UV-Vis Spectroscopy: Practice Questions

6 UV-Vis Spectroscopy Practice Problems

In the past few posts we’ve covered some of the key points of UV-Vis spectroscopy. Now comes the key question: how do we apply it? How can it be useful to us, especially in determining the structure of unknown compounds?

It’s one thing to “know” something because you’ve read it somewhere. It’s altogether a different thing to take that knowledge and apply it. Hence, tests.

Let’s practice applying the key principles of UV-Vis spectroscopy through a series of questions –  questions, by the way, of the type that may appear on your exams. We’ll begin with the simplest of concepts we have covered, and then build to incorporate some concepts from the rest of a typical course. [Answers to all at the bottom].

Table of Contents

1. Question One: Hexa-1,3-diene and Hexa 1,4-diene
2. Question Two: Alcohol vs. ketone
3. Question Three: Steroids
4. Question Four: Reactions
5. Question Five: Phenolphthalein
6. Question Six: Structure Determination
7. Answers

The first two questions should be gimmes if you’ve read the previous two posts (especially this one).

Question One

1. Which molecule absorbs at the longest wavelength, 1,3-hexadiene or 1,4-hexadiene?

Question Two

2. What about these two molecules?

These should have been pretty straightforward. Generally, the longer conjugation length, the higher the λ_max. 

Question Three

3. This one might initially look strange, but follows the same concepts as those above:

Question Four

4. Here’s a variant of those questions that connects to reactions you’ve likely seen if you’ve covered alkenes and alcohols:

Question Five

5. This question might seem like a bit of a reach, but it is not significantly more difficult than any of the previous questions. The bonus question below might take some thought.

Question Six

6. Here’s an application from structure determination. Say you’ve isolated a compound that you know has the formula C₂₀H₃₂O. One of your co-workers proposes the structures below.  Which do you think is most likely?

What differentiates questions like 4, 5, and 6 from 1, 2 and 3 is that they ask that you integrate what you’ve learned about UV spectra with material in the rest of the course.

So while 1, 2 and 3 are common “exercises” you may find in textbook sample problems, questions 4, 5 and 6 are more representative of what you’ll see on an exam.

It also illustrates why it’s difficult to memorize your way through organic chemistry. Once it gets sufficiently complex, the number of ways you can be asked to integrate different concepts is nearly infinite. However, the number of important concepts is relatively small. 

So ends our foray into UV-Vis spectroscopy, which has, unfortunately, only scratched the surface. No time to go deep into the Woodward rules, to discuss solvent effects, vibrational bands or other extensions.

We will, however, return to this topic one day – once we get around to writing about sigmatropic rearrangements.

In the next post we’ll move on to a more frequently used spectroscopic technique: infrared (IR) spectroscopy.

Notes

Answers

Answers:  1. 1,3 hexadiene (conjugated). 2. B (the ketone participates in conjugation, while the carbon with the alcohol does not). 3. B (three pi bonds) should absorb at a higher wavelength than A (2 pi bonds). Molecule A (ergosterol) absorbs UV light and undergoes a transformation to molecule B (ergocalciferol)  -> this is the first step in Vitamin D biosynthesis. Hence, this is why sunlight is needed to produce vitamin D.

4. i)Hydrogenation removes the alkene, not the C=O. The product absorbs at shorter wavelength. ii) this elimination reaction (E2) results in a double bond, which results in the two benzene rings being conjugated with each other. Therefore, the product will absorb at longer wavelength. iii) oxidation of the alcohol by PCC will result in a ketone, which will be in conjugation with the alkene and thus absorb at a longer wavelength.

5) Only in B is there conjugation between all three aromatic rings (examine that central carbon). Having the largest conjugated system it therefore absorbs at the longest wavelength and is the most highly coloured. Structure A) is the neutral form present up to pH 8.2 . Under highly basic conditions (pH above 13)  hydroxide ion will react with B at the central carbon, resulting in structure C.

6) The structure C₂₀H₃₂O has 5 degrees of unsaturation. Structure A might be reasonable (the C=O absorbance) but we would expect NaBH₄ to reduce the ketone to an alcohol. Since NaBH₄ does not affect lambda max, we can rule it out.

Likewise we would expect the ketone in structure B) to be reduced by NaBH₄ and since it is conjugated with the aromatic ring, this would affect lambda max [benzene rings are unaffected by Pd/C and H₂ under normal conditions].

We can rule out C) because its UV absorbance should be affected by hydrogenation with Pd/C and H₂. This leaves us with D, which has 5 degrees of unsaturation and no reactions that would affect its absorption maximum.