How to solve any NMR question
Solving NMR questions is easier than you think. All you need is a step-by-step process to help guide you through each question. And here it is…
Most NMR questions on an exam involve determining a specific structure rather than memorizing and repeating various NMR values. Typically, you will be given an NMR spectra and a molecular formula (sometimes an IR spectra will be provided). I have put together a few ideas that might make this process a bit easier. I am in the process of putting together a more concise document than this as a study aid. This post is meant to walk you through the thought process of how to tackle this type of problem. The description is a bit long (….so hold on!), but once you get it, you can just use the algorithm to solve your NMR problems. Here are some reference values and a couple of proton NMR spectra:
Proton NMR Reference Values
(cem.msu.edu)
(mhhe.com)
(process-nmr.com)
(1H NMR of Taxol; unknown source)
(Our example 1H NMR spectra for this post; unknown source)
Start with an algorithm to get you on track
When staring an NMR question, you can use the following algorithm to help guide you through the thought process:
(above should say C2H5Cl = C2H6)
Calculate Degrees of Freedom
We notice the first thing says calculate degrees of unsaturation…what is that? This allows us to determine if there are any double bonds or rings (cyclic structures) in the compound. Let’s look at an example; the formula is C4H8O2. Now, we need to compare this formula with the formula of a completely saturated hydrocarbon (all single bonds…no double bonds):
CnH2n+2
This formula tells us how many hydrogens we need to have a carbon compound with NO double bonds or rings. Let’s look at our example…
If I have a compound with 4 carbons, this 4 refers to n. Now, if we plug this in the formula, we get
C4H2(4)+2 = C4H10
This says that if we have a compound with only 4 carbons, we need 10 hydrogens to have a compound with no double bonds or rings. If we look at our example, we have C4H8O2. For now, all we need to look at is the C4H8 when dealing with degrees of unsaturation (we will discuss what to do with heteroatoms a bit later). What we now want to do is subtract the hydrogens in our example (C4H8) from the saturated formula (C4H10):
C4H10 – C4H8 = 2 hydrogens
This leaves us with a value of 2. Now, the last things we do to get our degrees of unsaturation is divide this number by 2:
2/2 = 1
How to use the degrees of unsaturation to get the answer
This leaves us with 1; therefore, we have 1° of unsaturation. So, what does this mean? Each degree of unsaturation equates to a double bond or ring. Here are a few examples to further clarify:
1° of unsaturation = 1 double bond or 1 cyclic structure
2° of unsaturation = 2 double bonds; 1 alkyne; 1 double bond and 1 cyclic structure; 2 cyclic structures
3° of unsaturation = 3 double bonds; 2 double bonds and 1 cyclic structure; etc…
When you see 4° of unsaturation, think benzene; 3° of unsaturation for the 3 double bonds and 1° of unsaturation for the ring.
In our example, it means we have one double bond or one cyclic structure in our compound. Let’s look at few ring systems:
The larger the ring, the more stable the ring (with this series). Three and four-membered rings are rare. Usually, you will hear more about 5- and 6-membered rings. Since this is the case, we more than likely have a double bond (but never rule out a ring until you have looked at the NMR spectra).
Next, if we look at the algorithm, we need to consider the other atoms (other than carbon and hydrogen) in our formula…oxygen. Since we have degrees of freedom…think carbonyl. That is all we can do for now with our algorithm. Let’s now look at our spectra and see if we can start to determine the structure from the peaks:
Interpreting the spectra – splitting patterns
We have three peaks: a quartet around 4 ppm; a singlet around 2 ppm; and a triplet around 1 ppm. At this point, let’s review singlet, doublet, triplet, quartet, and multiplet:
What do these peaks refer to and how to we get the specific peak pattern? Well, we use the n+1 rule to figure out the pattern:
From here, it is now a matter of trying to put the pieces together into the desired structure. Let’s review what we have figured out about our structure:
Putting the fragments together
Once you have your fragments, it is a matter of figuring out how to put them together. By looking at the spectra and where the peaks show up (ppm), you can figure out how the fragments go together.
EXTRA:
Click NMR pictures to see the images as a PDF.
Don’t forget IR Spectroscopy!
Here is a simple guide showing you the most common IR values