14.1: Nuclear Magnetic Resonance Spectroscopy

NMR-active Nuclei

The basis for NMR is the phenomenon that some atomic nuclei spin about their axes and as a result generate their own magnetic field, or < strong>magnetic moment, therefore these nuclei are called NMR-active. Not all nuclei have a magnetic moment, though; only those nuclei with an odd number of protons and/or neutrons have. Fortunately nuclei that are important for organic compounds, such as the ¹H isotope of hydrogen, the ¹³C isotope of carbon, the ¹⁴N isotope of nitrogen, ¹⁹F and the ³¹P are all NMR-active and therefore can be observed by NMR. Other nuclei, such as the common ¹²C isotopes of carbon and ¹⁶O isotope of oxygen, do not have magnetic moments, and cannot be directly observed by NMR.

Figure \(\PageIndex{1}\): The magnetic field associated with (a) The spinning Earth (b) A spinning Nucleus

Many kinds of nuclei behave as if they were spinning about an axis, somewhat as the Earth spins daily. Because they’re positively charged, these spinning nuclei act like tiny magnets and can interact with an external magnetic field, denoted B₀. Not all nuclei act this way, but fortunately for organic chemists, both the proton (¹H) and the ¹³C nucleus do have spins. The more common ¹²C isotope, however, does not have nuclear spin. (In speaking about NMR, the words proton and hydrogen are often used interchangeably, since a hydrogen nucleus is just a proton.) Let’s see what the consequences of nuclear spin are and how we can use the results.

Spin State and Magnetic Resonance

In the absence of an external magnetic field, the spins of magnetic nuclei are oriented randomly. When a sample containing these nuclei is placed between the poles of a strong magnet, however, the nuclei adopt specific orientations, much as a compass needle orients in the earth’s magnetic field. A spinning ¹H or ¹³C nucleus can orient so that its own tiny magnetic field is aligned either with (parallel to) or against (antiparallel to) the external field. The two orientations don’t have the same energy, however, and aren’t equally likely. The parallel orientation is slightly lower in energy by an amount that depends on the strength of the external field, making this spin state very slightly favored over the antiparallel orientation (Figure \(\PageIndex{2}\)).

Figure \(\PageIndex{2}\): (a) Nuclear spins are oriented randomly in the absence of an external magnetic field but (b) have a specific orientation in the presence of an external field, B₀. Some of the spins (red) are aligned parallel to the external field while others (blue) are antiparallel. The parallel spin state is slightly lower in energy and therefore favored.

If the oriented nuclei are irradiated with electromagnetic radiation of the proper frequency, energy absorption occurs and the lower-energy spin state “flips” to the higher-energy state. When this spin-flip occurs, the magnetic nuclei are said to be in resonance with the applied radiation—hence the name nuclear magnetic resonance.

The exact frequency necessary for resonance depends both on the strength of the external magnetic field, the identity of the nucleus, and the electronic environment of the nucleus. If a very strong magnetic field is applied, the energy difference between the two spin states is larger and higher-frequency (higher-energy) radiation is required for a spin-flip. If a weaker magnetic field is applied, less energy is required to effect the transition between nuclear spin states (Figure \(\PageIndex{3}\)).

Figure \(\PageIndex{3}\): The energy difference ΔE between nuclear spin states depends on the strength of the applied magnetic field. Absorption of energy with frequency ν converts a nucleus from a lower spin state to a higher spin state. (a) Spin states have equal energies in the absence of an applied magnetic field but (b) have unequal energies in the presence of a magnetic field. At ν = 200 MHz, ∆E = 8.0 × 10^–5 kJ/mol (1.9 × 10^–5 kcal/mol). (c) The energy difference between spin states is greater at larger applied fields. At ν = 500 MHz, ∆E = 2.0 × 10^–4 kJ/mol.

In practice, superconducting magnets that produce enormously powerful fields up to 23.5 tesla (T) are sometimes used, but field strengths in the range of 4.7 to 7.0 T are more common. At a magnetic field strength of 4.7 T, so-called radiofrequency (rf) energy in the 200 MHz range (1 MHz = 10⁶ Hz) brings a ¹H nucleus into resonance, and rf energy of 50 MHz brings a ¹³C nucleus into resonance. At the highest field strength currently available in commercial instruments (23.5 T), 1000 MHz energy is required for ¹H spectroscopy. These energies needed for NMR are much smaller than those required for IR spectroscopy; 200 MHz rf energy corresponds to only 8.0 × 10^–5 kJ/mol versus the 4.8 to 48 kJ/mol needed for IR spectroscopy.

¹H and ¹³C nuclei are not unique in their ability to exhibit the NMR phenomenon. All nuclei with an odd number of protons (¹H, ²H, ¹⁴N, ¹⁹F, ³¹P, for example) and all nuclei with an odd number of neutrons (¹³C, for example) show magnetic properties. Only nuclei with even numbers of both protons and neutrons (¹²C, ¹⁶O, ³²S) do not give rise to magnetic phenomena (Table \(\PageIndex{1}\)).

Table \(\PageIndex{1}\): The NMR Behavior of Some Common Nuclei

Magnetic nuclei	^1H	^13C	^2H	^14N	^19F	^31P
Nonmagnetic nuclei	^12C	^16O	^32S

 

Exercise \(\PageIndex{1}\)

The amount of energy required to spin-flip a nucleus depends both on the strength of the external magnetic field and on the nucleus. At a field strength of 4.7 T, rf energy of 200 MHz is required to bring a ¹H nucleus into resonance, but energy of only 187 MHz will bring a ¹⁹F nucleus into resonance. Calculate the amount of energy required to spin-flip a ¹⁹F nucleus. Is this amount greater or less than that required to spin-flip a ¹H nucleus?

Answer

• 7.5 × 10⁻⁵ kJ/mol for ¹⁹F
• 8.0 × 10⁻⁵ kJ/mol for ¹H

Exercise \(\PageIndex{2}\)

Calculate the amount of energy required to spin-flip a proton in a spectrometer operating at 300 MHz. Does increasing the spectrometer frequency from 200 to 300 MHz increase or decrease the amount of energy necessary for resonance?

Answer

1.2 × 10⁻⁴ kJ/mol