit in the equitorial positions. The axial equitorial 2,3-acetonation is the kinetic product.  It puts more strain on the α-anomer allignment, takes the 6-ring out of the chair conformation destabilizing the system.  ofc it depends on solvent, but the desired ring formation will happen under the equilibration conditions.  There are only 3 hydroxyls availible when the α,6 hydroxyl positions are protected, so you will get a mixture, but it is controllable.  example provided:

[Rutherford]

I've read in Modern Organic Synthesis by Zweifel that trans OH groups on rings don't form five membered cyclic acetals that easily, while cis do. Seems like it's a mistake. Do you have any reference or any book recommendation, as I would like to explore this further (protecting groups in carbohydrate chemistry).

[Dan]

I've read in Modern Organic Synthesis by Zweifel that trans OH groups on rings don't form five membered cyclic acetals that easily, while cis do. Seems like it's a mistake. Do you have any reference or any book recommendation, as I would like to explore this further (protecting groups in carbohydrate chemistry).

Your book is correct, phth is mistaken on this one. Mannopyranosides preferentially form the 5-membered cis-fused ketal with acetone or 2,2-dimethoxypropane, this is well documented in the literature. As you have numbered your original compound, this would be the 3/4 position. Note however that in carbohydrate nomenclature position 1 of an aldose is the anomeric centre (and the primary alcohol is position 6). Acetonation of methyl D-mannopyranoside gives methyl 2,3-O-isopropylidene-D-mannopyranoside in high yield, this one:



As for books, I will have a look in my collection when I get home from holidays and recommend one or two (PM me if I forget). They're general carbohydrate chemistry texts and I can't remember which ones cover protection in the most depth.

This one is very useful, but expensive: http://eu.wiley.com/WileyCDA/WileyTitle/productCd-0471133396.html

I need to see if I still have my copy of it.

[Rutherford]

Thanks Dan for clarifying this. Any thoughts about the initial problem?

[Dan]

Thanks Dan for clarifying this. Any thoughts about the initial problem?

This should give you some ideas: http://onlinelibrary.wiley.com/doi/10.1002/anie.199422721/abstract

[Rutherford]

I cannot access it.

[Dan]

Also:

http://onlinelibrary.wiley.com/doi/10.1002/anie.199422901/abstract
http://pubs.rsc.org/en/content/articlelanding/2008/cs/b717902b

I'm struggling to find free pages...

Edit: Try this.

[Rutherford]

So the usage of dispiroacetals will selectively protect O2 and O3 leaving O4 unprotected and ready to participate in methylation?

[Rutherford]

2,2,3,3-tetramethoxybutane (TMB) can be used, too to afford butane 2,3-bisacetal. The O(2) and O(3) atoms are selectively protected in excellent yield.

[Dan]

Yes, I would start with BDA (butane diacetal) protection. How do you proceed from there?

[Rutherford]

Deprotection of O5 with PBr₃ and subsequent methylation would give smaller than 50% yield I guess (statistical formation of products).

The alternative: Protect O4 with PMB (MPM) group, then deprotect O5 with BBr₃, then protect O5 with Bn group. Deprotect O4 with DDQ, methylate with NaH/MeI and finally deprotect O5 using hydrogenation. All these steps combined would probably give an ever lower yield than the first approach.

[Dan]

PBr₃ will generate HBr and cleave the BDA group. Remember, O5 (your numbering) is part of an acetal (as is the BDA group) so you can remove them both at the same time in the presence of an ether.

How are you dealing with O1 (your numbering)?