In this post I wanna talk about one interesting synthesis question a student brought to me.
So, without a further ado, here’s the question:
Propose the reasonable synthetic sequence to accomplish the following transformation—
While this synthesis may look intimidating, let’s break the whole process into small steps. This will help us see how we can figure out the necessary steps to get from the starting material all the way to the final product (we also call it a target molecule).
Basics of the Retrosynthetic Analysis
This is my first post on synthesis, I’ll also do a brief introduction to the retrosynthetic analysis.
So, what’s going to be our first step? Generally, you want to look at your starting material and the target molecule and see if you can find the difference. Find what has changed. Did you make new C-C bonds? Did you make any functional group transformations? In this case, we have added a whole new side to the molecule by making two new C-C bonds.
Figuring out the position of the new bonds helps us to come up with the key intermediates that we need to have in this synthetic scheme.
Alright, we have figured out the position of the new bonds, what now? The next step is to come up with what we call a “synthon.” Synthons are the molecular entities which don’t have to be “real” molecules but rather are scaffolds of molecules with the correct charges on atoms. There are two ways how you can find those charges: normal polarity and the reversed polarity method.
In a nutshell, the normal polarity works by “assigning” the negative charge to a heteroatom like “O” or “N” and alternating the charges down the chain. Likewise, the reversed polarity works by alternating the charges, however, you will start by assigning the positive charge to your heteroatom.
Breaking It Down for Our Molecule
Let’s try a normal polarity approach for this problem and see what we get.
By assigning the negative charge to the oxygen atom and alternating it from atom to atom we get the following picture.
Now, by breaking up the molecule at the connection points where we want to make our bonds, we get the following synthons:
And once we have our synthons, it’s time to convert those into actual reagents or the “synthetic equivalents:”
Synthetic equivalents are the actual molecules that have the same polarity on the atoms that the synthons have. Or, alternatively, they can be easily converted to reactive intermediates with the required charges.
For instance, if you have a plus charge on a C, that equates to a molecule with a halogen or other leaving group on that carbon, or a carbonyl such as an aldehyde, ketone, or a carboxylic acid or a carboxylic acid derivative. Essentially, if you have a plus, it’s an electrophile, so you should be thinking about what sort of a molecule you can come up with that has the same skeleton and is electrophilic.
Likewise, if you have a minus on your C, you’re dealing with some sort of a nucleophilic species. Typically, when we are thinking about the C-nucleophiles, we’re talking about various organometallic compounds, or enols/enolates. While there are other possibilities, these are the most typical examples you’ll normally see in your course.[transition]
So, looking at our synthons and thinking a little about all different possibilities, we can come up with the two molecules: a carbonyl compound (which happens to be our starting material) and an α,β-unsaturated ketone. These two molecules can react with each other under basic conditions in a reaction we know as Robinson Annulation. While you could possibly come up with something else or other alternatives, these two are, perhaps, the easiest and most straightforward synthetic equivalents which won’t require you to come up with exotic procedures, protecting groups, or other hoops you’ll need to jump through to get your chemistry going.
Putting It All Together
Now, when we have figured out the reagents (synthetic equivalents), it’s time to put it all together. First, we’ll need to make a good nucleophile out of our starting material. Since it’s a ketone, we can use a strong base to convert it into an enolate:
Once we have an enolate, we can add the other reagent to it to go through the Michael Addition:
Finally, once we have our Michael addition product, we need to finish this Robinson Annulation cascade by doing an intramolecular aldol condensation:
This sequence assembles the carbon skeleton for our molecule. You could also perform the Robinson annulation in acidic conditions as a single pot reaction. For the sake of practice, I encourage you to write this entire mechanism in acidic conditions and let me know if you got it in the comments below.
Alright, now, when we have assembled the molecular skeleton for our compound, we have one last thing left: we need to perform functional group changes.
In this case, we’ll need to reduce the C=C and the C=O double bonds. There are a couple of different ways how we can do it. For the sake of simplicity, I’ll kill both birds with the same stone and reduce both with molecular hydrogen at high pressure using a heterogeneous catalyst like platinum or palladium. You could also do this reduction with complex hydrides like NaBH4 or LiAlH4, which tend to completely reduce the conjugated carbonyls like the one we have. The complex hydrides are, however, less regioselective and occasionally don’t work for a complete reduction like the one we want to perform, so I’m choosing a “foolproof” way here, although it does require harsher conditions than you would expect in a regular reduction with hydrogen gas.
The final reduction gives us the target molecule!
Did you like this synthesis problem? Was it hard or were you able to see the solution right the way? Let me know in the comments below!
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