Second Most Important Reactions of Alkynes: Lindlar – Na/NH3
Today, let’s talk about what (in my opinion) are the second-most important reactions of alkynes.
Alkynes have two Pi bonds. As you know by now, Pi bonds can be hydrogenated to form C-H bonds and break the C-C Pi bond.
If you treat an alkyne with just Pd/C and H2, you’ll end up adding hydrogen not just to one of the pi bonds, but both of them. In other words, you get an alkane.
However, there’s a way to stop the reduction halfway through, so that you just reduce the alkyne. That’s what today is about.
There are two ways to do this, and stereochemistry is really important here.
The first way is to take a modified version of Pd/C that isn’t as reactive as normal. We call this a “poisoned” catalyst. Pd/C can be poisoned with the use of a little bit of lead (Pb) and an amine (such as quinoline) to make what we call “Lindlar’s Catalyst”.
Lindlar’s catalyst reduces alkynes to alkenes, to give the Z (“Cis”) alkene.
The second important partial reduction of alkynes involves sodium (Na) and ammonia (NH3). This ends up reducing the alkyne to give the trans alkene. And it stops there.
[A lot of students confuse this with NaNH2/NH3 – they’re not the same thing! NaNH2 is a base. Na is a reducing agent.
Some instructors use Li instead of Na for this. ]
Stereochemistry is super important because alkynes are kind of a blank canvas – they can be “decorated” in all kinds of different ways.
Thanks for reading – James
P.P.S Reagent Friday – Sodium