hey guys for today's problem we've got a little bit of a synthesis oriented
challenge what we want to do is we've got our we've got our trans butene and
we want to get to cis butene in three steps and so using the techniques you
know let's see if you can get from this to this in three steps
okay so let's talk about how we're gonna do this we've got our trans 2-butene we want to get to
our cis 2 butene now obviously this is not very practical question these are
both readily available so no one would actually do this in a lab but it's a
nice little thought experiment to test us and see if we can comb through our
our toolkits of of reactions and see if we can figure out a strategy a way to do
this and so we might look at the product and say alright well we want to get a
cis alkene, we want to get a Z alkene now what ways do I know how to do that well
I know that I can do a selective hydrogenation of an alkyne right we know
how did how to do how to do hydrogenation of an alkyne to
specifically get a Z alkene and so now we're left with the task of how do we
get to an alkyne so that we can do that okay well what methods do I know in
order to get an alkyne well we know that we can generate pi bonds through
elimination reactions but we need we would need to do two of them so maybe
we're starting to piece things together we can take a look at this and we can
say all right well what am I gonna do to set myself up for a situation where I
can get an alkyne so what if we do a dihalogenation right what if we do let's
just use bromine and we probably use maybe CCl4 right there and
so what's gonna happen there well we would get we would get we would get a
dibromo product right there right so we know that we can do dihalogenation
and we're gonna get to there so why is that useful because we know another
technique whereby we can get from this kind of molecule to the alkyne by doing
two successive eliminations right so what what method do we know that will
help us do that well we would need a base we need a pretty strong base let's
say we use NaNH2 and and we need two equivalents because we're gonna do two
eliminations right and we might we might run that in mineral oil and so we are
going to go right NH2 minus, a very very strong base that's going to go and grab
a proton do one elimination then it is going to go grab the other proton do the
other elimination and we are going to get to that alkyne right we're gonna get
to that alkyne that we wanted right and so we knew we were going to do that
because we had this strategy in mind we we that was the first thing we thought
of we thought the the best way to get that Z alkene is by hydrogenation and if
we remember what is the what is our method for doing that we might remember
that we're going to use lindlar catalyst right lindlar catalyst is going to
selectively hydrogenate the alkyne to the Z alkene so this is an excellent
little problem in terms of looking at something that looks like we haven't
even really done very much but then having to sift through our toolkit of
reactions and try to figure out a strategy so we've got our die
halogenation and then elimination twice to get the alkyne and then lindlar
catalyst to get us to the Z alkene.