The reduction of alkynes is one of the many reactions you will encounter as an organic chemistry student. Understanding the nature of reactants and intermediates will help you determine their products. In this article I’ll help you identify if your alkyne is reduced to an alkane, or perhaps a cis or trans alkene.
The first thing to ask yourself is ‘what is downsizing?
You can use the chemistry mnemonic of LEO the lion says GER
Where LEO = Loss of Electron Oxidation
and GER = Electron Reduction Gain
But this is too much thought for organic chemistry. In orgo you have to think about atoms and bonds.
So try this:
electron + proton = hydrogen atom
Therefore reduction = addition of hydrogen
If we are reducing an unsaturated molecule, reduction means breaking the pi bonds and adding hydrogen.
The standard reaction uses H2 gas in the presence of a metal catalyst. The catalyst grabs the molecule, breaks the pi bonds, and inserts hydrogen into each carbon that previously had a pi bond.
This takes your alkyne directly to an alkane
However, if you interfere with the hydrogen addition process, you end up breaking only one pi bond, resulting in a cis or trans alkene.
Let’s start with trans.
When an alkyne reacts with metallic sodium in liquid ammonia, the resulting reduced product is a trans alkene. Let us summarize the reaction without delving into the mechanism.
This reaction proceeds with a radical intermediate. When the pi bond in the alkyne is broken, the radical and lone pairs of electrons repel each other, pushing each other as far apart as possible. Since the farthest they can go is trans, their end product follows as a trans alkene.
now for cis
Reduction with H2 gas in the presence of a Lindlar catalyst results in a cis alkene.
Think of Lindlar’s catalyst as a weakened metal catalyst. This tries to reduce the alkyne but gets ‘stuck’ in the middle. Since the metal tends to grab carbon atoms from the same side, the 2 hydrogen atoms add without, or from the same side.