The UV-Vis Spectroscopy - Online Tutor, Practice Problems & Exam Prep

The UV-Vis Spectroscopy Concept 1

Hey, everyone. So let's take a look at UV-based spectroscopy. But before we do that, let's first take a look at the electromagnetic spectrum. Now recall that it represents a continuum of electromagnetic radiation containing all wavelengths and frequencies. And the part of the electromagnetic spectrum that we're going to focus on are the visible light spectrum and the UV region, because again this is UV Vis spectroscopy.

Now, when we say visible light region, we're talking about the portion where we can see without the aid of any instruments. We can see this with our eyes. And next to it is our UV region. Now, the visible light region, we're going to say ranges from 800 nanometers to approximately 400 nanometers. And the UV region picks up from there.

It goes from 400 nanometers to 200 nanometers. Now, the trend here is we're going from left to right. We're going from the radio waves, AM, and FM, towards gamma rays. A way to remember the order is our memory tool here, that "rude Martians invented very unusual x-ray guns." So if we take a look at this, "rude" is for radio waves, "Martians" is for microwaves, "invented" is for infrared, "very" is for visible, "unusual" is for UV, "x-ray" is for x-rays, and "guns" is for gamma rays.

So we're talking about the trend is we're going from radio waves to gamma rays. Remember, frequency uses the Greek letter mu, its units are typically in Hertz, and we're going to have lambda here represent wavelengths. Now, typically, we have them in meters, but when it comes to UV Vis spectroscopy, we're going to be looking at them in terms of nanometers. And the trend is as we go from left to right, as we go from radio waves to gamma rays, we would see that our frequency would increase and our energy would increase. They both are directly proportional to each other.

One goes up, the other one goes up. Now, frequency and energy are inversely proportional to wavelength. That means that if frequency and energy are going up, then wavelength has to go down. So again, there's this inverse relationship, this opposite relationship between wavelength versus frequency and energy. Alright.

So keep that in mind when we're talking about the electromagnetic spectrum. It's vast with all these types of electromagnetic radiation,We are only concerned with these two here. We are concerned with the visible and the UV sections. Okay?

So, again, we're going to pay attention to the range in which they operate. The visible light region is from 800 to around 400 nanometers, and then UV is about 400 nanometers to 200 nanometers. Alright. So first we look at the electromagnetic spectrum before we get into our real UV Vis spectroscopy ideas.

The UV-Vis Spectroscopy Example 1

Everyone, let's take a look at the following example. Here it says, from the choices provided, which kind of electromagnetic radiation contains the lowest amount of energy? Alright. So here we're talking about energy.

Here we have our memory tool up above to help us remember the order of the different types of electromagnetic radiation. So rude Martians invented very unusual X-ray guns. Remember, the general trend is that as we head this way, our energy increases. So if we take a look here from the choices provided, we have microwaves, which is "Martians." We have X-rays, which is "X-ray."

Gamma rays is the "guns." Ultraviolet is "unusual," and infrared is "invented." If we look at the ones that we circled, we can see that the one most to the left, and therefore, with the lowest amount of energy, would be "Martians," which is connected to microwaves. So, in this example, our final answer would be option A.

The UV-Vis Spectroscopy Concept 2

Hey, everyone. Let's take a look at the UV Vis spectrum. Now here we're going to say that UV spectroscopy is used to examine the electronic transitions of compounds with conjugated pi bonds. Now these conjugated pi bonds, that just means we have pi bonds, single bond, pi bond, which could be double bond, single bond, double bond, or maybe a triple bond's involved where you have double bond, single bond, triple bond. Now here we're going to say a spectrophotometer uses a light source that emits UV light at all wavelengths.

So this represents a spectrophotometer, this whole apparatus here. We're going to say monochromator filters, so we have our light source here, it goes into our monochromator, it filters the desired wavelengths needed by the solution for absorption. So it's kind of honing in and more focusing this wavelength. It's filtering it through to get it more specific. And it's going to go through our solution here.

We're going to say the absorption a is determined and plotted on a UV Vis spectrum. So here, it gets focused through the monochromator, passes through my solution, and as it passes through the solution it now hits our detector. Our detector creates this UV Vis spectrum. Here we're going to say if you look at it, our UV Vis spectrum, we say that this is our output, the reading that we've gotten from our light source. We're going to say on the y-axis, we have our absorption.

On our x-axis, we have our lambda, which is our wavelength in nanometers. Now here we're going to say its prominent feature is λmax, which is the wavelength of maximum absorption. So whatever this species, this conjugated species or a compound or system that we're examining, it's giving us this image. This top part represents its λmax, the area of maximum absorption for this particular conjugated compound or system. Right?

So this is how we utilize this spectrophotometer, which again is this whole apparatus, to basically give us a reading, a layout, a wavelength in terms of maximum absorption for this particular conjugated system.

The UV-Vis Spectroscopy Example 2

We take a look at the following example question: it asks which of the following compounds would produce a UV-Vis spectrum via a spectrophotometer. Now remember, we said that UV-Vis spectra are created through the use of a spectrophotometer, but for conjugated pi bond systems. So, we have to examine each of these and see which one has conjugation involved. Here, we have 1,4-cyclohexadiene. "Cyclo" indicates a ring; "hex" means 6 carbons.

We count: 1, 2, 3, 4, 5, 6. Here, "1" means a double bond starts at carbon 1, and "4" denotes another one starting at carbon 4. We can see that this compound or system is not conjugated. We don’t have a double, single, double sequence. There are gaps between the two double bonds, so this wouldn't work.

Next, we consider 1,3-cyclobutadiene, "cyclo" again for a ring and "buta" for 4 carbons. Then, 1, 2, 3, 4. The sequence of double bonds at positions 1 and 3 represents a conjugated system because it's double, single, double. So this would produce a UV-Vis spectrum.

Moving on to 1,4-cycloheptadiene. "Cyclohepta" means 7 carbons. So we say here, 1, 2, 3, 4, 5, 6, 7. Again, this is not conjugated, so it wouldn't work. Lastly, we have cyclooctene.

"Oct" indicates 8 carbons, and it only says "ene," so there is just one double bond. There isn't even another double bond where a possibility of conjugation could exist. So, this would not work either. Out of our choices, only option b, 1,3-cyclobutadiene, is a conjugated system and therefore is the only one that would produce a UV-Vis spectrum through the use of a spectrophotometer.

Alright, so just keep that in mind when we're talking about UV-Vis spectra; it only works on conjugated pi systems.

The UV-Vis Spectroscopy Concept 3

Hey, everyone. Let's take a simple look at conjugation and our lambda max. Now remember, only organic molecules with conjugated pi bonds will absorb a UV Vis spectrum. And the trend is the greater your conjugation, then the greater your lambda max. Remember, conjugation is just alternating pi bonds and single bonds.

Now if we take a look here, we have 1,3-butadiene, and we would say that this is a conjugated system because it goes pi bond, single bond, pi bond. Here, we'd say its theoretical lambda max value is going to be 217 nanometers. Now, if we go over to 1,3,5-hexatriene, this is also conjugated and it has even more conjugation because here it was pi bond, single bond, pi bond. Here, it's pi bond, single bond, pi bond, but then also additional conjugation, single bond, pi bond. So there's even more conjugation, and we said the trend is the higher the conjugation, then the higher lambda max.

Here, its theorized lambda max value is 258 nanometers. So we can see that the number has gone up. And this is the trend we need to remember: UV-Vis spectroscopy looks at these conjugated systems. The greater the conjugation that we have, the higher the lambda max will be. Later on, we'll go into further detail looking at this more theoretically, as well as doing calculations where we could figure out an estimate of our lambda max.

For now, just remember this general trend: The higher the conjugation, the higher the lambda max value.

The UV-Vis Spectroscopy Example 3

Hey everyone. So for this example question, it says, which of the following compounds would you expect to show UV absorptions in the 200 to 400 nanometer range? Alright. So we know that this range ties into UV-Vis spectroscopy, and we know that when we're talking about this topic, it deals with conjugated systems.

So the ones that are conjugated are the ones that could fit within this given range. So, if we take a look at our options, we have pi bond, single single pi bond, or pi bond, single single single pi bond. This is not conjugation, so there's no chance of it falling within this range. For the next one, we have pi bond, single bond, pi bond, single bond, pi bond. So we have a portion of it that's conjugated.

Therefore, it should fall within this range. So this is an answer. For the next one, we have pi bond, single bond. That group there is a nitrile. Remember, nitriles have pi bonds.

It's a carbon triple bonded to nitrogen. So although we can't see it, we need to remember that it would still be pi bond, single bond, pi bonds. So, this would have some level of conjugation, so this would be included. For this one, we don't even need to do much. We can see that there's a benzene ring here, which is a conjugated system.

It's pi bonds and single bonds alternating, so this is another one that fits. Then finally, this last one here we have pi bond, single bond, pi bond, single bond, pi bond. A portion of this molecule is conjugated. Therefore, it could fall within this range. So from our options, we can see that 2, 3, 4, and 5 could be expected to have a value that falls within this given range.

So here, out of our options, it will be answer e. E says 2, 3, 4, and 5. Again, one doesn't work because one is not conjugated at all, so it's not gonna fall within this range. Again, remember, when we're talking about UV-Vis spectroscopy, we're looking at conjugated pi bonds. Where there's conjugation, there's a chance of it falling within these ranges.

Which of the following compounds would you expect to show absorption in the portion of the electromagnetic spectrum that we can be visually seen without the use of equipment?

• A. 1,3,5-hexatriene (λ_max = 256 nm)
• B. Quinoline (λ_max = 313 nm)
• C. Lycopene (λ_max = 476 nm)
• D. Acetone (λ_max = 274 nm)
• E. 3-methylene-1-cyclohexene (λ_max = 230 nm)

A compound with the formula C₁₁H₁₄O has a λ_max of approximately 244 nm. Based on the given information, which of the following compounds is the likely identity of the compound?

1. IR Spectroscopy revealed a signal of 1710 cm^-1. 

2. Reduction of the functional group at 1710 cm^-1 created a chiral molecule.

3. Reduction of the functional group at 1710 cm^-1 caused no change in the λ_max of the compound. 

• A. I only
• B. II and III
• C. I and III
• D. III and IV
• E. II only