Introduction to NMR and TMS Standard Revision Notes | A Level AQA Chemistry

1. Why NMR Spectroscopy Matters

Imagine you have synthesised an organic compound in the lab and you need to confirm its structure — without destroying the sample. Nuclear magnetic resonance (NMR) spectroscopy is the tool that makes this possible, giving you a detailed map of where specific atoms sit within a molecule.

NMR works by placing a sample in a powerful magnetic field and detecting how different nuclei respond to radiofrequency radiation. For A-level chemistry, you need to know about two types: $^{13}$C NMR, which gives information about the positions of carbon atoms, and $^{1}$H NMR, which gives information about the positions of hydrogen atoms.

Both techniques rely on the same core principle — that nuclei in different molecular environments absorb energy at slightly different frequencies — but they produce spectra with very different levels of complexity. That difference is worth understanding before you read a single spectrum.

2. The Delta Scale and Chemical Shift

Here is the central idea: not all nuclei of the same element resonate at exactly the same frequency. A hydrogen atom attached to a carbonyl group behaves differently from one attached to a carbon in an alkyl chain, because the electrons surrounding each nucleus shield it from the applied magnetic field to a different degree. This difference in resonance position is called the chemical shift.

The delta ($\delta$) scale is used because the actual frequency difference between signals depends on the strength of the spectrometer's magnetic field — and different labs use different instruments. Expressing chemical shift in ppm makes values comparable across all instruments. A signal at $\delta = 2.0$ ppm means the same thing whether the spectrum was recorded on a 300 MHz or a 600 MHz spectrometer.

On the delta scale, TMS sits at $\delta = 0$ ppm on the right-hand side. Signals from most organic compounds appear to the left of TMS — at higher ppm values — because their nuclei are less shielded than those in TMS. The further left a signal appears, the more deshielded that nucleus is.

3. Tetramethylsilane (TMS) as the Standard

Every measurement needs a reference point, and in NMR that reference is tetramethylsilane, $Si(CH_3)_4$, universally abbreviated to TMS. Its signal is defined as $\delta = 0$ ppm, and all other signals in the spectrum are reported relative to it. The question the AQA specification asks you to answer is: why TMS specifically?

TMS has twelve equivalent hydrogen atoms and four equivalent carbon atoms, all in the same highly shielded environment. This means TMS produces a single, sharp, intense signal in both $^{1}$H and $^{13}$C NMR — straightforward to identify and straightforward to use as a baseline.

• Single sharp signal: All twelve $^{1}$H atoms (and all four $^{13}$C atoms) are equivalent, so TMS gives one clear reference peak rather than several overlapping ones.
• Signal at $\delta = 0$ ppm: The silicon atom donates electron density to the surrounding methyl groups, making the $^{1}$H and $^{13}$C nuclei highly shielded. This places the TMS signal well away from the signals of most organic compounds, which appear at $\delta > 0$ ppm.
• Chemically inert: TMS does not react with the sample being analysed, so it does not alter the compound's structure or produce additional signals from reaction products.
• Low boiling point: TMS boils at approximately 27°C, so it can be removed from the sample easily after the spectrum has been recorded, leaving the original compound intact.
• Soluble in common NMR solvents: TMS dissolves readily in the organic solvents and deuterated solvents used in NMR, ensuring it mixes uniformly with the sample.

Taken together, these properties make TMS the ideal internal standard: it is non-interfering, straightforward to identify, and its signal sits in a region of the spectrum that is free from virtually all organic compound signals.

4. $^{13}$C NMR vs $^{1}$H NMR: Comparing Complexity

Both types of NMR give information about the molecular environment of atoms, but $^{13}$C NMR produces noticeably simpler spectra than $^{1}$H NMR. Understanding why helps you interpret both types confidently.

In $^{13}$C NMR, each chemically distinct carbon atom in the molecule gives one signal. Because carbon-13 is only about 1.1% naturally abundant, the probability of two adjacent carbons both being $^{13}$C is very low. This means carbon–carbon coupling is negligible, and each signal appears as a single line. The result is a clean spectrum where the number of peaks directly tells you how many different carbon environments are present.

In $^{1}$H NMR, hydrogen atoms on adjacent carbons interact with each other through a process called spin–spin coupling (splitting). This causes signals to split into multiplets — doublets, triplets, quartets, and so on — making the spectrum considerably more complex. At A-level, you need to be aware that $^{1}$H NMR spectra are more complex than $^{13}$C spectra; the detailed interpretation of splitting patterns is covered when you study $^{1}$H NMR in full.

5. Chemical Shift Depends on Molecular Environment

The key principle underpinning all NMR interpretation is this: the chemical shift of a nucleus depends on its molecular environment. Electrons surrounding a nucleus shield it from the applied magnetic field. The more electron density around a nucleus, the more shielded it is, and the lower (further right) its chemical shift value. Remove electron density — by attaching an electronegative atom nearby, for example — and the nucleus becomes deshielded, shifting its signal to higher $\delta$ values (further left).

Consider a carbon atom in an alkyl chain: it sits in an electron-rich environment and resonates at a relatively low $\delta$ value (typically 5–45 ppm in $^{13}$C NMR). Now consider a carbonyl carbon (C=O): the electronegative oxygen withdraws electron density, deshielding the carbon and shifting its signal to 160–220 ppm. That is a dramatic difference, and it is entirely explained by the change in molecular environment.

The same logic applies in $^{1}$H NMR. A hydrogen on an alkyl group resonates at $\delta$ 0.5–1.5 ppm. A hydrogen on a carbon adjacent to a carbonyl group is deshielded and appears at $\delta$ 2–3 ppm. An aldehyde hydrogen (CHO) is strongly deshielded and appears at $\delta$ 9–10 ppm. Each shift reflects the electron density — and therefore the shielding — experienced by that particular nucleus in its specific molecular environment.

6. Solvents for $^{1}$H NMR: Deuterated Solvents and $CCl_4$

To record an NMR spectrum, the sample must be dissolved in a solvent. Here is the problem: if you use an ordinary solvent such as water or ethanol, the enormous number of hydrogen atoms in the solvent would produce signals that completely swamp the signals from your sample. The solution is to use a solvent that contains no ordinary hydrogen atoms.

AQA specifies two types of solvent for $^{1}$H NMR. The first is a deuterated solvent — a solvent in which all hydrogen atoms have been replaced by deuterium ($^2$H, often written D). Deuterium has a different nuclear spin from $^{1}$H and does not produce a signal in the $^{1}$H NMR spectrum. Common examples include deuterated chloroform ($CDCl_3$) and deuterated water ($D_2O$). The second option is tetrachloromethane ($CCl_4$), which contains no hydrogen atoms at all and is therefore completely silent in $^{1}$H NMR.

Both choices achieve the same goal: the solvent produces no $^{1}$H NMR signal, so the spectrum you record contains only signals from your sample. This is essential for accurate interpretation.

7. Exam Technique and Mark-Winning Strategies

NMR questions on AQA papers tend to fall into two categories: those that ask you to explain a feature of NMR (such as why TMS is used or why deuterated solvents are needed), and those that ask you to interpret a spectrum. The explanation questions are entirely predictable — and entirely learnable.

Worked example: Answering 'Explain why TMS is used as a standard' [4 marks]

A student is asked to explain why TMS is a suitable substance to use as a standard in NMR spectroscopy. Write a mark-winning response.

1. TMS gives a signal at $\delta = 0$ ppm, which is outside the range of signals from most organic compounds. — This earns a mark for identifying the reference position and its practical advantage. [B1]
2. The silicon atom donates electron density to the surrounding methyl groups, making the $^{1}$H and $^{13}$C nuclei highly shielded. — Linking the signal position to shielding and the role of silicon shows conceptual understanding. [B1]
3. All twelve hydrogen atoms (and all four carbon atoms) in TMS are equivalent, so TMS produces one sharp, easily identified reference peak. — Equivalence of nuclei is the reason for a single peak — this is a distinct marking point. [B1]
4. TMS does not react with the sample. Its low boiling point (~27°C) means it can be removed from the sample after the spectrum is recorded. — Inertness and low boiling point are two separate properties — either earns the fourth mark. [B1]

Answer: TMS gives a single sharp signal at $\delta = 0$ ppm because all its nuclei are equivalent and highly shielded by the electron-donating silicon atom. This places the reference signal outside the range of most organic compound signals. TMS is chemically inert and has a low boiling point, so it does not interfere with the sample and can be removed easily after the spectrum is recorded.

Notice how each sentence in that model answer targets a distinct marking point. Examiners award marks for specific, precise statements — not for general waffle about NMR being 'useful'. Practise writing responses where every sentence earns its place.

Every spec point in this lesson connects back to one central idea: NMR signals depend on molecular environment, and every practical choice — the delta scale, TMS as a standard, deuterated solvents — exists to make that environmental information as clear and reproducible as possible. Keep that thread in mind and the individual facts become much easier to recall and apply.

In AQA questions, the important point is that the position of a peak tells you about the chemical environment of a proton. The number of peaks tells you how many different proton environments are present, and the splitting pattern gives information about neighbouring hydrogens.