professor Dave here, let's talk about hydrogenation of alkynes.
so when we first talked about addition reactions one of the first things that we talked
about was hydrogenation, and we learned about some different ways that hydrogen
atoms can add to a pi bond, and so the first way that we looked at to do that
was just simple hydrogen over palladium. we saw that two hydrogen atoms would add
syn to the alkene and we saw that that was an example of hydrogenation, and so
for all the addition reactions we looked at all kinds of ways that we can do
chemistry with alkenes, but sometimes we want to hydrogenate alkynes, and so
there's a few different ways to do this. now the first is just the example we
just said, the same way that we can go from the alkene to the alkane with
hydrogen over palladium or some similar metal, is going to
hold here. it's gonna do the same thing, right, hydrogens can attach to, can
interact with any pi bond and so we're gonna have that from go from the alkyne
all the way to the alkane, so it's not going to be able to stop at the alkene
functional group level, it's gonna continue because we know that hydrogen
will interact with alkenes as well so we're gonna go from the alkyne
all the way to the alkane. so if you ever see an alkyne and these conditions we
know that we're just gonna get rid of all the pi bonds, we're gonna go straight
to the fully saturated alkane. but there may be instances especially in synthesis
right for a synthetic technique we may want to go just one step down, we may
want to go from the alkyne and stop at the alkene, and not go all the way to the
alkane, and so there's a couple ways to do this. the first one looks like this
this is this is a situation right with this this is called lindlar catalyst and
so what it is, is it's hydrogen and it's some it's always it's a metal it's
palladium but it's been poisoned and so what happens is that actually limits its
ability to hydrogenate. it is only able to hydrogenate alkynes and not alkenes
and so it's going to be able to interact with this alkyne and produce an alkene
but it is specifically going to produce the Z alkene right so remember when we
have an alkene we can have two different kinds of stereochemistry we could have
an E or a Z alkene and if you remember my tutorial on this we talked about how
to use the cahn-ingold-prelog Convention to assess whether an alkene is E or Z and
so in this case it's Z because if you remember zee zame zide, right so these
groups are on the same side of the plane containing the double bond so this is a
Z alkene and so when we see these conditions now there there are a few
variations, you may see it written a different a few different ways
oftentimes you will just see the words lindlar catalyst or simply lindlar and
so what this is telling us is we know we're going from the alkyne to the Z
alkene okay so that is one thing that we could do very deliberately in a
synthetic pathway, if we have an alkyne and we want to get a z alkene to then do
some other chemistry. now obviously if we have a way to get the Z alkene we want
to have a way to get the E alkene as well depending on what what we're trying
to do in our synthetic pathway and so there's another way. and let's start with
that same alkyne one more time and what we're seeing is that if we have this
situation we could see it like this with lithium and ethyl amine and ammonium
chloride, an alternate way that you will commonly see a sodium and ammonia so
really the key here is one of these metals lithium or sodium one of those
one of those alkali metals that is going to do this chemistry it is going to
again in a limited way it's not going to be able to reduce it's not gonna be able
to hydrogenate the alkene so it is gonna stop at the alkene functional group
level, it's gonna take the alkyne and it's gonna go to the
alkene but it will specifically go to the E alkene, okay so these are three
things that we want to remember in terms of hydrogenation of alkynes we have
three ways that we can go, we can go all the way to the alkane with simple
hydrogenation over palladium metal or we can use lindlar catalyst and go to the Z
alkene or we can use one of these two situations and go to the E alkene.
so let's put this into context let's say we're looking at this molecule this
alkyne and we are going to attempt three different kinds of hydrogenation let's
make sure that we understand what the product is going to be in each of these
cases so we've got an alkyne this is a 1 2 3 4 5 carbon alkyne really quickly a
lot of times students encounter some confusion when depicting alkynes, it does
have to have this geometry and these are carbons right there
remember that carbons participating in triple bonds are sp hybridized and
therefore exhibit linear geometry so this is the right way to draw that right
we can't have alkynes in a zigzag fashion because that that doesn't depict
the geometry accurately. we do have a carbon here and then 2 & 3 are triple
triple bound and then 4 & 5 so this is a 5 carbon this is this is 2-pentyne would be
the IUPAC name and now let's say that we're gonna do these 3 different kinds
of hydrogenation here let's say we do hydrogenation over palladium and so what
are we gonna get well we know that when we do this we are not able to stop at
the alkene right this is going to hydrogenate it's going to react with
both of those pi bonds and we're gonna go all the way to the alkane so we're
gonna get a 5 carbon alkane we're gonna get pentane right there. so that's what's
going to happen if we use the those conditions for hydrogenation. now let's
look over here now what is this we need to be able to look at conditions and as
best we can sort of memorize what they do so that on a test we know what to
draw and we should recognize that this is lindlar catalyst or once again we
might just see lindlar right there and what we want to associate in our minds
is that lindlar catalyst is always going to stop at the alkene and it
specifically will stop at the Z alkene so we're gonna get a five-carbon alkene
right so we're gonna have a 1/2 right this is gonna be and so 1 2 3 right it's
going to connect these two but they're gonna be Z it's gonna it's
gonna be a Z alkene right so this necessarily has to have this orientation
we need to have these alkyl groups projecting in the same direction because
zee zame zide, so 1 2 3 4 5 that is a five carbon alkene this is in the correct
position and we also have the correct stereochemistry at that location and so
we've got the Z alkene. that leaves our last situation here we could add these
conditions or once again you may often see sodium and ammonia and so when you
see something like that again we're looking for this alkali metal there
lithium or sodium that's what's going to be doing the chemistry and we are gonna
get again the five carbon alkane but now we are going to get the E alkene right
so the E alkene has the alkyl groups projecting in opposite directions so
whereas over here right these are the hydrogens right there those are the
ones that got on there so the hydrogenss added to the same side we've got Z same
side over here we added these two opposite sides okay so we've got the
alkane we've got the Z alkene and we've got the E alkene and hopefully that
helps us understand these three different sets of conditions for
hydrogenation of alkynes.