Proton NMR

Proton NMR

Lessons

In this lesson, we will learn:

  • To understand how NMR works as an analytical process.
  • To understand the descriptive terms when interpreting NMR spectra.
  • To understand how to predict splitting patterns when interpreting proton NMR spectra.
  • To apply principles of NMR to determine molecular structure.

Notes:

  • 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
    Carbon NMR

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

  • There are a number of differences to carbon NMR that make analyzing a proton NMR spectrum more complicated. Proton NMR spectra show splitting patterns in the peaks.
    Proton environments ‘couple’ with proton 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 result of the split depends on the number of adjacent protons (n), plus one. For example:
    • In a molecule with the chain R-CH2CH3\, a signal due to the -CH3\, protons will couple to the -CH2CH3\, protons, of which there are two. Using the n + 1 rule, 2 + 1 = 3, so the singlet due to -CH3\, 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, because the adjacent tertiary carbon has no protons attached to it. Using the n + 1 rule, 0 + 1 = 1, so the signal remains a singlet.
    See the table for the names of the split signals:

    N + 1 value

    Splitting pattern

    Example

    1

    Singlet

    -C(CH3)3

    2

    Doublet

    -CH(CH3)2

    3

    Triplet

    -CH2CH3

    4

    Quartet

    -CH(Cl)CH3

    5+

    Multiplet

    -CH2CH2CH3


  • 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

  • Below is a step by step outline to working out a 1H NMR spectrum using a worked example (guide in italic) if you are given 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)

    Splitting

    Integration

    0.8

    Triplet

    ~3

    1.2

    Multiplet

    ~2

    1.5

    Multiplet

    ~2

    2.2

    Triplet

    ~2

    11.8

    Singlet

    ~1


    STEP ONE: With a known molecular formula, look for a fit to a general formula. You are trying to find the hydrocarbon backbone that all organic molecules have! For example:
    • Alkane (CnH2n+2).
    • Alkene (CnH2n).
    • Aromatic compounds (C:H ratio generally nearer to 1:1 due to the C6H6 benzene ring).
    In this C5H10O2\, example, we have C5H10\, which would resemble an alkene, however there are also two oxygen atoms present, and we need to consider how they are connected in the molecule. So far, we could say we have either an alkene or alkane chain with one or more functional groups containing oxygen.

    STEP TWO: Look at the number of peaks in the spectrum compared to the formula. You are looking for signs of symmetry in the molecule!
    For 1H NMR, integration values help a lot. Alkyl chain branching and some aromatic rings will produce many protons in the same environment.
    • When a molecule is symmetrical, the same proton environments produce the same NMR signal.
    • Symmetry helps build the backbone of the molecule!
    We have five signals for ten protons over five carbon atoms: this is quite straightforward and could just be a straight carbon chain with the oxygen functional group(s) on one carbon atom. This would make a different hydrogen environment on each carbon atom in the chain. Five different signals works for a carboxylic acid, but the arrangement of an ester means the ten protons would have to be over four carbon atoms (try and draw a five-carbon ester!). The number of signals strongly suggests we do not have an ester molecule.

    STEP THREE: Try to identify the functional groups present from the rest of the formula. This is completing the rest of the molecule! For example:
    • A carboxylic acid or ester will have two oxygen atoms, and an aldehyde or ketone will have one. For all three, due to the C=O bond, any molecule with these groups will have two less hydrogens than an alkane analogue.
    • An alkene will have two less hydrogens than the alkene analogue because of the double bond to carbon that was otherwise making C-H bonds.
    From here, we have options for the two oxygen atoms in the molecule. Functional groups containing oxygen atoms include:
    • Aldehydes (RCHO) and ketones (R2CO), but these are only one oxygen atom each.
      We’d need two of them to get two oxygens as per our C5H10O2 molecular formula and that would give us less hydrogen atoms than we know we have.
      In short, there is no evidence that an aldehyde or ketone is present in the molecule.
    • Alcohols (R3C-OH) which we could have two of. This is unlikely though; their range is very broad and alcohol signals in 1H NMR are unpredictable as reactions with the solvent occur. Geminal diols (where both –OH groups are other same carbon) are also generally unstable. It is not the best idea to rely on 1H NMR for evidence of alcohols in a molecular structure.
    • Carboxylic acids (RCOOH), one of which would explain the two oxygen atoms and the two less hydrogens that we have in our formula, compared to the alkane.
      This is a possibility for the structure.
    • Esters (RCOOR’), which would explain the two oxygen atoms and the two less hydrogens that make the C:H ratio ‘alkene-like’. This is a possibility going forward.

    STEP FOUR: Use the information you’ve gathered from the molecular formula to pick out the appropriate signals – you should be able to use an NMR absorption table and your molecular formula to know where to expect signals.
    We have carried three possibilities for the structure into this step of our working:
    • An alkene with two alcohol groups attached.
    • A carboxylic acid.
    • An ester.
    We can now effectively rule out an ester now. There is a distinctive singlet peak at 11.8 ppm which an ester cannot account for. An ester would also cause deshielding (higher ppm) in protons on the sp3^{3} carbon adjacent to an ester group – the four other signals are all in a fairly ordinary alkyl chain range.
    In short, there is no evidence in our NMR spectra that specifically supports an ester, and there is specific evidence against it:
    • A carboxylic acid.
    • An ester.
    The distinctive singlet at 11.8 ppm is the range where the proton of a carboxylic acid group resonates. Carboxylic acid protons also predictable occur with singlet splitting. This is substantial evidence supporting a claim for a carboxylic acid.
    We can now assign part of our predicted structure to this NMR signal. We can also ‘strike off’ part of the formula.
    From C5H10O2, removing an acid COOH leaves the remaining molecule fragment C4H9.

    We have four signals left: 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 FIVE: Draw out your predicted structure with the NMR peaks assigned to the specific carbon atoms. See the picture below:


    1 Source for 1H NMR data: https://www2.chemistry.msu.edu/courses/cem251/SS13_HOVIG/Spectroscopy%20tables.pdf
  • Introduction
    Using 1H NMR.
    a)
    Splitting patterns

    b)
    Integration in 1H NMR

    c)
    Using 1H NMR to find structure - example.