Identifying NMR unknowns using the "integration first" approach

Description of the approach

Initial general evaluation

Before carrying out a detailed analysis of the ¹H NMR spectrum, some general analysis of the structure is advisable:

• If the formula is given, you should work out the no. of "elements of unsaturation" (sometimes called "double bond equivalents"). This is given by the formula

EOU = 1 + # carbons - 1/2(# hydrogens + halogens)

A nitrogen is counted as half a carbon, but oxygen and sulfur are not counted at all. A triple bond is equivalent to two double bonds. Note that a benzene ring has four elements of unsaturation- three double bonds + one ring.

• If an IR spectrum is given, assess what functional groups are present; particularly useful are O-H or N-H bands (around 3400 cm^-1) and C=O (around 1700 cm^-1); for more detail, see Introduction to IR spectroscopy.
• If a ¹³C NMR spectrum is given, count the number of peaks, which indicates how many types of carbon are present. Unless sp carbons are present (these are relatively uncommon), the carbon types can be characterised as those above (to the left of) 100 ppm, which are sp² carbons, and those below (to the right of) 100 ppm which are sp³ carbons. If there are any peaks above 170 ppm, these are likely to be C=O carbons.

Analysis of the ¹H NMR spectrum

STAGE 1: Integration

• Use the integration information to work out the number of hydrogens associated with each peak or set of peaks in the spectrum.
  • If the integration information is given as numbers x, y, z, use this to calculate the number of hydrogens at each position. If you know there are n hydrogens in total, then the no. of hydrogens at each position is given by:

(proportion of total integration) x (total no. of Hs)

i.e., for x, y and z peaks we have <math>\frac{x.n}{x + y + z}</math>,  <math>\frac{y.n}{x + y + z}</math>,  <math>\frac{z.n}{x + y + z}</math>

• • If the integration information is given only as an integration line, then measure (with a ruler) the vertical distance on that line that corresponds to each peak. Any unit can be used, but millimetres are recommended, as some precision is needed.
• Check that the peaks are distinct; in more complex molecules, groups may overlap, in which case the analysis is more difficult.

File:HanddrawnC6H4.png

benzene ring with 2 unknown attached groups

• If you have four or more EOUs, and you have peaks in the aromatic region (6.5-8.5 ppm), then you have an aromatic ring. The integration here tells you how many Hs are on that ring, and if this is a benzene ring you indirectly know how many groups are attached to the ring. For example, if you have 4Hs on the benzene ring, then you must have (6-4), i.e., two, groups attached to the ring.
• For peaks that show multiples of three hydrogens (3, 6, 9, 12), these are very likely to indicate CH3 groups (methyls).
  • If there are 6H, 9H, 12H in the same peak, these must be from multiple equivalent methyls, suggesting some form of symmetry.
• For sharp peaks that show two or four hydrogens, these are fairly likely to indicate CH3 groups (methylenes); however, the following exceptions should be noted:
  • If these 2 or 4H peaks occur in the aromatic region (6.5-8.5 ppm), these are combinations of aromatic C-Hs. A common pattern is a pair of doublets, each of 2Hs, indicating a para-disubstituted benzene ring.
  • If the 2H peak is broad, it could be an NH₂ or two OHs; this can be confirmed if an IR is available showing the characteristic N-H or O-H stretch (usually around 3400 cm^-1.
• If you have leftover 1H peaks, these usually indicate individual C-H ("methine") groups. Because these must be attached to at least three other atoms, these peaks are often multiplets. If they are broad singlets, then they are likely to be O-Hs or N-Hs, especially if this is confirmed in the IR spectrum.
  • If there are complex or indistinct peaks showing 2Hs or more, these are likely to be different types of H that just overlap, and they cannot be treated as simply as the above cases. For example, a complex peak that integrates to 3Hs is likely to be 2H + 1H; one that shows 4Hs may be 2H + 2H or 3H + 1H. If you have an unusual number such as 5H or 7H, these are almost certainly either from overlapping peaks, or else you have made an error in your integration calculations.
• Write down on a piece of paper all the groups that you have found.

STAGE 2: Build up the groups

You should now have most or all of the "pieces of the jigsaw". You need to begin putting the pieces together to make bigger groups, then finally work out how those groups are connected.

• Begin by adding "bonds" or points of attachment to your groups. For an aromatic ring, show the appropriate no. of "bonds" (= 6 - no. of Hs on the ring) - be sure to show these as not clearly attached to specific carbons (unless on a simple phenyl). For a CH₃, add one "bond"; for a CH₂, add two "bonds", and for a CH add three "bonds".

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You will now try to connect the bonds to each other, like fitting adjacent jigsaw pieces together. This is done initially using multiplicity, but chemical shift then becomes necessary in most cases.

• All groups with only one point of attachment (such as methyls or simple phenyl (C₆H₅) groups) must be on the end of the structure. These are analogous to the "edge pieces" in a jigsaw, and are best tackled first.
• If there are any single CH₃s (methyls) present, begin with one of these, and work through all of them. Start with any 3H peaks that are not singlets:

File:HanddrawnEthyl.png

Ethyl

• • If the methyl is a triplet, it is attached to a CH₂, which means you have an ethyl group (CH₃CH₂). Try to locate the particular CH₂ (look at the other 2H peaks); if it is a simple quartet then the ethyl group is isolated; if it is a higher multiplet then it must be attached to more C-Hs. Either way, attach the CH₂ to the CH₃ and show one "bond" coming off the CH₂.

File:HanddrawnEthylidene.png

Ethylidene

• • If it is a doublet, it is attached to a C-H. Try to locate the particular 1H peak that corresponds to this C-H; this will be difficult if you have multiple C-Hs, but if you have only one that must be it! If you have a C-H that is a clear quartet, then it is very likely to be the C-H attached to your CH₃, meaning that you have a CH₃CH or ethylidene group. Draw the C-H attached to your CH₃, with two bonds coming off the C-H.
• If you have a 3H singlet, it is an isolated CH₃, and you should use chemical shift to identify where it is likely to be:
  • Around 0.9-1.3 ppm, it's on an alkane chain, attached to a simple C that has no Hs on it.
  • Around 1.6-2.4 ppm, it's attached to a C=C (usually below 2 ppm), or a C=O (usually 1.9-2.3 ppm) or an aromatic ring (usually > 2.2 ppm).
  • Around 3.3-4.0 ppm, it's attached to an oxygen. Closer to 3.3 suggests a simple ether R-O-CH₃; closer to 4 ppm suggests an ester (RCO-O-CH₃) or an aryl methyl ether (ArOCH₃).

File:HanddrawnIsopropyl.png

Isopropyl

• If you have a 6H peak, corresponding to two single CH₃s, then many of the above rules apply, but with the additional constraint that the methyls must be equivalent.
  • If the peak is a triplet, you have two equivalent ethyl groups; there should be a 4H quartet elsewhere corresponding to the two equivalent CH₂s.
  • If the peak is a doublet, this could be two equivalent CH₃s attached to the same C-H, indicating an isopropyl group or (CH₃)₂CH- with one "bond" off the central CH carbon; this is very likely if you find a C-H that has a multiplicity of seven (septet). However, you should be careful; this pattern would also be seen if you have two similar but non-equivalent CH₃ singlets.

File:HanddrawnIsopropylidene.png

Isopropylidene

• • If the peak is a singlet, then you have two equivalent CH₃ groups with no Hs on the neighboring atoms. If these methyls are around 1 ppm there is a good chance they are an isopropylidene group or (CH₃)₂C, with two bonds off the central carbon.

File:HanddrawnTertButyl.png

TertButyl

• • If you have a 9H singlet, this almost certainly indicates a tert-butyl group (CH₃)₃C-, which has only one "bond" coming off the central carbon.
• Pick out remaining groups that are obvious from chemical shift.
  • Alkenes have a chemical shift around 4.6-5.0 ppm if terminal (R₂C=CH₂, and at 5-6 ppm if they are internal (R₂C=CHR), and the integration can again be a guide to how many alkene Hs are present. Remember that cis/trans isomers are possible when C=C double bonds are present. If electron-withdrawing groups at nearby, the peaks will be moved downfield, possibly into the aromatic region.
  • Aldehydes show a sharp peak around 9.5-10 ppm (or slightly higher if next to an aromatic ring). If the peak is a multiplet, this indicates how many Hs are on the carbon neighbouring to the aldehyde carbon.
  • Carboxylic acids show a broad singlet for the O-H at a variable chemical shift, usually 10-13 ppm. Sometimes the peak may be too broad to be seen at all.

STAGE 3: Finding missing pieces

By this time, you should have several molecule fragments, but some pieces have yet to be identified, so you will need to do an atom inventory. If you know the overall formula, subtract the atoms in your identified fragments, and you may find some "missing pieces". The ¹H spectrum shows only hydrogens, so you have hopefully identified all hydrogen-containing groups, but other fragments without hydrogen will still need to be accounted for. These commonly include:

• Ether oxygens in R-O-R. These can often be inferred from seeing peaks around 4 ppm.
• Carbonyl (C=O) groups. If there is a missing EOU as well as a missing oxygen, this is a likely choice; the presence of a strong IR peak around 1700 cm^-1 confirms this. If a C=O and an "ether" are both present, consider the possibility of an ester. The ¹³C NMR spectrum is especially useful in such cases; a ketone carbon is normally found around 195-220 ppm, whereas an ester carbon appears at 165-180 ppm.
• Halogens (F, Cl, Br, I) are usually singly bonded. Fluorine pulls neighbouring Hs far downfield, and it is also NMR-active; it causes splitting of the H signal via coupling. Chlorine, bromine and iodine do not cause noticeable splitting, but do move neighbouring Hs downfield.
• Nitrogen can form a wide variety of compounds, and in all cases it will move nearby Hs downfield (for example, a CH²N appears around 2.4-3.0 ppm). If N-H groups show, these will normally be broad singlets (like O-H) at variable chemical shift, and they can be confirmed via IR.
• Sulfur likewise moves neighbouring Hs downfield (CH²S also appears around 2.4-3.0 ppm), but S-H peaks do couple to Hs on neighbouring carbons. Sulfur also often forms sulfoxides R2S=O and sulfones R2SO₂, which produce a stronger downfield shift.

STAGE 4: Putting the pieces together

By this time, you should have a complete set of "jigsaw pieces" - molecule fragments that can be put together in different ways to produce a complete molecule.

• Label the pieces as A, B, C, D, with the appropriate number of "bonds" or connections coming off each. Putting the "edge pieces" (groups with only one "bond") on the outside, you should find there is only a handful of ways to organise the pieces into a final structure. Draw all of these final structures, which should be isomers of one another.
• If possible, determine which isomer is the correct one. This is not always possible without the use of more advanced techniques (e.g., coupling constants), especially if the isomers are very similar (e.g., cis vs. trans). Usually, however, it can be done using:
  • Multiplicity. If you have joined two H-containing carbons together, these will show coupling. If this coupling (splitting) is absent, then it means that these fragments cannot be together.
  • Chemical shift. If one fragment is joined to an electronegative atom or a pi-system, it will see nearby Hs moved downfield; the resultant chemical shift must match. For example, with an ester, any C-Hs next to the C=O should come around 2-2.5 ppm, whereas any C-Hs next to the -O- will come around 4-4.5 ppm.

You should now have your final structure!