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  1. Using 1H NMR.
  2. Splitting patterns
  3. Integration in 1H NMR
  4. Using 1H NMR to find structure - example.
Topic Notes

In this lesson, we will learn:

  • How to apply the principles learned in 13C NMR for use in 1H NMR.
  • How to predict and understand the splitting patterns and integration of 1H NMR signals.
  • How 1H NMR is used to determine the structure of organic molecules.


  • Nuclear magnetic resonance (NMR) spectroscopy is an extremely versatile tool to determine chemical structure. The details of how NMR spectroscopy works was covered in Carbon NMR

  • NMR for hydrogen nuclei is called proton (1H) NMR. This is the major isotope of hydrogen which is spin active.

  • All of the essentials you learned in 13C NMR apply to 1H NMR:
    • Signals are produced in an NMR spectrum which are related to ‘environments’, or how the H atoms in a molecule are bonded to the rest of the structure.
    • These environments are measured in ppm with TMS as a reference.
    • There are regions of the 1H NMR spectrum which reflect ‘proton environments’. Some include:
      • 0-2 ppm is the alkyl proton (R3CH, R2CH2, RCH3) region, with -CH3 closest to the 0 ppm reference.
      • 5-6 ppm is the alkenyl proton region (R2C=C(H)R).
      • 6.5 – 7.5 ppm is the aromatic proton region.
      • 9-11 ppm is generally the acid/carbonyl proton region. Carboxylic acid protons appear at 10-11 ppm, while aldehyde ones closer to 9-10 ppm. Ketones do not have any protons directly attached to them; a 9-10 ppm signal in the 1H NMR spectrum will settle if you have a ketone or an aldehyde!
      This and more can all be found in a 1H NMR absorption table.

  • Proton NMR spectra are a bit more complicated than 13C NMR. Proton NMR spectra show splitting patterns in the peaks.
    This is when proton environments ‘couple’ with other environments on carbons adjacent to them. This causes the signals to ‘split’ in a way that is sometimes called the n + 1 rule because the type of split depends on the number of adjacent protons (n), plus one. For example:
    • In a molecule with R–CH2CH3, the –CH3 protons will couple to the –CH2CH3 protons: there’s two of them. Using n + 1, 2 + 1 = 3, so the –CH3 signal becomes a triplet signal. This ‘triplet’ is known as a splitting pattern. .
    • In a molecule with the fragment R–C(CH3)3, a signal due to the –C(CH3)3 protons will not split with any other signal, because the adjacent carbon is tertiary – it has no protons attached to it. Using n + 1, 0 + 1 = 1, so the C(-CH3)3 signal remains a singlet.

    See the table for the names of the split signals:

    N + 1 value

    Splitting pattern

















  • Another difference from carbon NMR to proton NMR is the assigning of integration values to peaks. Integration values are found from the area under the peaks, and roughly relate to the number of protons that produce this signal. This is different from the splitting pattern!
    • For example, the protons -CH(OH) CH2CH3 would have an integration of 2.
    • The protons -CH2CH2CH3 would have an integration of 3.

    These ‘differences’ once you understand them are really just more information, or stronger evidence for the structure of a molecule – 1H NMR is more useful than 13C NMR for a lot of organic molecules!

  • Worked example: Find the structure of a molecule using 1H NMR spectra and the molecular formula.
    Determine the structure of a molecule with the formula C5H10O2.
    There are five signals in the 1H NMR spectrum:

  • Chemical shift (ppm)


















    This format of question can come with a lot of other information (such as mass spectrum, the 13C NMR or some test-tube reaction results) to make it easier, or it might not. The less information you’re given, the more you can give the examiner to get marks!

    STEP ONE: Just like we did with 13C NMR, start by using the molecular formula to spot functional groups!
    Recall some general formulae:
    • CnH2n+2 for an alkane. Think of this as your starting molecule or a blank slate with no functional groups to create obvious NMR signals.
    • CnH2n+2O for an alcohol. Alcohol groups won’t remove any H from the molecule because the oxygen has effectively ‘inserted’ between a C-H bond.
    • CnH2n for an alkene.
    • CnH2nO for a carbonyl e.g. ketone or aldehyde. Each carbonyl group will replace two hydrogens from the molecule, compared to if the carbon was just a -CH2- unit instead.
    • CnH2nO2 for an ester or acid. Both these groups have a carbonyl bond that has replaced two hydrogens.

    We have the formula C5H10O2, so two H atoms or C-H bonds have been replaced by the two oxygen atoms now present. This only reasonably matches a carbonyl (C=O) group as part of an ester or carboxylic acid functional group. We have some evidence for an ester or acid.

    STEP TWO: Compare the number of peaks to the molecular formula. You are looking for clues about symmetry!
    For 1H NMR, integration values help a lot because alkyl chain branching and aromatic rings can produce many protons in the same environment.

    We have five proton environments for ten protons over five carbon atoms. Apart from the CH3 chain ends, carbon atoms without any functional groups attached are just -CH2- so we have five signals for five “-CH2-” units. This just means each CH2 or CH3 section of the chain is a unique environment.
    It could be a straight carbon chain with the oxygen functional group(s) on one carbon atom. This makes every proton environment different (one is attached to the functional group, one is one carbon away, another two away, another three...).

    Five different signals works for a carboxylic acid, but not an ester. An ester would need the ten protons to be spread over only four carbons because the ester carbon won’t be bonded to any hydrogens at all. The number of signals is strong evidence that we don’t have an ester.

    STEP THREE: Now we can look for the expected peaks in the NMR spectrum, starting with the functional groups:

    A carboxylic acid is the major functional group that still fits both the formula and the number of expected proton signals. Carboxylic acid 1H NMR peaks are obvious singlet signals at around 10-12 ppm. We have a signal in exactly that range! It is a singlet, too. This is the -COOH proton environment.

    From here, we have assigned COOH of our molecule. What else do we have left? Try and ‘subtract’ COOH from the molecular formula: C5H10O2 – COOH = C4H9. So we have C4H9 of an organic molecule left – that’s just a four-carbon butyl chain.

    We have four signals left too: 0.8 ppm; 1.2 ppm; 1.5 ppm and 2.2 ppm, to assign to four proton environments. These are all in the region where saturated carbon is found – a typical alkane chain. The deshielding effect of the carboxylic acid group (producing higher ppm) gets weaker the further away from the group you go. These four signals are the four remaining carbon atoms in a straight chain; the higher the ppm, the closer it is to the carboxylic acid group.
    • The triplet at 2.2 ppm is due to the –CH2 protons of the carbon adjacent to the COOH group. The closest protons to the deshielding effect would predictably make them resonate further downfield (higher ppm) than the rest of the protons in the chain.
    • The two multiplets – coupling both sides of themselves in the straight chain – are predictably the middle peaks of the spectrum. They are more shielded than the adjacent carbon, the triplet at 2.2 ppm, but less shielded than the terminal carbon.
    • The signal at 0.8 ppm, the terminal carbon attached to a CH2 group, would couple for a triplet. This would be assigning the rest of the molecule.

    STEP FOUR:Draw out your predicted structure with the NMR peaks assigned to the specific carbon atoms. See the picture below:

    Notice that by seeing the formula and the number of NMR peaks we have a good idea about what our molecule is before we even look at the actual ppm values.
    You have seen a lot more molecular formulae and general formulae/functional groups than you have seen NMR spectra, so start with what you know best! There are millions of organic molecules but by starting with ruling out possible functional groups you will quickly narrow down your search for what the structure is, and that is important in a timed exam paper.

  • Worked example: Compare the 1H NMR spectra of two molecules:
    Below is the structure of two molecules. Describe and explain the differences expected in the 1H NMR of the two compounds.

  • In this question format, you know the structure already. What are the obvious differences? Start with those and try to add as much detail as possible:
    • The molecule on the left contains a ketone (RCOR’).
    • The molecule on the right contains an aldehyde (-CHO).
    • The molecule on the left will have an extra -CH3 group that the aldehyde does not.

    Think about these differences – how do they change the NMR spectrum?

    • The aldehyde will have a sharp singlet around 10-11 ppm with an integration value of 1. This corresponds to the proton attached to the carbonyl carbon. This signal will simply not be there in the ketone.
    • The ketone will have an extra -CH3 carbon environment around 1.5-2 ppm. This will also be a singlet with an integration of 3, whereas the other alkyl proton signals (from the ethyl chain attached to the benzene ring) will be a triplet or quartet. This singlet alkyl proton signal won’t be present in the aldehyde.

    Because you were given the structure of the molecules already, you need to give as much detail as possible on the actual signals. Don’t waste time on the signals they both have – those aren’t going to be used to identify the compounds are they? Always give:

    • The expected ppm values of the peaks that the molecules don’t have in common.
    • The splitting pattern (if it is 1H NMR).
    • The integration (if it is 1H NMR).

1 Source for 1H NMR data: https://www2.chemistry.msu.edu/courses/cem251/SS13_HOVIG/Spectroscopy%20tables.pdf