Sapling Chapter 15 #14

Sapling Problem

Chapter 15 #14

Draw the structure of the compound that is consistent with the 1H NMR data below. (Assume that long-range coupling is not observed.)

Chemical formula	Chemical Shift (ppm)	Relative Integration	Multiplicity
C5H9Br	1.06	3	Triplet
	2.00	2	Multiplet
	2.26	3	Singlet
	5.87	1	Triplet

With these kinds of problems, it is always helpful to start our calculating the degrees of unsaturation in the molecular formula.

DOU=(2(C)+2+N-X-H)/2

Here we just plug in the number of carbons, the number of hydrogens, and don’t forget to subtract your halogen!

DOU=(2(5)+2-1-9)/2=1

One degree of unsaturation means that there is a double bond somewhere in our formula (or a ring).

The next step is to look at the integration and multiplicity. I like to draw out basic structures next to my chart. Then I put them together at the end to get my final structure. The multiplet is the tricky atom here, so we will not draw that and see if we could get a multiplet from any of our final structures.

Remember that integration is a signal based on the number of hydrogens on a certain carbon, and integration is the “splitting” of that signal based on the number of hydrogens on adjacent carbons PLUS ONE!

Also, notice that there are only four signals in this NMR and five carbons in the formula. This probably means that there is a carbon somewhere that has no hydrogens on it (meaning it probably contains that one degree of unsaturation and is bonded to two other things)

We should also look at the chemical shift. The closer a carbon is to an electronegative atom, the more chemical shift it has. This means that the triplet at 5.87ppm is the closest to the bromine atom. It also means that the lower the shift, the further that hydrogen (on a carbon) is from the bromine. If the structure was a ring, all of the hydrogens would be experiencing some kind of pull from the bromine. However, we see a pretty big difference in shift, so we can conclude that our structure is linear.

Now we arrange our pieces that we drew to the side of the table into one structure. We can eliminate a structure piece based on our need for a degree of unsaturation

 Structures A,B,C,and D are all potential candidates for the NMR. The next step would be to construct charts for each structure, showing integration and splitting.

We see that structures A, C, and D do not give us the 1 integration hydrogen with the triplet splitting. They can be eliminated.  C and D also have all five signals instead of the four that are seen. Therefore, we can conclude that structure B is the correct one. It fits all of the splitting criteria and contains one degree of unsaturation, and fits the molecular formula.