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How Stereochemistry matters

I often say that the main theme of organic chemistry 1 is “Stereochemistry”. Today I want to try to make it a little clearer why that’s the case.

It starts with alkynes.

Earlier we went through different reductions of alkynes – the Lindlar, which gives you cis alkenes, and then Na/NH3, which gives you trans alkenes.

Alkynes are kind of like a blank canvas. You can do all the reactions of alkynes mentioned earlier, but by reducing them with Lindlar or Na/NH3, you can subsequently do all the reactions of alkenes.

When you combine the fact that a lot of reactions of alkenes have different stereochemical restraints (remember the 3 buckets?) this can lead to a lot of complexity. Very testable complexity.

It’s going to look simpler in the forward direction, so let’s do two examples.

Let’s start with a simple alkene, like 2-butyne. Depending on whether you use Lindlar or Na/NH3, you’ll get the cis or the trans alkene. Now, subject each of these alkenes to a typical reaction of alkenes – say, epoxidation with mCPBA. This will form epoxides of each of these two alkenes. Note how they are diastereomers of each other (stereoisomers but not enantiomers).

You can do the same thing with a different reaction – say, bromination with Br2. Repeating the same process gives you a mixture of diastereomers, although here bromination always gives you the trans product (it goes through the 3-membered ring!).

This should hopefully be understandable in the forward direction. But here’s the real test. Can you recognize it in the backwards direction? Because instructors love giving these types of problems for synthesis questions.

Here’s an example. Can you do it?


Thanks for reading! James