Leah here from leah4sci.com and in this video I'm going to solve question 1 of the Organic
Chemistry 2 Final Practice Exam. You can download the exam and solutions on my website leah4sci.com/Orgo2Final.
In question 1 we're told to determine the products of this reaction and at first it
looks scary, but the key is to figure out what we're looking at. Anytime you see heat
and no reagent or solvent and realize that you're reacting pi bonds, think of the diels-alder
reaction. The first molecule has a diene so that will be half of the diels-alder and this
one having an electron withdrawing group on it has to be the dienophile. We know the reaction
but it's kinda tricky to figure out the next step given the rings and all the substituents
unless you put it into the standard form trick that I teach you in the diels-alder series
and that's to set the molecules as follows.
The diene gets the pattern 1,2,3,4 and the dienophile gets A and B. We know the product
of this reaction is a cyclohexene where we have 1,2,3,and 4 fused to A and B with the
pi bond between 2 and 3. Any other substituents on the molecules will just follow the numbers.
So let's start with the diene. We know that this is the portion that will act as the diene
so we'll number it 1,2,3,4 and then set it up over here, 1,2,3,4. Then we have to complete
the ring with five and six but I don't want them to be in the way, I want to draw them
out of the way as you do with any bridgehead carbons just to exaggerated the bond from
4 to 5 to 6 and back to 1 which is fine as long as 4 is attached to 5 which is attached
to 6 and back to 1. Let's not forget the OCH3 on carbon number 2. The dienophile is pretty
straightforward. We have a pi bond that'll be A and B. The question is, do we label it
A and B from the carbonyl down? or do we label it A and B towards the carbonyl? For this
you have to look at what groups we have and how that impacts the electronegativity and
the resonance of the molecule. The diene has an electron donating Oxygen from carbon number
2. If these electrons were to resonate onto the pi bond, then these electrons of the pi
bond would resonate towards carbon 1 making carbon 1 partially negative.
Now let's see how the dienophile can resonate so we know where we have a more negative or
more positive carbon. The electrons on the pi bond will resonate towards the carbonyl
carbon causing those electrons, the pi bond between carbon and Oxygen to collapse onto
Oxygen. This makes the lower carbon, meaning the one farthest away from the carbonyl partially
positive and the Oxygen partially negative. If we have a partially negative carbon on
the diene, we want to line up the more partially positive carbon of the dienophile next to
that and so we draw it as follows: We have the pi bond, this would be our partial negative
and that means the carbonyl is all the way down here. Now that we have it setup, we'll
proceed with the reaction, we have a concerted mechanism where all the electrons move at
the same time giving us a cyclo addition. Drawing the product looks scary so don't even
look at your started molecules when you're first drawing the product. Instead we'll just
draw a cyclohexane because that is the product of the diels-alder reaction. 1,2,3,4,A, and
B. Now let's look back at what we have and let's not forget the A and B over here. Starting
on the diene, I have an OCH3 attached to carbon 2 and since carbon 2 is pi bound to carbon
3 it's SP2 hybridized, it is flat trigonal planar so the OCH3, the methoxy group doesn't
come out of the page or go into the page instead it's in the plane of the page.
What we also have are 5 and 6 coming up as a bridge, as another fused ring so I can draw
it as I see it or I can just pop them in here as 5 and 6 and just show it like that, as
a bridge coming up out of the page. Then we move on to the dienophile. A and B are on
the reactive carbons but let's not forget the ring form with C, D, and F where F has
carbonyl so we just add that in. C,D,E and F, F binds back to A and F has a carbonyl.
One more thing to keep in mind here is that when these add, they add to the top or bottom
face of this molecule, meaning it'll have the substituents on A and B pointing up or
down. But because we have a bridge here we want them to be opposite each other so there's
no steric hindrance or no interference with the molecules.
You can see these molecules written in a lot of different ways. One of the most common
ones is the bread basket so let me show you the product in that form. We have the cyclohexene
product drawn not quite like a chair conformation but drawn where you actually see the side
angles of the bonds. Don't forget we still have the 1,2,3,4, A, and B with the pi bond
between 2 and 3, an OCH3 flat or trigonal planar with a carbon at 2 so we're not showing
it up or down, we're showing it straight. Carbons 5 and 6 are shown kinda like a handle
to the bread basket, that's why it's called the breadbasket form coming up as a bridge
out of this page. And then we'll show the ring attached to A and B, we want to show
them down so it's the opposite orientation, that's C,D,E, and we already have F. And let's
not forget that carbonyl on carbon F.
Be sure to join me in the next video where we solve question 2. You can find this entire
Orgo 2 Final Exam with the solutions on my website leah4sci.com/Orgo2Final.