Problem 662

Problem # 662

The ¹H and ¹³C NMR spectra of a compound with chemical formula C₁₀H₁₄O are shown below. The compound's IR spectrum shows a broad peak at 3,300 cm^-1. Determine the structure of this compound.

Solution

Let's go through the steps you should take to solve any NMR structure elucidation problem.

1. Are there any hints?

Yes. A broad IR peak around 3,300 cm^-1 tells you this compound contains an alcohol.

2. How many IHD are there?

(indices of hydrogen deficiency are also called degrees of unsaturation or double bond equivalents, depending on the textbook.)

The formula is C₁₀H₁₄O. That's the same as C₁₀H₁₄ (oxygens don't change IHD count). A fully saturated compound has formula C_nH_2n+2, so a C₁₀ molecule should have 22 hydrogens (2 x 10 + 2). The difference between C₁₀H₂₂ and C₁₀H₁₄ is 8 hydrogens, which corresponds to 4 IHD.

3. Draw some C₁₀H₁₄O structures with 4 IHD and eliminate, learn, repeat.

This isn't shown on the image below, but if you don't have an idea of the structure of the molecule, just start drawing structures! Never look at a blank page- just start drawing structures with the correct number of IHD (in this case, 4 IHD).

The peaks around ~7 ppm on the ¹H NMR tell you its probably aromatic (also, benzene has 4 IHD) so that's a good place to start. Draw a few benzene candidate structures with formula C₁₀H₁₄O, but then eliminate structures that don't fit the data. How? I like to go through this check list:

Eliminate by number of signals: Do the candidate structures you drew give the proper number of ¹H and ¹³C NMR signals? If not, eliminate!
Eliminate by multiplicity: Do your structures have the correct ¹H NMR multiplicities? If not, eliminate!
Eliminate by integration: Do your structures have the correct ¹H NMR integrations?If not, eliminate!
Eliminate by chemical shifts: Would the structures you drew have chemical shifts that are about the same as the chemical shifts in the ¹H spectrum given in the problem?

That's the order in which I usually eliminate candidate structures- fastest method (determining the number of expected NMR signals) to slowest method (predicting chemical shifts).

Some things we know so far about this molecule:
Form the IR we know It contains an alcohol.
The gigantic singlet with integration of 9 screams tert-butyl
The total integration of the aromatic region (~7 ppm) is 4, which means this molecule is disubstituted benzene. They're both doublets with integration of 2 which points to a para-substitution pattern.

As you elminate incorrect structures you will learn what fits the data, and be able to draw better candidate structures. Then you can repeat this process. Once you have a structure that passes these four problems, you probably have the correct structure.

Notice how we didn't even really need the ¹³C NMR to answer this problem. The ¹H NMR is usually enough.

I can't stress how important it is to just draw something! Even if you have no idea what the answer might be, don't even leave a blank page for an NMR problem. Just starting drawing out structures with the proper formula and IHD count!