6.3.2 (b,c) 1H NMR Spectroscopy

(b) analysis of a high resolution proton NMR spectrum of an organic molecule to make predictions about:

(i) the number of proton environments in the molecule

(ii) the different types of proton environment present, from chemical shift values

(iii) the relative numbers of each type of proton present from relative peak areas, using integration traces or ratio numbers, when required

(iv) the number of non-equivalent protons adjacent to a given proton from the spin– spin splitting pattern, using the n + 1 rule

(v) possible structures for the molecule

{In examinations, NMR chemical shift values will be provided on the Data Sheet.}

{Restricted to functional groups studied in the A level specification.}

{Learners will be expected to identify aromatic protons from chemical shift values but will not be expected to analyse their splitting patterns.}

(c) prediction of a Carbon-13 or proton NMR spectrum for a given molecule

1H NMR (or Proton NMR) has the advantage that 1H is the primary isotope of Hydrogen.

Because we are analysing different nuclei the chemical shifts for proton NMR are much smaller than for 13C NMR.

Low resolution proton NMR works in much the same way as 13C NMR, although you can rely on peak areas to represent the ratio of equivalent Hydrogen environments.

The peak areas are often shown as integration traces (looking like a much-stretched S) - which you can measure with a ruler if necessary, although numbers are usually drawn on by the board

In the molecule above there are:

a) 6 Hs in environment a

b) 3 Hs in environment b

c) 3 Hs in environment c

b) 2 Hs in environment d

It would be possible to "guess" the identity of most of the Hydrogen environments by looking at the heights of the peaks but don't say "height" in an exam - say peak area.

The largest peak would have to be environment a.

The smallest peak would have to be environment d.

The two other peaks don't look the same size, but their integration traces are the same height.

Notably their integration traces are half as big as for environment a.

The 1:2 ratio reflecting the respective number of Hydrogens in each environment.

You could distinguish between these peaks only by looking at their chemical shifts.

Sometimes, as above, the board doesn't even print an integration trace.

They simply write the numbers from the ratio above each peak.

So in this case the middle peak is easily identified from the number of Hydrogens in the environment being 2.

The others, with peak areas of 3 would only be distinguished by referring to their chemical shifts.

High resolution H NMR sees individual peaks split due to "spin-spin coupling" which we don't really need to understand.

As far as A level is concerned this only happens in H NMR.

Generally, an A level examiner will accept "multiplet" for anything bigger than a quartet as it becomes very difficult to distinguish between them.

The amount of splitting is due to the number of Hydrogens in different neighbouring environments.

Ethane would give us 1 singlet - all the Hydrogens are in the same environment so they cannot split each other.

Propane has two environments.

One environment contains 6 H's, the other contains 2.

Splitting is governed by the n+1 rule where n is the number of neighbouring H atoms

The environment containing 6 Hs has two neighbours (n=2) and so is split in a triplet (n+1 =3)

The environment containing 2 Hs has 6 neighbours (n=6) and so is split in a multiplet (6+1 =7)

You would need an integration trace here because "height" really is little guide to area.

At high resolution the splitting emerges.

The environment C containing 6 Hs has two neighbours (n=2) and so is split in a triplet (n+1 =3)

It may look like it has 4 neighbours but the two sets of CH2 are equivalent to each other.

The environment A containing 4 Hs has 5 neighbours (n=5) and so is split in a multiplet (5+1 =6)

Three environments.

The Methyl group on the right has no H atom neighbours and remains a singlet.

The CH₃ on the left has two neighbours and is therefore split into a triplet.

The CH₂ on the left has three neighbours and is split into a quartet.

The problem with Hydrogen atoms that are attached to Oxygen atoms is that they have a very wide range of chemical shift (due to Hydrogen bonding) and produce small peaks.

The same is true of N-H hydrogen atoms.

They can appear almost anywhere, even within another multiplet, causing confusion.

Fortunately, they can be removed.

In the case of Ethanoic acid there really isn't any confusion

But running the H NMR again in D₂O (heavy water, water containing Deuterium, ²H) removes this signal because these H atoms are replaced by the D atoms which have an even mass number.

CH₃COOH + D₂O --> CH₃COOD + DHO

Why this happens is anyone's guess but the usual technique is to run the sample with and without heavy water so that it is clear which (if any) peaks disappear.

Here the sample was being run in a deuterated solvent (CDCl₃) to avoid signals from the solvent.

A drop of D₂O and only one peak disappeared, meaning only one OH (or NH) environment.