IR Spectroscopy: Some Practice Problems

By itself, Infrared (IR) spectroscopy isn’t a great technique for solving the structure of an unknown molecule. However, we’ve seen that IR spectroscopy can be a great technique for identifying certain functional groups in an unknown molecule – especially functional groups containing OH or C=O.

In this post we’re going to go through four (simple) practice problems where you’ll be provided with an IR spectrum and the molecular formula, and are then charged with the task of figuring out which molecule best fits the spectrum.

Let’s begin.
(answers to each problem, along with analysis, are at the bottom of the post. Don’t see them until you’ve given each problem an attempt).

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Problem #1: Unknown molecule with molecular formula C₅H₁₀O. Which of these five molecules is it most likely to be?

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Problem #2: Unknown molecule with molecular formula C₆H₁₂O.

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Problem #3: Unknown molecule with molecular formula C₆H₁₄O .

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Problem #4: Unknown molecule with formula C₄H₈O₂  (Also, smells like vomit)

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Answers

Problem 1: 

• You’re given the molecular formula, which is C₅H₁₀O. This corresponds to an index of hydrogen deficiency (IHD) of 1, so either a double bond or ring is present in the molecule. This immediately rules out d) whose IHD is zero and thus has a molecular formula of C₅H₁₂O.
• Looking at the spectrum we see a broad peak at 3300 cm^-1 and no dominant peak around 1700 cm^-1 (That peak halfway down around 1700 cm^-1? It’s too weak to be a C=O. )
• That broad peak at 3300 tells us that we have an alcohol (OH group). The only option that makes sense is e) (cyclopentanol) since it has both an OH group and an IHD of 1. It can’t be b) since that molecule lacks OH.
• a) and c) are further ruled out by the absence of C=O ; B is ruled out by the presence of the OH at 33oo

Problem 2: 

• A molecular formula of C6H12O corresponds to an IHD of 1 so either a  double bond or ring is present in the molecule.
• There is no strong OH peak around 3200-3400 cm^-1 (that little blip around 3400 cm^-1 is too weak to be an OH). We can immediately rule out a) and e) .
• However, we do see a peak a little above 1700 cm^-1 that is one of the strongest peaks in the spectrum. This is a textbook C=O peak. We can safely rule out b) which lacks a carbonyl.
• The only option that makes sense is d) (2-hexanone) since c) doesn’t match the molecular formula (two oxygens, five carbons).
• Note also that the C-H region shows all peaks below 3000 cm^-1 which is what we would expect for a saturated (“aliphatic”) ketone.

Problem 3:

• A molecular formula of C₆H₁₄O corresponds to an IHD of zero. No double bonds or rings are present in the molecule.
• Using this we can immediately rule out d) and e) since their structures cannot correspond to molecular formula (they are both C₆H₁₂O)
• There is no OH peak visible around 3200-3400 cm^-1. We can rule out a) and b) .
• This leaves us with c) . It’s an ether.
• Useful tip: ethers are “silent” in the prominent parts of the IR spectrum; this functional group is best identified through a process of deduction. Seeing an O in the formula but no OH or C=O peaks, the only logical selection is c) .
• Final note: e) is a cyclic ether called an “epoxide”. The important clue to distinguish c) and e) was the fact that we were given the molecular formula. In the absence of that information it would have been difficult to tell the difference without a close consultation of an IR peak table.

Problem 4: 

• The immediate giveaway is the smell of puke. That’s butyric acid for sure!
• More seriously: the formula of C₄H₈O₂ corresponds to an IHD of 1. We can immediately rule out c) .
• Looking at the IR spectrum we see a huge peak in the 3300-2600 cm^-1 region that blots out everything else.  This seems like a textbook “hairy beard” typical of a carboxylic acid, but let’s look for more information before confirming it. We can at least rule out a) , which has no OH peaks.
• We also see a strong peak a little above 1700 cm^-1 which is typical of a C=O. We can safely rule out e) which lacks carbonyl groups entirely.
• This leaves us with two reasonable choices: b) (the carboxylic acid) and d) the ketone / alcohol. How to choose between the two? The “hairy beard” is diagnostic. Alcohol OH peaks don’t fill up 600 wavenumber units the way that carboxylic acid peaks do. [Go back and look at a few examples from the previous post if you’d like confirmation]  A more subtle way to distinguish the two might be the position of the carbonyl peak, but carboxylic acids (1700-1725 cm^-1) show up largely in the same range as do ketones (1705-1725 cm^-1).

You might recognize that in each of these four examples we followed a simple procedure:

1. Since we were given the molecular formula, we calculated the index of hydrogen deficiency. This is a quick calculation and gives us useful information. We were able to use it to “rule out” a few answers which you might classify as “trick questions”.
2. Next, we examined the hydroxyl region around 3200-3400 cm for broad, rounded peaks (“tongues”) typical of OH groups . The most important question we want to answer is: “is there an OH present”?
3. Then, we looked at the carbonyl region from about 1650 – 1830 cm for  sharp, strong peaks (“swords”) typical of C=O groups. Here we want to quickly know if there are any C=O groups present.
4. Using these three pieces of information we could then rule out various options that were given to us, narrowing down the possible options.  Granted, these were relatively simple examples (only C,H, and O), but the thought process is what’s important.

Notes and Conclusion

• It’s nice to be able to get a positive ID on a functional group, but ruling things out can be valuable too. The absence of an OH or C=O peak (or both) is still helpful information! We used this in Problem 3 to infer the existence of an ether by the absence of an OH or C=O.

• A related point: information you get about a molecule from various sources (e.g. molecular formula, UV-Vis, IR, mass spec, 13-C and 1H NMR)  is self-consistent and should not contradict. 

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