Saturday, March 19, 2016

How To Approach NMR Problems

There is no perfect way to approach an NMR problem. Often students don't even know where to start. I'd like to offer my thoughts on how to approach figuring out a structure when given an NMR spectrum.

There are 3 main things you must consider (not necessarily in this order):
chemical shift (ppm value)
integration (how many H that make up the peak)
peak shape (singlet, doublet, triplet, etc)

Chemical shift
The ppm value of a peak gives you many hints as to what sort of H group it is. 
Insert graph or figure
Look for the obvious peaks such as aromatics, aldehydes, and carboxylic acids

This tells you how many H's make up a particular peak. This doesn't tell us what type of H it is, but it tells us how many, and this will help us figure out the overall structure of the molecule.

Peak shape
The shape of the proton peak tells us how many protons are on adjacent carbons. Remember the n+1 rule? 

How to approach a problem
Now that we understand the three most important characteristics of the spectrum, we can start to figure out the structure of the molecule. Again, there's no perfect way to do this. But I suggest using a fragmentation approach. In other words, figure out the fragments of the molecule first, and then try to piece them together and see if it all agrees with your spectrum. Let's work through a problem so you see what I mean.

We'll start with a simple one. Often you are given other data, such as IR, mass spec, molecular formula, or at least the proton integration. We'll make it easy for this first one and provide the molecular formula which is C4H8O.

The integration of the peak at 2.5 ppm is 2H
The integration of the peak at 2.2 ppm is 3H
The integration of the peak at 1.1 ppm is 3H

How do we solve it? So first off, there's no aromatic protons, no aldehyde peaks no broad singlets (so no alcohols), so we can rule out all these groups. Can't be carboxylic acid or ester either because these require 2 oxygens. Since there is 1 oxygen based on the molecular formula, there may be an ether or ketone.

Let's look at peak shape. The first one that stands out is the singlet at 2.2 ppm. This means that it's "isolated" (in other words this group is not adjacent to a carbon that has C-H bonds). It also has an integration of 3H. So most likely, this is an isolated CH3 group.

Let's look at the peak at 1.1 ppm. It is a triplet which means it neighbors a carbon (or carbons) that is bound to a total of 2 protons. This peak also has an integration of 3H, so it is likely a CH3 group. Since it's a triplet, it must neighbor CH groups on either side, or it neighbors a CH2 group on one side. 

The peak at 2.5 ppm is a quartet (neighbors carbon(s) bound to 3 protons total). The peak has an integration of 2H, so likely a CH2 group. We saw previously that we have a CH3 group that could perhaps neighbor a CH2 group, so perhaps this is the CH2 group that neighbors the CH3 group.

So what do we know so far? We likely have a CH3 and CH2 groups that neighbor each other and then an isolated CH3 group:


So we've accounted for 3 out of the 4 carbons and 8 out of the 8 hydrogens. This means that the 4th carbon is bound to oxygen. Since we've already accounted for all the hydrogens, we can assume that this last carbon isn't bound to any hydrogens, so it must be bound to oxygen through a double bond (so it's a ketone).


Does the structure make sense? Work through the peak shape and integration data to confirm that this structure is consistent with the spectrum (you will find that it is).

What about the chemical shifts? In this case, I didn't find the chemical shifts too valuable or necessary to solve the structure. This will not always be the case, and others may find that the chemical shift data makes more sense to them, so let's discuss briefly. 

Remember that for any given proton peak, its chemical shift value is dependent on the electron density of the adjacent groups. The closer the proton peak to electron dense groups, the higher the ppm value. For example, the CH3 on the left of the structure is adjacent to CH2 (not electron dense), so this group has the lowest ppm value (1.1 ppm). Compare this to the other CH3 group that's on the right of the ketone, it is adjacent to a C=O group which is fairly electron dense due to the double bond and oxygen. Because of this, the CH3 group on the right is shifted more DOWNFIELD (higher ppm value, in this case 2.2 ppm). The CH2 group is the farthest downfield (highest ppm value) because it's a secondary carbon and it also neighbors the ketone. 

With a more complex structure, the chemical shift data becomes very important because it tells us how to put all the fragments together. 

More complicated examples to come!

No comments:

Post a Comment