# Retrosynthesis of a lactone

We have been asked to do a retrosynthesis of (6⁠S)-6-[(2⁠S)-2-hydroxy-4-phenylbutyl]oxan-2-one J from 3-phenylpropan-1-ol K:

My first instinct would be to do a functional group interconversion of the alcohol to a ketone, to then do a Baeyer-Villiger oxidation.

The key to this synthesis should be — in my opinion — how to generate the two hydroxy functions with anti stereoselectivity starting from at least one achiral reagent. Most of the time you would attempt building up chirality de novo using chiral auxiliaries, catalysts or reagents as these methods are often more cost-efficient than acquiring a specific chiral reactant — the obvious exception being if a compound exists in the natural chiral pool that you can easily use (which is not really the case here). Having identified our most important problem, we can turn to the other ones:

• we need to generate a δ-lactone
• we need to form a $\ce{C-C}$ bond, preferably in a chiral manner, to a $\ce{C6}$ body (the bit of the final product that is missing in the reactant).

Scheme 1: General retrosynthesis idea.

In my opinion, the δ-lactone is a non-issue. If you have an open-chain ester such as 3, you can probably form the six-membered lactone simply by $\ce{K2CO3}$-mediated transesterification. Practically all literature methods to synthesise esters should also predominantly give the δ-lactone especially since the unfavourable ζ-lactone is the alternative. (In the unlikely case that it is formed, it can easily be transesterificated as noted.) I would put this step last.

Scheme 2: Transesterification step.

We now have 3, an anti-configured 1,3-diol. When I see this, my mind immediately jumps to the Evans-Tishchenko asymmetric reduction. Thus, we can arrive there from either of the two possible β-hydroxyketones 4 or 5. For this to work best, I would want the carboxy group to be masked as a protected alcohol here, cf. 6.

Scheme 3: Evans-Tishchenko step. $\ce{X}$ would be $\ce{H}$ for the newly-formed hydroxy group and $\ce{CH3CH2CO-{}}$ for the inducting one.

Both of these β-hydroxyketones could be generated from aldol reactions; however, only ketone 4 suggests that it can easily be reached from the desired reactant 2. If we wanted to synthesise 5, we would need an extra carbon atom on 2 which we would need to introduce in an additional reaction. However, the route to 4 only requires selective oxidation to the aldehyde — my personal favourite is Dess-Martin–oxidation, but you should probably check all the others that suggest themselves (Swern and relatives, TEMPO, Ley-Griffith, etc.) as one could be superior over the others due to your lab spirits liking it.

Finally, this also tells us that the aldol addition to generate 4 from 7 and 8 should proceed in a stereoselective manner giving the (S)-configured product predominantly. Again, many methods are known and I would suggest a Paterson aldol reaction. Analysis of the relevant literature (unfortunately, I did not find a website, so a citation it is) reveals that $\ce{(-){-}Ipc2BOTf}$ and Hünig base should give the (S) product preferentially.[1]

Scheme 4: Final aldol disconnection.

This leads us to the following forward synthesis:

$$\ce{\mathbf{2} ->[DMP] \mathbf{7} ->[(-){-}Ipc2BOTf][ (iPr)2NEt, \mathbf{8}] \mathbf{4} ->[SmI2, EtCHO] \mathbf{6} ->[1) TBSOTf; 2) PPTS][3) DMP; 4) {Pinnick}] \mathbf{3} (R = H) ->[DCC, DMAP][K2CO3] \mathbf{1}}$$

Reference:

[1]: A. S. Franklin, I. Paterson, Contemp. Org. Synth. 1994, 1, 317–338. DOI: 10.1039/CO9940100317.

The first idea that comes to mind is an aldol, as you pointed out too:

The only thing left to figure out is what aldol conditions to use to achieve enantioselectivity and the enantioselective reduction that is needed afterwards.

The Bayer-Villiger I don't understand, as:

I guess the regioselectivity should fit, with the alcohol-substituted side having the higher migrating group abilities, but I don't see how you would achieve the donating group to be in 3-position in the resulting cyclopentanon.

Hope that helps...

• After the aldol you'd still need to reduce the beta-hydroxyketone to a diol, I think? You'd have to make the aldol enantioselective and the reduction diastereoselective (probably not a difficult task, since there are tons of procedures out there). – orthocresol Feb 22 '17 at 14:15
• yeah right ; ) then its probably bad with two enantioselective transformations needed... – logical x 2 Feb 22 '17 at 14:19
• Only the aldol is enantioselective, the reduction can be controlled by the 1st stereocentre. chem.uky.edu/courses/che538/ac/2006.fall/problems/grad.01/… (section 4.1.2) I agree with your overall idea anyway, aldol seems the most straightforward choice for a 1,3-dioxygenated compound. – orthocresol Feb 22 '17 at 14:21