Organic Chemistry · Structure Determination: Nuclear Magnetic Resonance Spectroscopy

The Nature of NMR Absorptions

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Nuclear magnetic resonance (NMR) spectroscopy works because certain atomic nuclei behave like tiny bar magnets. A nucleus has a property called spin, described by a spin quantum number I. Only nuclei with nonzero spin are NMR-active — and of those, the ones with I = 12, such as ¹H and ¹³C, give the cleanest signals. When an NMR-active nucleus sits in a strong external field B₀, its two possible spin orientations have different energies, separated by ΔE = γℏB₀, where γ is the gyromagnetic ratio, a constant of the nucleus itself. Irradiating the sample with radio-frequency (RF) energy whose frequency exactly matches this gap flips spins from the lower (aligned) state to the higher (opposed) state — that absorption of energy is the NMR signal.

Two facts dominate the rest of this chapter. First, the energy gap is tiny, so only about one nucleus in tens of thousands sits in the lower state — NMR is inherently insensitive, which is why samples are concentrated and scans are averaged. Second, the field a nucleus "feels" is modified by its own electrons, so identical nuclei in different environments absorb at slightly different frequencies — the origin of chemical shifts and the reason NMR is the most informative experiment for determining organic structure.

Why this matters

NMR is the single most powerful tool for determining the structure of an unknown organic compound — how many kinds of protons and carbons it contains and how they connect.
The same physics underlies magnetic resonance imaging (MRI): hydrogen nuclei in water and fat absorb and re-emit RF energy with relaxation times that differ between tissues.
Pharmaceutical and food-industry QC uses NMR to verify that a synthesized product matches its intended structure and purity.
For exams, spin states, absorption frequency, and relaxation explain why 1H NMR shows certain signals and why 13C NMR needs signal averaging (a later topic).

The college version

Core Concepts

Which nuclei are NMR-active

Whether a nucleus has spin depends on its mass number and atomic number:

Odd mass number → half-integer spin. ¹H, ¹³C, ¹⁹F, ³¹P, and ¹⁵N all have I = 12 and are strongly NMR-active.
Even mass number and even atomic number → I = 0. ¹²C, ¹⁶O, and ³²S have no spin and give no NMR signal. Conveniently, carbon NMR therefore "sees" only the 1.1% of carbon atoms that are ¹³C.
Even mass number and odd atomic number → integer spin. ²H (deuterium) and ¹⁴N have I = 1; they are NMR-active but give more complex signals.

For an I = 12 nucleus there are exactly two spin states, α (aligned with B₀) and β (opposed to B₀).

The energy gap and the resonance condition

In zero field the α and β states are degenerate. In a field B₀, the gap between them is

$Δ E = γ ℏ B 0$

and since ΔE = hν, the frequency at which absorption occurs is

$ν = γ B 0 2 π$

This is the : a nucleus absorbs RF energy only when the applied frequency matches its . The γ value sets how "strong" a nucleus appears: ¹H has γ= 2.675 × 10⁸ rad s^-1 T^-1 (42.58 MHz per tesla), so a 7.05 T magnet gives a 300 MHz proton spectrometer; ¹³C has about one quarter of that γ, resonating near 75 MHz in the same magnet.

Why the signal is weak: Boltzmann populations

The α and β states are populated according to the :

$N α N β = e Δ E / kT$

where k is the Boltzmann constant and T the absolute temperature. Because ΔE is minuscule compared with kT at room temperature, the α excess is tiny — about 1 part in 40,000 for protons at 300 MHz (Worked Example 2). Signal intensity is proportional to this population difference, which is why NMR is far less sensitive than infrared or mass spectrometry. Higher-field instruments (600, 900 MHz) increase ΔE and the signal — why modern NMR magnets are superconducting.

Relaxation: the pathway back to equilibrium

After absorbing, a spin sits in the β state. If nothing returned spins to α, the population difference would vanish and absorption would stop — . Two relaxation processes restore equilibrium:

Spin–lattice (longitudinal) relaxation, time T₁: excess spin energy is released to the surroundings as heat. Short T₁ gives sharp lines and allows rapid re-acquisition of scans.
Spin–spin (transverse) relaxation, time T₂: energy is exchanged among neighboring spins without net loss. Very short T₂ broadens lines.

Relaxation also matters in MRI, where T₁- and T₂-weighted images distinguish tissues. Without relaxation, spectra could never be recorded — the sample would stay saturated.

Common Confusions

Do Not Confuse	With	Difference
NMR-active nuclei	All nuclei	Only nonzero-spin nuclei absorb; 12C, 16O, 32S are NMR-silent
Spin I = 12 nuclei	Spin I = 1 nuclei	1H/13C/19F/31P give simple two-state spectra; 2H/14N give more complex patterns
The resonance frequency	The chemical shift	Frequency is ν= γB₀/2π for the whole experiment; chemical shift is the small environment-dependent variation (next topic)
Weak NMR signal = bad instrument	Weak signal = intrinsic low population excess	Even a perfect instrument sees only ~1 extra spin per 40,000 because ΔE ≪ kT
T₁ spin–lattice	T₂ spin–spin	T₁ releases energy to surroundings (scan repetition); T₂ exchanges energy among spins (linewidth)

Eli explains

The same idea, in plain words

Explain it like I’m 10

Every atom's core (nucleus) is a tiny spinning magnet, but only some cores are magnetic enough to notice. If you put those magnetic cores in a huge magnet and "ping" them with just the right radio wave, they flip over and absorb the ping — that's the NMR signal. Because only about 1 in 40,000 cores is in the right state to flip, the signal is weak, and each core needs a moment to settle back down before it can be pinged again.

Worked example

Example 1: Calculating the resonance frequency of 1H in a 7.05 T magnet

A "300 MHz NMR spectrometer" uses a 7.05 T magnet. Verify this with the proton resonance frequency. Given: γ(¹H) = 2.675 × 10⁸ rad s^-1 T^-1, π= 3.1416.

Step 1 — Write the formula first:

$ν = γ B 0 2 π$

Step 2 — Substitute values with units:

$ν = (2.675 \times 108 rad s -1 T -1)(7.05 T) 2(3.1416)$

Step 3 — Cancel units and evaluate. The radian is dimensionless and T^-1 · T = 1, leaving s^-1 = Hz:

$ν = 1.886 \times 109 s -1 6.2832 = 3.00 \times 108 s -1 = 300 MHz$

Sanity check: 42.58 MHz/T × 7.05 T = 300 MHz. "300 MHz NMR" refers to the proton frequency; higher-field magnets (14.1 T → 600 MHz) give proportionally larger gaps and stronger signals.

Example 2: Energy gap and population excess for 1H at 300 MHz

For protons at 300 MHz and 298 K, find (a) the energy gap ΔE and (b) the fractional excess of α over β spins. Given: h = 6.626 × 10^-34 J s, k = 1.381 × 10^-23 J K^-1.

Part (a) — Frequency form of the energy equation:

$Δ E = h ν$

$Δ E = (6.626 \times 10 -34 J s)(3.00 \times 108 s -1) = 1.99 \times 10 -25 J$

Part (b) — Thermal energy first:

$kT = (1.381 \times 10 -23 J K -1)(298 K) = 4.12 \times 10 -21 J$

Then the gap-to-thermal ratio:

$Δ E kT = 1.99 \times 10 -25 J 4.12 \times 10 -21 J = 4.83 \times 10 -5$

Because this ratio is ≪ 1, the population ratio is approximately 1 + ΔEkT = 1.000048, and the fractional excess of lower-state spins is about half that:

$N α - N β N α + N β \approx Δ E 2kT \approx 2.4 \times 10 -5$

Roughly 2.4 extra α-spins per 100,000 — about 1 in 40,000. This tiny excess is why NMR needs concentrated samples and signal averaging, and why raising the field (bigger ΔE) is the main route to better sensitivity.

Key takeaways

NMR-active nuclei have nonzero spin: odd-mass nuclei like ¹H and ¹³C have I = 12; ¹²C and ¹⁶O are invisible.
Energy gap: ΔE = γℏB₀; resonance frequency: ν= γB₀2π.
¹H resonates at 42.58 MHz per tesla; ¹³C at about one quarter of that.
Absorption occurs only at the resonance condition — RF frequency must match the Larmor frequency.
Boltzmann population excess is ~10^-5 at room temperature — the reason NMR is insensitive and needs many scans.
Relaxation (T₁ spin–lattice, T₂ spin–spin) prevents saturation; slow relaxation broadens lines.
The environment-dependent frequency variation of identical nuclei is the basis of chemical shifts.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

Which of the following are NMR-active: ¹²C, ¹³C, ¹H, ¹⁶O, ³¹P, ²H?
Show answer
¹³C, ¹H, ³¹P, and ²H are NMR-active. ¹²C and ¹⁶O have I = 0.
Write the resonance-frequency formula in terms of γ and B₀. What happens if the field is doubled?
Show answer
ν= γB₀2π. Doubling the field doubles the frequency (300 → 600 MHz).
Calculate the ¹³C resonance frequency in a 7.05 T magnet, given γ(¹³C) = 6.728 × 10⁷ rad s^-1 T^-1.
Show answer
ν= (6.728 × 10⁷)(7.05)2π = 4.743 × 10⁸6.2832 = 7.55 × 10⁷ Hz = 75.5 MHz — about one quarter of the proton frequency.
Why is the NMR signal so weak compared with IR or mass spectrometry signals?
Show answer
The population excess of lower-energy spins is only ~1 in 40,000 at room temperature; signal intensity is proportional to that tiny excess.
What would happen if spin relaxation did not occur during an NMR experiment?
Show answer
Without relaxation, populations would equalize (saturation) and net absorption — the signal — would stop.
A proton absorbs at 300 MHz in a 7.05 T magnet. At what frequency would it absorb in a 14.1 T magnet?
Show answer
600 MHz, because ν= γB₀2π scales linearly with field.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Spin quantum number (I): Intrinsic "amount of spin" of a nucleus; must be nonzero for NMR
Gyromagnetic ratio (γ): Nuclear constant relating magnetic moment to spin; sets the resonance frequency
Resonance condition: The match between applied RF frequency and the nucleus's Larmor frequency
Larmor frequency: Precession frequency of a spin in a field, ν= γB₀/2π
Boltzmann distribution: Population ratio of spin states at equilibrium, e^ΔE/kT
Spin–lattice relaxation (T₁): Return of excited spins to equilibrium by releasing energy to surroundings
Spin–spin relaxation (T₂): Exchange of spin energy among neighboring nuclei
Saturation: Equalized α/β populations, so no net absorption

Sources & references

openstax.org — Organic Chemistry

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