DETERMINATION OF THE STRUCTURE FROM SPECTRAL DATA

IB Chemistry

Example 1 C7H16O
MW 116 First calculate the degree of unsaturation: the result is 0. The compound will not have a ring or a double bond.

IR Spectrum
The compound has an oxygen, but no double bonds, so it must either be an alcohol (-OH) or an ether (-O-). Alcohols are quite distinctive in IR, showing up with a strong, broad band in the region 3500-3250. Note the broad band at 3391: this indicates that the compounds is indeed an alcohol. You can also look for the C-OH stretch at 1260-1050.

Think of possible structures

Now we know that the compound has an -OH group, but there are lot of ways that this seven-carbon molecule could be put together. It could be straight-chained or it could be branched; the -OH group could be (theoretically) anywhere in the molecule. Let's look at the NMR to get an idea of how many different hydrogens are in the molecule so that we can narrow down the number of possible structures.

Proton NMR Spectrum

The NMR of this compound is below:

There are only four different kinds of protons, and 12 of them (at 0.9 ppm) are next to a carbon that has only one hydrogen (because it is a doublet). This indicates that the molecule has two of the sub-structures below:

These two sub-structures account for 6 of the 7 carbons in the molecule. Therefore the "something" in each must be the same carbon. Let's put them end-to-end and replace the "something" with a carbon, then sketch in the proper number of bonds:

Now we need to place an -OH group in the molecule. It can't be on either of the carbons that are connected to the two methyl groups, because the position of each hydrogen in blue is dictated by the fact that it splits the 12 hydrogens at 0.9 ppm into a doublet. Therefore, it must be on the middle carbon, like this:

In the NMR, protons on the same carbon as an -OH group - the hydrogen in purple above - will show up from 2-4 ppm. Sure enough, there is a peak constituting one proton at 3.1 ppm; furthermore, this peak is a triplet indicating 2 protons (blue) on adjacent carbons.

The peak at 1.7-1.9 ppm has two protons and is an octet (although it's kind of hard to see all 8 peaks); this peak corresponds to the blue protons, which are split by both the red and purple protons (a total of 7 hydrogens on neighboring carbons). The hydrogen on the oxygen (green) shows up as a singlet at 1.4 ppm. Although hydroxyl protons usually show up from 2-4 ppm, the exact position is not always predictable. Hydroxyl protons are usually not split.

Summary
Example 1 is 2,4-dimethyl-3-pentanol:

Example 2 C5H10O
MW 86 Calculate the degree of unsaturation: the answer is 1; it can have a carbon-carbon double bond, a carbonyl, or a ring.

IR Spectrum
Since the degree of unsaturation indicates that the compound could have a carbonyl, let's look for that first. There is a band at 1726, suggesting a saturated aldehyde (1740-1720). Whenever you suspect a compound is an aldehyde, look for bands in the region 28302695. They often appear as shoulder-peaks, just to the right of the C-H alkane stretch peaks. In the spectrum below, note the two bands in this region, suggesting that the compound is indeed an aldehyde.

Proton NMR Spectrum

Since the IR spectrum indicates an aldehyde, look for this functionality in the NMR spectrum. The aldehydic proton appears in the NMR from 9-10, usually as a small singlet.

The spectrum above shows a small singlet corresponding to one proton at 9.2 ppm, confirming that the compound is an aldehyde. Protons on the carbon adjacent to the aldehyde carbonyl will show up at 2-2.7 ppm; this is the triplet peak of 2 protons at 2.4 ppm on the above spectrum. Thus, so far we know that there is an aldehyde group next to a methylene group which is next to a carbon that has two hydrogens:

This accounts for 3 of the 5 carbons in the molecule. The un-colored hydrogens in the above structure could correspond to the peak of 2 hydrogens centered at 1.6 ppm; this peak is a pentet indicating that these protons are adjacent to carbons with a total of 4 hydrogens. The peak centered at 1.35 ppm has two hydrogens and is a sextet, indicating it is next to carbons that have a total of 5 hydrogens. Finally, the peak at 0.9 ppm has 3 hydrogens and is a triplet, indicating it is a methyl group adjacent to a carbon that has 2 hydrogens. Therefore, it looks like the molecule is a straight-chain of 5 carbons with the aldehyde group at one end:

Note that the closer a group is to the carbonyl function, the further downfield it is shifted. Here is how the NMR correlates to the structure:

Summary
Example 2 is 1-pentanal:

1-pentanal 6

Example 3 C4H8O
MW 72 First calculate the degree of unsaturation: the answer is 1. This means that the compound has four carbons and an oxygen, it can have a carbon-carbon double bond, a carbonoxygen double bond - a carbonyl, or a ring.

IR Spectrum
The IR spectrum for Example 1 is below. Since the degree of unsaturation indicates that the compound could have a carbonyl, let's look for that first, since carbonyl bands are strong and distinct. Carbonyls show up in the region 1760-1665, and specifically, saturated aliphatic ketone close to 1715. Sure enough, there is a band at 1718 indicating a saturated aliphatic ketone.

Think of possible structures

Now we know that the compound has a carbon-oxygen double bond, but there are still a few ways that this four-carbon molecule could be put together. Examples are below:

The second and third structures above are saturated aliphatic aldehydes, which show up at 1740-1720. While in the above IR spectrum the band at 1715 might be close to the range of saturated aliphatic aldehydes, an aldehyde would also show a distinct band for HC=O stretch in the region 2830-2695, so it is not likely that it is one of these structures. That leaves the first compound, which is 2-butanone. 7

Proton NMR Spectrum

A table of characteristic NMR shifts is available online: click on the link below. Note that this chart is also linked to in the frame to the left. Before you look at the NMR spectrum, think about what the spectrum of 2-butanone should look like. There are three different types of protons:

The 3 protons in green will be a singlet and show up from 2-2.7 ppm. The 2 protons in blue will be split to a quartet by the protons in red; they will show up from 2-2.7 ppm. They will be further downfield (have a higher ppm value) than the protons in green because they are shielded both by the carbonyl and by the red methyl group. The 3 red protons are the farthest from the carbonyl and are split into a triplet by the blue protons. Let's look at the NMR and see if this is what we see.

Sure enough, here is how they correlate with the structure:

Note the pattern of the ethyl group -CH2CH3 in the above NMR spectrum. Whenever you suspect an ethyl group in a molecule, look for a quartet of 3 protons and a triplet of 2 protons, with the methylene (-CH2-) group further downfield than the methyl group (-CH3).

Summary
Example 3 is 2-butanone:

Example 4 C4H8O2
MW 88 Calculate the degree of unsaturation: the answer is 1. The molecule can have a carboncarbon double bond, a carbonyl, or a ring.

IR Spectrum
If you have a double bond and an oxygen in the molecule, always look for a carbonyl, because carbonyl bands are strong and easy to find. Sure enough, the band at 1713 indicates a carbonyl. Another prominent band is the characteristic wide O-H stretch at about 3150; this band looks more like a carboxylic acid O-H stretch than an alcohol O-H stretch (see example 3 for IRs of alcohols). C-O stretches show up from 1320-1000.

NMR Spectrum
In the NMR, since we suspect a carboxylic acid, look for a broad singlet in the region 1013.2 ppm.

Indeed, the broad peak at 11.8 indicates a carboxylic acid proton: -CO2H. The triplet at 1 ppm corresponds to a methyl group next to a methylene (-CH 2-) group; the sextet at 1.7 ppm is a methylene group next to a total of 5 hydrogens; the triplet at 2.3 ppm is a methylene group next to a methylene group. Therefore, it looks like the molecule is straight-chain:

The structure correlates with the NMR spectrum like this:

Summary
Example 4 is butanoic acid:

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Example 5 C7H8O
MW 108 Calculate the degree of unsaturation: the answer is 4. The molecule probably has an aromatic ring.

IR Spectrum
The most prominent band is the O-H stretch at 3335. Also look for aromatic bands from 1600-1585 and 1500-1400: they are present.

NMR Spectrum

The 4 protons from 6.6-7.2 ppm indicate an ortho- or meta-disubstituted aromatic ring. The phenol proton, -OH, shows up anywhere from 4-12 ppm; the singlet at 5.2 ppm indicates this proton. Since the ring is disubstituted, the singlet at 2.3 ppm of 3 hydrogens must indicate a methyl group on the ring.

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Summary
Example 5 is 3-methylphenol:

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Example 6 C6H15N
MW 101 Calculate the degree of unsaturation: the answer is 0. The molecule has no double bonds or rings.

IR Spectrum
Since the molecule has a nitrogen, look for a band in the region 3400-3250 - there is a single small band at 3384, which probably indicates the N-H stretch of a secondary amine. (Recall that tertiary amines will not show a band in this region because they do not have any N-H's to stretch.)

NMR Spectrum

Amine protons show up from 0.5-3.0 ppm if the amine is not on an aromatic ring; the small "buried" peak at 1 ppm indicates a secondary amine peak:

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There are only two other types of protons in the molecule: the doublet at 1 ppm indicates 12 hydrogens adjacent to one hydrogen and the septet at 2.9 ppm indicates 2 hydrogens adjacent to 6 hydrogens. The only way the molecule can be "put together" is to have each R group coming off the nitrogen to be the same, and to be -CH(CH3)2.

Summary
Example 6 is diisopropylamine:

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Example 7 C3H5O2Br
MW 153 Calculate the degree of unsaturation: the answer is 1. The molecule has one double bonds or rings.

IR Spectrum
Since the molecule has an oxygen and one degree of unsaturation, look for a carbonyl band - there is one at about 1700. There is also a wide O-H stretch band centering about 3000; this is characteristic of a carboxylic acid.

NMR Spectrum

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The singlet at 11.6 ppm supports the IR suggestion that the compound is a carboxylic aci, R-CO2H. The triplets of two hyrogens each at 3 and 3.6 ppm indicate two different methylene groups, both adjacent to a carbon that has two hydrogens. Since the molecule only has 3 carbons, these must be next to each other. A methylene group which has a bromine on the same molecule will be from about 2.5-4 ppm and a methylene group adjacent to a carboxylic acid will be from about 2-2.7 ppm. The following molecule satisfies these criteria:

Summary
Example 7 is 3-bromopropionic acid:

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Example 8 C6H10
MW 82 Calculate the degree of unsaturation: the answer is 2. The molecule has two double bonds and/or rings.

IR Spectrum
The IR spectrum below doesn't show much of interest. We would expect two carboncarbon double bonds (1680-1640), or one carbon-carbon triple bond (2260-2100). Could it be a ring, or two rings? The C-H stretching vibrations are all under 3000, indicating that there are no hydrogens on a carbon-carbon double bond or on a carbon-carbon triple bond - it is possible that the compound is a internal alkyne.

NMR Spectrum

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The above spectrum indicates two identical -CH2CH3 groups. The degree of unsaturation of 2 both hints at an alkyne, although the carbon-carbon triple bond stretch 2260-2100 is not seen. This does not rule out the compound being an alkyne, because the normally weak carbon-carbon triple bond stretch is not observed in a symmetrically substituted alkynes. If we put the carbon-carbon triple bond in the middle of a straight carbon chain molecule, we get this:

This compound has two equivalent ethyl groups and a symmetrically substituted carboncarbon triple bond. The C-H stretch is not seen in the IR because it is not a terminal alkyne (and therefore does not have a C-H on the triple-bonded carbon. This is how it correlates with the NMR spectrum:

Summary
Example 8 is 3-hexyne:

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Spectroscopy Tutorial: Degree of Unsaturation

A calculation of the degree of unsaturation is a good way to start a spectroscopy problem. It tells you how many rings and double bonds are in the molecule; thus, you know if you should look for a carbonyl or a carbon-carbon double bond, or a ring or an aromatic ring. The degree of unsaturation can be calculated readily from the molecular formula of all compounds containing carbon, hydrogen, oxygen, nitrogen, sulfur, or the halogens, by applying the following rules:

Rule 1: Replace all halogens in the molecular formula by hydrogens. Rule 2: Omit oxygens and sulfurs. Rule 3: For each nitrogen, omit the nitrogen and one hydrogen.

Applications of these rules reduces the molecular formula in question to the molecular formula of the hydrocarbon which has the same degree of unsaturation. The degree of unsaturation of a hydrocarbon is easily deduced if one remembers that a saturated hydrocarbon has the formula CnH2n + 2. Thus for the formula CnHm,

Applying these rules to the molecular formula C8H8NOBr.

Rule 1, replace halogens with hydrogens: C8H9NO Rule 2, omit oxygens: C8H9N Rule 3, omit the nitrogen and one hydrogen: C8H8

Therefore,

This means that the molecule has five rings and/or pi bonds (or any combination of the two). A degree of unsaturation greater than or equal to 4 doesnt demand but should suggest the possibility of an aromatic (benzene) ring.

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