Introduction to NMR: Carbon-13 (13C) NMR

Introduction to NMR: Carbon-13 (13C) NMR

Lessons

In this lesson, we will learn:

  • How NMR is used as evidence for the structure of organic molecules.
  • How to read 13C NMR spectra and identify different carbon environments.
  • How to predict a 13C NMR spectra for simple organic molecules.

Notes:

  • Nuclear magnetic resonance (NMR) spectroscopy is a very useful tool to help chemists understand the structure of a molecule.
    In a short sentence, NMR spectroscopy tells chemists how the atoms in a molecule are attached to one another. For atoms like carbon and hydrogen, it will give you strong evidence for particular ‘environments’ in the molecule like -CHO, COOH and other groups/bonds.

  • The longer explanation of how this works is below:
    • Some nuclei have a property called spin – it is not literally spinning but for now you can think of it this way. Hydrogen-1 (1H) and carbon-13 (13C) both have this property - very useful for organic chemists.
    • NMR machines have a magnetic field that ‘flips’ the spin of a nucleus - think of it as now spinning the other way (clockwise \, \, anticlockwise). When this ‘flips back’ to its original spin state it releases radio-frequency energy which is measured by the NMR spectrometer.
    • Electrons around the nucleus interfere with the machine’s magnetic field; the electrons are shielding the nucleus.
    • As a result, there is a difference between the machine’s field and the nucleus’ field it experiences. This chemical shift is seen by the change in frequency (Hz) of energy the nucleus emits.
    • Because the electrons cause the chemical shift (symbol δ \delta), NMR just reveals what the electrons are ‘doing’ around a nucleus – different shielding means different chemical shift. What bonds are the electrons making?
    • NMR spectrometers run at a variety of frequencies (their field strength) so the units on an NMR spectrum is parts per million (ppm) . Ppm is just a fraction (parts per million!) of the nuclei’s resonance frequency compared to the resonance of a reference sample called tetramethylsilane (TMS) defined as 0 ppm . This TMS reference is just like how the freezing point of water was chosen to be 0°C and we measure everything else compared to that. For 1H and 13C the shifts are positive ppm values, which means the larger the ppm, the less shielding of the nucleus.

    Using NMR over time, chemists have built up a large catalogue of chemical shifts that are evidence for a nucleus’s environment – what it is bonded to – so we learn the structure of the molecule around these nuclei. This catalogue or database is called an NMR absorption table and they are widely available online and in most chemistry textbooks. This data is how NMR is used to help find a molecule’s structure.

  • NMR can be run for any nucleus that has non-zero spin, so conveniently for organic chemists, NMR works for carbon AND hydrogen because they both have a ‘spin active’ isotope. 1H is the major isotope of hydrogen and is spin-active, and 13C is a minor (1.1%) spin-active isotope of carbon.

  • An NMR spectrum shows an absorption peak for every environment, not every atom!
    To find a ‘carbon environment’, tell a story of how the specific carbon atom is connected to the rest of the molecule. If the story is different to any other carbon atom, it is a unique environment and will have a unique NMR peak/absorption. You can have every atom in a unique environment, but it’s not always the case.
    • Symmetry is your friend when analysing NMR! If you have a symmetrical molecule (such as tetramethylsilane, TMS) some nuclei will produce the same NMR spectrum because they’re in the same environment as something else. This is how NMR spectra can show ‘less peaks’ than the number of atoms in the molecule being analyzed. The reference TMS is an example of this:
      • TMS, (CH3)4Si, has four identical methyl groups. Because there is only one carbon environment, there is only one peak in the 13C NMR spectrum. It has twelve hydrogens; all of them are in one of four identical methyl groups, so there is only one peak in the whole 1H spectrum too.
    • When you are mapping peaks on an NMR spectrum to a molecule, different peaks will usually be given labels like Ca, Cb or Ha/Hb etc.

    See below for this example using TMS. It will show only one signal in a 1H NMR spectrum and only one signal in a 13C NMR spectrum – both are defined as 0 ppm for reference.

  • A 13C NMR spectrum has absorption ‘regions’ where certain carbon environments are generally found:
    • Between 0-100 ppm, saturated carbon signals will be found. This region can be split into two further parts:
      • 0-50 ppm: saturated alkanes. From lower to higher ppm, this includes terminal carbon atoms (-CH3), secondary carbon (-CH2-) and tertiary carbon. Most of the halogens (R3C-X) are also found here.
      • 50-100 ppm: saturated carbon bonded to oxygen, such as ethers (R3C-O-R’) and alcohols (R2C-OH). Alkynes (C \equiv C) are also found here.
    • Between 100-200 ppm, unsaturated carbon signals will be found. This region can also be split into two parts:
      • 100-150 ppm: unsaturated hydrocarbon environments. This includes alkenes (R 2 C=CR 2 ) and aromatic carbon.
      • 150-200 ppm: unsaturated carbon bonded to oxygen. This includes ketones (R2C=O) aldehydes (RCHO), esters (RCOOR’), amides (RCONR2) and carboxylic acids (RCOOH).

    These values reflect how ‘deshielded’ the nucleus is; the larger the ppm, the more deshielded a nucleus is. Deshielding is more prevalent in double bonds and bonds to electronegative atoms.

  • Worked 13C NMR example: pentan-3-one.
    Below is the structure of pentan-3-one, C5H10O, where we will go through explaining why it has the NMR spectrum that it does.
    How many carbon environments can you see? Remember symmetry!

  • Start by telling the story, from one end of the carbon chain to the other.
    • The first carbon is at the end of a hydrocarbon chain bonded to three hydrogen atoms – it’s a CH3 carbon. Next to it is a -CH2- group followed by a carbonyl, another CH2 and finally a -CH3 chain end. We can call this Ca.
    • The second carbon is part of a CH2 ‘unit’ bonded to a carbon atom on either side. One is a CH3 chain end, and one is a central carbonyl carbon. This story is already not the same as the first carbon – they’re two different carbon environments and will have different 13C NMR peaks. We can call this Cb.
    • The third carbon is in the exact middle of the carbon chain. It has no hydrogen atoms bonded to it. This story is already different from the first two so it is unique. We can call this Cc.
    • The fourth carbon is a -CH2- carbon bonded to a -CH3 chain end carbon and a carbonyl carbon which is in the middle of the molecule. Beyond that is another CH2 group and beyond that is the CH3 chain end. This is the same story as the second carbon. These are identical carbon environments and they will produce the same signal in a 13C NMR spectrum. This is Cb.
    • The fifth carbon is a -CH3 chain end carbon. Next to it is a -CH2- group followed by a carbonyl, another CH2 and then the CH3 chain end. This is the same story as the first carbon. They are identical carbon environments - Ca.

    If you were to run a 13C NMR spectrum for pentan-3-one then, you will see three carbon environments despite the molecule having five carbon atoms.
    Where will they be? Look at the point above and think about what type of carbon environments we have:
    • The first and fifth carbon atoms are -CH3 which is a saturated alkane/hydrocarbon, mostly occuring between 0-50 ppm. Remember that higher ppm is caused by deshielding, near electronegative atoms and double bonds. There is no double bond or electronegative atom here nor on the CH2 next door, so this will likely be a very low ppm, around 0-20 ppm.
    • The second and fourth carbon atoms are -CH2- which are saturated alkane/hydrocarbon as well, but they are one carbon away from a carbonyl carbon which is generally much higher ppm. These will be in the 0-50ppm range, but probably on the higher end such as 30-50ppm.
    • The third carbon atom is a carbonyl carbon. These are amongst the most deshielded (highest ppm) absorption peaks, in the 150-200ppm range and probably close to 200.

    The 13C NMR absorption peaks for pentan-3-one are ~8ppm (Ca), ~35 ppm (Cb) and ~210 ppm (Cc).1 See the image below:

  • Exam questions on NMR tend to give you the NMR spectrum (or the peaks with their ppm values) and the molecular formula.
    Going straight to the NMR spectrum when you don’t know what you’re looking for is going to waste your time in an exam - a molecular formula is a HUGE clue to which peaks to expect and where in the NMR spectrum. Use your general formulae:
    • CnH2n+2 is an alkane – your peaks will all be around the 0-50ppm and you can almost certainly rule out peaks outside the range.
    • CnH2n is possibly an alkene – look for peaks corresponding to C=C environments (100-150ppm). It could also be a cyclic alkane.
    • CnH2nO could be a ketone or aldehyde. Look for an obvious peak around 180-200ppm – aldehydes and ketones are extremely clear in NMR spectra.

    Train yourself to spot functional groups from the molecular formula. Here’s an example:
    • C5H10O2 – which functional groups do we know with oxygen atoms?
      • Ketones and aldehydes – we could have two to account for the two oxygens, but that would ‘cost’ four less hydrogen atoms than a normal five-carbon molecule so our molecule cannot have the formula C5H10O2, it would have to be C5H8O2. It can’t be a dione or dial.
      • Alcohols – an -OH group won’t reduce the number of hydrogens on the chain so you would still expect CnH2n+n in the formula. It’s unlikely to be an alcohol.
      • Ester – this will have two oxygens and compared to a standard alkane you have lost two hydrogen atoms for the C=O bond. This is possibly an ester – look for ester peaks (150-200ppm).
      • Carboxylic acid – this is two oxygen atoms and because of the -OH it only loses two hydrogens compared to an alkane’s CnH2n+n formula. This matches – you could have a carboxylic acid and carboxylic acid NMR peaks are very obvious, amongst the highest ppm for 13C and 1H NMR.

      Knowing what functional groups are possible from your molecular formula is extremely helpful before looking at the NMR spectrum so you know what you are going to be looking for!

  • Worked example: C3H6O
    The compound with molecular formula C3H6O has three peaks in its 13C NMR spectrum: a peak at ~5 ppm, ~37 ppm and ~200ppm. Suggest the structure of this molecule.

    Let’s start by working on the molecular formula.
    • C3H6O matches the formula for a carbonyl group in a hydrocarbon chain – an aldehyde or ketone.
    • We have three peaks for only three carbon atoms.
      That means there are no equivalent carbon environments, so the carbonyl can’t be in the middle of the chain otherwise the two chain ends would be the same signal. That makes it a carbonyl on a chain end – an aldehyde.

    So without even looking at the actual ppm values we have been given, we would expect:
    • A carbonyl carbon signal (~200ppm).
    • Two alkyl chain signals: one next to the carbonyl chain (-CH2CHO) which will probably be closer to 50 ppm and one on the end of the chain (H3CCH2CHO) which is closer to 0-20 ppm.

    These are our signals! 5ppm is the -CH3 carbon, 37 ppm our -CH2 middle of the chain, and -CHO is our carbonyl carbon up at 200ppm.
    With this said we have accounted for all three peaks/signals shown and have a molecule – propanal – that accounts for the molecular formula. See the image below:

    References:
    1 https://www.chem.wisc.edu/areas/reich/nmr/c13-data/cdata.htm - large database of 13C and 1H NMR data values for organic molecules.
  • Introduction
    Introducing NMR
    a)
    How does NMR work?

    b)
    How to read an NMR spectrum.

    c)
    NMR absorption regions.

    d)
    Worked example: Predicting NMR environments.

    e)
    Using molecular formula in NMR exam questions.

    f)
    Worked example: Predicting structures from NMR.


  • 1.
    Suggest possible structures for compounds, given the NMR peaks and their molecular formula.
    Below is some NMR data and a molecular formula for some compounds. Suggest a structure for them:
    a)
    Formula C6H12O with five 13C NMR signals: 205 ppm, 52.7 ppm, 30 ppm, 24.7 ppm, 22.6 ppm.

    b)
    Formula C7H6O with five 13C NMR signals: 192 ppm, 137.7 ppm, 129.9 ppm, 134.7 ppm.

    c)
    Formula C4H8O3 with three 13C NMR signals: 179 ppm, 72 ppm, 28 ppm.