Chapter 13: Spectroscopy |

**Interpretting ^{1}H-NMR
Spectra **

Let's summarise what can be obtained from a ^{1}H
NMR spectrum:

How many types of H ? |
Indicated by how many groups of signals there are in the spectra |

What types of H ? |
Indicated by the chemical shift
of each group |

How many H of each type are there? |
Indicated by the integration (relative
area) of the signal for each group. |

What is the connectivity ? |
Look at the coupling patterns. This tells
you what is next to each group |

- The chemical shift is the position on the d scale (in ppm) where the peak occurs.
- Typical d /ppm values for protons in different chemical environments are shown in the figure below.
- There are two major factors that influence chemical shifts (a) deshielding due to reduced electron density (due electronegative atoms) and (b) anisotropy (due to magnetic fields generated by π bonds).

Note that the figure shows the typical chemical
shifts for protons being influenced by a *single group*. In cases where
a proton is influenced by *more than one group*, the effects are essentially
cumulative.

- The area of a peak is proportional to the number of H that the peak represents
- The integral measures the area of the peak
- The integral gives the
of the number of H for each peak*relative ratio* - Don't under estimate what the integral can tell you!
- Summing the integrals can give you the empirical number of H and can be related to the molecular formula

- The proximity of other "n" H atoms on neighbouring carbon atoms, causes the signals to be split into "n+1" lines.
- This is also known as the
*multiplicity*or*splitting*of each signal.

An example of an H NMR is shown below.

Based on the outline given above the four sets of information we get are:

5 basic types of H present in the ratio of 5 :
2 : 2 : 2 : 3.

These are seen as a 5H "singlet" (ArH), two 2H triplets, a 2H quartet and a
3H triplet. Each triplet tells us that there are 2H in the adjacent position,
and a quartet tells us that there are 3H adjacent.

(Think of it as the lines you see, L = n + 1, where n = number of equivalent
adjacent H)

This tells us we that the peaks at 4.4 and 2.8 ppm must be connected as a CH_{2}CH_{2}
unit.

The peaks at 2.1 and 0.9 ppm as a CH_{2}CH_{3} unit. Using the
chemical shift charts, the H can be assigned to the peaks as below:

7.2ppm (5H) = Ar**H**

4.4ppm (2H) = C**H**_{2}O

2.8ppm (2H) = Ar-C**H**_{2}

2.1ppm (2H) = O=CC**H**_{2}CH_{3 }and

0.9ppm (3H) = CH_{2}C**H**_{3}

© Dr. Ian Hunt, Department of Chemistry |