Since the effect being measured involves the measurement of spin states of a nucleus, the values of ΔΔ size 12{Δ} {}E will be affected by the local magnetic field of a nucleus being examined.

The local magnetic field is, in turn, affected by the chemical environment of the nucleus. ΔΔ size 12{Δ} {}E thus becomes a measure of the chemical environment of the nucleus. Hydrogen atoms bonded to sp3sp3 size 12{ ital "sp" rSup { size 8{3} } } {} carbon atoms are found in different regions of the NMR spectrum from hydrogen atoms attached to alkene sp2sp2 size 12{ ital "sp" rSup { size 8{2} } } {} carbon atoms, alkyne sp carbon atoms, and aromatic sp2sp2 size 12{ ital "sp" rSup { size 8{2} } } {} carbon atoms, oxygen, nitrogen, metals, etc.

We can then use ΔΔ size 12{Δ} {}E to give us some information about the nature of the atoms to which a particular hydrogen atom is attached.

The position of a signal in the NMR spectrum is known as its chemical shift. It has become conventional to reference the NMR signals for hydrogen atoms to the signal found for TMS (tetramethylsilane, Si(CH3)4Si(CH3)4 size 12{ ital "Si" \( ital "CH" rSub { size 8{3} } \) rSub { size 8{4} } } {}), which has been assigned a value of 0. This scale gives the chemical shifts for other atoms in terms of δδ size 12{δ} {} which has frequency units of parts per million (ppm). These values are constant regardless of the field strength of the magnet being used. Most hydrogen atoms attached to organic molecules will resonate in the region δδ size 12{δ} {} = 0 to 10 ppm. Often, the more acidic the proton is, the more positive this value becomes (acid protons in carboxylic acids will often be seen at >+10 ppm). This means that the proton is experiencing less electron density around it, a phenomenon known as deshielding. The less electron density there is about a given hydrogen atom, the more deshielded it is. On the opposite extreme, the more electron density a hydrogen atom experiences, the more shielded it is. While most hydrogen atoms are found in the NMR spectra at positive δδ size 12{δ} {}values, some may be found at negative δδ size 12{δ} {}values. This often occurs when protons are attached to metals (metal hydrides). In these cases, it is typical to find hydrogen signals with δδ size 12{δ} {} = 0 to -30 ppm. Chart 1 gives a general guideline for where certain types of protons are found in NMR spectra. 

Chart 1. NMR Chemical Shift Assignments 

Figure 3

One of the most useful aspects of NMR spectroscopy is that the signal intensities are directly proportional to the amount of a particular type of H atom present. This can be used in two ways.

Figure 4

Thus, if we have an ethyl group, there are three protons from the methyl (CH3)(CH3) size 12{ \( ital "CH" rSub { size 8{3} } \) } {} group and two protons from the methylene (CH2)(CH2) size 12{ \( ital "CH" rSub { size 8{2} } \) } {} group. We expect to see two signals - one for the methyl and one for the methylene - with relative peak areas of 3:2. The process of determining the peak areas is known as integration and the resultant peak areas are known as the integral of the spectrum.

In order to use this information, you need to determine what types of atoms are the same and what types are different within a molecule. Atoms may appear the same for two reasons.

We can also use integration to find the relative amounts of substances present. For example, if we have a mixture of iodomethane and chloromethane dissolved in solution, the ratio of the two peak areas will give the relative amounts of these two compounds. This is useful for determining purity and product yield. We can introduce a known amount of one sample and use its concentration to determine the concentration of another species in solution. The one caution here is that the relative peak areas need to be normalized for the number of protons that produce the signal. In the example of iodomethane and chloromethane, this is easy since they both have methyl groups and the signals are both due to three protons. However, if we were to determine the concentration of dichloromethane versus chloromethane, we would need to take into account that one molecule has three H atoms whereas the other only has two.

Figure 5

Butane

Figure 6

Pentane

The two methyl groups are equivalent to each other, as are two of the two methylene groups. The middle methylene group is unique. In principle, we would expect to see three signals in a 6:4:2 or 3:2:1 ratio.

Ethyl acetate

Figure 7

Ethyl acetate has three types of H atoms. There are two methyl groups and one methylene group. We would expect to see three signals in a 3:3:2 area ratio.

Coupling is another very important aspect of NMR spectroscopy that gives a great deal of information about structure. Coupling results in a signal being split into more signals due to neighboring atoms which also have a nuclear spin. This arises frequently between protons on adjacent carbon atoms. One important aspect of this phenomenon is that equivalent nuclei do not couple to each other.

The reason coupling occurs can be described as follows. For example, let us consider the case of a molecule containing two carbon atoms, each with a single proton and other atoms which do not have a nuclear spin coupling to the protons. One such molecule is 1,1-dibromo-2,2-dichloroethane.

Figure 8

Figure 9

We need to consider a couple of other cases in order to have enough information on coupling patterns to understand common problems. There are cases where there is more than one proton on adjacent carbon atoms.

Let us first consider the case where one or more protons on one carbon atom (let's call it Carbon A) "see" two identical protons on a neighboring carbon atom (called Carbon B). What types of magnetic fields will be seen by the protons on Carbon A? To sort this out, we need to consider the different possible spin combinations of the protons on Carbon B. This is done purely by probability. There are four possibilities:

Figure 10

These can be described by the spin numbers: (+1/2, +1/2), (+1/2, -1/2), (-1/2, +1/2), (-1/2, -1/2). It should be easy to see that the energies of the two combinations (+1/2, -1/2) and (-1/2, +1/2) will be equal. We can order these possibilities according to their expected energies in the presence of a strong external field:

Figure 11

The splitting of the protons on Carbon A will be into three signals in a 1:2:1 ratio, the 2 arising because that energy level is twice as probable.

The case for three protons on an adjacent carbon atom is worked out in a similar fashion. Again, the splitting seen by the protons on Carbon A attached to Carbon B (a methyl group) would be as follows:

Figure 12

There are 8 possible combinations of spin states which divide into a 1:3:3:1 ratio. Either all spins are up, two up and one down, two down and one up, or all up. A proton or protons on one carbon atom adjacent to a methyl group will, therefore, split into a quartet with area ratios of 1:3:3:1.

Ethyl

If we have a CH3CH2−CH3CH2− size 12{ ital "CH" rSub { size 8{3} } ital "CH" rSub { size 8{2} } rSup { size 8{ - {}} } } {} group, as in chloroethane, we would expect to see two peaks in a ratio of 3:2. The methyl group signal will be split into a triplet (with relative areas of 1:2:1) by coupling to the methylene protons. The methylene protons are split into a quartet (with relative areas of 1:3:3:1) by coupling to the methyl protons. Therefore, we expect the spectrum of an ethyl group to look something like…

***SORRY, THIS MEDIA TYPE IS NOT SUPPORTED.***

Notice that the chemical shift of a peak split by coupling is defined as the center of the peak pattern. As mentioned earlier, the distance between the peaks of the CH3CH3 size 12{ ital "CH" rSub { size 8{3} } } {} group (the coupling constant) will be the same as the distance between the peaks of the CH2CH2 size 12{ ital "CH" rSub { size 8{2} } } {} group. Also note that the total intensity of the peaks due to the CH3CH3 size 12{ ital "CH" rSub { size 8{3} } } {} group is 1.5 times the size of the total intensity for the peaks of the CH2CH2 size 12{ ital "CH" rSub { size 8{2} } } {} group.

Butane

In the molecule CH3CH2CH2CH3CH3CH2CH2CH3 size 12{ ital "CH" rSub { size 8{3} } ital "CH" rSub { size 8{2} } ital "CH" rSub { size 8{2} } ital "CH" rSub { size 8{3} } } {}, the methyl groups are equivalent and the methylene groups are equivalent. As described earlier, we would expect to see two signals in a 3:2 ratio. If we now consider coupling, we see that the methyl groups are attached to methylene groups. The CH3CH3 size 12{ ital "CH" rSub { size 8{3} } } {} protons will thus be split into triplets by the methylene protons. Likewise, the methylene protons will be split into quartets by the protons on the methyl groups. However, notice that because the methylene groups are equivalent, they do not couple to each other. Therefore, we expect to see the same pattern as that predicted for the ethyl group shown above.

More Complex Coupling Patterns

If a hydrogen atom is attached to a carbon atom to which nonequivalent carbon atoms with different numbers of protons are attached, then a more complex pattern is observed. One arrives at the total number of signals expected by multiplying the individual couplings. For example, consider 1,2-dichloropropane with three carbon atoms having different protons attached:

Figure 13

Quartet of triplets

Figure 14

Triplet of quartets

Figure 15

You will be given a packet of several spectra. Using the chart and the information on chemical shifts, coupling patterns, and integrated intensities given above as a guide, identify the unknown organic compounds. Answer some review questions to refresh your mind. We will work on number 2 and 5 with you.

Lab Revision Questions: NM

(Total 20 Points)

On my honor, in preparing this report, I know that I am free to use references and consult with others. However, I cannot copy from other students’ work or misrepresent my own data.

…………………………………………………………….. (signature)

Print Name: _________________________________________

1. The atomic weight of Br is listed as 79.909 on the periodic table, but we use 79, not 80, when calculating the molecular weights of Br-containing compounds for mass spectrometry – why? (2 points)

2.

(a) Draw trans-3-bromo-1-butylcyclohexane in planar form (no chair). Be sure to indicate the stereochemistry in your drawing. (2 points)

(b) Now draw both chair conformations of trans-3-bromo-1-butylcyclohexane. (2 points)

(c) Circle the conformer above that is lower in energy. (1 point)

3.

(a) How many degrees of unsaturation does C5H7ClOC5H7ClO size 12{C rSub { size 8{5} } H rSub { size 8{7} } ital "ClO"} {} have? (2 points)

(b) Draw an isomer of C5H7ClOC5H7ClO size 12{C rSub { size 8{5} } H rSub { size 8{7} } ital "ClO"} {} that contains a ketone (There are many correct answers.). (2 points)

(c) Draw an isomer of C5H7ClOC5H7ClO size 12{C rSub { size 8{5} } H rSub { size 8{7} } ital "ClO"} {} that does not contain a ketone. (Again, there are many correct answers.). (2 points)

4. Explain why 3-Chloropentane shows four different absorbencies in the 1H1H size 12{ {} rSup { size 8{1} } H} {} NMR spectrum. (3 points)

Figure 16

5. Write the structure from the data given. (4 points)

Molecular formula CxHyNBrCxHyNBr size 12{C rSub { size 8{x} } H rSub { size 8{y} } ital "NBr"} {}

IR:

Figure 17

1H1H size 12{ {} rSup { size 8{1} } H} {} NMR:

Figure 18

13C13C size 12{ {} rSup { size 8{"13"} } C} {} NMR:

Figure 19

Report QuestionsNMR Laboratory

(Total 70 Points)

On my honor, in preparing this report, I know that I am free to use references and consult with others. However, I cannot copy from other students’ work or misrepresent my own data.

…………………………………………………………….. (signature)

Print Name: _________________________________________

1. Explain each of the following observations. (3 X 2= 6 points)

(a) The compound CH3CH2CH=CH2CH3CH2CH=CH2 size 12{ ital "CH" rSub { size 8{3} } ital "CH" rSub { size 8{2} } ital "CH"= ital "CH" rSub { size 8{2} } } {} shows an IR absorbance around 1550 cm–1, but its structural isomer CH3CH=CHCH3CH3CH=CHCH3 size 12{ ital "CH" rSub { size 8{3} } ital "CH"= ital "CHCH" rSub { size 8{3} } } {} does not-why?

(b) The 1H1H size 12{ {} rSup { size 8{1} } H} {}-decoupled 13C13C size 12{ {} rSup { size 8{"13"} } C} {} NMR spectrum of CHF3CHF3 size 12{ ital "CHF" rSub { size 8{3} } } {} shows a single peak that is split into a quartet.

2.

(a) Draw trans-4-(1-ethylpropyl)cyclohexanol in planar form (no chair). Be sure to indicate the stereochemistry. (2 points)

(b) Now draw trans-4-(1-ethylpropyl)cyclohexanol in its lowest energy conformation. (2 points)

(c) Now draw cis-4-(1-ethylpropyl)cyclohexanol in its lowest energy conformation. (2 points)

(d) Considering your answers to (b) and (c), which is lower in energy, cis or trans-4-(1-ethylpropyl)cyclohexanol? Explain your answer briefly (one sentence). (2 points)

3. In the following compound, label symmetry-equivalent H atoms with the same letter and nonequivalent H atoms with different letters, as we did in class. Then predict the 1H1H size 12{ {} rSup { size 8{1} } H} {} NMR spectrum of the compound, indicating the approximate chemical shift, integration, and multiplicity for each resonance that you expect to see. (20 points)

Figure 20

Figure 21

4. Below are the IR spectra of two isomers having the formula C7H7ClC7H7Cl size 12{C rSub { size 8{7} } H rSub { size 8{7} } ital "Cl"} {}. (Total 8 points)

1) What is the unsaturation number of these isomers? _____ (2 points)

2) Why is there such a difference in the intensity of the C-H stretching and C-H bending region between these two isomeric compounds? (2 points)

3) What structure would you assign to Compound A? Indicate on the spectrum which IR bands support your structure assignment. (4 points)

4. What structure would you assign to Compound B? Indicate which IR bands support your structure assignment. Note: NUJOL is a paraffin oil used to prepared the IR sample. (8 points)

Figure 22

Figure 23

5. Find the structure of the compound and explain each spectrum to support your answer.

(a) Compound contains C, H, and O. (10 points)

MS:

Figure 24

IR:

Figure 25

1H1H size 12{ {} rSup { size 8{1} } H} {} NMR:

Figure 26

13C13C size 12{ {} rSup { size 8{"13"} } C} {}NMR:

Figure 27

(b) Compound contains C, H, and O. (10 points)

MS:

Figure 28

IR:

Figure 29

1H1H size 12{ {} rSup { size 8{1} } H} {} NMR:

Figure 30

13C13C size 12{ {} rSup { size 8{"13"} } C} {} NMR:

Figure 31