1.12. How To Interpret NMR Spectra and Generate Plausible Molecular Structures
It is possible to analyze a 1H NMR and/or 13C NMR spectrum and generate a structure (or a small set of structures) that would match the data. This combines all previous skills and often requires moderate-to-significant practice. There are two common ways of asking students to perform this task: interpret spectra and choose from possible structures or interpret spectra and draw the structure(s) that fit the data.
This task may be considered the main goal of this chapter. Depending on the spectra and information given, the difficulty can range from trivial to challenging.
All structural determination questions are logic puzzles. In long answer questions part marks are awarded for showing the structural fragments and logic/rationale behind your interpretation; even if you struggle to generate a complete answer you can still achieve marks by showing your work/thought process and getting some of the pieces.
As a general guideline for the order of steps:
1. Use any extra information given.
- Remove solvent signals if needed.
- If you are given a molecular formula, determine the Degree of Unsaturation.
- If you are told there is a certain functional group, draw this as a fragment and make a note of which atoms from the molecular formula have been accounted for.
- If you are given the 1C NMR and/or DEPT spectra, use the chemical shift and/or phase to look for obvious groups (signals that must be certain functional groups, CH2’s in the DEPT-135, etc.) and make a note of them.
2. Identify and label all distinct signals in the 1H NMR spectrum. Going left-to-right is often simplest. Circling the signals and giving them labels (A, B, C, … ; Signal 1, Signal 2, Signal 3, … ; etc.) typically works well.
3. Determine the integrations for all distinct signals in the 1H NMR spectrum. Often this is given in the modern format, with numerical integrations written on the spectrum.
- Check if the total integration accounts for all hydrogens in the molecular formula. If it does not, can the total number be multiplied to account for them all (i.e. is symmetry in the molecule resulting in the signals accounting for multiple chemically equivalent sets of hydrogens).
4. Determine obvious and/or plausible functional groups. Some chemical shifts are very characteristic. This is normally aided by considering integrations (e.g. a signal with an integration of 1 around 10 ppm is probably an aldehyde; a signal with an integration of 3 around 1 ppm is probably a methyl (CH3); etc.).
- Broad signals represent hydrogens in O-H and/or N-H bonds. Depending on the chemical shift and other data it MIGHT be possible to determine which kind and/or what functional group (e.g. a carboxylic acid’s chemical shift is distinct; if the molecular formula has no nitrogen then it cannot be an amine/amide/etc.).
5. Draw the obvious/plausible functional groups. Subtracting these elements from the molecular formula might help to determine what functional groups the remaining signals are (e.g. if there are only two Hs unaccounted for, then the remaining signals cannot be CH3’s; if there is only one oxygen atom, then there is not an ester functional group; etc.). Draw any new fragments determined this way.
6. Attempt to determine connectivity, especially for signals with simple multiplicities (s, d, t, q). Connect any obvious or necessary pairings. Redraw these larger fragments beneath your original work. Do NOT ERASE OR WRITE OVER ANYTHING unless you determine that you made a mistake in a previous step. This helps save time if you find an error (instead of starting over from scratch, you can start in the middle).
7. Check if complex multiplicities can be explained. Would combining the fragments in a certain way give rise to complex multiplicities for the signals that have them?
8. Generate possible structures. Typically only a few (1-3) will be possible from the fragments generated. A moderate amount (4-6) can still proceed to the next step but will be time consuming (as an alternative, quickly check previous steps and see if anything was missed to rule out more options).
9. Fact check and verify. Rule out structures until only one remains, then CONFIRM THE REMAINING STRUCTURE MATCHES THE DATA. This step is by far the most important. For each candidate structure, check…
- Does the structure have the correct chemical formula?
- Does the structure fit the DoU?
- Quickly consider the 1H NMR spectrum of the structure. Comparing it to the spectral data you have, would the structure theoretically have…
- …the correct number of signals?
- …the correct multiplicities, chemical shifts, and integrations for signals?
Because this task is the primary purpose of this chapter three worked through examples are given.
Example 1:
1. Use any extra information given.
No solvent signals indicated.
DoU = [2 + 2∙C + N – X – H]/2 = [2 + 8 + 0 – 0 – 10]/2 = 0
Logic: If there are no rings and no π bonds, then there are no functional groups with π bonds (no alkenes, no aldehydes, no ketones, etc.). Do not consider these as options.
No specific functional groups indicated.
No 1C NMR and/or DEPT spectra.
2. Identify and label all distinct signals in the 1H NMR spectrum.
3. Determine the integrations for all distinct signals in the 1H NMR spectrum.
Integrations are given in the spectrum. The total integration is [1 + 2 + 1 + 6] = 10. There are 10 Hs in the chemical formula, so all are represented in the spectrum.
4. Determine obvious and/or plausible functional groups.
Signal A is broad.
Must be O-H or N-H.
There are no N’s, so must be O-H.
There is only one oxygen in the molecular formula.
Probably an alcohol functional group.
Signal B integrates to 2 and is around 3.4 ppm.
There are no halogens. There is only one oxygen.
Probably a CH2 next to an oxygen.
Signal C integrates to 1 and is around 1.3 ppm.
The integration represents only one hydrogen.
Probably a CH.
Signal D integrates to 6 and is around 0.9 ppm.
No one group can have an integration of 6.
Probably multiple chemically equivalent groups.
Probably 2 chemically equivalent CH3’s.
5. Draw the obvious/plausible functional groups.
6. Attempt to determine connectivity, especially for signals with simple multiplicities (s, d, t, q).
7. Check if complex multiplicities can be explained.
Only one structure was possible.
In the generated structure the hydrogen labelled C would be a triplet of septets.
In the given spectrum Signal C is complex/multi-coupled.
Signal C could be a triplet of septets.
No other signals are complex.
Complex couplings are plausibly explained.
8. Generate possible structures.
All fragments have been combined. No other structures can be generated.
9. Fact check and verify. Rule out structures until only one remains, then CONFIRM THE REMAINING STRUCTURE MATCHES THE DATA.
The structure has molecular formula C4H10O. The given molecular formula is C4H10O. Match.
The structure’s formula has DoU = 0 and there are no rings or π bonds in the structure. The given molecular formula has DoU = 0. Match.
The structure would generate 4 signals. There are 4 signals in the spectrum. Match. The structure would have a broad signal integrating to 1 (chemical shift variable), a doublet integrating to 2 (around 3.3-4.0 ppm), a complex signal integrating to 1 (around 1.4-1.9 ppm), and a doublet integrating to 6 (around 0.9-1.2 ppm). Match.
The structure of the unknown compound in Example 1 is:
Example 2:
1. Use any extra information given.
No solvent signals indicated.
DoU = [2 + 2∙C + N – X – H]/2 = [2 + 10 + 1 – 0 – 11]/2 = 1
No specific functional groups indicated.
If there are only 3 signals in the 13C NMR spectrum, then there are multiple chemically equivalent carbons.
2. Identify and label all distinct signals in the 1H NMR spectrum.
3. Determine the integrations for all distinct signals in the 1H NMR spectrum.
Integrations are given in the spectrum. The total integration is [4 + 4 + 3] = 11. There are 11 Hs in the chemical formula, so all are represented in the spectrum.
4. Determine obvious and/or plausible functional groups.
There are no broad signals.
There are no O-H or N-H.
Signal A integrates to 4 and is around 3.8 ppm.
No one group can have an integration of 4.
Probably multiple chemically equivalent groups.
There are no halogens. There is only one oxygen.
Probably NOT 4 chemically equivalent CH’s next to one oxygen (VERY unusual).
Probably two chemically equivalent CH2’s next to one oxygen.
Signal B integrates to 4 and is around 2.7 ppm.
No one group can have an integration of 4.
Probably multiple chemically equivalent groups.
There are no halogens. The DoU is 1, so there are no alkynes or aromatic rings. Outside standard range of being next to a carbonyl.
Probably NOT 4 chemically equivalent CH’s next to one nitrogen (VERY unusual).
Probably two chemically equivalent CH2’s next to one nitrogen.
Signal C integrates to 3 and is around 2.4 ppm.
There are no halogens. The DoU is 1, so there are no alkynes or aromatic rings.
Could be CH3 next to carbonyl.
There are already two CH2 groups next to an oxygen (Signal A). Signal A had only one plausible option. The oxygen is probably already accounted for, so a CH3 next to a carbonyl is possible but unlikely.
Could be CH3 next to nitrogen.
5. Draw the obvious/plausible functional groups.
6. Attempt to determine connectivity, especially for signals with simple multiplicities (s, d, t, q).
7. Check if complex multiplicities can be explained.
No signals are complex.
8. Generate possible structures.
All fragments have been combined. No other structures can be generated.
9. Fact check and verify. Rule out structures until only one remains, then CONFIRM THE REMAINING STRUCTURE MATCHES THE DATA.
The structure has molecular formula C5H11NO. The given molecular formula is C5H11NO. Match.
The structure’s formula has DoU = 1 and there is one ring and no π bonds in the structure. The given molecular formula has DoU = 1. Match.
The structure would generate 3 signals. There are 3 signals in the spectrum. Match. The structure would have a triplet integrating to 4 (around 3.3-4.0 ppm), a triplet integrating to 4 (around 2.3-3.0 ppm), and a singlet integrating to 3 (around 2.3-3.0 ppm). Match.
The structure of the unknown compound in Example 2 is:
Example 3:
This example is intentionally very difficult to showcase limitations at an introductory level.
1. Use any extra information given.
No solvent signals indicated.
DoU = [2 + 2∙C + N – X – H]/2 = [2 + 20 + 1 – 0 – 13]/2 = 5
If the signal at 4.2 ppm is not a triplet, it must be a complex multi-coupled signal that coincidentally looks like a triplet.
No 1C NMR and/or DEPT spectra.
2. Identify and label all distinct signals in the 1H NMR spectrum.
3. Determine the integrations for all distinct signals in the 1H NMR spectrum.
Integrations are given in the spectrum. The total integration is [1 + 2 + 2 + 2 + 1 + 1 + 1 + 3] = 13. There are 13 Hs in the chemical formula, so all are represented in the spectrum.
4. Determine obvious and/or plausible functional groups.
Signal A is broad.
Must be O-H or N-H.
There is only one option.
Probably a carboxylic acid functional group.
Signal B is broad and integrates to 2.
Must be O-H or N-H.
Both oxygens are already accounted for.
Must be N-H.
Probably NH2 of an amine.
Signal C integrates to 2 and is around 7.1 ppm.
No position in an aromatic ring can have an integration of 2.
Probably multiple chemically equivalent groups.
Probably 2 chemically equivalent CH’s of an aromatic ring.
Signal D integrates to 2 and is around 7.0 ppm.
No position in an aromatic ring can have an integration of 2.
Probably multiple chemically equivalent groups.
Probably 2 chemically equivalent CH’s of an aromatic ring.
Signal E integrates to 1 and is around 4.2 ppm.
NOT POSSIBLE. Cannot be CH of an ester, both oxygens already accounted for.
Probably part of a lower range set pulled to a higher ppm by nearby electronegative atom(s).
Cannot be CH attached to oxygen, both oxygens already accounted for. Cannot be CH attached to halogen, no halogens in formula. Cannot be CH of alkyne, DoU = 5 and already accounted for (1 π bond in carboxylic acid; 3 π bonds and 1 ring in aromatic ring; 1 + 3 + 1 = 5).
Possibly CH attached to N and close to electronegative atom(s).
Nearby electronegative atoms would be oxygens of carboxylic acid.
Possibly CH attached to aromatic ring and close to electronegative atom(s).
Nearby electronegative atoms could be nitrogen of amine or oxygens of carboxylic acid.
Signal F integrates to 1 and is around 3.4 ppm.
NOT POSSIBLE. Cannot be CH attached to oxygen, both oxygens already accounted for. Cannot be CH attached to halogen, no halogens in formula.
Probably part of a lower range set pulled to a higher ppm by nearby electronegative atom(s).
Cannot be CH of alkyne, DoU = 5 and already accounted for (1 π bond in carboxylic acid; 3 π bonds and 1 ring in aromatic ring; 1 + 3 + 1 = 5).
Possibly CH attached to N and close to electronegative atom(s).
Nearby electronegative atoms would be oxygens of carboxylic acid.
Possibly CH attached to aromatic ring and close to electronegative atom(s).
Nearby electronegative atoms could be nitrogen of amine or oxygens of carboxylic acid.
Signal G integrates to 1 and is around 3.2 ppm.
NOT POSSIBLE. Cannot be CH attached to oxygen, both oxygens already accounted for. Cannot be CH attached to halogen, no halogens in formula.
Probably part of a lower range set pulled to a higher ppm by nearby electronegative atom(s).
Cannot be CH of alkyne, DoU = 5 and already accounted for (1 π bond in carboxylic acid; 3 π bonds and 1 ring in aromatic ring; 1 + 3 + 1 = 5).
Possibly CH attached to N and close to electronegative atom(s).
Nearby electronegative atoms would be oxygens of carboxylic acid.
Possibly CH attached to aromatic ring and close to electronegative atom(s).
Nearby electronegative atoms could be nitrogen of amine or oxygens of carboxylic acid.
Signals F and G have very similar chemical shifts, are both CH’s, and both have complex multiplicities.
Probably a pair of diastereotopic H’s on the same carbon.
Signal H integrates to 3 and is around 2.2 ppm.
Could be CH3 attached to aromatic ring.
Cannot be CH3 next to carbonyl (combining carboxylic acid fragment and methyl fragment makes a complete structure without the other atoms of the formula). Cannot be CH3 attached to alkyne, DoU = 5 and already accounted for (1 π bond in carboxylic acid; 3 π bonds and 1 ring in aromatic ring; 1 + 3 + 1 = 5). Cannot be CH3 attached to alkene, DoU = 5 and already accounted for.
Probably a CH3 attached to an aromatic ring.
5. Draw the obvious/plausible functional groups.
6. Attempt to determine connectivity, especially for signals with simple multiplicities (s, d, t, q).
7. Check if complex multiplicities can be explained.
In the left generated structure:
The hydrogen labelled E would be a doublet of doublets.
In the given spectrum Signal E is complex/multi-coupled.
Signal E could be a doublet of doublets.
The hydrogen labelled F would be a doublet of doublets.
In the given spectrum Signal F is complex/multi-coupled.
Signal F could be a doublet of doublets.
The hydrogen labelled G would be a doublet of doublets.
In the given spectrum Signal G is complex/multi-coupled.
Signal G could be a doublet of doublets.
No other signals are complex.
Complex couplings are plausibly explained.
In the right generated structure:
The hydrogen labelled E would be a doublet of doublets.
In the given spectrum Signal E is complex/multi-coupled.
Signal E could be a doublet of doublets.
The hydrogen labelled F would be a doublet of doublets.
In the given spectrum Signal F is complex/multi-coupled.
Signal F could be a doublet of doublets.
The hydrogen labelled G would be a doublet of doublets.
In the given spectrum Signal G is complex/multi-coupled.
Signal G could be a doublet of doublets.
No other signals are complex.
Complex couplings are plausibly explained.
It is not possible to rule out one of the structures using the complex multiplicities.
8. Generate possible structures.
Two structures can be generated.
9. Fact check and verify. Rule out structures until only one remains, then CONFIRM THE REMAINING STRUCTURE MATCHES THE DATA.
Both structures have molecular formula C10H13NO2. The given molecular formula is C10H13NO2. Both match.
Both structures’ formula has DoU = 5 and there are 1 ring and 4 π bonds in each structure. The given molecular formula has DoU = 5. Both match.
Both structures would generate 8 signals. There are 8 signals in the spectrum. Both match. The structures would both have a broad signal integrating to 1 (chemical shift variable but high), a broad signal integrating to 2 (chemical shift variable), a doublet integrating to 2 (around 6.5-8.5 ppm), a doublet integrating to 2 (around 6.5-8.5 ppm), and a singlet integrating to 3 (around 2.2-.5 ppm).
The left structure would have a doublet of doublets integrating to 1 (pulled a moderate amount above 2.2-3.0 ppm, close proximity to electronegative oxygens) and two doublet of doublets both integrating to 1 (pulled slightly above 2.2-2.9 ppm, moderate proximity to electronegative oxygens).
The right structure would have a doublet of doublets integrating to 1 (pulled slightly above 2.2-3.0 ppm, moderate proximity to electronegative oxygens) and two doublet of doublets both integrating to 1 (pulled a moderate amount above 2.2-2.9 ppm, moderate proximity to electronegative oxygens).
In the given spectrum, the signal for the CH has been shifted downfield (deshielded, higher ppm) by 1.3 ppm units. The signals for the diastereotopic hydrogens of the CH2 have been shifted downfield by 0.3 and 0.5 ppm units.
The CH is shifted downfield more.
The left structure is more likely.
The structure of the unknown compound in Example 3 is most likely:
