All right, there we go. Now we're recording. As I said on Monday, your active assignment six is open and available, and then I'm going to post quiz six today.
It kind of depends on if we get as far as I would like. Quiz six is a little heavy on concepts. that we're covering Friday.
So I'm hoping to get a jump on that, getting into that a little bit earlier today to help you out there so it's not a mad dash over the weekend to get all that content down. In the meantime, I am working on grading exam one. I hope to have that by the end of the week. end of the week.
I'll let you know on Friday how that's looking. If it's not already done at that point, probably won't be done on Friday. Probably it'll be over the weekend that I finish that up. But for today, we're going to continue our journey in conjugated pi systems. So I left you last time with, we still need to go over the mechanism for halogenation and halohydrins, the halohydrin reaction for our conjugated dienes.
So I have that mechanism queued up for you. And then we're going to move into a molecular orbital description of why dienes behave so differently from other, when they're conjugated so differently from cumulated or isolated dienes, and how that's going to extend beyond just dienes, like trienes, tetraenes, et cetera, how that goes. And then that molecular orbital theory description really pertains to reactions, but the simple case of understanding or proof of what we know about that really comes in. and UV-Vis spectroscopy. So we'll spend a couple minutes talking about how that works with organic chemistry before we get into one more reaction for this chapter, which is a paracyclic reaction called the Diels-Alder cycloaddition.
So that is what's coming. Before I launch into our content, are there any questions hanging from Monday's content? Not so much.
So just one reminder for Friday's class, we're going to be talking about variations on the Diels-Alder Cycle Edition that are best understood by being able to see it in three dimensions. So I'm going to bring my model kit down to show you a couple of examples but if you have have a model kit that you like to use, bring it with you to class or at least bring a handful of pieces if you don't want to lug the whole thing with you. You'll need at a minimum a cyclohexane ring and another two carbons beyond that.
So you'll need like eight to ten carbons should get the job done. Plus some colored markers and things like that. So bring model kits on Friday if you want to play along with me.
Before we get into the new stuff, I just want to start with a quick recap of what I went over last time. So let me zoom this out a little bit because I added a little bit more information to this reaction. coordinate diagram that we used for explaining the thermodynamic versus kinetic conditions that lead us to 1, 4 versus 1, 2 addition products respectively. Bless you.
So what you saw last time was this reaction coordinate diagram, which splits at the the intermediate going through two different transition states to two different products. And we had a good rationale on these products and why they had different energies. It was based on where the alkene was. So that alkene stability really dictated the energy of the final product. So that's what the note is here in green.
But then I did not go through why we have this difference in transition state energy that is kind of opposite of the stability of the final product. So So I wanted to make sure you had a description of that as well. So I drew in the transition states for both processes.
So when we look at what's happening getting to our more stable thermodynamic product where our iodine is adding to the end here, the transition state for that has the iodine adding to that end carbon, and that puts a partial positive on a primary-like position. As opposed to when we're making what turns out to be our less stable product, when the iodine adds in... into the allylic cation.
It's adding into this more substituted side, so our partial positive charge is in a more substituted or more stable position, okay? So the partial positive charge that we have coming from what was our allylic cation intermediate is what's dictating the difference in energy between these two transition states, okay? That happens to be different than where our ultimate energies are because the carbocation versus the alkene Stabilities are opposite in these specific cases.
So hence that little inversion of who goes where in the end, crisscross our paths a little bit. So that kind of sums up and gives you your little reminder. of when we add 1, 2, so hydrogen and halogen adding two adjacent carbons, that was our kinetic set of conditions. That's the less stable situation.
The 1, 4 addition is our thermodynamic set of conditions that gives us the more stable product and stability dictated by position of alkene. But 1, 2 versus 1, 4 is the other thing you want to remember. And then what I didn't give you, I gave you some examples of... What happens when we do a halogenation as opposed to a hydrohalogenation?
So this is addition of a halide across your conjugated pi system. We can also get a 1, 2 or a 1, 4 product out, but I didn't show you how that works because there's a little bit of a difference between that and what we were doing with the hydrohalogenations. So, draw out the microns before you for just bromination of 1, 3-butadiene. It's going to start out just like any halogenation.
reaction starts out where we're going to have our molecular bromine adding across one of the double bonds of our diene. So these are the same arrows that you pushed for halogenation in CHEM 235. We're going to wind up making a bromonium ion here from our bromine through these two arrows, alkene adding into bromine, bromine adding back into alkene, and then we're kicking out a bromide. That'll come back into play later, right? But when we get to this bromonium cation, it is in a position where the alkene that didn't immediately react actually can rearrange our backbone a little bit.
So it's actually that alkene that will shuffle over and break open that ring-strained bromonium ion, right? That three-membered ring is looking to get ring opened so that alkene can facilitate it, which puts your bromine out on one end here, moves our double bond over to the middle, and our positive charge... winds up out on the other end. That's another allylic cation.
Once you have an allylic cation, you are automatically into resonance with this version of the allylic cation. And that shows you that you have two electrophilic sites that your halogen can now add back into. So the halide ion will add in either in the 4 position, getting to our 1,4 product, or in the 2 position, getting to our 1,2 product. And the differentiation between those...
Those two would fall on a very similar reaction coordinate diagram to what we saw for hydrohalogenation. So you're still going to get thermodynamic higher temperature conditions leading to 1,4 and then kinetic lower temperature conditions leading to 1,2 addition. And then the twist on our halogenation reactions was always that halohydrin formation reaction. That's when we just have water as the solvent, and then it outcompetes the bromide ion as the nucleophile in the second step. So I have the mechanism for that.
laid out here for you as well. Well, most of the mechanism of that. So it starts out exactly the same. Alkyne reacts with bromine, makes bromonium, bromonium opens up into the resonance pair of allylic cations, but now.
Now instead of adding your bromine in as the nucleophile, we're going to add water in as the nucleophile either in the four position or the two position. Again, high temperature versus low temperature for the thermodynamic versus kinetic pathway. Your water winds up four or two relative to the one of bromine and then you just need to deprotonate that in the final step to get to your halohydrate.
So nothing really new there except for that like ring opening by the adjacent double bond leading to the allylic cation. So allylic cation you had already seen forming the halonium you had already seen it's just this step that's something new. And again that one is not in your textbook.
Questions now that you've had that refresher of the things that were in Monday's class meeting? Ready to move on then? Okay, so the key theme for today, I'm going to go back to some basics about conjugated pi systems. And the idea of them being more stable than other types of pi systems. And it's not just the stability we're going to talk about here.
I want to talk about overall structure along with the reactivity. Because one other feature of our conjugated pi systems that I didn't talk about on Monday, let's just redraw our 1,3-butadiene. When we look at 1,3-butadiene, diene. When you think about bond lengths, which you haven't thought about since, oh, way back at the beginning of Chem 235, that's fine. Just relative values, longer versus shorter.
What's longer, a single or a double bond? Single is longer. What is interesting about our conjugated pi systems, our double bonds are kind of standard double bond length, but when we look at the bond length on the single bond here. It's actually much shorter than you would expect. It's not quite double bond short, but it's not as long as you would expect a single bond to be.
And this is something we're going to see again when we talk about benzene structures in the coming weeks. But what is happening here, I already alluded to on Monday when I said, when we look at this conjugated pi system and start thinking about the hybridization of every carbon atom within that system, everything's sp2, right? So one thing that's happening, this shorter bond length that we're getting between the two carbons that we draw out as being single bonded, its single bond truly is a shorter single bond.
When we think about what... what's happening with that bonding, if we had an sp3 hybrid orbital making the sigma bond between carbons two and three, we're talking about something that's shaped like this, right, our kind of lopsided dumbbells that were our sp3 orbital shapes. We're talking about overlapping two sp3 orbitals to make the standard sigma bonds that you know and love.
Oops, that one is much smaller than the other one. There we go. So when we think about making that sigma bond, we think of this kind of an overlap.
But if we're now not overlapping an sp3, but rather an sp2 pair of hybrid orbitals, the sp2 orbitals ... Well that's super exaggerated, that's a little more sp. They're a little shorter and a little fatter than our sp3 hybrid orbitals, right? So they look a little bit more like this. These are still going to overlap to form a sigma bond, but in this case it's going to be a little bit shorter than it would have been for overlapping sp3s.
So when you think about making bonds that way, believe it or not, you're thinking about molecular orbital theory. And I want to expand on this a bit more because if we look at the pi systems, this begins to explain a lot of the things that we see in terms of the structure and the reactivity of our conjugated pi systems. Thinking about those pi molecular orbitals, the pi bonds are what we consider to be our double bonds, right?
But remember, a double bond really is a sigma bond inside of a pi bond. So every single bond in our diene has these sp2 hybrid orbitals. orbitals overlapping to make the sigma bond backbone of this.
So that's your single bond. Sigma is single. When you have a double bond, it's that single bond plus a pi bond on top of it.
Triple bonds would have two pi bonds on top of it. Right. What is a pi bond?
The pi bonds come about when we take when you have an SP2 hybridized orbital. Well, we started with an S and three P's to make that hybrid orbital. Two of them went into the hybrid to make the SP2. That third P orbital is unhybridized.
Right. So our pi bonds are coming from the unhybridized P orbitals and how they overlap. So anytime you see something that we're calling an sp2 carbon, the main thing we're going to wind up thinking about is the fact that it's not that this is sp2 hybridized, it's that there is an unhybridized p orbital within that, which can do different things for us. So our unhybridized p orbitals, these are going to be our dumbbells, and I'm drawing these vertical instead of horizontal for a reason.
When we take two unhybridized p orbitals and combine them, that's when we start to make our pi bonds. Our pi bonds really are kind of a hot dog bun around the hot dog of a sigma orbital. I want to add a little bit to this idea because this should be familiar right from Chem 235, bring two p orbitals together make that hot dog bun situation.
These are the nodes, looks like a crazy smiley face or something, but really these are the nodes that sit in the middle of our dumbbells, sort of where the nucleus of the atom would be. It's a little bit more complicated than we just bring these together and we form the hot dog. Because when you get into P-Chem, for those of you who are chemistry majors, you're going to learn all about wave functions. You might have heard a little bit about this in Gen Chem.
Depending on who you had, you might have heard more. about it in Chem 235, but you have a sign for your wave function that describes the orbital and it's positive or it's negative. In organic chemistry what we like to do is we say you have shaded and unshaded because it lets us draw this with a really clear picture. So if you bring your p orbitals together so that their signs are aligned, that their positive and negative ends or their shaded and unshaded ends are aligned, then you wind up with a new molecular orbital that we consider to be a bond. We call that a bonding orbital.
However, It's possible that your two p orbitals could come together misaligned. In fact, 50% of the time, they're going to come together so that they don't line up, in which case they make this kind of repulsive situation where they don't overlap and don't give us a new bond. So we have a bonding orbital when they overlap properly, but then we have an anti-bonding orbital.
when our overlap does not occur. This is referred to as being out of phase. Whereas when we had the bonding occur, we were in phase. And this is all terminology that comes from the fact that our electrons have that wave-like nature.
So we're talking about sine waves that either are constructive or destructive. So constructive leads to bonding. Destructive leads to this antibonding situation. Both of these things are equally likely to occur. So anytime we're thinking about orbitals, we're thinking about both the bonding and the antibonding type situation can occur.
We usually depict this in what we refer to as a molecular orbital diagram. also known as just an MO diagram. A molecular orbital diagram will show all of the different bonding, anti-bonding, non-bonding possibilities that result when we combine two orbitals.
And furthermore, we're going to plot these relative to their energy. So back to this idea of just a simple alkene, which is going to have two sp2 carbons, so two unhybridized p orbitals that come together to make that pi bond. We're going to think about that pi bond coming together.
in either a constructive our orbitals overlap making that hot dog which we tend to show not as the hot dog because those get ugly but just as two bonds or two orbitals that are next to each other and in phase or they can come together out of phase which we will show that way. Let me slide my entire depiction of that down just a little bit. So our relative energy increases as we go up, and we have two possible orbitals available for this pi system.
We refer to them as the pi orbital, that's our bonding orbital, and the pi star orbital, that star means anti-bonding. And like I had in our cartoon of what happens when we bring things together in phase versus out of phase, out of phase, what we wind up with in between the orbitals, we have the pi orbitals. is actually a node.
So we make a new node. That node is going to be present when we're out of phase in that molecular orbital diagram as well. Sometimes you'll see it drawn in, sometimes it's just implied by the fact that you have this flip in phase.
So molecular orbital diagrams take into account what our orbitals can look like, all possibilities, but then we populate them with a number of electrons that will occupy those orbitals, and that tells us a little bit about the stability of any given system just based on its structure and its electronics. So when we look at this simple alkene, we're going to have two electrons in the pi system. So any double bond is going to contribute two electrons into the pi system. So when we go to populate our molecular orbital diagram, we want to fill in in these blanks that I put between the orbitals and the name pi or pi star.
These blanks are for whether or not we have electrons. in these orbitals. Just like when you were doing simple electron configurations for atomic orbitals, like think about it.
I asked you to start to populate your S and P orbitals for carbon. You start by putting your electrons in the lower energy first and work your way. way up and kind of half fill things before you fully fill things and move on, that's exactly what we're going to be doing here.
We're going to take these two pi electrons, we can put two electrons per orbital and start at our lowest energy orbital and then start to fill them up and see how high they go. In the case of this simple alkene, our first electron will go into our pi bonding orbital and our second electron, we're going to fill that energy level before we go any further and just pair it up. It spins, but they're paired up in that orbital. That orbital is full, and we've used up all of our electrons.
This is a stable situation. Everything's in a bonding orbital. Nothing has to go into antibonding. Life is good for that molecule.
We'll talk about how we start to destabilize things, because that's really where reactions occur. That's coming down the road. Right now, we just want to work on how we depict the energy levels. The next thing I want to show you is what happens when we start with the conjugated system. Because this molecular orbital diagram is going to hold for any isolated alkene.
Well, once we start putting together conjugated systems, our pi orbital picture will change. Because we actually have... Remember sp2 orbitals all the way through all four of the carbons involved in that diene.
We're gonna have four unhybridized p orbitals that need to come together. So we're talking about combining four different p orbitals. Sometimes in phase, sometimes out of phase. Oh my god, this is going to get complicated, right?
But it won't. So one thing that's going to happen when we start thinking about the ways that these p-orbitals can come together, whether they're coming together in phase, out of phase, two in phase, two out of phase, is that we're going to have to think about how we're going to do it. phase, one in phase, three out of phase. We're going to stick with a high level of symmetry and we're going to wind up with a total number of molecular orbitals that's equivalent in number to the atomic orbitals that are making it up. So if we're putting in 4p orbitals, we're going to use those to make 4pi molecular orbitals in our diagram.
So I'm going to draw these out similarly to what I did for a simple alkene. I'm going to make kind of a string of four p orbitals that are all linked together kind of at their centers. That's my basic framework for what a pi molecular orbital will look like.
There's going to be four of them so those of you who have tablets copy that and paste it. Oops that's the wrong thing to paste I meant to copy. Copy.
And what I'm going to do is just paste four of these kind of evenly spaced, horizontally relative to each other. And then we're going to figure out on this energy diagram where the lowest energy is at the bottom, what our different structures will be. When you're doing a molecular orbital diagram, your lowest energy molecular orbital, every single p orbital that comes together in that MO is going to be in phase. You will have zero nodes.
I am going to shade the tops of each of those orbitals to indicate our all-in-phase situation. That is always our lowest energy ground state molecular orbital, regardless of what. atoms and how many of them are making up the molecule.
When we go to our next energy level, we're going to start adding nodes in. We're going to add in one node at a time. So our next energy level will have just one node and you want to place it symmetrically within your molecule.
So I'm going to just put it smack down the center. So kind of between the second and the third. the third molecular orbitals.
And then I'm going to figure out what phases it took to come together to create a system that has just that one node. So I'm going to start on the left. I'm going to shade the top just like I did with the bottom set of molecular orbitals. But then when I get to that node, I'm going to flip my phase and I'm going to shade the bottom.
And then I'll keep everything in phase with that until I get to another node, which we won't have because we used up our one node already. When I go to my next energy level, I'm going to have two nodes. I'm going to place those symmetrically as well, so they're between one and two and three and four.
My first orbital I'm shading the top of and I'm immediately going to flip my phase so that first p orbital is not in phase with anything else. It's just the middle two that are in phase with each other and then we have to flip again to get to the fourth one which will be out of phase with the middle as well. And then when you get to your highest energy level, you're going to have the maximum number of nodes possible for that, which is usually, well, in an acyclic system, it'll be number of orbitals minus, or number of p orbitals going into the orbital minus one.
So we're going to have three nodes here. So we literally have a node between every single thing within this molecule. So we're just going to switch the phase between every single.
p orbital to get into this system. So we like to name these. I was calling things pi and pi star when we had just a simple alkene. This is more complicated than a pi system, so we're going to take it to the next letter in the Greek alphabet and call these psi orbitals. So we have psi 1, psi 2, psi 3, and psi 4 for the names of these molecular orbitals.
And each of those orbitals, once again, can hold two electrons. Now, in our diene system, how many pi electrons do we have to populate our molecular orbital? Four, yeah.
Two double bonds. So it's going to be four pi electrons. And then when we start populating our orbitals, again we're going to start down to the bottom, lowest energy, psi 1, we'll get our first two electrons. Psi 2, we'll get our next two electrons.
And you might be thinking, ooh, that's not stable, right? Well, here's the other funky thing that happens, is that we don't actually consider every last one of these molecular orbitals beyond that ground state to be antibonding. What's antibonding in any given molecular orbital diagram, I'm going to go back up to our simple alkene, is just the orbitals that are above the halfway point.
Okay, so if we draw a line in between, like just right smack down the middle of the energy levels of all of the different orbitals in order, anything that is on the lower part of the diagram would be considered a bonding orbital. Anything in the upper part would be considered an anti-bonding orbital. So if we take that concept back down to our four molecular orbital system, our antibonding orbitals really are just psi 3 and psi 4. And therefore, our bonding orbitals are where all of our electrons reside. So this is still a stable situation.
Now there's one other feature I want to point out on these molecular orbital diagrams because we care a lot about that boundary between where we do and we don't have electrons. So the molecular orbital that has the highest energy and is occupied is known as the highest occupied Molecular orbital, abbreviated as the HOMO. Even if there's just one electron in it, it would be considered occupied. So you look for the one that's highest energy and has electrons. The one that has the lowest energy and is not occupied, that would be pi star in our original alkene, would be referred to as the lowest unoccupied molecular orbital.
and that is abbreviated as the LUMO. And the distance between those is actually an important feature that we will come back to shortly. So in our diene molecular orbital diagram, our HOMO is psi 3, or psi 2, sorry.
And then our LUMO, lowest unoccupied, is psi 3. Once again, this is where reactions happen, and we'll see this as being a consequence for spectroscopy as well when we talk about UV spectroscopy in just a few minutes. Kind of getting the hang of this? We can do the same thing with an extended number of orbitals. I'm going to just quickly draw out what happens if we had six molecular orbitals.
Remember, you can grab the notes after class, so you don't need to keep up with my drawings right now. I mostly want you to see what happens in a relative sense here. So with six molecular orbitals, and we're going to string those six together, those six p orbitals, and I'm going to copy and paste that.
Okay. So there's the six skeletons of our molecular orbitals for a triene system. So we're going to have size one through six. Oops. Did I draw too many?
Yes, I did. Take that first one out. All right, so there I've got all six skeletons of my molecular orbitals to put together. And then I'm just going to build them real quickly. Our lowest one, remember, has zero nodes.
We'll have one centrally located node for psi 2. So there's one phase flip. and everything's in phase on either side of that. Psi 3 will need two nodes, once again placed symmetrically.
We need three nodes in Psi 4. Notice the symmetrical placement even though we wind up with a couple of like orbital one and six wind up not in phase with anything. Let's do four nodes for psi five. Bless you.
Like this. And then finally, size six will just have a node between every last one of them. And we've constructed that molecular orbital diagram short of filling in our electrons. How many electrons are going to populate these orbitals?
Six, correct. So parent 1, parent 2, parent 3. Notice we look at where our bonding versus anti-bonding is. That all of our orbitals are occupying bonding, sorry, all of our electrons are occupying bonding orbitals.
And then our homo orbital, highest occupied. is psi three and then the lowest unoccupied the lumo is psi four and i just realized that i have the whole point of me going through all of these is i wanted to give you a comparison of what happens one versus two versus three bonds and i did not grab that picture one second i'm going to pull it out of some old notes where do i have it i think it's this day Spoilers on what's coming here. This is what I want.
How do I copy this over? How did I do that before? Okay, I'm just going to do it on these notes and then I'll copy it over after class. So what we just built were simple alkene, diene, and triene molecular orbitals. If we were to put them all on the same relative energy scale, Then the two things I want you to notice are that we have a ground state energy, so the energy of the lowest orbital that gets increasingly lower as we increase our conjugation.
So better and better stability. So isolated versus conjugated versus more conjugated, more stable. The other thing that's notable, notice that the HOMO and LUMO orbitals are noted for each of these. The difference in energy between our HOMO and LUMO orbitals, known as the HOMO-LUMO gap, gets smaller.
Like I said, this does have implications on reactions. We will talk about that in the future. Today what I want to talk about is something that's easier to see that difference in, which is spectroscopy, and specifically UV spectroscopy. So slight change of gears here. And a little bit of a disclaimer, we don't use UV spectroscopy a ton in organic chemistry, as you'll see.
It doesn't give us the broad kind of structural information that we get out of IR or NMR. So it is useful in various... specific situations as I will explain. So when we talk about UV spectroscopy, where are we in the electromagnetic spectrum? Because we've dealt with infrared, which is way out here.
This is relatively low energy. We've dealt with things a little further out this way, x-rays, gamma rays, not so much, right? We've even dealt with radio waves, NMR. We're really probing out in this very, very low energy area. So I want you to think about what's happening with energy in the electromagnetic spectrum, and we're going to focus on what's happening in the UV invisible range now, which is higher energy than we've dealt with before.
So I want you to remember your equations for how energy relates to your frequency. Planck's constant times frequency, direct proportion to energy. Our wavelength Planck's constant times the speed of light divided by wavelength, we have an inverse proportion between our wavelength and our energy. So when we're thinking about wavelengths, the higher the wavelength, the lower the energy.
So increasing wavelength going this way, this is actually lower energy down at the right-hand side of the spectrum. whereas our higher energy is on the left-hand side of the spectrum. So we're going to be talking today about UV visible spectroscopy, where we're dealing with typically for visible spectroscopy between 400 and 700 nanometers. So the higher that frequency, the lower that energy is. Energy, as we were just talking about on our molecular orbital diagrams, the smaller the homolumo gap, the lower energy we're going to be.
That's what we're going to see reflected perfectly with UV-Vis spectroscopy. So when we do UV bispectroscopy on organic molecules, a lot of them don't give you a lot of very good information. But here's a handful, well in fact here are ones that we just were looking at.
Notice that our simple alkene is not on this list. So diene has a UV-Vis absorption, which we measure as lambda max, which this is the kind of data you get out of UV-Vis spectroscopy. We're looking for, when we're plotting absorbance versus our wavelength, we're looking for the maximum absorbance that we get. We refer to that as our lambda max.
That is our telltale frequency that we're using here. For a simple diene, our lambda max is at 217 nanometers. Back at our UV-Vis spectrum, this is going to be way down in the far, getting closer to the far UV. This is not in our visible spectrum at all. But if we add another alkene to that and go to a triene, Our wavelength where our lambda max occurs is getting longer and longer and longer.
So it goes up to 258. If we start putting four of them together, we get up to 290. We're still not getting into the 400 nanometers that is the violet end. of our visible spectrum. But you can imagine that if we threw in more double bonds, like we have in this molecule, we will get there. This molecule is carotene.
Carotene has a lambda max of 440. 440 nanometers, that's absorbing in the violet. Wait, you guys know carotene. It's not purple, is it?
It absorbs in the purple, but remember how your color wheels work? If we're absorbing in the purple violet, what we're really seeing is over here in the orange and yellow. Okay, so it absorbs violet and this has that orange appearance. Another example, let's add a few more double bonds to this. Oh my goodness, look at the length of that molecule.
How many double bonds are in conjugation there? That's a whole lot, right? This is lycopene. Lycopene has a lambda max of 502. We look at where that falls in our absorbance. 502 is in the green.
So we're on the green bordering on cyan, so we're over around here. And that is going to be pretty red. Tomatoes, right? So that increase of wavelength that we're seeing is because of that decrease in that HOMO-LUMO gap energy. What's really happening in UV-Vis spectroscopy is we're promoting an electron from the HOMO, the highest occupied level.
up to the luma. So the easier we can bridge that gap, the less energy it takes to do that, the more likely it is to wind up falling in the visible range for us. Not all spectroscopy works like that. UV is kind of destructive in that way, because we're then generating radicals that can do things. Luminol experiment at the end of the semester, you get to have great fun with that concept.
I do have just a couple more examples queued up here that anthocyanins these are things that are red pink or colorless until you deprotonate them and I just wanted to point out that in their red pink or colorless stage we're lacking some conjugation here. If we protonate them and get to this phase then we extend the conjugation and we start to get into blue or purple. Usually the really dramatic thing is colorless to blue or purple with these.
You can read about those all kinds of places online. And there's some spectra that go along with the chlorophylls and some of the anthocyanins that go along with those that wind up looking very, very purple. And with just two minutes left, I don't think I have time to introduce our Diels-Alder reaction.
So I'll just give you the preview of the concept that we're going to talk about on Friday. So on Friday, we're going to talk about pericyclic reactions. So, paracyclic reactions are reactions that can occur by either making or breaking bonds using nothing but orbital overlap.
So we're going to come back to molecular orbital theory to describe these reactions so you can see how that occurs. I'm just going to give you a very quick preview of the type of reaction we'll be doing, the Diels-Alder reaction, which in its most basic form takes a diene. We're going to react it with what we refer to as a dienophile. Don't worry about these terms. We'll get into them on Friday.
Notice, no obvious electrophile, no obvious nucleophile. Yet, we're going to use these to make two new bonds to get to a product like this. So that is our Deals Alder Cyclo-Edition, totally different type. of mechanism, simple orbital overlap, no ions involved, nothing polar involved to get us to this point.
So it'll be a pretty packed day, but less molecular orbital theory, more arrow pushing and thinking about structures. Bring your model, please.