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Organic chemistry
Reduction of alkynes
Different methods for reducing alkynes to alkenes or alkynes. Created by Jay.
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at about6:25, why doesn't the Na bond to the C? Why instead does the C get an H from the NH3?(11 votes)
It does bond to the C, but the bond is ionic rather than covalent. The electronegative difference between C and Na is great enough so that the bond is essentially ionic.
The carbanion is a very strong base. It can easily remove a proton from NH3 to form the much weaker base NH2-.(13 votes)
At1:40, Jay introduces liquid ammonia as a reagent. If i remember correctly, ammonia is a gas at room temperature and at normal pressure. Does that mean that this reaction needs to take place at either a low temperature or a high pressure (or both)?
Will this reaction occur with ammonia as a gas? Perhaps the ammonia is aqueous or in some other solution? ... or am I missing something?(8 votes)- That is correct. The boiling point of ammonia is -33°C. If the flask is immersed in a mixture of dry ice and acetone (-78°C), the ammonia will condense to a liquid. High pressure is not needed. There must not be any water present.(11 votes)
- Why does using Lindlar Pd as the catalyst reduce it only to an alkene while using Pt as the catalyst allows it to be reduced all the way to an alkane? What is the stopping mechanism with the Lindlar Pd catalyst?(7 votes)
- Lindlar’s catalyst is a Pd catalyst that is "poisoned" with traces of lead and quinoline. This reduces its activity so that it can reduce alkynes but not alkenes.(13 votes)
- In Na/NH3 system, Is there any possibility that the alkene products are further reduced to form alkane or polymerized like free radical polymerization of vinyl monomers?(7 votes)
No, due to the radical mechanism, the pi electrons in the alkene will not split like in the alkyne. The Na/NH3 system is similar to the poisoned catalyst system where the alkyne will stop after the formation of the alkene.(4 votes)
What happens to the sodium atoms when they donate that one electron?(3 votes)
I believe they just become Na+ ions. They no longer affect the reaction.(5 votes)
- Hi since NH3 is a catalyst ,it should be conserved after reactions ends and product is formed but we are only left with NH2 not NH3 and in the reaction you did not mention Any role of hydrogen(1 vote)
- NH3 is not used as a catalyst, it is a reagent. Lindlar Pd is a catalyst because it is not used.(5 votes)
I get it that you can do cis or trans hydrogenation, but why is each reaction restricted to form either form? What makes the Lindlar Pd one give out cis alkenes and what makes Na / NH3 give off trans alkenes?(1 vote)
The hydrogen is absorbed onto the metal catalyst. When the alkyne approaches a hydrogen molecule absorbed onto the catalyst, the hydrogen atoms are both on the same side of the triple bond, leading to a cis alkene. With sodium and ammonia, as explained in the video the sodium donates an electron to one carbon which causes one of the pi bonds of the alkyne to split, with one carbon ending up as a radical (with a single unbonded electron) and the other as a carbanion (with a pair of unbonded electrons). The electrons of the radical and the carbanion repel each other by moving to opposite sides of what is now an alkene bond - this leads to a trans product.(4 votes)
How exactly is palladised charcoal formed?
and how is quinoline a poison for alkynes?(1 vote)- Add PdCl₂ and HCl to a warmed, aqueous suspension of activated charcoal. Then add formaldehyde (this reduces Pd²⁺ to Pd). Neutralize with NaOH. Filter off the catalyst. Wash it with distilled water, and dry it over KOH.
Quinoline is a poison for the catalyst, not the alkyne.
The nitrogen lone pair in quinolone binds chemically to the activated sites on the surface of the catalyst.
This reduces the number of active sites
Poisoned sites can no longer accelerate the reaction the catalyst was supposed to catalyze.
The activity of the catalyst is reduced.
Thus, a poisoned Pd/C catalyst will reduce an alkyne to the corresponding alkene, but it cannot reduce the alkene product to the corresponding alkane.(3 votes)
- Here NH3 is used to convert a alkene to trans alkene, I need to know that what if HCl is used instead of NH3 (HCl can serve the same function as NH3 i.e, donating a proton). Are the products formed be same in both cases?
Is this reaction an ideal one (if HCl is added)?(2 votes)- HCl in its pure form is a gas and unable to act as a solvent. HCl as a solution will react with the sodium and prevent the reduction from occurring.(1 vote)
- Why is the Na,NH3 reduction trans? Why can it not be cis?(1 vote)
- The radical electrons want to be as far apart from each other as possible, so they arrange to be on opposite sides of the molecule. Jay mentions this at3:50.(3 votes)
6:25
Video transcript
In this video, we're going
to take a look at two ways to reduce alkynes. The first way is a
reaction we've seen before. This is the
hydrogenation reaction. And we saw it before when
we hydrogenated alkenes to form alkanes. Here we're going to hydrogenate
an alkyne to form an alkene. And to do a
hydrogenation reaction, we need some hydrogen gas,
so some H2 right here. And then a metal
catalyst, so we're going to use Lindlar
palladium, which is a special type of catalyst. It will catalyze the
reduction of the alkyne on the left, the
alkene on the right. However, the reduction of
the alkene to the alkane down here is slow. So, so slow that we can
stop it if our goal is to just make an alkene. So this reaction will
form a cis-alkene. And it was a syn edition
of our hydrogens. So we're going to get
the two hydrogens adding on to the same
side, and this has to do with the mechanism of
a hydrogenation reaction. So you can check out the
earlier video on hydrogenation of alkenes to see more details. So Lindlar palladium,
a poison catalyst, it will reduce an
alkyne to an alkene. It will produce a cis-alkene. All right, so that's how
to make a cis-alkene. Let's take a look at how
to make a trans-alkene. So how do we reduce an alkyne
to make a trans-alkene. So here is our alkyne So we
have our triple bond like that. And we're going to
add sodium metal, and we're also going to add
liquid ammonia like that. So we're going to
form a trans-alkene. So I'm going to put-- this
time my two hydrogens are going to be on opposite
sides of each other. So this is formation of
a trans-alkene like that. And it does this by an anti
addition of hydrogens, right? So these are adding from
opposite sides like that. Let's take a look
at the mechanism to form a trans-alkene. So I start with my alkyne. So I'll go ahead and
put in my carbons there and put an R
group on the left side. And I'll make this
an R-prime group to distinguish it from
the R group over there. So we start with sodium, which
we know, being in Group 1, has one valence
electron like that. And in the first step
of the mechanism, this sodium atom is going to
donate its valence electron to the alkyne. So when we're showing the
movement of one electron, we use a half-headed arrow. So I'm going to show this
electron moving over here, but it's only one
electron so I'm only going to do a half-headed
arrow like that, not a full-headed arrow. So one of these bonds
here between the carbons is going to break. And one of the
electrons is going to move over here to
this carbon like that. And one of the
electrons is going to move over to the
carbon on the left. So let's go ahead
and draw the result of all those electrons
moving around. So we have an R group here. And we had a triple bond,
but now we only have a double between our two carbons, and
then we have R-prime over here. So the carbon on the right
picked up an electron from sodium, and it also
picked up an electron from the breaking of
that one bond there. So now it has two
electrons around it like that, which
gives us a negative 1 formal charge on this carbon. So it's a carbanion. It's an anion here. The carbon on the left
picked up one electron for the breaking of
that bond like that. So that's a radical
that's something we haven't talked about before. So we actually form what's
called a radical anion here. So let's go ahead
and write that. This is a radical
anion, so radical because there's an
unpaired electron there. And then it also has a carbanion
in the same molecule like that. So we have these electrons
that are pretty close together, at least how I've
drawn them, right? So we know that electrons
are all negatively charged, so all these electrons are
going to repel each other. So this isn't the most
stable way for this molecule to have in terms
of a conformation. These electrons
are going to repel, and they're going to
want to try to be as far away from each other
as they possibly can. So what's going to
happen is, we have our two carbons right here. And let's say that these
two electrons stay over here on this side. This one electron's going to
go over to the opposite side. They're going to try to get
as far away from each other as they possibly can. And same thing with these
R group here, right? So this R group is going
to try to get as far away from this R-prime
group as it possibly can. So this trans conformation
is the more stable one. So this is our negatively
charged carbanion right here. So in the next step
of the mechanism, we remember ammonia is present. So let's go ahead and
draw an ammonia molecule floating around like that. So here is our ammonia molecule. And the carbanion is
going to act as a base, and it's going to take a proton
from the ammonia molecule. So this lone pair
of electrons is going to form a new
bond with this proton, and these electrons are going
to kick off onto the nitrogen. So let's go ahead
and draw the results of that acid-base reaction. So now we have our two
carbons, with an R group right here, R-prime right here. And now this carbon on the
right is bonded to a proton. It bonded to a
hydrogen like that. And then we still have
our radical down here, so there is one electron
on that carbon as well. All right, so the next
step of our mechanism? Well, there's plenty
of sodium present. So here's a sodium atom
with one valence electron. The sodium is going to donate
this electron to this carbon. So just use a
half-headed arrow to show the movement of one electron. So if that sodium atom donates
that one valence electron to that carbon, let's go ahead
and draw the results of that. So we have two carbons double
bonded, an R group over here, a hydrogen, and an R-prime. And this carbon had
one electron around it. It just picked up one
more from a sodium atom. So it's like that,
which would give it a negative 1 formal charge. So this carbon has a
negative 1 formal charge. So let's go ahead and draw
that negative 1 formal charge. It's a carbanion. And once again, ammonia
is floating around, so let's go ahead
and draw ammonia right here, so NH3 like that. And the same thing is going to
happen as did before, right? The negative charge is
going to grab a proton. It's going to act as a base. And these electrons are going
to kick off onto the nitrogen here. And so we protonate
our carbanion, and we have completed our
mechanism because now we have our two R groups
across from each other. And we added on two hydrogens
across from each other as well like that. So we formed a trans-alkene. All right, so
that's the mechanism to form a trans-alkene. Let's look at a few examples. Let's start with this
alkene right here. OK, so carbon triple
bonded to another carbon, and we'll put a methyl group
on each side like that. OK, so let's do a few
different reactions with the same substrate here. So our first reaction will
just be a normal hydrogenation with hydrogen gas, and let's
use platinum as our catalyst. So this is not a
poison catalyst. This is a normal catalyst. So what's going to
happen is, first you're going to reduce
the alkyne to an alkene. And then since there's
no way of stopping it, it's going to reduce
the alkene to an alkane. So this is going to
reduce the alkyne all the way to an alkane. So if we go back up here
to beginning, remember, we said that a poison catalyst
will stop at the alkene, but if it's not a
poison catalyst, it's just going to hydrogenate
your alkene to an alkane down here. So this reaction is going
to produce an alkane. Let's go ahead and
draw the product. So we know that there are
four carbons in my starting materials. There's going to be four
carbons when I'm done here, so these two carbons in the
center here are going to turn into CH2's. And then on either side,
we still have our CH3's. So this is going to form
butane as the product. All right, this time
let's use a hydrogen gas, and let's use a
Lindlar palladium here. This is our poisoned catalyst. So it's going to reduce
our alkyne to an alkene, and then it's going to stop. And you have to think, what
kind of alkene will you get? You will get a cis-alkene. So if we draw our two hydrogens
adding on to the same sides, so now we have our methyl
groups going like that. So our methyl groups
will be going-- this and this would be our
product, a cis-alkene. All right, let's do one
more, same starting material. So this one right
here, except this time we're going to add
sodium, and we're going to use ammonia
as our solvent. And remember, this will reduce
our alkyne to an alkene, but it will form a
trans-alkene as your product. So when you're drawing
your product down here, you want to make
sure that your two hydrogens are trans
to each other. So they add on the mechanism,
and then your two methyl groups would also be on the
opposite side like that. So look very closely as to what
you are reacting things with. Is it a normal
hydrogenation reaction? Is it a hydrogenation reaction
with a poison catalyst, which would form a cis-alkene? Or is it reduction with
sodium and ammonia, which will give
you a trans-alkene.