14.13: Constructing Partial Structures

When you look at an IR spectrum, you immediately see little chunks of the structure, because you see individual bonds. You know at a glance that the compound contains a C=O bond or an O-H bond. That can be very reassuring, because you can quickly imagine what you are dealing with.

NMR spectra often take more work. You may have to put pencil to paper to come up with a structure. The work pays off, because you can get a much more detailed picture of the structure.

Step by Step

Let's start with a ¹³C NMR spectrum. Suppose you had peaks in the spectrum at 200, 35, and 15 ppm. We might assign these three peaks as follows:

shift (ppm)	partial structure
200	sp^2 C=O
35	sp^3 C-C=O
30	sp^3C-C=O
15	sp^3 C-C

We are saying that the first carbon is in the sp² (trigonal planar) region and that it is so far downfield because of a double bond to oxygen. The other two carbons are in the sp³ or tetrahedral region. One of them isn't very far downfield; it is probably just attached to another sp³ carbon. The other two, at 35 and 30 ppm, are both a little further downfield. That's roughly the right place for a tetrahedral carbon attached to a trigonal-planar carbon; that is, these carbons are each attached to either a double bond or a carbonyl.

In the partial structure, we always bold or underline the carbon that corresponds to the peak we are discussing. If you don't do that, it isn't clear whether the peak at 30 comes from a carbon next to the carbonyl (C=O), or the carbon in the carbonyl itself. Also, at the peak at 15, we want to make it clear that we are talking about a single carbon atom; leaving the partial structure as C-C implies that this spectroscopy observes bonds, but it does not. IR spectroscopy observes bonds. ¹³C NMR spectroscopy observes carbon atoms.

Now, suppose we look at the ¹H NMR spectrum for the same compound. Maybe we will see three peaks this time. There is a quartet integrating for 2H at 2.3 ppm, a singlet integrating for 3H at 2.1 ppm, and a triplet integrating for 3H at 1.1 ppm. We enter those characteristics in a table. This time, there are three features to explain for each peak.

shift	integ.	multipl.	partial structure
2.3	2H	quartet	CH3-CH2-C=O
2.1	3H	singlet	CH3-C=O
1.1	3H	triplet	CH3-CH2

First, we need to explain the shift. All of these peaks are in the upfield end of the spectrum (below 5 ppm), so they are likely from hydrogens on sp³ or tetrahedral carbons. The first two are slightly downfield, just past 2 ppm. That suggests that the sp3 carbons they are attached to may, in turn, be attached to sp2 carbons: either double bonds or carbonyls. We already know there is a carbonyl from the ¹³C spectrum, so let's assume that's what is causing the shift near 2 ppm. The third peak, at 1.1 ppm, is in the normal range; this hydrogen is on a tetrahedral carbon, likely attached to other tetrahedral carbons.

To demonstrate what the integration is telling us, we just show the correct number of hydrogens. There are two hydrogens responsible for the peak at 2.3 ppm. Three others are responsible for the peak at 2.1 ppm, and another three give rise to the peak at 1.1 ppm.

Finally, we need to explain the multiplicity. The peak at 2.3 ppm is a quartet, so by the "n+1" rule it must be next to a CH₃ group. The peak at 1.1 ppm is a triplet, so it must be next to a CH₂ group. (It does not take long to figure out that these two peaks represent hydrogens that are next to each other.) Finally, the peak at 2.1 ppm is a singlet. It has no hydrogen neighbors at all.

Notice that we do not need to know the structure to fill in these partial structures. We are just writing down what the data is telling us. From there, it isn't very far to determine the overall structure.

Exercise \(\PageIndex{1}\)

Fill in partial structures for the following peaks.

a) 10.1 ppm, 1H, triplet b) 3.4 ppm, 1H, septet c) 7.3 ppm, 2H, triplet

d) 5.4 ppm, 1H, quartet e) 1.4 ppm, 2H, sextet f) 8.0 ppm, 1H, singlet

g) 2.1 ppm, 3H, singlet h) 6.8 ppm, 2H, doublet i) 0.9 ppm, 6H, doublet

Answer

Aromatic (benzene etc) peaks are labeled "Ar" to distinguish from alkene peaks that show up further upfield (lower shift). Also, some peaks may be in two symmetric positions and are labeled with "x2".

a) 10.1 ppm, 1H, triplet, CH₂-CH=O b) 3.4 ppm, 1H, septet, O-CH(CH₃)₂

c) 7.3 ppm, 2H, triplet, CH=CH-CH x 2 (Ar) d) 5.4 ppm, 1H, quartet, CH₃-CH=C

e) 1.4 ppm, 2H, sextet, CH₃-CH₂-CH₂ f) 8.0 ppm, 1H, singlet, C=CH-C (Ar)

g) 2.1 ppm, 3H, singlet, CH₃-C=C or CH₃-C=O or CH₃-N; need context to choose

h) 6.8 ppm, 2H, doublet, CH=CH-C x 2 (Ar) i) 0.9 ppm, 6H, doublet, CH-CH₃ x 2

Exercise \(\PageIndex{2}\)

Identify the errors in the following partial structures:

a) 3.6 ppm, 2H, triplet, CH₂-CH₂ b) 2.1 ppm, 2H, singlet, CH₃-C=C

c) 7.4 ppm, 2H, doublet, CH=CH₂-C d) 1.8 ppm, 2H, quintet, CH₂-CH₄

e) 7.8 ppm, 1H, triplet, -CH=CH₂ f) 1.7 ppm, 1H, nonet, NH₂-CH(CH₃)₂

Answer

Identify the errors in the following partial structures:

a) 3.6 ppm, 2H, triplet, CH₂-CH₂ the first carbon must be attached to O to have a shift at 3.6 ppm

b) 2.1 ppm, 2H, singlet, CH₃-C=C the integral says only 2H, not 3H

c) 7.4 ppm, 2H, doublet, CH=CH₂-C the shift implies aromatic, so there can only be one H per carbon; must be symmetry

d) 1.8 ppm, 2H, quintet, CH₂-CH₄ there can't be four hydrogens on one carbon; must be some hydrogens on each side

e) 7.8 ppm, 1H, triplet, -CH=CH₂ the shift implies aromatic, so there can only be one H per carbon; must be one one each side

f) 1.7 ppm, 1H, nonet, NH2-CH(CH3)2, an attached nitrogen would shift this hydrogen past 2 ppm; also, coupling is rarely seen across O or N, so the two neighboring H on the left are probably on a carbon.