Nuclear magnetic resonance spectroscopy

In a nutshell

Nuclear magnetic resonance (NMR) spectroscopy tells you where the carbon and hydrogen atoms sit in a molecule, so it is the most powerful single tool for working out an unknown structure.

It joins mass spectrometry and infrared spectroscopy as the third analytical technique: used together, they let chemists confirm the structure of a new compound. The exam tests whether you can read the data to solve a structure, not the physics of how the machine works.

Assumed knowledge: Organic analysis (mass spectrometry and infrared spectroscopy), Introduction to organic chemistry (structural and skeletal formulae).

Core content

Why chemists use NMR

No single technique proves a structure on its own. Chemists have built up a range of analytical techniques which, used together, confirm the structure of a new compound:

• Mass spectrometry gives the relative molecular mass (from the molecular ion) and fragments.
• Infrared spectroscopy identifies the bonds and functional groups present.
• NMR gives information about the position of the ¹³C or ¹H atoms in the molecule, which is the carbon-hydrogen skeleton the other two cannot show.

There are two kinds you need: ¹³C NMR (looks at carbon atoms) and ¹H NMR, also called proton NMR (looks at hydrogen atoms).

Chemical environments: one signal for each environment

This is the idea the whole topic rests on.

Atoms of the same element can sit in different chemical environments depending on what is bonded around them. In NMR:

• each set of atoms in the same environment gives one signal (peak);
• so the number of peaks tells you the number of different environments.

Atoms are in the same environment (they are equivalent) if swapping them over leaves the molecule unchanged. A molecule's symmetry therefore reduces the number of peaks.

Still don't get it? · equivalent environments

Think of a cinema. Two seats are "the same kind of seat" if you could swap the two people sitting in them and nobody could tell the difference. A front-row aisle seat and a back-row middle seat are different kinds; the two aisle seats in the same row are the same kind.

Atoms work the same way. Two hydrogen atoms are in the same environment if you could swap them and end up with exactly the same molecule, drawn the same way. In propanone, CH₃COCH₃, the two CH₃ groups are mirror images across the C=O, so swapping them changes nothing: they are equivalent and count as one environment, not two.

The exam point: count the number of different environments, then look hard for symmetry that makes atoms equivalent. One environment gives one peak, so a symmetrical molecule gives fewer peaks than you might first expect.

The chemical shift (δ) scale

The horizontal axis of every NMR spectrum is the chemical shift, given the symbol δ and measured in parts per million (ppm).

• δ measures how far a signal lies from the signal of the standard, TMS, which is defined as δ = 0.
• Chemical shift depends on the molecular environment of the atom: electron-withdrawing groups nearby (such as O, or a C=O) pull electron density away and push the signal to a higher δ (further left, called downfield).

So the position of a peak (its δ value) tells you the kind of environment, and you match it against the chemical shift table in the Chemistry Data Booklet. You are given that table in the exam, so the skill is reading and matching it, not memorising numbers.

Tetramethylsilane (TMS): the standard

A small amount of tetramethylsilane, Si(CH₃)₄, is added to the sample as the reference point against which every shift is measured. The same standard is used for both ¹H and ¹³C NMR.

Why TMS is a suitable standard (each is a separate creditable reason):

• All 12 hydrogen atoms are equivalent (and all 4 carbons equivalent), so it gives a single, strong peak that is easy to see with only a little added.
• Its peak lies upfield (to the right) of almost all other organic signals, at δ = 0, so it does not overlap the peaks you are trying to read.
• It is chemically inert, so it does not react with the sample.
• It is volatile (low boiling point), so it is easily removed afterwards to recover the sample.
• It is non-toxic.

Exam caution: if a question states that TMS is inert and non-toxic, those two cannot then be your answer. Give the independent reasons instead (one peak, all protons equivalent, upfield of other peaks, easily removed).

Solvents for ¹H NMR

A solvent that contained ordinary hydrogen atoms would swamp the spectrum with its own huge ¹H peaks. So ¹H NMR samples are dissolved in solvents that are proton-free:

• Deuterated solvents, most commonly CDCl₃ (deuterium, ²H, absorbs well outside the normal ¹H region, so it is not seen), or
• tetrachloromethane, CCl₄, which contains no hydrogen atoms at all.

The reason to state is simply that they have no ¹H atoms, so they give no signal to confuse the spectrum.

¹³C NMR

¹³C NMR gives simpler spectra than ¹H NMR: the peaks are not split, so you read it as a straight count.

• The number of peaks = the number of different carbon environments.
• The δ value of each peak identifies the kind of carbon, read off the Data Booklet (for example a carbonyl carbon in an aldehyde or ketone appears far downfield at about 190 to 220 ppm; an ester or acid C=O at about 160 to 185 ppm; a C-O carbon at about 50 to 90 ppm; a plain alkane carbon at about 5 to 40 ppm).

Below is the ¹³C spectrum of methyl propanoate, CH₃CH₂COOCH₃. Its four carbons are all in different environments, so there are four peaks:

¹H NMR: integration and the number of protons

A ¹H spectrum carries an extra layer of information that ¹³C does not:

1. Number of peaks = number of different proton environments.
2. Chemical shift (δ) of each peak = the kind of environment (match to the Data Booklet).
3. Integration: the area under each peak (shown as a step on the integration trace) is proportional to the number of ¹H atoms in that environment. So the integration gives the relative numbers of equivalent protons, as a ratio.

Integration gives a ratio, not absolute numbers: a 3 : 2 : 1 trace could mean 3 : 2 : 1 protons or 6 : 4 : 2.

Here is the ¹H spectrum of ethanol, CH₃CH₂OH. Three environments give three peaks, with integration 3 : 2 : 1:

Spin-spin splitting and the n + 1 rule

In a high-resolution ¹H spectrum each peak may be split into a cluster of smaller lines by the protons on the neighbouring carbon atom. The pattern tells you how many protons are next door.

The rule is:

A signal is split into (n + 1) lines, where n is the number of equivalent protons on the adjacent carbon atom(s).

AQA limits you to aliphatic compounds and to these patterns:

Two further points the examiners insist on:

• Equivalent protons do not split each other. Protons in the same environment (for example the three H of one CH₃, or two identical CH₂ groups) produce a singlet, not a split pattern.
• O-H and N-H protons usually appear as a singlet and do not split the protons next to them, so an OH or NH often shows up as a lone singlet.

Still don't get it? · the n + 1 rule

Why should the hydrogens on the carbon next door change the shape of my peak?

Picture each neighbouring proton as a tiny bar magnet that can point either with the big magnetic field or against it. The proton you are watching feels the field plus whatever its neighbours' little magnets add.

• One neighbour: it points up or down, two equally likely options, so your proton feels two slightly different fields and its signal splits into two lines (a doublet).
• Two neighbours: the combinations are (up, up), one-up-one-down (which can happen two ways), and (down, down). That gives three field strengths in a 1 : 2 : 1 ratio, so you get a triplet.
• Three neighbours give four combinations in a 1 : 3 : 3 : 1 ratio, a quartet.

Count the combinations and you always get one more line than the number of neighbours. That is the exam shortcut: lines = (number of adjacent equivalent H) + 1. So a CH₃ sitting next to a CH₂ (2 neighbours) is a triplet, and that CH₂ next to the CH₃ (3 neighbours) is a quartet. A neighbouring CH₂CH₃ is the classic quartet-and-triplet pair.

Worked examples

Example 1: counting the peaks in a ¹³C spectrum (using symmetry)

How many peaks does each compound give in its ¹³C NMR spectrum?

Propanone, CH₃COCH₃:

• The molecule is symmetrical, so the two CH₃ carbons are equivalent (one environment).
• The C=O carbon is a second environment.
• 2 environments, so 2 peaks.

Propanal, CH₃CH₂CHO:

• The CH₃, the CH₂ and the CHO carbons are all different, with no symmetry.
• 3 environments, so 3 peaks.

The lesson: always check for symmetry before you count, because equivalent carbons share a peak.

Example 2: full structure from ¹H NMR (shift + integration + splitting)

A compound of molecular formula C₄H₈O₂ gives these three peaks. Deduce the structure.

Work through the three peaks:

• Peak a: δ ≈ 3.7 is in the 3.7 to 4.1 range for a proton on a carbon bonded to O (an ester O-CH); integration 3 means a CH₃; a singlet means no H on the adjacent carbon. So this is -O-CH₃.
• Peak c: δ ≈ 1.1 is in the 0.7 to 1.2 range for a plain CH₃; integration 3; a triplet means 2 H on the adjacent carbon (a CH₂). So a CH₃ next to a CH₂.
• Peak b: δ ≈ 2.3 is in the 2.1 to 2.6 range for a CH next to C=O; integration 2 means a CH₂; a quartet means 3 H on the adjacent carbon (a CH₃). So a CH₂ between the C=O and the CH₃.

Assemble the fragments CH₃CH₂-, a C=O, and -OCH₃, using up C₄H₈O₂:

$CH_3CH_2{-}CO{-}O{-}CH_3$

The compound is methyl propanoate. (Notice how splitting separates the two CH₃ groups that both integrate for 3: one is a singlet, one is a triplet.)

Example 3: turning an integration trace into a ratio

The integration trace for ethanol's three peaks measures 4.5, 3.0 and 1.5 units.

• Divide each by the smallest value (1.5):

$4.5 : 3.0 : 1.5 \;=\; 3 : 2 : 1$

• So the environments contain protons in the ratio 3 : 2 : 1, matching CH₃ : CH₂ : OH.

Always reduce the trace to the simplest whole-number ratio, exactly as you would an empirical formula.

Common exam mistakes

• Forgetting symmetry, so predicting too many peaks. Equivalent atoms share one peak. Propanone gives 2 ¹³C peaks (not 3) because the two CH₃ are equivalent, and a para-disubstituted benzene gives fewer peaks than its carbon count suggests.
• Saying propanone has two ¹H peaks. Both methyl groups are equivalent, so propanone gives one ¹H peak.
• Vague splitting statements. "The triplet is due to the CH₂" is not enough and loses the mark. You must say the signal due to a particular group is split into a triplet by the (2 equivalent) protons on the adjacent carbon.
• Explaining splitting without the number of neighbours. Just writing "the n + 1 rule applies" does not score; state how many hydrogens are on the adjacent carbon.
• Predicting a split for equivalent protons. Two identical CH₂ groups, or an isolated CH₃ with no neighbouring H, give a singlet, not a doublet or triplet.
• Quoting "inert" and "non-toxic" for TMS when the question already states them. Give independent reasons (one peak, all protons equivalent, upfield of other peaks, volatile/easily removed).
• Giving the wrong solvent, or D₂O as the general solvent. State a proton-free solvent, CDCl₃ or CCl₄, and the reason that it has no ¹H atoms.
• Reading integration as absolute numbers. It is a ratio: a 3 : 2 trace could be 6 : 4 protons.
• Giving TMS as a displayed or structural formula when the molecular formula Si(CH₃)₄ was asked for (a repeated examiner complaint).
• Misreading the δ axis for ¹³C, for example calling a peak at 200 ppm an ester (160 to 185) when it is an aldehyde or ketone (190 to 220).

Key definitions

• Chemical shift (δ) - the position of an NMR signal measured in parts per million (ppm) relative to the signal of TMS, which is set at δ = 0.
• Chemical environment (equivalent atoms) - atoms are in the same environment, and give a single signal, if they can be exchanged without changing the molecule; the chemical shift depends on the molecular environment.
• Integration - the area under a ¹H peak, which is proportional to the number of ¹H atoms in that environment and so gives the relative numbers of equivalent protons.
• The n + 1 rule - a signal is split into (n + 1) lines by n equivalent protons on the adjacent carbon atom(s).
• Standard (TMS) - tetramethylsilane, Si(CH₃)₄, added to calibrate the δ scale; it gives one strong peak upfield of other signals, is inert, volatile and non-toxic.

Specification

• I can state that chemists use a range of analytical techniques (mass spectrometry, infrared and NMR) which together confirm the structure of a new compound.
• I can state that NMR gives information about the position of ¹³C or ¹H atoms in a molecule, and that ¹³C NMR gives simpler spectra than ¹H NMR.
• I can explain the δ scale for chemical shift, that chemical shift depends on the molecular environment, and that integrated ¹H spectra give the relative numbers of protons in different environments.
• I can state that ¹H NMR samples are dissolved in deuterated solvents or CCl₄ and explain why.
• I can explain why TMS is a suitable standard.
• I can use ¹H and ¹³C spectra with chemical shift data from the Data Booklet to suggest possible structures or part-structures.
• I can use integration data to find the relative numbers of equivalent protons.
• I can use the n + 1 rule to deduce doublet, triplet and quartet splitting of adjacent, non-equivalent protons in aliphatic compounds.

Related notes

• Organic analysis (mass spectrometry and infrared, the other two techniques NMR is combined with)
• Introduction to organic chemistry (structural, displayed and skeletal formulae)
• Aldehydes and ketones (3.3.8): the carbonyl carbon seen far downfield in ¹³C spectra
• Carboxylic acids and derivatives (3.3.9): esters and acids, a common source of NMR unknowns
• Organic synthesis (3.3.14): confirming the product of a synthetic route