NMR spectrum chemical shift confusion

Recently started on NMR spectrum, and I'm slightly confused about chemical shifts and their connection to the resonance frequency and magnetic field.

From what I know, the "new" way of doing NMR means keeping the radio-frequency constant and increasing the external magnetic field in small increments.

Now, I believe a greater electron density around the protons (more shielded) will protect the positive nucleus from the external magnetic field and thus, the positive nucleus experiences a lesser net magnetic field. Hence, there is a lower resonance frequency which should result in a smaller chemical shift (the magnetic field will determine the energy differences between the spins).

From what I understand, because of experiencing a lesser net magnetic field, we increase the external magnetic field so that we can get a reading (this apparently gives rise to the idea of high field on the more shielded side). However by increasing the external magnetic field, won't we increase the energy differences between the spins and cause the resonance frequency to be higher which conflicts with the idea that we have a lower resonance frequency?

• It looks like you are describing continuous wave (CW) and not pulsed NMR. The latter is generally considered the "new" way of doing NMR. – Buck Thorn May 7 at 10:15

It looks like you are describing continuous wave (CW) and not pulsed NMR. The latter is generally considered a more modern way of doing NMR as it was developed after the CW technique. The pulsed technique was also accompanied by the development of FT processing which resulted in a Nobel price being awarded to Richard Ernst.

In CW NMR you usually "sweep" the field - increase the external field $$B_0$$ in small increments, as you explain - while keeping constant the frequency generated by a "read" RF coil. Coincidence between the read frequency and the Larmor frequency $$\omega = \gamma \left(1-\sigma \right)B_0$$ - the resonance condition - results in a signal registered with a pickup coil. Here $$\sigma$$ is the chemical shielding which reduces the effective magnitude of the applied field $$B_0$$ at the nucleus.

by increasing the external magnetic field, won't we increase the energy differences between the spins and cause the resonance frequency to be higher which conflicts with the idea that we have a lower resonance frequency?

This is exactly the point of increasing the field. The Larmor frequency is related to the splitting in the energy levels, and you need a greater field to bring about a splitting in a more shielded nucleus, so that the resonance condition is satisfied.

The resonance frequency in such a CW experiment is always $$\omega = -\gamma B_1$$, but as the field at the shielded nucleus is $$\left(1-\sigma \right)B_0$$ and the shielding opposes the applied field, it is necessary to apply a larger field ("go upfield") to excite a more shielded nucleus.

Normally the external magnetic field is held constant and an r.f. pulse is applied to generate the nmr signal. The electrons in the molecule allow the external field and the magnetic dipole due to nuclear spin to interact with one another. This interaction causes a local magnetic field that can add to or oppose the external field and so a particular nucleus has a resonant frequency that is slightly different to one in a different environment, for example methyl protons vs aromatic ones. This difference in frequency with respect to a standard compound (such as tetramethysilane TMS) is the chemical shift. The shift is often measured in parts per million change in frequency and a positive ppm corresponds to de-shielding and in this case the nucleus experiences a larger magnetic field than the reference TMS which is taken to be 0 ppm. In a de-shielded nucleus the local magnetic field due to local motion of the electrons adds to the external field, and opposes this when shielded causing a negative chemical shift.

(The chemical shift $$\displaystyle \delta_{ppm}=\frac{v-v_{ref}}{v_{ref}}\cdot 10^6$$. The $$v$$ are frequencies, e.g. $$v_{ref}=60\cdot 10^6$$ Hz. The reason for using a ratio is so that the chemical shift from nmr machines of different frequency can be compared, 100 vs 300 MHz for example).

The r.f. pulse contains many frequencies so can be used measure lots of different nmr transition frequencies.)