hey guys so for this practice problem we have a little two-step sequence here
we've got this substrate with two different kinds of carbonyl
functionality and we are gonna run this through two reactions one with strong
base and then two with these conditions here so try to think critically and see
if you can get the right answer here.
okay so let's take a look at this we've got this six carbon substrate right one
two three four five six, six carbons we have an aldehyde functional group and a
ketone functional group and we are doing this the first reaction is simply strong
base we've got some hydroxide and so ideally the first thing that we think of
is enolate chemistry right we know that with an aldehyde or a ketone in the
presence of strong base like hydroxide those are your classic conditions to do
some aldol chemistry and so we know that we're gonna make an enolate now the
thing is we've got alpha protons here and here and here but there's one clue
here there are two carbonyls so we know that we are opening up the possibility
of some intramolecular chemistry right because anytime it is possible for
something like an enolate in this case to attack another portion of the
molecule that is going to be tremendously kinetically favorable
because it doesn't then have to find something else in solution to collide
with the the other functional group that's gonna interact with is right
there ready to go so that certainly is going to happen much faster now the
thing is ring size is gonna dictate the position of enolization
because if we analyze say here right what what size ring is that gonna allow
us to get well this carbon would have to attack this so that'd be a one two three
four membered ring that's not so good for membered rings
are a little bit too small to be favorable likewise if we analyze here
one two three four it would attack that carbon and another four membered ring
not so great but what if we analyze here right we've got some protons there that
are enolizable and if we go ahead and and and get one of those guys so let's
let's take our hydroxide and let's have that go up here so let's make our
enolate now we're gonna we're in a much better position here because we've got
this carbon is going to attack this carbon and so this is a 1 2 3 4 5 6
membered ring that we're gonna make because this carbon right the alpha
carbon there is going to attack this carbonyl carbon so let's show that
happening we're gonna reform that carbonyl this is going to go over here
and attack that and so we get well let's just go ahead and draw the six membered
ring so you know what let me do that let's get a little more room we're gonna
go over here so let's draw the six membered ring and now all we have to do
is this so let's say we call this number one right we can just do this like this
we'll just number them we know that we're making a six membered ring so
let's just make the six membered ring now carbon one is the one that attacked
carbon six right so that is our new bond right there now what else is there well
on carbon two we reformed the carbonyl right on carbon two this pi bond came
back down so let's draw that and then three four five nothing and then on six
six should have that oxyanion now right so this was the this was the carbon
where we analyzed there is the carbon that did the attacking it
made the bond from 1 to 6 that's our new sigma bond and then 6 was a carbonyl
now oxyanion right so that's that and then let's just pick up a proton from
water and protonate that real quick and
we'll get our aldol addition product and then we know that from the aldol
addition product we can just go ahead and get the aldol condensation product
so let's take some more base and let's go ahead and get that proton and kick
this off we're in strong base so another hydroxides not going to hurt and we'll
do this so that is the product of number one number one is getting us to this six
membered ring where we have the carbonyl reformed we've got that pi bond that's
the product there where was this intramolecular aldol condensation now
number two what does it look like we're gonna do here well let's draw these next
to each other and in a very particular way and maybe it will become more
apparent
when I draw them this way it may become a little more apparent what's going to
happen this is a classic this is a great diene and therefore this is a dienophile
and we know that we're going to be doing some diels-alder chemistry and
this is possible because of this pi bond in the aldol condensation product so
let's go ahead whenever we're doing diels-alder we can draw these dashed
lines just to guide us because we know we're going to be making a new six
membered ring if you don't get a six membered ring when you're doing
diels-alder something went wrong because it will always make a new six membered
ring and all we have to do is shuffle these pi electrons around this pi bond
can become the new pi bond this pi bond can become a new sigma bond this pi bond
can become a new sigma bond and we are going to get this we know that this is
the new six membered ring this is the old six membered ring this is still
there we've got that new pi bond and there has to be a bridgehead carbon
there okay so that is gonna be our product just to
really quickly clarify what we did here these are the new bonds right we've got
these bonds here we made two new carbon-carbon sigma bonds in doing the
diels-alder right that's always going to be the case we're gonna make a new the
the four carbons that were participating in our sorry the six carbons that we're
participating in pi bonds are now in a six membered ring together so that is
these six carbons but the thing is this there was this fifth carbon on the
diene and that's not going anywhere right that better still be in your
product so that's gonna be a bridgehead carbon right there because it's not part
of the new six membered ring but the the initial five membered ring still exists
so we have this five membered ring in addition to this new six membered ring
and then the rest of this molecule should also still be there nothing
happened to the rest of that so when you're doing diels-alder like this
whether whatever you're doing with your carbons you definitely want to make sure
you don't gain or lose any carbon material so say if you didn't have
bridgehead carbon because you forgot about it you could count your carbons
and say oh I've got 12 but over here or you know or sorry yeah
10 but over here we had 11 because 6 and 5 so that would tip you off you'd say
well I I must have lost something and then you could remember that bridgehead
carbon so that's the deal with the diels-alder reaction and so once again
this was a little bit of a clever one we had to be a little bit insightful we had
to say I know that I have this compound with two different carbonyl
functionality with two different kinds of carbonyl functionality reacting with
strong base we knew that we were going to be doing some enolate chemistry some
aldol condensation specifically we're gonna do intramolecular aldol
condensation and that guided us as to the s to the location where enolization
took place because it had to be the location that would allow for a stable
ring to form so four membered rings no good but we analyzed here and we were
able to cyclize and get a six membered ring and that's how we got all the way
to the aldol condensation product for number one so number one was our
intramolecular aldol condensation and then just reacting with this diene and
some heat those are perfect conditions for diels-alder we had our diene and our
dienophile and this was our diels-alder product so this is the
product of this two-step sequence.