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Diels-Alder: intramolecular
How to analyze the product of an intramolecular Diels-Alder reaction.
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- At3:02, the end product shows the stereochemistry of carboxylic acid as dashed- wouldn't it be a wedge since it is on the right side of the dienophile?(3 votes)
- I thought that was intentionally done because they mention that the end product is the endo product, therefore that means they rotated that carboxylic acid around for the endo approach. Correct me if I'm wrong...(4 votes)
- At3:12the carboxylic acid group is on the right of the dienophile. However, at5:18the carboxylic acid group is on the left side of the dienophile. I understand we switched the orientation, but why are we allowed to do this? How do we know how to look at it to determine what is right and what is left?(4 votes)
- He was explaining that we could not obtain an endo product from the first reaction, and basically wanted us to work backwards as in, "Which steriochemistry will give us the endo product?" He rotated the dienophile portion of the molecule so that the carbonyl groups were inward in order to obtain the endo product. Otherwise, it would have been an exo product.(1 vote)
- At4:12, they have the
wrong
product as the oxygen of ester is replaced by a -CH2 group(3 votes) - I still don't understand WHY atoms on the right side rotate a certain way while the atoms on the other side rotate a different way (wedge vs. dash). Is there a way to know this WITHOUT memorization (maybe conceptually). I don't understand this for the Diene and the Dienophile.(2 votes)
- Samuel, first of all we must consider that this drawings are representations of the molecules and do not are the molecules. Therefore, the real molecules are three dimensional and our drawings are two dimensional. So, we need some form of representation in the 2D paper of the 3D molecules. This is the wedge and dashes. The 'body' of the molecule is represented as in the plane of the paper, the dashes are going behind this plane, and the wedges its coming at you. That said, when he says that the part right of the molecule relative to the line is going in a certain direction, that is another standard of the model, but is not exactly what really happens in the real word. IS a model of visualization for us. However, some reactions are more favored than others, but this is another kind of problem. Hope that helped!(2 votes)
- So basically in this molecule, we can just change whether the COOH group is on the left or the right because one of the sigma bonds (attached to the ether Oxygen) can rotate. The endo product is naturally favored, so the sigma bond will rotate to accommodate that since sigma bonds aren't normally fixed in space?
I know I put that as a statement; but I'm really I'm just asking to make sure I'm not just entirely off.(2 votes)- Yeah the molecule can do single bond rotations to form different conformations which would make a Diels-Alder possible.(1 vote)
- Why do inside substituents on the diene go up? What even determines "up"? Would they be going down if I looked from the opposite direction? So, why this direction and why up?
Furthermore, what is with the line through the dienophile, determining the same thing?(1 vote) - Does adding pressure change bond lengths?(1 vote)
- Not really, they're unchanged no matter the pressure exerted on them. The things which change bond lengths are the amount of electrons composing the bond and the types of atoms involved.
Hope that helps.(1 vote)
- Is there is a second possible product (the enantiomer of the one shown) for this reaction?
I think we can get a second product by having the dienophile portion of the molecule approach the diene portion from above (rather than below as Jay shows3:30).(1 vote)
Video transcript
- [Narrator] If a diene and a
dienophile are both contained in the same molecule,
that molecule can undergo an intramolecular Diels-Alder reaction. This molecule on the left
undergoes an intramolecular Diels-Alder reaction to form
the product on the right. And notice how we form
two rings for our product, so this is a pretty cool reaction. The first thing we need to do is find our diene and our dienophile. Our diene is over here on the left, so let me highlight the pi electrons. These pi electrons in red
are part of the diene, and so are these pi electrons in magenta. Our dienophile is over here on the right. These pi electrons, let me make them blue, these pi electrons are
part of our dienophile. First let's show how those two rings form, and we won't worry about
stereochemistry right now. Let's just think about
moving those six pi electrons to form our two rings. Our pi electrons in red move into here to form a bond between these two carbons. Our pi electrons in blue move into here to form a bond between these two carbons. Obviously they'd have to
be a lot closer together in space in reality, and our pi electrons in
magenta move into here. Let's draw our cyclohexene ring. Here's our cyclohexene ring. Our electrons in red formed this bond, our electrons in blue formed this bond, and the electrons in
magenta formed this bond. Alright, let's look at what's
attached to those carbons. This carbon right here is
this one in our product, and so there must be a
CH2 and then an oxygen. So there is a CH2 and then an oxygen coming off of that carbon. Next let's look at this carbon. Maybe I should change colors here. Let me make this blue. This carbon right here is this one. And we know we have a carbonyl
coming off of that carbon, and then we hit this oxygen. That's the same oxygen as before. We have a carbonyl coming
off of that carbon, and then that goes straight to the oxygen. Already we see those two rings. The next carbon is this
one. I'll make it green. And we can see we have a methyl group and a carboxylic acid
coming off of that carbon. Again, I'll just draw these in
without any stereochemistry. There's our methyl group, there is our carboxylic acid, CO2H. And then finally, our last carbon, and I'll just make this yellow here, this carbon has a methyl
group coming off of it. There's our methyl group. Now we can see that the
formation of these two rings. But the next thing we
need to do is to account for the stereochemistry that
we see in our product here. And notice we're told that
this is the endo product. The endo product needs
to have an endo approach, and we've talked about
this in earlier videos. If this is our dienophile, these
carbonyls on the dienophile need to point towards the diene. Let's go to a video so we can see how to use the model set to better
visualize the endo approach. Here's our molecule,
and I'm going to rotate about this bond, just
so we can make our diene and our dienophile
approach each other easier. Now we have our carbonyls
pointing towards the diene, so this is the endo approach, and now if I hold it like this, you can see the diene and the dienophile. On the left is a picture of
what we saw in the video, and we know that a bond
forms between this carbon and this one, so I'll draw
in a dotted line here. On the right is a picture of our product, so that bond that forms is
between this carbon and this one. And let me put it in on the drawing too, so we're talking about this bond. We know that another bond forms between this carbon and this one, so I'll draw in a dotted line here. And that's a bond between
this carbon and this one, so I'll draw in that bond. And then on this drawing we're talking about this bond in blue. Next let's think about the
stereochemistry of the diene. Here is our diene, and let's
put in these two hydrogens. These two hydrogens are
inside substituents. We know that inside substituents go up. If we find those two
hydrogens on our picture, here they are, and for our product, those two hydrogens are going up in space. If we are staring down at the
molecule from this direction, those two hydrogens are coming out at us. In our drawing, here's
one of the hydrogens coming out at us on a wedge, and at this carbon would be the other one. Let me go ahead and draw in this wedge. Here's our other hydrogen
coming out at us in space. The outside substituents go down, so let me use blue for this. This methyl group is
an outside substituent. So is this CH2. On our picture, here is our methyl group and here is that CH2. And for our product, those
two are going away from us. It's a little bit hard to see, but this methyl group is
going away from us in space, and so is this CH2, so for our product, here is our methyl
group going away from us and here is our CH2 going away from us. Next let's think about the stereochemistry of the dienophile. I like to draw a line right here, and we analyze everything on
both sides of that double bond. First let's look at
what's on the left side. We have a carboxylic acid on
the left side of that line and we have this carbonyl. It's hard to see the carbonyl, but it's right back here,
and then this carboxylic acid I've represented with this red atom here just to make it easier to
work with in the model set. Those two are gonna end
up on the same side, so these two are gonna
end up on the same side, and they're to the left
of this line that I drew. We know those two are
going to end up down. If we find them in our product, here is the carboxylic acid. It's actually going down, away from us, and then this bond to that carbonyl is going away from us too. Let's find those on the
drawing of our product. Well, here is the carbonyl,
and you can see it's going away from us, it's
on a dash for our drawing, and here is the carboxylic acid. It's also going away from us on a dash. So those two end up on
the same side of the ring, in this case, down in space. Let's look at what's on the
right side of our double bond. We know there's a hydrogen here, so let me change colors so
we can see these things. There's a hydrogen and
we have a methyl group on the right side of our double bond. And here is the methyl group, which I've used as a yellow atom, and here is the hydrogen. The stuff on the right
ends up on the same side, and it goes up for our final product. If we look at our product here, this hydrogen is actually
going up in space, and it's really hard to
see for this methyl group, but it's actually going up if
you reorient this molecule. So these two are going up in space. Here is that hydrogen and
here is that methyl group. So the stuff to the right side
of the double bond goes up, and the stuff to the left side, and again, this is the endo approach, ends up going down.