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Organic chemistry
Course: Organic chemistry > Unit 14
Lesson 2: UV/Vis SpectroscopyConjugation and color
Why extensively conjugated compounds absorb light in the visible region and thus have color. Created by Jay.
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Hello, I actually came here looking for explanations regarding Bathochromic and Hypsochromic shifts explanations (red and blue shifts, not the physics side but the chemistry side) since they're pretty hard to find in textbooks. This video was helped in explaining the theory of UV-Vis and I got some grasps on colours, however does anyone know where else I should go to understand the colour shifts better?(4 votes)
Here are two links that might get you started.
https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Spectrpy/UV-Vis/spectrum.htm
http://www.uobabylon.edu.iq/eprints/publication_11_8282_250.pdf(6 votes)
When we have more than one double bond, I get that we have 2 pi MO and 2 pi* MO;
but why are the 2 pi MO placed on top of each other? Don't they have the same amount of energy?
Isn't each pi Molecular Orbital representing one double bond? (or am I understanding this incorrectly?)(2 votes)
Actually, there are pi orbitals with different levels of energy, but it's not one double bond vs. the other double bond; instead it's comparing the different bonding orbitals that are achieved when molecules with pi bonds come together in either an in-phase or out-of-phase formation; a good explanation can be found here:
https://chemistry.stackexchange.com/questions/8910/why-does-the-energy-gap-for-%CF%80-%CF%80-transitions-shrink-with-the-size-of-the-pi-co(4 votes)
I see, conjugation allows the colour. But if the compound containing "COOH", do we have n-pi* transition? This transition allows some colour?(2 votes)- So I get that an electron in a π bond can absorb energy from a specific light wavelength to get from homo to lumo. But the bonding doesn't break, so the electrons have to get back to homo, releasing the absorbed energy. In what form is this energy released? Because if that would be in the form of light, it would be in the same wavelength as the absorbed wavelength and the molecule would still be colorless.(2 votes)
The energy can be released in the form of light, but when the conjugation is sufficiently extensive then the energy difference between the HOMO and the LUMO will be comparatively small leading to a larger wavelength. Therefore light absorbed and emitted will be in the visible region and have a color we can observe.(1 vote)
If an electron absorbs a photon with a specific color of light, shouldn't it reemit that colored photon back out when it falls back to its ground state?
For example, I've read that a black colored object absorbs all wavelengths of light and converts it into heat? How come the electrons of a black colored object don't reemit those absorbed colors as photons, thus reemiting all wavelengths of light?(1 vote)- So it's good to keep in mind that electron transitions are more complicated in reality. So in general if an electron is excited to a higher energy level it is less stable compared to it at the ground level. The electron can reach this excited several ways and isn't just limited to photon absorption; it could also simply absorb heat and cause an excitation.
When it falls back to the ground state, or relaxes, it can do so with two main processes; radiative and non-radiative methods. Radiative would be where a photon is emitted and proceeds via fluorescence or phosphorescence. Both emission methods emit a photon of lower energy (and higher wavelength) compared to the absorption photon. Fluorescence is the more energetic of the two. The actual reason for the difference gets too involved in energy states so I won't overdo it.
Non-radiative processes would be where the electron relaxes without the emission of a photon. These would include processes like vibrational relaxation, internal conversion, and intersystem crossing. These processes manifest themselves as higher kinetic energy and heat for the atom/molecule.
So if an atom's electron absorbs a photon of light with a certain energy/wavelength/frequency it could reemit that photon via fluorescence or phosphorescence, but not at the original energy/wavelength/frequency because they are coupled with non-radiative processes that produce a little heat too. Additionally it's important to note that that the emitted photon is not always in the visible range and could be in the UV range which would be unseen by our eyes and appear clear/colorless to us. That's why a lot of organic solutions appear colorless; because they're emitting in the UV range which we can't see because of the limitations of our eyes.
A black object relaxes its electron back down the ground via non-radiative processes primarily releasing heat. This is why black objects like the asphalt of a road are noticeable warmer than objects of different colors. But again it's not an all or nothing deal, black objects still emit some photons.
Hope that helps.(2 votes)
- Why does a plant appear green?(1 vote)
Do a google image search for "chlorophyll absorbance spectrum". You will see that chlorophyll absorbs blue and red light. This leaves green/yellow as the main wavelengths of light that are reflected (and subsequently received by the eyes).(2 votes)
Why isn't one of the COO−-Oxygens being protonated when giving acid to the basic form of phenophtalein? Is it due to steric hindrance?(1 vote)- How "extensive" does the conjugation need to be for the molecule to show color?(1 vote)
- I wouldn't say there is an exact number of pi bonds needed, it really depends on the specific molecule. A molecule like β-carotene needs 11 pi bonds to emit visible light, but then a molecule like cyanidin only needs 8 pi bonds to do so too. I would cautiously estimate that 8 pi bonds at least are needed to emit light with a higher wavelength than ultraviolet (But I'm almost certain there are examples of molecules who can do it with 7 pi bonds or less). In general though the greater number of conjugated pi bonds in a molecule, the greater the chance for it to display visible color.
Hope that helps.(1 vote)
- so we can say that this is one of applications of Molecular orbital theory. i need to know some real life i mean practical life applications of molecular orbital theory.(1 vote)
- I feel like some great key has been revealed. Is all color due to extensive conjugation? Like the color of one's eyes?(1 vote)
Video transcript
- [Voiceover] Most organic molecules don't have any color at all. An example of that would
be ethene, or ethylene. Ethene has two carbons, and each of those carbons
is sp2 hybridized. So, each of those carbons has a p orbital. We have a total of two p orbitals, or two atomic orbitals. Those two atomic orbitals
are going to recombine to form two molecular orbitals. So, one bonding molecular orbital, and one antibonding molecular orbital. The bonding molecular orbital has a lower energy. This represents the bonding molecular orbital, down here. The antibonding molecular orbital is higher in energy, so this is the antibonding molecular orbital. Ethene has two pi electrons. Let me highlight those two pi
electrons in magenta, here. Those two pi electrons go into the bonding molecular orbital. So, that orbital is occupied. We could also call this orbital the highest-occupied molecular
orbital, or the HOMO. This would be the lowest, unoccupied molecular orbital, or the LUMO. There's a difference in energy between the HOMO and the LUMO. That difference in
energy is very important because that difference in energy corresponds to a wavelength of light. Energy is equal to Planck's constant, times the speed of light,
divided by the wavelength. So, energy ... Energy and wavelength are inversely proportional to each other. A certain amount of energy corresponds to a certain
wavelength of light. This energy difference ... This energy difference between the HOMO and the LUMO, corresponds to a certain
wavelength of light. That wavelength of light turns out to be approximately 171 nanometers. When ethene absorbs light, at a wavelength of 171 nanometers, that corresponds to the
proper amount of energy between the HOMO and the LUMO. That's enough energy for
one of those pi electrons to jump from the HOMO to the LUMO in a pi-to-pi star transition. We talked a lot about
pi-to-pi star transitions in the first video, on
UV/Vis spectroscopies. So, make sure to watch that video before you watch this one. That's the idea of what's
happening with ethene. It absorbs light at a
wavelength of 171 nanometers. Let's move on, to 1,3-Butadiene, which has four carbons. Each one of those carbons
is sp2 hybridized. So, each one of those
carbons has a p orbital. We have four p orbitals,
four atomic orbitals, which would recombine to form four molecular orbitals, two bonding, and two antibonding. The two bonding molecular orbitals are lower in energy than the two antibonding molecular orbitas. We have a total of four pi
electrons, four butadienes ... Here's two pi electrons,
and here are the other two, so four pi electrons. Those pi electrons occupy the two bonding molecular orbitals. Next, our job is to find the highest-occupied molecular orbital ... That's this one ... And the lowest unoccupied
molecular orbital. That's this one. The energy difference between the HOMO and the LUMO is what we're thinking about, here. Notice, that this energy difference ... This difference in energy is smaller than the difference in energy in the previous example. So, if you think about the equation that relates energy and wavelength ... If you decrease the energy ... Since they're inversely
proportional to each other, you must increase the wavelength. So, we must have a higher
wavelength than before. Instead of 171 nanometers, 1,3-Butadiene is going to absorb light at approximately a
wavelength of 217 nanometers. Finally, let's look at 1,3,5-hexatriene. If I look at the pi electrons, we have two, four, and six pi electrons. Those six pi electrons fill the three bonding molecular orbitals. Next, we find the highest-occupied
molecular orbital, and the lowest unoccupied
molecular orbital, and look at the energy
difference between them. Notice the energy difference
has gotten even smaller. If we, once again, decrease the energy, we increase the wavelength
of light that's absorbed. The wavelength of light must be even higher than this. It goes up to approximately
258 nanometers. So, hexatriene absorbs light at approximately 258 nanometers. That's still in the UV region of the electromagnetic spectrum. So, hexatriene doesn't have a color. In order for something to have a color, it has to absorb light
in the visible region. That is only accomplished by thinking about conjugation. Here, we have hexatriene,
which is conjugated. Double-bond, single-bond, double-bond, single-bond, double-bond. But, it's still absorbing
light in the UV region. Imagine a molecule that has much more conjugation than hexatriene. Increased conjugation means an increased wavelength
of light that's absorbed. Notice our trend, here. As we increase in the
amount of conjugation, we increase the wavelength
of light that's absorbed. If we have an
extensively-conjugated molecule, we could absorb light at
an even higher wavelength. If we get past approximately
400 nanometers, we're into the visible region. Those molecules should have color. That's the idea. Extensive conjugation leads to color. Let's look at beta-Carotene, which is what we talked
about in the previous video. We say that beta-Carotene is orange. So, we have an orange molecule, here. Once again, the key is conjugation. Look how conjugated beta-Carotene is. Look at all the alternating
single, and double, bonds. The extensive conjugation means that this molecule can absorb light at a longer wavelength. It absorbs light in the visible region of the electromagnetic spectrum. Beta-Carotene absorbs blue light, and reflects orange, which is why we said carrots and the molecule
looks orange, here. It's conjugation that
you have to think about. Let's move on to one more example, phenolphthalein. All right, so, phenolphthalein is probably the most famous acid-based indicator. If you've taken a general chemistry lab, I'm sure you've used phenolphthalein as an indicator for your
acid-based titrations. You know, at a low pH,
in an acidic environment, phenolphthalein is colorless. But, if you add base ... Here, we'll talk about adding base, here. You increase the pH, you see a pink or magenta color. Let's see if we can explain why. First, let's add some sodium hydroxide. Let me add some sodium hydroxide, here. There's a hydroxide anion. I'm drawing in another
hydroxide anion, here. So, a negative-one charge on the oxygen. Hydroxide anion is going
to function as a base. It's going to take this proton, which we could take these electrons, and move them into here, which pushes these electrons over to here. These electrons move in, here, and these electrons come
off, onto the oxygen. The other hydroxide
would take this proton, and these electrons would end up on the oxygen, here. So, let's follow those electrons. Electrons in magenta move
into here, like that. Let's use blue for the next one, here. These electrons, in blue, move into here. Let's go for green. These electrons, in green, move into here. Then, finally, these electrons, in red, move off onto this oxygen. So, we give that oxygen a negative-one of formal charge. If this hydroxide anion takes this proton, then these electrons
end up on this oxygen, giving that oxygen a
negative-one formal charge. We form an ion, here. This ion has a pink, or
magenta, color to it. If we look closely at it, we can see why. Look at the conjugation that's present. There's all kinds of alternating single-, and double-bonds, right? So, double bond, single bond, double bond, single bond, double bond, single bond, double bond, single bond,
double bond, and so on. Pretty much the entire ion is conjugated. We have this extensive conjugation which allows the ion to
absorb in the visible region. That's why it appears to
be a pinkish color, here. We could go back the other direction. If we add some acid, we could turn this back
into the colorless form of phenolphthalein, here. The reason why this is colorless is because this carbon ... Let me go in and highlight
this carbon, right here. This carbon is sp3 hybridized. It's sp3 hybridized. Therefore, it doesn't have a p orbital. We have a little bit of a conjugation in the benzine rings. We have these alternating single- and double-bonds, here. But, this conjugation is disrupted when we get to the central carbon. So, we don't have the conjugation throughout the entire molecule, here. We don't have enough conjugation, and so it doesn't absorb
in the visible spectrum. That's why it appears to be colorless. If we look at that carbon,
over here, on the right ... Let's use red, here. This carbon, right here ... This carbon is sp2 hybridized. So, it has a p orbital, which allows delocalization
of these electrons. We have extensive conjugation, and so we get the color. Hopefully, this just helps you understand how important conjugation is for color.