Organic Chemistry I & II
Organic Chemistry I & II

Organic Chemistry I & II

Lead Author(s): Steven Forsey

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Organic Chemistry I & II is designed for instructors who want an active, dynamic, and understandable approach to support their own efforts in the classroom. This ever-evolving textbook includes auto-graded questions, videos and approachable language in order to make difficult concepts easier to understand and implement.

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Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

Up to 40-60% more affordable

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Carey & Giuliano, “Organic Chemistry”, 10th Edition

$219

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Solomons et al., “Organic Chemistry”, 12th Edition

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David R. Klein, “Organic Chemistry”, 3rd Edition

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Always up-to-date content, constantly revised by community of professors

Constantly revised and updated by a community of professors with the latest content

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

In-book Interactivity

Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

Customizable

Ability to revise, adjust and adapt content to meet needs of course and instructor

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

All-in-one Platform

Access to additional questions, test banks, and slides available within one platform

Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

About this textbook

Lead Authors

Dr. Steven Forsey, Ph.D.University of Waterloo

Steven Forsey is currently a Professor at University of Waterloo, teaching a variety of organic chemistry courses to Chemistry, Science, Chemical Engineering, Nanotechnology and distance education students. He received his Ph.D. (2004) for Synthetic Organic Chemistry from University of Waterloo, Ontario. He is a recipient of the Excellence of Science Teaching Award and has acted as the Teaching Fellow for the Department of Chemistry since 2016.

Contributing Authors

Felix NgassaGrand Valley State University

Neil GargUCLA

Jennifer ChaytorSaginaw Valley State University

Greg DomskiAugustana College

Christian E. MaduCollin Community College

Christopher NicholsonUniversity of West Florida

Franklin OwEast Los Angeles College, UCLA

Robert S. PhillipsUniversity of Georgia

Grigoriy SeredaUniversity of South Dakota

Simon E. LopezUniversity of Florida

Brannon McCulloughNorthern Arizona University

Jason JonesKennesaw State University

José BoquinAugustana College

Stephanie BrouetSaginaw Valley State University

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Chapter 17: Nuclear Magnetic Resonance

An MRI scan of a head shown in five separate slices. ​MRI machines utilize the concept of nuclear magnetic resonance in order to provide images of organs and structures inside the body. [1]


Contents

Learning Objectives

  • Understand the theory behind nuclear magnetic resonance.
  • Deduce molecular structure using NMR and other spectroscopic data.
  • Predict a molecule’s splitting pattern in 1H-NMR.
  • Be able to explain and interpret a proton or a carbon signal’s chemical shift.

17.1 Introduction

In the previous chapter, we learned how IR and Mass Spectrometry techniques could be used to determine the structure of a compound. However, major limitations exist. IR only shows the presence of certain functional groups. Mass spectrometry can suffer from fragments having the same m/z ratio. In this section, we’ll see how Nuclear Magnetic Resonance (NMR) spectroscopy is an essential tool , giving detailed information on an unknown’s structure. 

17.1.1 Theory of Nuclear Magnetic Resonance

All matter exhibits a measurable magnetic field. That’s right! The clothing you are wearing, the screen on which you are reading this chapter, and even your own body—each of these are tiny generators of a magnetic field! While this may be difficult to believe, an atom’s magnetic behavior is the core of NMR spectroscopy.

From physics, we know that an electric current is generated whenever there is a flow of electrons. If the electrons flow in a circular trajectory, then a magnetic field is generated perpendicular to the direction of the electric field. If you’ve taken physics, this is the “left-hand rule.” From quantum mechanics, we know that electrons exist outside the nucleus and are in constant flux. Hence, each nucleus generates its own tiny magnetic field. The following illustration for an aggregate sample shows how the direction of these tiny magnetic fields are completely random.

Figure 17.1. Direction of nuclear magnetic moments is random in the absence of an applied magnetic field

However, in the presence of an external magnetic field, the direction of these tiny magnetic fields narrow to only two possibilities: parallel and anti-parallel. When an external magnetic field is applied, alignment with the field will be slightly favored energetically over alignment against the field. The energy difference between the two spin states is very small and there will be a small excess of molecules that are aligned with the field.

Figure 17.2. In the presence of an applied magnetic field, nuclear magnetic moments align either with or against the direction of the external field.

In 1952, two physicists, Felix Bloch and Edward Mills Purcell, discovered that when radio waves were used to excite such a sample, nuclei absorbed the incident radiation and as a result were excited to a higher energy state. In addition, each nucleus experienced a nuclear spin inversion. In other words, nuclei that were initially antiparallel to the external magnetic field became parallel and vice versa.

Figure 17.3. When a specific radio frequency irradiates a molecule, nuclei absorb the incoming radiation and undergo a spin flip in their magnetic fields.​

Not all wavelengths within the radio frequency region would excite nuclei. The incident energy had to match or be in resonance, with the energy difference between the ground and excited states. When a nucleus experiences the right combination of magnetic field and electromagnetic radiation a proton will flip its spin to the higher level. The proton is said to be in “resonance” and its absorption of energy is detected by the NMR spectrometer. Spin flipping occurs when electromagnetic radiation equals exactly the energy difference between the two spin states.

The strength of the magnetic field is a huge factor on the resonance frequency. The energy gap between the two nuclear spin states is a function of the applied magnetic field Bo, ΔE=γBo, where γ is a constant for a given nuclei. If a very strong magnetic field is used, the energy gap between the two nuclear spin states is larger. Hence, a higher frequency is required to induce the spin-flip or resonance. When a weaker magnetic field is used, the energy gap is smaller, and a lower frequency is required (Figure 17.4).

Figure 17.4. As the strength of the external magnetic field from the instrument increases, the frequency required to induce a nuclear spin flip increases proportionally.

Not all nuclei are NMR active. The quantum mechanical requirement is for a nucleus to have either an odd number of protons or an odd number of neutrons. In other words, nuclei containing an even number of both protons and neutrons are found to be NMR inactive. NMR active nuclei include 1H, 13C, 2H, 14N, 19F, 11B and 31P. NMR inactive nuclei include 12C, 16O, and 32S.

Why are different frequencies required? If you recall from IR spectroscopy, different frequencies would excite various functional groups. Similarly, in NMR spectroscopy, it’s all about the plethora of hydrogen atoms! It turns out that different types of hydrogen atoms absorb different frequencies of radio-waves. The next section discusses what exactly we mean by different types of hydrogen atoms.

In NMR spectroscopy, one signal is observed for each type of carbon and hydrogen atom. Hence, NMR spectroscopy delineates the carbon-hydrogen framework of an organic molecule.

17.2 Determining the Number of Signals: Sets of Equivalent Protons

In IR spectroscopy, a signal is generated for each type of molecular motion of a specific functional group. In mass spectrometry, a signal would be generated for each type of fragment formed. In 1H-NMR spectroscopy, a signal is generated for each set of chemically equivalent protons.

What does it mean for protons to be chemically equivalent? NMR differentiates proton nuclei by their chemical environments. Basically, what atoms or groups of atoms are near the proton? These other atoms exert their own magnetic fields on the proton nuclei. Hence, different atoms influence the magnetic fields of proton nuclei differently. Chemically equivalent protons have identical chemical environments. In terms of molecular structure, chemically equivalent protons have the same number and type of nearby substituents.

To determine which protons are equivalent, we want to first identify any planes of symmetry in a molecule that are perpendicular to an internuclear axis. If such a plane exists, then one side of the plane is a mirror image of the other side (Figure 17.5).

Figure 17.5. In 3-methyl-2,4-pentanedione, a plane of symmetry bisects the molecule at carbon #3.

Second, hydrogens attached to the same carbon atom are almost always considered identical to each other. In other words, they are chemically equivalent to each other and represent one set. We’ll discuss exceptions later in this chapter.

17.2.1 Aliphatic Hydrocarbons

Let’s take a look at methane. Based on the previous guidelines, are the four protons equivalent?

Figure 17.6. Structure of methane


Q17.1 - Level 1

Are the four protons equivalent in methane?

A

Yes, all four protons are equivalent.

B

No, the four protons are unique from each other.


Let’s now examine ethane. Do the six protons experience the same magnetic environment?

Figure 17.7 Structure of ethane


Q17.2 - Level 1

Are the six protons in ethane equivalent?

A

Yes, all six protons are equivalent.

B

No, the six protons are unique from each other.

C

The three protons on the left carbon are different than the three protons on the right carbon.


Let’s now examine propane. How many sets of chemically equivalent protons do you think there are?

Figure 17.8 Structure of propane


Q17.3 - Level 2

How many set of chemically equivalent protons does propane have?

A

1 set

B

2 sets

C

3 sets

D

8 sets


Aliphatic alkanes are an example of organic molecules that contain homotopic protons, which are all chemically equivalent. A simple test can be performed to determine if protons are homotopic. Select any proton and substitute it with a different atom or group. Then, select another proton on the original molecule and substitute it with the same atom or group chosen before. Compare the two new images produced. If these images are the same, then the protons are considered homotopic (Figure 17.9).

Figure 17.9. Substitution test to determine if any two protons are homotopic

What if the substitution test results in molecules that are stereoisomers of each other instead?

Consider butane. Substitution of any two protons results in a pair of images, which are enantiomers of each other. These protons are said to be enantiopic. In 1H-NMR, enantiopic protons are considered chemically equivalent (Figure 17.10) because they exist in the same chemical environment. Protons that experience different chemical environments are called diastereotopic protons and are considered chemically nonequivalent.

Figure 17.10. Substitution test to determine if any two protons are enantiotopic

Finally, consider (R)-1-fluoropropane-2-ol. Assuming that X has a lower priority than F; substitution of Ha and Hb leads to the R, R and R, S products respectively, which are a pair of diastereomers. Ha and Hb are said to be diastereotopic. In 1H-NMR, diastereotopic protons are considered chemically nonequivalent; they will produce distinct signals (Figure 17.11).

Figure 17.11 Substitution test to determine if any two protons are diastereotopic.


Q17.4 - Level 3

Match each compound with the corresponding number of sets of equivalent protons.

question description
Premise
Response
1

A)

A

3

2

B)

B

6

3

C)

C

5

4

D)

D

4

E

2

F

1


17.2.2 Cycloalkanes

Cycloalkanes exhibit restricted rotation. Because of this, the protons are relatively more fixed in position than in the case of protons in aliphatic hydrocarbons. For example, in the case of chlorocyclopropane, the 1H-NMR spectrum displays three distinct signals. The following analysis shows why.

Figure 17.12. Structure of chlorocyclopropane showing the three different sets of protons Ha, Hb, and Hc

The protons marked Ha are all cis to the chloro group, while the protons marked Hb are all trans to the chloro group. Because of restricted rotation, their distances to the chloro group are relatively fixed. It follows that Ha experiences a different magnetic field strength than Hb. Hence, in cycloalkanes, a pair of methylene protons attached to the same carbon can be chemically different from each other. So, Ha and Hb give rise to two separate signals. Hc is unique unto itself, as it is the only proton attached to the carbon containing the chloro group. This cis-trans analysis can be very useful in predicting the number of 1H-NMR signals for substituted cycloalkanes.


Q17.5 - Level 2

Determine the number of sets of equivalent protons for cis-1,2-dichlorocyclohexane.

question description
A

2

B

3

C

4

D

5

E

6


17.2.3 Alkenes

Alkenes are another class of compounds that have restricted rotation about the C-C double bond. Using an identical analysis, answer the following question.


Q17.6 - Level 1

How many 1^1H-NMR signals do you expect for chloroethene?

question description
A

1

B

2

C

3


17.3 A Typical 1H-NMR Spectrum

We’ll next examine the different components in a typical 1H-NMR spectrum. Additionally, we’ll explore the theoretical aspects of a signal’s appearance and location on a spectrum.

17.3.1 The Chemical Shift

The NMR spectrometer must match the energy of proton to make it resonate. To do this, the spectrometer can be run in two ways: 1) varying or sweeping the magnetic field over a small range while observing the radio frequency (rf) or 2) sweeping the frequency of rf radiation while observing the external field constant.

Below is the 1H-NMR spectrum of methanol. The x-axis represents the “chemical shift” and is symbolized with the Greek letter δ. NMR spectra are calibrated such that 1 δ is equivalent to one part per million (1 ppm) of the NMR spectrometer frequency. For example, in a 400 MHz spectrometer, 1 δ equals one millionth of 400,000,000 Hz or 400 Hz. The chemical shift is calculated from the following equation:

where(νobs - νref) is the number of Hz away from a reference signal, and νspect is the operating frequency of the spectrometer in MHz. The advantage of reporting chemical shifts as relative values is that it allows for comparison of spectra taken on different NMR instruments of varying operating frequencies and magnetic field strength.

The x-axis in the NMR spectrum is the chemical shift which is proportional to the observed frequency. Higher energy values are downfield, or to the left and lower energy values are upfield or towards the right on the spectrum. It takes more energy to resonate deshielded protons; thus they appear downfield. It takes less energy to resonate shielded protons; thus they appear upfield. The NMR spectrum of methanol has two types of protons which experience different magnetic environments.

17.13A.png
Figure 17.13a . A typical 1H-NMR spectrum. The y-axis is signal intensity, while the x-axis is the chemical shift. At 0 ppm is the internal standard TMS.​​


Figure 17.13b. Calculation of chemical shift.

0 ppm is set as the internal standard, usually tetramethylsilane (TMS). In other words, all chemical shifts are relative to the single absorption from TMS. In both 1H and 13C NMR, TMS is the traditional standard as it only gives one signal that is usually very far away from absorptions of other organic molecules. Some spectra will show a single small line at 0 ppm, while in others the TMS signal is excluded. 

Figure 17.14. The structure of tetramethylsilane

A range of only 14 ppm is very small and can be a disadvantage in 1H-NMR. Signals that occur too closely together have a chance of overlapping each other, making the analysis of the spectra difficult. This is where the operating frequency of the spectrometer plays a huge role. For example, two signals that are 6 Hz apart on a 60 MHz spectrometer will be 40 Hz apart on a 400 MHz instrument. The 400 MHz instrument will have much higher resolution of individual peaks.

Figure 17.15. Effect of operating frequency on NMR resolution. Higher frequency NMR spectrometers tend to offer higher resolution spectra.

Chemical shift is proportional to the frequency of radiowaves that is being absorbed by a specific proton. In other words, it is proportional to the energy that is required for the nuclear spin flip to occur. Typically, frequency increases with the strength of the external magnetic field a proton's nucleus is exposed to.

Why do protons have different chemical shifts? Protons have different chemical shifts because the magnetic field from a nucleus’s electrons can interfere with the external magnetic field. This occurs when the electrons’ induced magnetic field circulates in a direction that opposes the instrument’s magnetic field. In a C-H bond, the opposing molecular field would make it resonate upfield relative to a hypothetical naked proton. The strength of the induced field also depends on the electron density near the hydrogen atom in the sigma bond. The greater the electron density, the greater the shielding, and vice versa.

17.16.png
Figure 17.16. If an electric current travels in a circular fashion, a magnetic field is generated perpendicular to the electric field.

Hence, if electron density around a nucleus is significant, the nucleus will effectively be shielded from the external field. It will need less energy to make the nuclei resonate, resulting in a relatively small chemical shift. For example, take a look at the 1H-NMR spectrum for propane.

Figure 17.17. Proton NMR spectrum of propane

Both signals are relatively upfield, being very close to 0 ppm. All protons in propane have electron-donating alkyl groups nearby, resulting in a shielding effect.

Now let’s look at the opposite effect. What if a compound contained atoms that could withdraw electron density away from a specific proton?


Q17.7 - Level 1

Select the atom that is the most electron-withdrawing (most electronegative).

A

N

B

Cl

C

Br

D

C

E

O


Figure 17.18. Protons close to electron-withdrawing groups are deshielded from the applied magnetic field. For resonance, they require a smaller applied magnetic field (Bo) (—higher chemical shift, farther away from TMS). Such protons absorb downfield from other protons.

Let’s take a look at 1-fluoroethane.

Figure 17.19. Different types of protons on 1-fluoroethane


Q17.8 - Level 2

Do you expect the methyl or methylene protons to resonate further downfield?

A

Methyl protons

B

Methylene protons

C

Both proton groups should absorb at the same chemical shift


Inductive effects also explain the slight differences between tertiary, secondary, and primary protons (~1.7, 1.2, 0.9 ppm, respectively). Carbon is slightly more electronegative than hydrogen. Tertiary protons have the largest number of nearby carbons and are therefore deshielded to a certain extent. Primary protons have the fewest number of carbons nearby and are relatively more shielded.

Table 17.1. Chemical shifts for alkyl halogens​

Another factor that can cause deshielding or shielding is induced magnetic fields cause by π electrons. Hence, protons that are near π bonds, sp2 or sp-hybridized carbon atoms, also tend to absorb radiowaves further downfield. (This will be discussed in further detail in the next section.)

The following diagram shows the characteristic chemical shifts for various protons:

Figure 17.20. Characteristic chemical shifts for various protons​


Q17.9 - Level 1

Identify the set of equivalent protons in the following compound with the 2nd^{\text{nd}} smallest chemical shift.


Q17.10 - Level 1

Are the red and blue methyl protons on ethyl bromide equivalent?

question description
A

Yes

B

No

C

Depends on the temperature


If a solution is cooled so that rotations are slowed down, different conformations can be seen. For example, cyclohexane at room temperature only gives one peak, but at lower temperatures, a complex splitting pattern is observed, because axial and equatorial protons are now resolved due to slow interconversion between chair conformations (Figure 17.21).

Figure 17.21. Temperature dependence of proton equivalency in cyclohexane

3D Molecule*: cyclohexane


Q17.11 - Level 3

Match the following compounds with the correct number of 1^1H-NMR signals expected at room temperature.

question description
Premise
Response
1

A)

A

3

2

B)

B

6

C

2

D

5

E

4


17.3.1.1 Diamagnetic Anisotropy

Why are vinylic and aromatic protons so much more downfield than alkyl or alkynyl (hydrogen bonded to a triple bond) protons? The proposed explanation has to do with the direction in which the induced magnetic field from nearby π electrons is circulating. This phenomenon is known as diamagnetic anisotropy, which means that a proton’s magnetic field is influenced by its location in space relative to the induced magnetic field from the π electrons.

Figure 17.22 shows a circulating electron that induces a magnetic field that opposes the applied field (Bo). The strength of the magnetic field felt at the nucleus will depend on where the proton is located in the induced magnetic field. If a proton is located in the induced magnetic field that opposes the applied field, it would be shielded and would appear upfield. If it was located in the induced magnetic field that is in the same direction as the applied field, the hydrogen would take less energy to resonate and would appear downfield.

Figure 17.22. Induced magnetic field by circulating electrons​

The molecular magnetic fields generated by ∏ electrons are directional and anisotropic. This means that the measurement varies depending on the direction in which it is taken.

Figure 17.23. Anisotropic effect for alkenes, aldehydes, benzene, and ethyne

The major anisotropic effect for alkenes and aromatic rings occurs when the molecular axis is perpendicular to the external magnetic field. In this orientation, the π electrons circulate in a direction that generates a magnetic field antiparallel to the external field above and below the π bond. The protons attached to the sp2 carbon are located in the magnetic field created by the circulating electrons that enhance the applied field.These protons are deshielded and resonate downfield  (Figure 23a). Of course, not all of the alkene molecules in a solution are lined up in the position shown in Figure 17.23a because molecules are always in motion. The figure depicts the maximum deshielding effect and a molecule on its edge would give the minimum deshielding effect. Thus, the observed chemical shift is an average of all the possible orientations.

Aldehyde protons resonant between 9-10 ppm and are heavily deshielded. This is due to the combined deshielding effect due to the π bond and the electron withdrawing ability of the electronegative oxygen.

Alkynyl protons, on the other hand, exhibit anisotropic behavior when the π bond axis is parallel to the external magnetic field. This results in an induced field from the π bonds also. However, the magnetic field of the attached protons is not enhanced. The proton nuclei are not within the volume of space covered by the inducted magnetic field from the π electrons. Hence, alkynyl protons absorb upfield from vinylic and aromatic protons. Shown in Figures 17.23a to 17.23d are the anisotropic effect for alkenes, aldehydes, benzene, and ethyne.

Ring current effects are elegantly displayed in the chemical shifts of the protons on [18] annulene. Two peaks are seen in the 1H-NMR spectrum, one at 8.9 ppm and the other at -1.8 ppm, which is upfield from TMS! [18] annulene is aromatic and the molecule has 18 protons: 12 pointing outside and six pointing into the aromatic ring (Figure 17.24). The inside protons are shielded by the induced magnetic field of the circulating electrons and thus opposes the applied field. This anisotropic magnetic field creates the unusually low chemical shift of these protons (Figure 17.24).

Figure 17.24. Anisotropic effects of [18] annualene


17.3.2 Signal Intensity

The y-axis of a 1H-NMR spectra is proportional to the total area under the signal curve and the area under each peak is proportional to the number of hydrogens generating that peak. This integration yields the relative number of equivalent protons giving rise to the signal. The larger the number of equivalent protons, the larger the integration value. Consider tert-butylmethylether.


Q17.12 - Level 1

What is the expected relative integration value of the methyl protons versus the t-butyl group protons?

question description
A

1:1

B

1:3

C

1:6

D

1:9


Figure 17.25. Proton NMR spectrum of tert-butylmethylether


Q17.13 - Level 1

Match the indicated proton sets with the expected relative intensities on a 1^1H-NMR spectrum.

question description
Premise
Response
1

A)

A

6

2

B)

B

2

C

4

D

3


17.3.3 Splitting of Signals by Neighboring Sets of Protons

Shown below is the 1H-NMR spectrum of diethyl ether. One can clearly see very specific, orderly signal patterns appearing. This is strongly different than the simpler 1H-NMR spectrum of tert-butylmethylether. The theory which gives rise to these unique signal patterns is known as spin-spin coupling.

Figure 17.26. Proton NMR spectrum of diethyl ether

If protons from one set of equivalent protons are near protons from a different set of equivalent protons, their magnetic fields can interact or “couple” with each other. The simplest type of spin-spin coupling occurs when different protons are on adjacent carbons. In other words, there are exactly three sigma bonds between them.

Let’s see what can happen when one proton Ha has exactly one nonequivalent neighboring proton Hb, where His chemically different from Hb. Remember how in NMR spectroscopy only two possible nuclear spin states exist? They were what we called parallel and antiparallel. Suppose that the nuclear spin of Ha is parallel to the external magnetic field. What are the possible spin arrangements that Hb can have? Well, there are only two possibilities. Hb can either be parallel or antiparallel, with respect to Ha’s nuclear spin. Because of these two possibilities, the signal of Ha gets split into two separate peaks. The chances of each spin configuration occurring is 50%; they have an equal chance of occurring. Hence, the two separate peaks will be of identical relative integration values. This splitting pattern is known as a doublet. The following splitting tree shows what happens as the number of neighboring protons Hb increases sequentially by one, with all nuclear spin configurations being shown.

17.27 The effect of spin coupling on the strength of the magnetic field felt at the nucleus, B.​

Spin-spin splitting or coupling is caused by the presence of nonequivalent vicinal or neighboring protons. The splitting in hertz between the two peaks is called the coupling constant and is shown with the letter J. Because coupling is caused entirely by internal forces, the magnitude of the coupling constant is not dependent on the magnitude of the applied field. Thus, coupling constants are reported in hertz and will be the same no matter what the magnitude of the applied field is, i.e. 60 or 400 MHz.

Figure 17.28. Depiction of the coupling constant J in a doublet

A simple mathematical relationship can be derived. A signal will be split into n+1 peaks, where n is the number of neighboring protons that are equivalent to each other but is different from the proton that is being observed.

A triplet is created when a proton couples with two magnetically equivalent protons. There are four possible spin states of Hb and Hb’: 1) both are aligned with the magnetic field or both are parallel; 2) Hb parallel and Hb’ antiparallel; 3) Hb antiparallel and Hb’ parallel; and 4) both Hb and Hb’ antiparallel. Notice that spin states 2 and 3 would have the same effect on Ha. Thus Ha would experience three possible electronic environments and a triplet would result. Integration of each peak in the triplet gives a ratio of 1:2:1.

Figure 17.29. Illustration of spin-spin coupling of two equivalent protons with one non-equivalent proton to produce a triplet showing 3 peaks of a 1:2:1 relative intensity. Using the formula, n + 1, where n is the number of neighboring equivalent protons, three peaks (triplet) should be observed for Ha: 2 + 1 = 3 peaks

Now consider another scenario of Ha coupling with three adjacent magnetically equivalent protons Hb, Hb’ and Hb’’. Once again, Hb protons will have an effect on the chemical shift of Ha. There are now four possible electronic environments, and a quartet would result. Integration of each peak in the quartet gives a ratio of 1:3:3:1.

Figure 17.30. Illustration of spin-spin coupling of three equivalent protons with one non-equivalent proton to produce a quartet showing four peaks of a 1:3:3:1 relative intensity. Using the formula, n + 1, where n is the number of neighboring equivalent protons, four peaks (quartet) should be observed for Ha: 3 + 1 = 4 peaks

Closer examination of the relative intensities reveals a surprising occurrence. Do you recognize the following diagram (Table 17.2)?

Table 17.2. Pascal's Triangle gives the relative intensities of peaks for each splitting pattern in 1H-NMR spectra.​​​

Pascal’s triangle is observed. In other words, the relative intensities within each splitting pattern can easily be predicted. Signals greater than a septet are often difficult to resolve. The term “multiplet” is often used for such signals, especially when the relative intensities of each peak are unclear.

Let’s go ahead and do some sample exercises together. Consider propane. The six methyl protons are one set, and the two methylene protons are another set.


Q17.14 - Level 1

What are the correct splitting patterns for each proton set in propane (CH3_3CH2_2CH3_3)?

A

6 CH3_3 protons = triplet; 2 CH2_2 protons = septet

B

6 CH3_3 protons = triplet; 2 CH2_2 protons = quartet

C

6 CH3_3 protons = doublet; 2 CH2_2 protons = triplet

D

6 CH3_3 protons = doublet; 2 CH2_2 protons = quartet


Figure 17.31. Proton NMR spectrum of propane

Let’s try another problem: butan-2-one.

Figure 17.32. Structure of butan-2-one


Q17.15 - Level 2

Assign each signal in the following 1^1H-NMR spectrum to the correct proton set in 2-butanone.

question description
Premise
Response
1

A)

A

1

2

B)

B

3

3

C)

C

2


Q17.16 - Level 1

Match the indicated proton set to the splitting pattern it will generate on a 1^1H-NMR spectrum.

question description
Premise
Response
1

A)

A

Singlet

2

B)

B

Quartet

C

Doublet

D

Triplet

E

Septet

F

Quintet

G

Septet

H

Sextet


17.3.3.1 Complex Splitting by Nonequivalent Sets of Protons

Tree diagrams are used to help think about the coupling between protons and the splitting patterns they produce. In a tree diagram, the different couplings are applied sequentially, from largest to smallest coupling constants. For example, with ethyl bromide, the Ha protons are adjacent to two Hb protons (Figure 17.33). If Ha did not couple with the adjacent protons, it would be a singlet. The next step in the tree diagram is to couple Hwith one Hb proton with a coupling constant of 7 Hz to produce a doublet. If the doublet is now coupled with the second Hb proton, the doublet would also be split into four. Since the coupling constant is the same for both Hb protons (7 Hz), only three peaks are seen because two middle peaks align with each other. Notice that this also gives the correct ratio 1:2:1. Thus the tree diagram gives the same pattern as spin-spin coupling diagrams of adjacent nonequivalent protons.