ok for this practice problem we're gonna look at some Diels Alder reactions so
this is a Diels Alder reaction in the forward direction and then we're going
to try to do a retro Diels Alder as well so if you're unfamiliar with this go
ahead and take a look at my tutorial on Diels Alder cycloaddition and we're
gonna try to draw the correct product for this Diels Alder and then given the
product of this Diels Alder try to draw the reactants because it's very
important to be able to think retrosynthetically in this manner, so give this a shot
ok so with Diels Alder reactions it's very important to understand that we're
always gonna be reacting a diene and a
dienophile, so here's our diene, meaning we have two double bonds and then a
dieonophile is something else that has at least one pi bond and so typically that's
going to be an alkene but we can also have an alkyne dienophile so the
easy part about the Diels Alder is that the mechanism is always the same. we're
always gonna create a six-membered ring between the four carbons that are the
that are participating in double bonds in the diene and then the two carbons
that are participating in the double bond in the dienophile that is involved in the
reaction, so we're going to push these pi bonds around and we are gonna
create a six-membered ring so here are the two new sigma bonds, in a Diels Alder
we are always going to create two new sigma bonds and one new pi bond so once again
it doesn't matter which way we cyclize these because they're actually not
moving in one direction or another, it's a concerted reaction but this pi pond has
gone here so that's this new pi bond right there and then this pi bond has
become this sigma bond and this pi bond has become this sigma bond. but the
interesting thing about using an alkyne as a
dieophile in a Diels Alder is that we're only utilizing one of these
pi bonds, the other pi bond is going to remain intact and so that's why we still
see a pi bond right there. if this had been an alkene like normal, we would not
see that pi bond there and that would be our, we have a product without that
pi bond there, but since it was an alkyne we do have an additional pi bond
these are now sp2, trigonal planar so those methyls are flat and in-plane so
that's the correct Diels Alder product for that reaction. now over here we
want to go retrosynthetically
it's very important to be able to look at a product and see the reactants that
it might have come from and so if in Diels Alder we always make a
six-membered ring, if we're doing retro Diels Alder we always break a
six-membered ring, and what we need to be able to do is identify the six-membered
ring and then see that since this pi pond is here right remember we're always
getting a pi bond then it must be these two sigma bonds that we break and
therefore pull apart and so what we can do is to even do retroactive electron
pushing arrows, why don't we make arrows that undo what we did, and so
what we get here is this sigma bond not literally because this is the reaction
in the forward direction but just for our own bookkeeping we can put this
sigma bond down here to generate that pi bond, this sigma bond will break and make
a pi bond there and make a pi bond there and so what we're understanding is that
this carbon here, this was not participating in the Diels Alder
reaction but it was part of the diene because the diene must have been this
five-membered ring, so because if we're breaking these bonds right there
boom boom and then pulling those fragments apart after were done
shuffling the pi electrons, we must have pi bonds there and there to create the
diene, and then on the dienophile, we must have had a pi bond there, and in terms of
stereochemistry because these were cis to one another
on the product we must have had a cis or Z alkene as opposed to E, if
this had been one dash and one wedge we would know that it would be an E
alkene instead as the dienophile, so this is the correct product for this
Diels Alder and these are the correct reactants for that Diels Alder product
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