Deuterated Solvents in NMR

Introduction

An NMR sample is mostly solvent: a few milligrams of compound dissolved in around half a millilitre of liquid. If that solvent contained ordinary hydrogen, its enormous ¹H signal would swamp the spectrum of the compound. The solution is to use deuterated solvents , in which hydrogen has been replaced by deuterium. Understanding these solvents explains several features you will see in real spectra, including small unexplained peaks that belong to the solvent rather than the sample.

Core explanation

The problem with ordinary solvents. A typical sample contains roughly a thousand times more solvent molecules than compound molecules. With a hydrogen-containing solvent such as CHCl₃ or H₂O, the solvent's signal would be vastly larger than any compound signal, overloading the receiver and hiding peaks nearby.

Deuterium is invisible in ¹H NMR. Deuterium (²H or D) has a nuclear spin, but its resonance frequency in a given field is about 6.5 times lower than that of ¹H. A ¹H spectrum is recorded in a narrow band of frequencies that does not include deuterium, so a fully deuterated solvent gives no ¹H signal.

The deuterium lock. Spectrometers also use the solvent's deuterium signal to stabilise the magnetic field. A separate channel continuously monitors the deuterium resonance and corrects any drift in the field, keeping peaks sharp during long acquisitions. The deuterium signal is also used for "shimming" — fine-tuning the field to make it uniform across the sample.

Residual solvent peaks. No deuterated solvent is 100% pure. Commercial CDCl₃ is typically 99.8% D, so about 0.2% of molecules are CHCl₃. Because the solvent is so concentrated, even this small fraction gives a visible signal. You must learn to recognise these peaks and not assign them to your compound.

Common deuterated solvents:

Solvent Formula Residual ¹H peak (δ) Typical use --- --- --- --- Deuterochloroform CDCl₃ 7.26 most organic compounds Deuterium oxide D₂O about 4.8 (HOD) water-soluble compounds, salts, sugars Dimethyl sulfoxide-d₆ (CD₃)₂SO 2.50 polar compounds, acids, amides Methanol-d₄ CD₃OD 3.31 polar compounds Acetone-d₆ (CD₃)₂CO 2.05 moderately polar compounds

Why CDCl₃ is the default. It dissolves most neutral organic compounds, is relatively cheap, is easily removed by evaporation so the sample can be recovered, and its single residual peak at δ 7.26 lies in a region that interferes only with aromatic signals. It has no exchangeable deuterium, so O–H and N–H protons of the sample remain visible.

Water peaks. Deuterated solvents absorb moisture from the air. Traces of water give a peak whose position depends on the solvent: about δ 1.56 in CDCl₃ and about δ 3.33 in (CD₃)₂SO.

Solvent choice changes the spectrum. In D₂O or CD₃OD, exchangeable protons in the sample are replaced by deuterium, so O–H and N–H peaks disappear — the same effect as a D₂O shake.

Step-by-step reasoning

To decide whether a small peak belongs to the compound:

1. Identify the solvent used for the spectrum. 2. Recall its residual peak position (for example δ 7.26 for CDCl₃). 3. Check the expected water peak position for that solvent. 4. Exclude these peaks from integration and assignment. 5. Assign only the remaining signals to the compound.

Visual explanation

In the simulation, record the spectrum of methylbenzene first in CHCl₃ and then in CDCl₃. In CHCl₃ a gigantic solvent peak at δ 7.26 dwarfs every other signal. In CDCl₃ the solvent peak shrinks to a small spike, and the methyl singlet near δ 2.3 and the aromatic multiplet near δ 7.2 become clearly visible.

Real-world analogy

Trying to record the ¹H spectrum of a compound in ordinary chloroform is like trying to hear a whisper at a rock concert. Switching to CDCl₃ is like turning off the band: the whisper can be heard, though a faint background hum — the residual solvent peak — always remains.

Real-world example

Metabolomics laboratories analyse urine and blood samples by NMR to detect disease markers. They add a small amount of D₂O to provide a lock signal and use special pulse sequences to suppress the huge remaining water signal, allowing dozens of metabolites to be measured at once.

Why?

Why does the CDCl₃ residual peak appear as a singlet even though the proton is next to a carbon? Neighbouring nuclei that would split it are chlorine atoms and ¹²C, which do not couple in the normal way, and coupling to the rare ¹³C isotope produces only tiny satellite peaks. The CHCl₃ proton therefore shows as a single line.

Common misconception

"A peak at δ 7.26 in CDCl₃ means the compound is aromatic." It may be only the residual CHCl₃ signal. Before concluding anything, check the integration and whether other aromatic signals are present.

Worked example

Question: The ¹H spectrum of butanone in CDCl₃ shows peaks at δ 1.05 (t, 3H), 1.56 (s, small), 2.14 (s, 3H), 2.45 (q, 2H) and 7.26 (s, small). Assign each peak.

Reasoning: Butanone, CH₃COCH₂CH₃, has three environments. The triplet, singlet and quartet match CH₃ of the ethyl group, CH₃CO and CH₂. The small peaks at 1.56 and 7.26 match water and residual CHCl₃.

Answer: δ 1.05 CH₃CH₂, 2.14 CH₃CO, 2.45 CH₂; δ 1.56 is water and δ 7.26 is residual solvent.

Quick check

1. Why does using D₂O as the NMR solvent make the O–H signal of a sugar disappear? Answer: The O–H protons exchange with deuterium from the solvent, and deuterium does not appear in the ¹H spectrum.

Exam focus

Give two reasons for using CDCl₃: it contains no ¹H to swamp the spectrum, and it dissolves most organic compounds. You may also mention that it is volatile and easy to remove. Examiners sometimes include a solvent peak to test whether you ignore it.

Advanced insight

In ¹³C spectra, CDCl₃ gives a triplet at δ 77.2 with equal line intensities (1 : 1 : 1). Deuterium has spin 1, so it has three spin states rather than two, and one attached deuterium splits the carbon signal into 2nI + 1 = 3 lines. The same rule explains the seven-line ¹³C signal of (CD₃)₂SO at δ 39.5.

Summary

NMR samples are dissolved in deuterated solvents because deuterium does not absorb in the ¹H range, so the solvent does not swamp the spectrum; its deuterium signal also locks the field. CDCl₃ is the usual choice. Incomplete deuteration produces small residual peaks, such as δ 7.26 for CHCl₃, and water gives extra peaks that must not be assigned to the compound.

Practice questions

1. Give two reasons why CDCl₃ is widely used as an NMR solvent. Answer: It has no ¹H atoms to produce a large solvent signal, and it dissolves most organic compounds; it is also volatile, so the sample can be recovered easily. 2. Explain why a small singlet at δ 7.26 appears in a spectrum run in CDCl₃. Answer: About 0.2% of the solvent molecules are CHCl₃, and because there is so much solvent this residual proton gives a visible signal. 3. Which solvent would you choose for an amino acid that is insoluble in CDCl₃, and what would happen to its NH₃⁺ and COOH signals? Answer: D₂O; the exchangeable NH and OH protons would be replaced by deuterium, so their signals would disappear. 4. What is the purpose of the deuterium lock? Answer: The spectrometer monitors the solvent's deuterium signal and adjusts the field to correct drift, keeping the spectrum stable and peaks sharp.