19B.2-19B.5 - Carbon And Proton NMR | Edexcel A-Level Chemistry

NMR reveals how atoms are grouped into chemical environments. Use carbon signals, proton chemical shifts, integration and splitting together to distinguish isomers and predict the spectra of a known structure.

What NMR Evidence Means

Two compounds can have the same molecular formula but different atom connections. For example, ethyl ethanoate and methyl propanoate both have the formula $\ce{C4H8O2}$ , so formula evidence alone cannot decide between them. NMR helps because atoms in different chemical environments absorb at different chemical shifts.

NMR spectra are read as evidence about environments:

Spectrum	Main evidence for this lesson	What it helps decide
${}^{13}\mathrm C$ NMR	chemical shift of carbon signals; number of carbon environments	which types of carbon are present and how many different carbon environments the molecule has
high-resolution ${}^{1}\mathrm H$ NMR	chemical shift, relative peak area, and splitting	which types of proton are present, their relative numbers, and how many adjacent non-equivalent protons they have

Chemical shift

The position of an NMR signal, measured in ppm relative to tetramethylsilane (TMS), which is assigned $0\,\mathrm{ppm}$ .

In many printed spectra the ppm scale decreases from left to right, so a signal further left is at higher chemical shift.

The useful reasoning chain is:

peak position → chemical-shift range → atom environment
number of signals → number of different environments
integration → relative number of protons
splitting → adjacent non-equivalent protons
combined evidence → possible structure

Environments, not atoms

NMR counts environments, not every atom separately. Equivalent atoms give the same signal, so two or three atoms may produce one peak if they are in the same chemical environment.

Equivalent protons form one proton environment and do not split one another. In ethane, the two carbons and all six protons are equivalent by symmetry; count each set once rather than applying neighbour splitting to equivalent protons.

Carbon-13 NMR And Carbon Environments

${}^{13}\mathrm C$ NMR spectroscopy gives information about the positions of carbon atoms through their chemical environments. Each distinct carbon environment gives one ${}^{13}\mathrm C$ NMR signal.

Do not use ${}^{13}\mathrm C$ peak height or area to count carbon atoms. The reliable A level move is to count the number of different carbon environments and use the chemical-shift position as environment evidence.

Draw or imagine the displayed structure.
Label carbons that are bonded to the same types of atoms and sit in the same symmetry position.
Group equivalent carbons together.
Predict one ${}^{13}\mathrm C$ signal for each group.
Use the Data Booklet chemical-shift range to support the type of carbon.

Compound	Carbon-environment reasoning	Expected ${}^{13}\mathrm C$ signal count
propanone, $\ce{CH3COCH3}$	the two methyl carbons are equivalent; the carbonyl carbon is different	2
butan-2-one, $\ce{CH3COCH2CH3}$	carbonyl carbon, acyl methyl, methylene, and terminal methyl are all different	4
2-methylpropane, $\ce{(CH3)3CH}$	three methyl carbons are equivalent by symmetry; the central carbon is different	2

The chemical shift then adds type evidence. A carbonyl carbon appears much further downfield than an alkyl carbon, while a carbon bonded to oxygen appears in a different region from an ordinary carbon-carbon alkyl environment.

For example, supplied carbon shifts of about $205$ and $30\,\mathrm{ppm}$ for propanone identify a ketone carbonyl and a methyl environment. There are three carbon atoms but only two environments: both methyl carbons contribute to the same signal. A carbon spectrum is normally proton-decoupled in these questions, so each carbon environment is shown as a single line; apply proton splitting rules to the proton spectrum, not to these carbon lines.

Common-error contrast

A molecule with four carbon atoms does not automatically have four ${}^{13}\mathrm C$ peaks. Ethyl ethanoate has four different carbon environments, so it does have four peaks; propanone has three carbon atoms but only two carbon environments because the two $\ce{CH3}$ carbons are equivalent.

Proton Environments And Integration

High-resolution ${}^{1}\mathrm H$ NMR gives information about the positions of hydrogen atoms in a molecule through their proton environments.

Proton environment

A set of hydrogen atoms with the same chemical surroundings; each environment gives one signal.

For ${}^{1}\mathrm H$ NMR, three pieces of evidence are used together:

Evidence	Meaning	Caution
chemical shift	type of proton environment, such as alkyl, next to oxygen, alkene, arene, aldehyde or acid proton	quote or use the supplied Data Booklet range rather than guessing from memory
relative peak area, often called integration	relative number of protons in each environment	it gives a ratio, not an absolute count unless other evidence fixes the formula
splitting pattern	number of adjacent non-equivalent protons	use the $(n+1)$ rule only for the adjacent protons causing the split

Integration is especially powerful because it links a spectral signal to a proton count. If three signals have relative areas $3 : 2 : 3$ , the molecule has proton environments in that ratio. It might be exactly $\ce{3H}$ , $\ce{2H}$ , and $\ce{3H}$ if the formula supports eight hydrogens in those environments.

Both possible structures have formula $\ce{C4H8O2}$ , but their proton environments differ.

Structure	Key proton-environment pattern
methyl propanoate, $\ce{CH3CH2COOCH3}$	an $\ce{OCH3}$ singlet with integration $\ce{3H}$ , plus an ethyl group next to a carbonyl
ethyl ethanoate, $\ce{CH3COOCH2CH3}$	a $\ce{COCH3}$ singlet with integration $\ce{3H}$ , plus an $\ce{OCH2CH3}$ ethyl group

Both have three proton environments, so the number of ${}^{1}\mathrm H$ signals alone is not enough. The chemical shifts, integration and splitting must be read together.

Ratio versus signal count

A peak area ratio of $3 : 2 : 3$ is not the same evidence as three signals: the ratio tells the relative number of protons under the signals. The number of signals tells the number of proton environments. These are different kinds of evidence.

Splitting And The n Plus One Rule

In high-resolution ${}^{1}\mathrm H$ NMR, a proton signal is split by adjacent non-equivalent protons.

The n + 1 rule

\text{number of peaks} = n + 1

$n$ is the number of neighbouring protons that are equivalent to one another but in a different environment from the protons giving the signal.

Adjacent non-equivalent protons, n	Splitting pattern	Common name
0	1 peak	singlet
1	2 peaks	doublet
2	3 peaks	triplet
3	4 peaks	quartet

Equivalent protons do not split their own signal. Protons on rapidly exchanging $\ce{O-H}$ or $\ce{N-H}$ groups may appear broad or may not follow a simple splitting pattern in ordinary A level data, so use the pattern supplied in the question.

For bromoethane, $\ce{CH3CH2Br}$ , there are two proton environments.

Signal	Adjacent non-equivalent protons	Splitting	Reason
$\ce{CH3}$ protons	2 on the neighbouring $\ce{CH2}$ carbon	triplet	$n+1 = 2+1 = 3$
$\ce{CH2Br}$ protons	3 on the neighbouring $\ce{CH3}$ carbon	quartet	$n+1 = 3+1 = 4$

The $\ce{CH2Br}$ signal is also shifted downfield compared with an ordinary alkyl $\ce{CH2}$ because the carbon is bonded to bromine. Splitting tells you about neighbours; chemical shift tells you about environment.

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19B.2-19B.5 - Carbon And Proton NMR