Leah here, from leah4sci.com and in this video we're going to look at alkyne reduction or
hydrogenation.
Where we reduce an alkyne to an alkane, a cis alkene or trans alkene.
Recall from general chemistry that reduction means gain of electrons.
We're gonna translate this to the gain of hydrogen atoms which also happens to include
the electrons.
Since an alkyne has two pi bonds, we have the potential to reduce both of them, meaning
break the pi bond and add hydrogen atoms.
An alkyne can be reduced in three different ways.
The first is a standard catalytic reduction using h2 gas and some metal like platinum,
palladium or nickel.
This reaction will attack both pi bonds giving us two sets of hydrogen atoms being added,
we'll show one at it for the first pi bond that we break and a second set of hydrogen
for the second pi bond that we break.
This is the most that we can reduce an alkyne, because we only have two pi bonds, but if
we go with the partial reduction and only attack one pi bond, we have the option of
forming a cis alkene or forming a trans alkene.
The cis reduction happens in the presence of a poison metal catalyst, the most common
one you'll see is the Lindlar's catalyst so you'll write it as h2 with Lindlar's.
This will break just one pi bond and add two hydrogen atoms sine to each other on the same
phase of the molecule, putting the two r groups sine to each other on the other phase of the
molecule for a cis product, and finally we have the dissolving metal reduction which
is dissolving a group one metal like sodium or lithium in a solution-like liquid ammonia.
This reaction will break one pi bond and give us two hydrogen atoms that add anti or opposite
from each other, pushing the r groups anti or opposite to each other for a trans product.
Let's go into each reaction in more detail.
Catalytic reduction of alkynes is very similar to what you saw from alkenes.
Recall from alkene reactions that catalytic reduction takes an h2 gas and puts it on to
the pi bond and breaks the pi bond adding to the same phase of the molecule.
The mechanism isn't fully understood but we have some idea as to what's going on.
Imagine that you have a metal catalyst that is holding hydrogen atoms, this is because
the h2 gas already broke apart and was grabbed by the metal and is just sitting on the surface
of the metal.
Now imagine that the metal is going to reach for the carbon atoms and when it does so,
it breaks that pi bond between them.
This is why the reaction is a sine addition because the metal grabbing the carbon, reaches
for both carbons at the same time from the same position and then the hydrogen atoms
that are also sitting on the metal will simply slide over and attach on to the carbon atoms,
they're adding from the same phase of a molecule.
This would happen to both phases of the pi bond so that your ultimate product is a carbon
chain with just a sigma bond between the two carbons and then two sets of hydrogen atoms
having added together at the same phase.
So we'll show green for the first pi bond that we broke, and blue for the second.
This is a fully reduced alkyne.
When you write out this reaction, you have to specify or assume that is h2 and excess
because to reduce an alkyne with two pi bonds, we're adding two sets of hydrogen atoms.
You'll it shown with the word "excess", just the letters xs or even two molar or two equivalents
to show that we're adding 2 mols.
But don't think that 2 moles means if I decide to only put 1 mol, then we'll only have one
pi bond broken, that's not the case because imagine a situation where you start with three
alkynes and you're thinking if I only use 3 moles of h2 gas, then each alkyne will have
one h2 added and I will get an alkene product.
That is not the case because it's not easy to control and the product will be random,
the green one might get zero reduction, giving you an alkyne as your final product, the blue
one might get a double reduction, giving you an alkene as your final product and the purple
might go halfway, giving you an alkene product.
This is not all alkene, it's nothing, complete, impartial.
Definitely not effective to get a product.
So what do you do if you want a catalytic reduction that stops at the alkene?
Question is, do you want cis or trans?
In order to get a cis alkene, we have to use a poison metal catalyst.
A poison metal catalysts still uses h2 gas.
That means we're still adding the hydrogen but being that the catalyst is poison, it's
a little slower, it's a little effective, and it's only capable of adding one rather
than two moles of h2.
Imagine that the metal, as sluggish as it is, works the same way, it grabs the carbon
from the same phase and adds the two hydrogen atoms to the same phase, showing this with
one pi bond still intact and the two hydrogen atoms having added to the same phase, the
two r groups get pushed down, get pushed away for a cis product.
The most common poison metal catalyst that you're going to see in organic chemistry is
the Lindlar's catalyst and I like to think of it as the lazy catalyst.
It's so lazy it does half a job and then it quits.
Most of you will only have to recognize Lindlar's without the details, but in case your professor
asks, the Lindlar's catalyst is a monstrosity that looks like this, quinoline, palladium
dissolved with barium sulphate in a methanol solution or we could just say Lindlar's, much
easier.
Don't memorize Lindlar's cis, understand why so the key why in this reaction is that a
metal grabs from the same phase, the hydrogen's get added sine to each other and that gives
us a cis product.
This is opposite of the dissolving metal reduction, where the key intermediate that determines
the why is a radical, and that radical intermediate will force a trans final product.
This reaction happens because we're using a group one element which in its neutral state
has just one valence electron.
For example, sodium or lithium.
This is dissolved in a cold solution, ammonia at room temperature is not liquid but if it's
really cold, it will be liquid with a radical intermediate.
Let's take a look at the mechanisms so you'll understand exactly why it goes trans.
This mechanism starts out with a sodium and solution, even though it's neutral, it should
be happy, it's not.
It has that one unpaired valence electron making it very unhappy and it doesn't want
the electron so it gives it away and it gives that electron to the carbon atom.
Notice for showing the mechanism with a single headed or a fish hook arrow, to represent
the movement of one rather than two electrons, making this a radical mechanism.
I also want you to follow the colors of every electron as we move along so you'll understand
exactly where things are moving and why they're doing what they do.
We'll show the pi bond as two red electrons breaking apart.
When carbon gets that one green electron from sodium, that electron does not want to be
by itself, so one electron from the pi bond with a fish hook arrow showing that just one
electron moves over to meet that green electron and form a lone pair, the other electron gets
kicked away.
It's a third wheel, there's nothing for it and so it breaks away on to that second carbon.
Our intermediate looks something like this, on the right we have a green and a red electron
with a lone pair and therefore a negative charge.
On the left we have a neutral radical.
This right here is where we determine that the product will ultimately be trans, soon
as we go from the triple bond which is linear to the double bond, the angle changes.
We have two options, on the left, we'll show the lone pair of electrons cis to the radical,
on the right we'll show the lone pair of electrons trans to the radical.
Don't forget these electrons have their personal space and the VSEPR (Valence Shell Electron
Pair Repulsion) theory tells us that they don't like being next to each other.
When we have this cis intermediate the two electron clouds are too close together, electrons
near electrons makes them very unhappy, very unstable and therefore not likely to form.
On the right, we have the two electron clouds as far away from each other as possible, having
the lone pair of electrons facing completely anti, completely away from that radical, they're
as far away from each other as possible, making this the absolute most stable intermediate
and since the molecule had to adopt a trans configuration, the ultimate product will also
be a trans configuration because of this intermediate right here.
But we have too many electrons going on and this is very unhappy.
Remember this reaction takes place is nh3 liquid which means we have ammonia floating
around everywhere.
Ammonia is a polar protic solvent with a partially negative nitrogen atom and a partially positive
hydrogen.
That partially positive hydrogen will attract the lone electron pair, the one that's made
from the green and red electrons.
These electrons will reach out for one of the hydrogen atoms, breaking the bond between
hydrogen and nitrogen to collapse those electrons back on to the nitrogen atom.
The green and the red electron are still attached to the molecule but now they're sitting as
a bond between carbon and the purple hydrogen we grabbed from ammonia, we also have an nh2
remaining in solution with a lone pair of electrons from the initial nitrogen and a
second lone pair from the bond between hydrogen and nitrogen that collapsed, giving this a
negative charge well guess what, that sodium, when it donated it's electron gave us an na+
and this is now a positive negative ion.
This gives us a balanced salt somewhere in solution that we don't really care about,
what we do care about is the fact that we still have a radical and that is not stable.
Lucky for this molecule, there is another unhappy sodium floating around the solution,
it is unhappy because despite being neutral, it has that one unpaired electron and it wants
to get rid of it.
It gives that electron to the carbon that already has one red electron, the one radical
allowing sodium to get rid of that electron, forming an na+ that dissolves in solution
and a second lone electron sitting on the left carbon atom.
This gives carbon on the left a lone pair negative charged.
But carbon is not quite satisfied because the negative charge on carbon still no good
so it reaches out for hydrogen for another ammonia floating around the solution, breaking
the bond between hydrogen and nitrogen, collapsing those electrons on to nitrogen, and this gives
us our final product.
We have the red electron and green electron, once a lone pair, now sitting as a single
bond between carbon and the hydrogen that we grabbed from ammonia, the rest of the ammonia
molecule is now nh2- floating around the solution, able to pair up with the sodium that we created
from that second radical step.
But these are just spectators, what we care about is the fact that we formed a neutral
trans alkene.
Be sure to join me in the next video where we look at the mechanism for the hydrohalogenation
of alkynes, you can find that video along with the entire alkyne video series, practice
quiz and cheat sheet by visiting my website, leah4sci.com/alkyne .