Sample Problem:

Here's a pretty typical example of the type of NMR problem you will encounter.

Start by gathering the six clues, one at a time.

Clue 1. Determine the degrees of unsaturation.

This is easy. Just plug the formula C₉H₁₂ into the equation for degrees of unsaturation:
Degrees of Unsaturation = [((# of Carbons x 2)+2) - # of Hydrogens]/2.
[((9 x 2)+2) - 12]/2 = [20-12]/2 = 4.

(For non hydrocarbon elements: treat nitrogens as ½ of a carbon in the above equation, halides as 1 hydrogen, and ignore oxygens.)

There are 4 degrees of unsaturation. With four or more degrees of unsaturation, you can often expect to find an aromatic ring in your unknown compound.

Clue 2. Look at the IR to find what functional groups are in the molecule.

You have a jumble of peaks between 1800 and 2000 cm-1 in the IR, indicative of an aromatic ring. You can confirm the presence of the ring by looking at the NMR, at the peaks around 7 ppm.

Clue 3. Find out how many hydrogens each set of peaks represents.

To get the relative ratios, it's best to use a ruler. Since that's difficult on a computer screen, I can tell you that the relative integration for the peaks are 5:1:6 for the three sets of signals. Since this adds up to the number of hydrogens in the molecular formula (12), the peaks at around 7 ppm represents 5 hydrogens, the peak at 3 ppm represents 1 hydrogen, and the peak at around 1 ppm represents 6 hydrogens.

Clue 4. Find all of the compound fragments.

To find the compound fragments, look at the NMR integration. First, you notice the benzene peaks at around 7.0 ppm, which is the piece you have already identified from the IR. It integrates for 5H, so from the integration you know that the benzene ring is substituted exactly once (6H for unsubstituted benzene - 1H), since one of the ring hydrogens is must be replaced by a substituent. Write down this fragment (if you haven't already).

fragment worth C₆H₅

Now take the molecular formula and subtract out the aromatic piece you just figured out to see what you have left. The benzene substituent is a C₆H₅ fragment, so if you subtract that from C₉H₁₂ you get C₃H₇ (note that the benzene ring contains all 4 degrees of unsaturation: one for the ring and three for the three double bonds. This means you can have no more rings and no more double or triple bonds).

C₃H₇ is what you have left. Now take a look at the other integrations. You have a set of peaks that integrates for 1H, so write that fragment down:

The bonds coming off marked as lines are simply bonds not attached to hydrogens. The other peak is a doublet that integrates for 6H. Trouble you say? A carbon can't have six hydrogens coming off of it? This is really no problem. It just means that there are two symmetric 3-hydrogen groups (methyls) that are chemically equivalent in the NMR, so draw those .

, symmetric

That's all of the fragments of the molecule. You can check to make sure all atoms are accounted for by subtracting out all of the fragments from the molecular formula. C₉H₁₂ - (Benzene fragment C₆H₅) - (CH fragment) - (2 CH₃ fragments) = C₀H₀. Good, all atoms are accounted for.

Clue 5. Piecing it all together.

At this point there is only one way that the pieces can go together, and that is as shown. The 4 fragments come together to make isopropyl benzene:

the four fragments come together to make

Clue 6. Checking the answer.

You know this structure is correct because there was only one way to put the pieces together, but let's just check the peak splitting to make sure. For the tertiary carbon with one hydrogen, you would expect a septet (multiplet) that integrates for 1H. Check! For the CH₃s you would expect a doublet that integrates for 6. Check! For the aromatic ring, you would expect several peaks at around 7 ppm that integrates for 5H. Check!