Carbon NMR Spectroscopy: Decoding Molecular Frameworks

Dive into Carbon NMR spectroscopy, a crucial analytical method in organic chemistry. Discover how it reveals carbon environments, identifies functional groups, and elucidates molecular structures.

In this lesson, we will learn:

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

• Nuclear magnetic resonance (NMR) spectroscopy is an extremely versatile tool to determine chemical structure. NMR works because of a property called nuclear spin:
  • If a nucleus has non-zero spin, it will interact with a magnetic field.
  • NMR applies a magnetic field that flips the ‘spin state’ of the nucleus. In this state the nucleus resonates due to applied radiation (radio waves) being absorbed. The frequency of this resonance can be measured.
  • The electrons around a nucleus are also affected by the NMR field. They set up their own magnetic field that interferes with the original NMR field; we say the electrons shield$\,$ the nucleus with their magnetic field.
  • As a result, the field the nucleus actually experiences is different – shifted – from the NMR field. This is seen in the change in frequency of energy the nuclei emit (see the second point in this list).
  • Depending on how the electrons are arranged around a nucleus – how it is bonding - the chemical shift, symbol $\delta$, will be a different value.
  • Because NMR machines come in and can run at a variety of frequencies (their field strength), the units on an NMR spectrum is parts per million (ppm). Ppm is a fraction of how far away the resonance frequency is from the resonance of a reference sample called tetramethylsilane (TMS), which is a signal chosen to be 0 ppm.
  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 is how NMR is used to find molecular 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. NMR for carbon detects the minor isotope $^{13}C$, or carbon-13, which is spin active (I = $\scriptsize1/2$ )


• A $\,$ $^{13}C$ spectra shows an absorption peak for every carbon environment – not necessarily every atom!
  To identify a carbon environment, “tell a story” of how the specific atom is connected to the rest of the molecule. If it is connected in a unique way, it is a unique environment.
  • NMR environments, when assigning peaks to atoms or vice versa will normally be given a unique label for each environment, e.g. $C_a$, $C_b$, $C_c$, $C_d$.
  • If a molecule has symmetry, look for atoms in symmetrical environments. Nuclei in symmetrical environments will produce the same NMR spectrum. This is how NMR spectra sometimes show ‘less peaks’ than the number of atoms in the molecule being analyzed.
  See below for an example:


• A $\,$ $^{13}C$ NMR spectra has absorption ‘regions’ where certain carbon environments are found. NMR absorption tables with specific ppm values are widely available online and in chemistry textbooks. Below is a general description:
  • 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 (R₂C=CR₂) 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).
  
  
• Below is a step by step outline to working out an NMR spectrum using a worked example, if you are given / know the molecular formula:
  Determine the structure of a molecule with the formula C₅H₁₀O₂
  
  There are five signals in the $\,$$^{13}C$ NMR spectrum: a signal at 180 ppm; 34 ppm; 27 ppm; 22 ppm and 13 ppm.
  
  STEP ONE: With your known molecular formula, look for a fit to a general formula. You are trying to find out the hydrocarbon backbone! For example:
  • Alkane (C_nH_2n+2).
  • Alkene (C_nH_2n).
  • Aromatic compounds (C:H ratio generally nearer to 1:1 due to the C₆H₆ benzene ring).
  
  In this C₅H₁₀O₂ example, we have C₅H₁₀ which resembles an alkene, however there are also two oxygen atoms. 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: Compare your formula with the number of peaks in the spectrum. You are looking for signs of symmetry. If there is not one signal for every carbon atom in your molecule, you have some form of symmetry! This could be alkyl chain branches or clues to where the substituents on the aromatic ring are.
  When a molecule is symmetrical, carbon environments become equivalent. The same carbon environments will produce the same NMR signal. This is how you may have less signals than you do carbon atoms in the molecule.
  We have five signals for five carbon atoms, so each carbon environment is unique. This likely means any functional group(s) are not attached in the middle of the chain. Alone though, this is not concrete evidence.
  
  STEP THREE: Try to identify the functional groups present from the rest of the formula. 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 alkane 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 molecular formula and that would give us less hydrogen atoms than we know we have.
    The information that we have means it is unlikely that an aldehyde or ketone is present in the molecule.
  • Alcohols (R₃C-OH), which we could have two of. We could then have an alkene double bond which explains the two less hydrogens than an ordinary diol would have. This is a possibility going forward.
  • Carboxylic acids (RCOOH) and esters (RCOOR’), 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 another possibility for the structure.
  
  STEP FOUR: Use the information you’ve gathered from the molecular formula to pick out the appropriate signals. This is where you rule out structure ‘candidates’ and find evidence for the remaining structure. You should be able to use an NMR absorption table and your molecular formula to know where to expect signals.
  We have carried two possibilities for the structure into this step of our working:
  • An alkene with two alcohol groups attached.
  • A carboxylic acid.
  • An ester
  There are no signals from 34 to 180 ppm. This leaves no evidence for a C=C bond (100-150ppm) or carbon atoms bonded to an alcohol group (around 50-75 ppm). The typical range that an ester carbon resonates at (160-170 ppm) also shows no peaks. The ester and alkene with alcohol groups can be ruled out.
  • An alkene with two alcohol groups attached.
  • Ester.
  • A carboxylic acid.
  We are left with the possibility of a carboxylic acid. The NMR spectrum contains a single signal at 180ppm, which is the range where carbon atoms with carboxylic acid groups resonate. This is substantial supporting evidence.
  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: 34 ppm; 27 ppm; 22 ppm and 13 ppm, to assign to four carbon atoms. 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 probably just the four remaining carbon atoms in a straight alkane chain; the higher the ppm, the closer it is to the carboxylic acid group. This would be assigning the rest of the molecule.
  
  Finally, draw out your predicted structure with the NMR peaks assigned to the specific carbon atoms. You should be able to account for all of the peaks – one for each carbon environment! See the picture below:
  
  
• Aside from just ‘ppm’, chemists use a few different descriptive terms when analyzing an NMR spectrum. Here is a look at the different NMR descriptions, compared to the ppm scale and the TMS reference you’ll see on every spectrum:
  • Chemical shift$\,$ just means the different frequency a nucleus resonates at compared to TMS, the reference at 0 ppm. We would say the reference TMS has zero chemical shift.
    • Higher ppm$\,$ is further from 0 ppm, so this is a larger chemical shift.
    • Lower ppm$\,$ is a smaller chemical shift.
  • Shielding$\,$ is talking about how a nucleus is ‘shielded’ from the NMR magnetic field by the electrons surrounding it. Compared to the vast majority of carbon environments, the carbon nuclei in TMS (the reference at 0 ppm) is shielded.
    • A signal at higher ppm reflects a deshielded$\,$ nucleus.
    • A signal at lower ppm reflects a shielded$\,$ nucleus.
  • The field$\,$ means the magnetic field strength that makes the nucleus resonate. The slightly electropositive silicon atom in the reference TMS makes the four carbons surrounding it more shielded by electrons; they need a stronger (higher) field than most carbon atoms to resonate.
    • A signal at higher ppm is further from 0 ppm, so a lower field strength made that nucleus resonate. This is a downfield$\,$ signal.
    • A signal at lower ppm is closer to 0 ppm, so a relatively higher field strength made it resonate. This is an upfield$\,$ signal.