Putting It Together

I have a lot of students ask me about how to do synthesis. Today, I’ll share with you  a 3-step method I suggest to them, and today we’ll start to go through it. It goes like this:

1. Analyze the starting material and the product – ask yourself, what bonds are formed and what bonds are broken?
2. When you’ve made your list, ask yourself – what reactions do I know that will form/break these bonds?
3. In what order do I do these reactions?

As an illustration, let’s use this synthesis question for alkynes.

Part 1. Asking “what’s new” ? 

A lot of students expect that they’ll be able to look at a synthesis problem and “just see the answer”. This is not how it works. A synthesis question is a bit like solving a Sudoku . There’s a method to it.  Before you can make headway, you have to do some arithmetic first.

Step 1 is where you do your arithmetic. It might really help if you draw out all the hidden hydrogens, because these can be hard to spot. You want to make as complete a list of the bonds that form and break as possible – including all the hydrogens.

Like this:

Part 2: Asking “what are some potential reactions?”

The next step is to ask yourself what reactions form and break these bonds. There’s no getting around the fact that you have to know what each reaction does. But there are a few general hints (that extend beyond alkynes):

• any time you’re breaking a C-C (pi) bond, it’s an addition reaction
• if you’re forming a C-C (pi) bond, it’s an elimination reaction. 
• substitution reactions form and break a bond on the same carbon
•  look for good leaving groups
• pay close attention to stereochemistry

For the reaction we’re talking about, we’re breaking two C-C (pi) bonds, which points to two addition reactions. We’re forming two C-H bonds and two C-OH bonds.

What are some addition reactions that will form these bonds? Here, you have to come up with a list of possibilities. We’re just “playing around” here.

• Hydroboration (form C-H, form C-OH)
• Hydration (form C-H, form C-OH)
• dihydroxylation (form C-OH, form C-OH)
• hydrogenation (form C-H, form C-H)

Next,  we’ll go through these possibilities and figure out which ones make sense and which don’t. 

Part 2

Today, let’s finish up a quick snapshot of how to break down a simple synthesis problem.

Last time I said that the process goes something like this:

1. Identify the bonds that break and form.
2. Identify reactions
3. Put them in the right order

The second step – identifying reactions – is really where the rubber meets the road with synthesis questions. There’s no way around it: you have to learn which reactions form and break certain bonds.

For the synthesis problem below, last time  we came up with the following possibilities

• Hydroboration (BH3, then H2O2/NaOH)
• Oxymercuration  (H2O, HgSO4)
• Hydrogenation (Pd/C, H2) or Lindlar’s catalyst/H2 or Na/NH3
• Dihydroxylation (OsO4)

There are two ways to test out whether these reactions would be appropriate for our synthesis. The first way is to start with the alkyne and work forward using these reactions. The second way is to start with the product and work backwards using these reactions. This is a short synthesis problem, so let’s take the slightly easier route, which is working forward from the alkyne.

If we apply each of these reactions to the alkyne, we’ll get the following results. As you should be able to see, only two of them are potentially useful. The other ones won’t get us to our desired product in a reasonable number of steps.

Looking at the results, it seems only that Lindlar and Na/NH3 are acceptable reactions to start on the alkyne.

Given that this is the case, let’s think about what the other reaction could be. If we’re hydrogenating, this will form two C-H bonds. So we’d need to form two C-OH bonds.

One way we know of forming two C-OH bonds is through dihydroxylation of alkenes with OsO4. This reaction proceeds in syn fashion, so working backwards, we would have to start with the following alkene (the trans).

If we’re starting from the trans alkene, then we must have started with Na/NH3 as our first step. That’s the only way to get the trans alkene from the starting alkyne.

So let’s put it together now: two steps.

I realize that this might seem like a simple example. And it is, in the grand scheme of things.  But in order to get good at anything you have to start somewhere. And by recognizing the bonds that form and break and the reactions that can get you there, at least you’ll have a method. You can start whittling away at a tough synthesis problem bit by bit instead of staring at it forever, expecting to just “see it”.

Thanks for reading!

James

P.S. Advanced readers only: Here, we did an “anti” hydrogenation followed by a “syn hydroxylation”. But did you notice that we could have obtained the same product through a syn reduction (Lindlar, H2) and an “anti”-dihydroxylation (mCPBA, followed by NaOH/H2O). Neat, huh?

Part 3