Introduction To UV-Vis Spectroscopy
Last updated: October 31st, 2022 |
Understanding UV-Vis Spectroscopy Will Make You More Fun At Parties
In today’s post we’ll discuss why most molecules are colourless, introduce the useful technique of UV-visible spectroscopy, and finally explain why molecules like chlorophyll and β-carotene are coloured.
We’ll even finish by showing you how to use a UV-Vis spectrum to predict the color of a molecule, which is a great trick to roll out at parties. No, seriously, it’s incredible. We’re talking about understanding the chemistry of colour, everyone! It doesn’t get better than this.
First, the summary. We’ll walk through the details below.
Table of Contents
- Converting Frequency Units Into Energy Units
- Ground State Electrons Can Be Promoted To Excited States Through The Absorption Of Light
- Case Study: The Molecular Orbital Diagram Of H2
- Why Most Molecules Containing Only Single Bonds Are Colorless
- Pi Bonds Absorb At Longer, More Energetically Accessible Wavelengths
- The UV-Vis Spectrometer
- How Does Conjugation Of Pi Bonds Affect Lambda Max?
- How Does Lambda Max Relate To The Color We Perceive?
- Conclusion: UV-Vis Spectroscopy
If you want someone to read something you’ve wrote, starting with an equation is generally a bad idea. However, we’re going to start with one of the most beautiful, amazing, and just plain useful equations in all of science. So if you drop off after reading this, really, that’s your problem.
From general chemistry, you may recall the immortal equation
E = hν
where E is energy, h is Planck’s constant (6.626 × 10-34 m2 kg / s ) and ν is frequency (in m-1) .
Why is this equation so frickin’ useful? Because it relates energy to the frequency of light.
To be more specific, when I say “light” I mean, “photon”, as in a carrier of electromagnetic radiation. For the purposes of today’s post, here’s the part of the electromagnetic spectrum we’ll be discussing today: the UV and visible frequencies.
Way back in general chemistry (Bohr model of the hydrogen atom, anyone?) you saw how an electron can be promoted from the ground state orbital to an excited-state orbital through the absorption of a photon of frequency:
ν = ΔE / h
[where ΔE is the difference in energy between the ground and excited states].
Since it is an electron that is being promoted from one energy level to another, we call these “electronic transitions“. Generally, the frequency of radiation required for electronic transitions is in the ultraviolet and visible portion of the electromagnetic spectrum.
This has practical importance in a great number of ways, but for our purposes, we’ll see that the most prominent is that light can promote electrons from bonding orbitals to anti bonding orbitals, and therefore potentially lead to the breaking of chemical bonds.
This will help us to understand:
- why molecules absorb UV light in the first place (sigma -> sigma* transitions)
- why UV radiation is extremely harmful (especially far-UV radiation)
- and ultimately, why certain molecules have color (pi –> pi* transitions)
Let’s start with the simplest molecule, hydrogen (H2) and build from there.
In the beginning of your organic chemistry course, you likely saw how atomic orbitals can overlap to form molecular orbitals.
In hydrogen (H2), for example, two 1s atomic orbitals overlap to form two sigma ( σ) molecular orbitals. The number of orbitals is always conserved: since we start with two atomic orbitals, we end up with two molecular orbitals.
Constructive overlap leads to the formation of the lower-energy sigma ( σ )orbital. Destructive overlap leads to the formation of the higher energy sigma star (σ*) orbital. These two orbitals differ by energy ΔE .
Since each hydrogen atom brings one electron to the party, we have two electrons to fill up our molecular orbitals with, and they will fill up the lower energy orbitals first – much like how nobody chooses to stand in the aisle of a city bus (higher energy) when an empty seat is present (lower energy).
This gives us a diagram that looks like this: (If you find this intimidating, just focus on that red ΔE).
That ΔE is important: by analogy to the Bohr model, if molecular hydrogen (H2) is exposed to light of frequency
ν = ΔE / h
an electron will be promoted from the ground state (sigma) molecular orbital (the highest occupied molecular orbital, or HOMO) to the excited sigma* molecular orbital (the lowest unoccupied molecular orbital or LUMO).
It’s a bit like how toddlers climb stairs, one step at a time: one foot ascends from the highest occupied step to the lowest unoccupied step, and the energy necessary to do this is determined by the difference in height between the steps.
Now, we can convert frequency to wavelength through the equation
c = ν λ (c is the speed of light, λ is wavelength).
For the H-H bond in H2, ΔE corresponds to a measured absorption wavelength maximum of 112 nm, which is deep, deep, deep in the UV region of the electromagnetic spectrum.
112 nm is not exactly the tanning-bed area of the UV. No, this is the death-ray part of the UV spectrum. That’s because UV radiation below 120 nm is also where common sigma bonds like C-H and C-C absorb, and hence being exposed to far-UV light would quickly fragment the bonds that make up the proteins, sugars, and DNA present in our bodies and turn us to goo.
Let’s make this more clear.
Most single (i.e. sigma) bonds such as C-C, C-H, O-H, and C-O have ΔE values that correspond to light in the deep UV part of the spectrum. They appear colourless to us because light in the visible region (400-700 nm) simply doesn’t have enough energy to excite their bonding electrons to an excited state.
This is why water is colourless. This is why ethanol is colourless. This is why diethyl ether, hexanes, chloroform, and a host of other molecules you encounter both in the lab and in everyday life are colourless: the ΔE for the bonding orbitals is too large for relatively low-energy photons of visible light to excite them.
[Insert Jeb Bush joke?]
In fact, UV in the region below 120 nm is so energetic, it is completely absorbed by atmospheric O2 and N2. In other words, if you want to measure the exact wavelengths that , say, ethanol absorbs at, you would need to place it in a vacuum chamber and expose it to hard UV light. This is of academic interest, sure, but outside the scope of a typical intro course. For our purposes, we won’t discuss sigma → sigma* transitions further.
Bottom line: we don’t generally observe sigma → sigma* transitions.
[One exception that we do encounter are dihalogens like Cl2 and I2 that can absorb visible light to generate free radicals through homolytic cleavage of sigma bonds . This is an example of breaking sigma bonds through absorption of UV or visible light, made possible because these bonds are quite weak, about 50-60 kcal/mol, and the corresponding ΔE is small ].
OK: we’re going to ignore σ→σ* transitions. So let’s talk about π-π* transitions instead.
This is where it gets interesting and relevant.
Let’s remind ourself of what we’re talking about by starting with a simple molecule. Here’s ethene (a.k.a. “ethylene”), the simplest alkene. A quick look at its structure reveals 5 sigma bonds and one pi bond.
- Recall that sigma bonds are the result of “end-on” overlap between s or spn (sp, sp2, or sp3) orbitals , whereas pi bonds result from the “side-on” bonding of adjacent p orbitals.
- All else being equal, π bonds are weaker than the comparable sigma bonds, because there is less orbital overlap.
- Weaker bonds mean that the energy gap ΔE between the π (HOMO) and π* (LUMO) will be correspondingly smaller. [In our analogy, a shorter step in a staircase requires less energy to climb].
- Since ΔE is smaller for a pi bond, this corresponds to a longer wavelength (of lower frequency) required to excite an electron from the pi to the pi* orbital.
For ethene, the wavelength absorption maximum for this pi-pi* transition is about 170 nm: still in the deep UV, but not as extreme as for the corresponding C-C sigma bond.
We’d expect ethylene to be colourless, and it is: it just absorbs a somewhat closer to the visible portion of the spectrum than, say, ethane.
Let’s illustrate this with a figure that often looks scary to students: an orbital energy diagram.
*Trigger Warning* All this figure is trying to show is that ΔE for the C-C π bond (in green) is smaller than ΔE for the C-C σ bond (in grey). If you get that, you’ve grasped the key point.*
Wait a second, you might say. How do we know that ethene absorbs light in the UV around 170 nm?
Glad you asked. Let’s introduce a very important device called a UV-Vis Spectrometer.
The basic idea behind UV-Vis spectroscopy is to shine light of varying wavelengths through a sample and to measure the absorbance at each wavelength. Only the wavelengths corresponding to the ΔE for an electronic transition will be strongly absorbed. [For a schematic of how the spectrometer works: check this excellent page, which is also an excellent alternative explanation of UV].
A UV-Vis spectrum plots absorbance (or its inverse, transmittance) of the sample versus wavelength. Here’s the spectrum for ethene. [In this case the wavelength is plotted versus transmittance, the inverse of absorbance (high absorbance = low transmittance, and vice versa). ]
Note that the wavelength of maximum transmittance is at 174 nm. We call this λmax , pronounced “lambda max”. Very little light passes through the sample at this wavelength, because the wavelength corresponds very closely to ΔE for the π to π* transition.
The UV-Vis spectrometer is a useful tool because it allows us to nail down exactly where samples absorb light, and thus quantify electronic transitions. For example, knowing that the λmax for ethene is at 174 nm allows us to calculate the energy gap ΔE , which turns out to be about 164 kcal/mol.
In our previous post on natural pigments, we noted that their large number of conjugated pi bonds was responsible for their colors.
UV-Vis spectroscopy helps us to understand exactly how conjugation relates to the λmax of a molecule – and thus, its color (or lack thereof).
For example, let’s look at what happens to λmax when we increase the conjugation length from 1 (ethene) to 2 (butadiene) to 3 (hexatriene).
As the number of conjugated pi bonds increases, the λmax increases as well!
Because longer frequency = smaller energy, this means that the energy gap ΔE between the highest-occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) decreases as the number of conjugated pi bonds increases.
For kicks, here’s what the bonding picture (roughly) looks like. Again, just focus on the ΔE: it gets smaller as the number of conjugated pi bonds increases.
As this trend continues past hexatriene toward molecules with longer conjugation length, λmax starts to creep into the visible region of the spectrum. It’s already at 258 nm for a conjugation length of 3. Color starts to appear when the conjugation length approaches 7 or so. [For instance, Amphotericin B has 7 conjugated pi bonds and absorbs around 403 nm, in the violet.]
Thus, the λmax of molecules are largely a function of the conjugation length: as conjugation number increases, so does λmax.
But one question remains:
One last piece of the puzzle. How does the wavelength of maximum absorbance (λmax) relate to the actual color?
First, a refresher from the last post. We see the complementary colour of the major color that is absorbed. A molecule that absorbs in the blue will appear orange, because we perceive the colors that are reflected, and orange is the complementary color of blue.
For example, this molecule, Rhodamine B [Note 3] absorbs at about 560 nm (green) and appears red , the complimentary color of green.
Knowing where a molecule absorbs visible light allows us to make predictions about its color. Interesting!
That is a pretty badass cocktail party trick, I think. Think about that the next time you look at a leaf, a tomato, a carrot, or yellow crayfish blood.
We could go on (and we will!) . But let’s leave it there for now.
BTW: A fantastic treatment of UV-Vis spectroscopy with less jibber-jabber has been written by Tim Soderbergh over at LibreText. Check it out.
Also, Reusch’s online textbook entry on UV-Vis spectroscopy is more in-depth and richer in deep detail on this topic than this post is.
This post has walked through some of the theory behind UV-Vis spectra.
In the next post, we’ll go into some of the more practical aspects of UV-Vis spectroscopy. What kind of questions can UV-Vis help us answer about an unknown molecule?
The λmaxisn’t solely a function of the number of pi bonds. If that were true, lycopene (11 conjugated pi bonds) and b-carotene (also with 11 conjugated pi bonds) would have the exact same color. They don’t, since λmax is also impacted by substituents such as attached alkyl groups, whether the pi bond is interior or exterior to a ring (endocyclic / exocyclic) and attached heteroatoms.
Back in the 1940s when UV-Vis was practically the only spectroscopic method available, Woodward and Fieser developed a set of empirical rules (the Woodward-Fieser rules) that aim to predict the λmax based on a number of structural factors. Variants of these rules for more complex conjugated systems also exist (Fieser-Kuhn rules).
Note 2. I would have liked to have used the UV-vis spectra of natural pigments like lycopene and chlorophyll as examples, but they contain more than one maximum. This isn’t easy to explain. Hence it isn’t as straightforward to translate into a color as Rhodamine B is. Here’s lycopene, for instance (source: wikipedia)
Note 3. Also missing here is why UV-Vis spectra are broad and not straight lines. For instance, if, say, rhodamine has a delta E that corresponds to light of 560 nm, why does it also absorb light at 559 and 561 nm ? Part of the answer is that the bonds are in constant vibration and this adjusts the value of delta E, so that a range of energies are absorbed.
Note 4. [How far can we take this? At the far extreme, you have graphene, which is just a big flat conjugated pi system with a near infinite number of conjugated pi bonds. In this situation pretty much every wavelength of light corresponds to some kind of pi-pi* transition, and the result is a black color .]