Organic Chemistry Study Tips

By James Ashenhurst

Thoughts on Synthesis

Last updated: September 23rd, 2022 |

Part 1 – Synthesis

Let’s start talking about synthesis. Guess where I’m going to start? You might think,  “what bonds form, and what bonds break?”.

Well… yes, but first, let’s start with arithmetic.

Back in first grade, your teacher might have given you questions like this:

4+ 2 = ?

3 + 5 = ?

And you got more and more comfortable with answering those questions. But then, one day, your teacher threw you a curveball.

3 + ? = 9

The first time you saw this it might have looked weird. But then you learned that you could solve them by moving the 3 over to the right hand part of the equation (making it negative) and then solving it that way.

?  = 9 – 3

Believe it or not, we can apply this lesson to organic chemistry questions.

NOW we start with “what bonds break, what bonds form?” (it’s inevitable, I know).

Every time they learn a new reaction, I tell my students to ask themselves  what the “pattern” of bonds forming and bonds breaking for a given reaction is. Here’s the Wittig reaction, for instance. (The same pattern applies for every Wittig reaction)

Now there’s a reason why I tell my students to focus on “bonds formed” and “bonds broken” a lot.   I do it not only because it’s a simplifying way of looking at reactions, but it also works in the reverse direction.

 

So once you’ve learned a reaction in the forward direction, reverse that pattern. This is how you’d go “backwards” from a product, to give you back the starting material. Every “bond formed” becomes a “bond broken” and vice versa.

 

Next, try apply this pattern to different products. In the example of the Wittig reaction provided below, try drawing different alkenes, and applying the “pattern” of the Wittig to give you back different ylides and aldehydes (or ketones).

The more you practice, the better you’ll get.

Part 2- What’s New

Every synthesis problem is different, but there are still some common problem solving techniques you can use to solve each synthesis problem in turn.

One thing you *don’t* want to do is try to look at a problem and see if you “just get it”. That strategy has a pretty low success rate. You have to have a method. 

Solving a synthesis problem is like eating a really big steak. If you tried to eat it whole, you’d probably choke. What you want to do is to cut it into tiny little pieces. Then it’s manageable.

My advice is to break it down into 3 questions. I’ll go through them in 3 separate posts.

Today’s question to ask is, “what’s different?“.

And by different, I mean – you guessed it – what bonds have formed, and what have broken. If you can’t answer that, you want to be as *specific* as possible about what has changed in the molecule. What new atoms are there? What atoms have disappeared? Every tiny little piece of information helps in solving the puzzle.

Let’s look at this example.

  • Here, it should be easy to spot the fact that we’ve lost an ester. Specifically, we’re missing OMe and C=O. One carbonyl is left behind.
  • Also, we’ve added some carbons. There are a methyl group and an ethyl group that weren’t there before. That means we’ve formed C-C bonds.
  • Here’s the most subtle tip. If you’re really good, you’ll be able to spot the loss of C-H bonds. Note how each of the carbons on the starting material are either C=O, CH2, or CH3, but our product has a tertiary carbon (C-H). It’s a good guess that our starting material lost a C-H somewhere.

Making a list of bonds that form and break focuses our attention. Think of it like a “to do” list.

The next step will be to make a candidate list of reactions that will accomplish these “tasks” on our to-do list.  More on that tomorrow.

Try applying this to the next synthesis problem you see and tell me how you do.

Thanks for reading! James

Part 3 – What Reactions? 

Getting comfortable with thinking about reactions in the backwards direction is going to be a really valuable skill for doing synthesis.

You might have noticed that there’s more than one way that the alkenes above could be made using the Wittig reaction. Make a note of these reactions for which this is true; it will come in handy when planning a synthesis.

In the above part I  said that the first question to ask yourself when you’re doing a synthesis problem is to say “what’s new”? You want to identify all the bonds that form and break, if possible, all the atoms that are new, and all the atoms that have left.

This is going to be the “to do” list.

Once you’ve got a list, the next question to ask is:

“What reactions do I know that will form or break these bonds?”

Note that answering this second question can be a lengthy process. If you’re stuck, go through your textbook. Look for examples of reactions that might be able to achieve what you’re trying to do.

I wish I could give you a short cut toward making this step efficient. But I can’t. There’s no formula. There’s no short cut. At some point, you just have to know what bonds are formed or broken by each given reaction. I say there’s no short cut, because this data represents experimental results that were discovered – and they’ve been placed in the course material for a reason, as exemplars  of different key types of reactions.

It’s like learning a language: when you learn English, at some point, you just have to learn that the word “sheep” refers to a certain type of animal with fluffy fur that says “baaaa!”. There’s just no getting around it.

Expanding the number of reactions you know is like expanding your vocabulary in a language. The more reactions you know, the more powerful you will be coming up with potential solutions to synthesis problems.

OK, let’s look at the problem from yesterday.

Here, we see that we’re losing an ester, we’re adding carbons, and we’re breaking at least one C-H bond.

Let’s come up with some ideas.

  • loss of ester – this could occur through decarboxylation. In order for this to occur we’d need a beta-keto carboxylic acid. We’re starting with a beta-keto ester, so that’s definintely a possibility.
  • formation of C-C bonds ; there aren’t too many reactions that form C-C, but one of significance is enolate alkylation. I say this and not the Claisen or Aldol because we seem to be adding carbons that don’t have oxygens on them. This would suggest a straight alkylation. Also, the loss of C-H is consistent with alkylation.
  • if we’re alkylating, what kind of alkyl halides would we need? We need to add methyl and ethyl groups. We can’t do this in just one step. So there must be *two* alkylation reactions.


These ideas came from looking at the patterns of different reactions, and recognizing those patterns. It will take some time to see the patterns. That’s why instructors harp on the importance of doing problems – because by doing problems, you’ll get much better at this.

The most fun part (for me anyway) is that there’s not always one right answer.  There can be multiple ways to solve these problems. It’s like traveling from one place to another. You can fly, you can take the Interstate, you can take the back roads. As long as you get to the final destination, t’s OK.
Next: let’s work on the timing.

Part 3 – Putting Things In The Right Order

When doing a synthesis question, once you’ve asked yourself : 1) what new bonds formed and broken? and 2) what reactions can I use to accomplish these tasks?, the third question to ask yourself is simply:

“In what ORDER do these reactions happen?”

The order in which certain reactions occur is extremely important in synthesis. Often the presence (or absence) of a certain functional group is critical for various reactions to occur. For instance, in the synthesis below, it’s important to convert the methyl ester to the carboxylic acid before trying to decarboxylate; and it’s important to do the alkylation before the decarboxylation, because the enolate is much easier to form.
Trying to figure out the order in which things are done is MUCH easier at this stage – when we’ve narrowed down so many variables – than it is at the beginning, when there’s so many possibilities.


By the time you get to this step, most of the order in which to do things should be straightforward. Although at this stage, small modifications to the synthesis can occur. Give your final synthetic route a last look-over to make sure there aren’t any problems that could crop up with your synthesis. If there’s some obvious incompatibilities, it might be necessary to re-order things, or even to add in one last class of modifications: protecting groups. More on that soon.

Part 4 – Application

I recently got an email from a reader with an advanced practice problem they were struggling with.

Redacted

Here’s the problem. I really hope this wasn’t for a take home exam or anything. I want to show an example of the thought process that goes into solving a synthesis problem by applying the “3 questions” method I teach.

Synth problem

So how do you actually apply the “3 questions” method? Here’s what I wrote back. 


The first thing I’d notice is that all the groups on carbons 1 through 4 (that you’ve labelled) are the same. What’s changing is C-5 and C-6.

What’s new?
– new methyl group on C-5
– c-6 attached to an amine (NH2)
– C6 attached to a carboxylic acid.

What reactions do we know that can do each of the following
– form a C-C bond (such as C-CH3)
– form a C-NH2 bond
– form a C- CO2H bond.

Now let’s think:
CO2H can come from hydrolysis of CN
C-NH2 can come from reduction of C=N
[together these 2 steps are called the Strecker synthesis]
C-CH3 can come from alkylation of an enolate

this means C-6 used to be C=O…

because we can easily turn C=O into C=N with NH2  [typo: should have written NH3]
and we can add CN to C=N  (this is the Strecker synthesis)
we can also form the enolate of C=O by treating  with LDA and then CH3-Br

So we need to turn that C5-C6 double bond so that there’s a C=O at C-6.
This requires anti-markovnikov hydroboration of an alkene (BH3 then H2O2)

Then oxidize that alcohol to an aldehyde (PCC)

Part 3 is putting all those reactions together in the right order.

I’m omitting one final detail: acetone is used here too [any idea why?! ]  But this shows the thought process.

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