Measurement of Group Electronegativity using Nuclear Magnetic Resonance Spectroscopy

Author: J. M. McCormick*

Last Update: July 21, 2008

Introduction

The concept of electronegativity is central to our understanding of bonding, the way in which atoms are held together in molecules and polyatomic ions. It was devised by Linus Pauling to explain why the bond energy of a polar bond was not simply the average of the bond energy of corresponding homonuclear species (i. e., why isn’t the bond energy of HCl not the average of the bond energies of H₂ and Cl₂?). Pauling realized that this was because the electrons involved in the bond were not being equally shared (as was the case in Lewis’s model), but were being pulled toward one atom in the bond. He coined the term electronegativity (symbol χ) as the ability of an atom to attract electrons in a bond to itself. From measured bond energies, Pauling was able to calculate the electronegativity of most of the elements (defining χ for fluorine to be 4.00) and predict bond energies that had not been measured.^1,2  Elements with high electronegativities (non-metals) have a greater propensity to draw electrons to themselves, while elements with low electronegativities (metals) do not. If two atoms in a bond have the same, or nearly the same, electronegativities the bond will be a covalent bond. But as the difference in electronegativity, Δχ, increases the bond’s electrons are pulled more toward the more electronegative atom.  This distortion of the electron density leads to a slight charge separation (build up a small positive charge on one atom and a small negative charge on the other) and gives a polar covalent bond. If Δχ is large enough, then the electrons are pulled totally to one atom and we have an ionic bond. Limiting values of Δχ have been arbitrarily defined for the various types of bonds, but it is better to consider electronegativity as a qualitative, rather than an absolutely quantitative, tool for discussing bonding.³

Since Pauling’s original work there have been a number of attempts to refine the concept of electronegativity. For example, Mulliken attempted to make it more quantitative by defining electronegativity as the average of an atom's electron affinity and its ionization energy.  Others have attempted to extend the concept of electronegativity to groupings of atoms.  These group electronegativities are is completely analogous to an atomic electronegativity, and are simply a measure of how much a group of atoms pulls electrons toward the group in a bond. In organic chemistry group electronegativities are very useful to explain chemical reactions based on certain atom groupings called functional groups. In this laboratory exercise we will be measuring group electronegativities in a series of organic molecules using nuclear magnetic resonance (NMR) spectroscopy.

NMR is the most powerful instrumental technique in modern chemistry. It can give a chemist information on the structure of a molecule and how it interacts with other molecules in solution. It is the latter which is used to produce the image in magnetic resonance imaging (MRI). NMR is based on the fact that some nuclei have a nuclear magnetic moment (i. e., spin), just like electrons.  A nuclear magnetic quantum number is given the symbol I and the different values it can have are m_I.  This is just like for an electron which has S = 1/2 and m_s = ±1/2, except for a nucleus I is not forced to be 1/2 and neither are the m_I.  When placed in a magnetic field states with different spin will split such that states having a negative m_I go down in energy while ones having a positive m_I go up in energy.  Such a situation is shown in Fig. 1 for a system with I = 1/2 and m_I =  ±1/2.  This is analogous to what bar magnets do when brought near each other: they align themselves so that their poles line up in a lowest-energy arrangement.  We can keep the bar magnets in their higher energy state (poles not aligned properly) by the input of energy (via our hands).  For the nuclei the energy to flip them from being aligned with the field (m_I negative) to being aligned against the field (m_I positive) is DE in Fig. 1, and we can make the nuclei flip from one state to the other if we exactly match DE with a photon of the same energy (E_photon = hn).

Figure 1.  Energy of levels with m_I = +1/2 and -1/2 as a function of increasing magnetic field strength.  Note that when the field is zero, the different m_I are degenerate (have the same energy).

Even in the very large magnetic fields required for NMR, DE for nuclei where I ¹ 0 is very small and the photon frequencies required are very low (radio frequencies).  And this presents a problem because at room temperature enough energy is available in the surroundings to flip the nuclei, which makes the flips hard to observe, but not impossible.  Two ways to observe the nuclei flipping are possible.  The first, old-fashioned way, is to hold the magnetic field constant and scan different radio frequencies until the one that matches DE for a particular nucleus is found, much in the same way you could find the frequencies of all the radio stations in the area by turning the tuning knob on a radio.  This continuous wave (CW) approach is slow, requires concentrated samples, and does not give good results in some cases.  Luckily, the second way of observing nuclear spin flips solves all of these problems and makes possible a range of experiments that are not possible with the CW approach. 

In this second approach, which is used in all modern NMR instruments, a short blast of radio waves is delivered to the sample and then re-emitted radiation by the sample is monitored over time.  Because frequency and time are related by the Heisenberg Uncertainty Principle (like energy and position are), if we know the duration of the radiation pulse precisely we will have many different frequencies present at the same time.  By judiciously choosing our pulse's duration we can hit our sample with all frequencies that the nucleus of interest can absorb.  The result, as you can imagine, is chaos, analogous to hitting all of the bells in a carillon at the same time.  There is a lot of noise, but we can pick out each individual bell by its distinct sound (frequency), if we listen attentively over time. We use the same method with the nuclei.  All of the nuclei are excited by the pulse, but then they began to "relax" and emit radio waves of an energy that matches their DE.  If we "listen" to the sample "ringing" over time, the result is a free induction decay (FID) like that shown in Fig. 2.  The complex FID pattern contains information on all the nuclei that were excited by our radiation pulse, and we can convert theses oscillations in time back to each nucleus' frequency by using a mathematical process called a Fourier transform.

Figure 2. Typical free induction decay (FID).

The frequency of radio waves that any nucleus will absorb obviously depends predominantly on the nucleus, but it also depends on the local environment in which the nucleus finds itself.  This is because each nucleus is surrounded by a cloud of electrons which modulates the magnetic field experienced by the nucleus.  If we pull electrons away from a nucleus (as when it forms a bond with a more electronegative atom) the nucleus is more exposed to the magnetic field. This in turn changes the frequency of radio waves it will absorb.  The opposite effect is observed if electrons are pushed toward a nucleus.  To simplify things a standard has been set for each NMR-active nucleus. The frequency change due to the differing chemical environment around a nucleus (called a chemical shift) is expressed relative to this standard.  In all cases the chemical shifts are very small, only a few parts per million (ppm), and so it is convenient to express chemical shifts in units of ppm relative to the standard at 0.000 ppm.

We need to consider on other effect in our discussion of NMR, and that is coupling.  Recall that when several bar magnets placed on a table they "know" about each because their magnetic fields interact. Nuclear magnetic moments can interact, or couple, too.  This coupling leads to a splitting of the signal (line) that arises from each distinct nucleus into a series of lines.  The way in which a single line is split, the spacing between the lines and their intensities can all be used to obtain information on the environment around an individual nucleus through its interactions with neighboring nuclei.  Coupling is most commonly seen with proton (¹H) NMR, because this isotope has a high natural abundance, and it is likely that a molecule will have many different hydrogen atoms that are near enough to another for coupling to occur.  For most other NMR-active nuclei their natural abundance is low and so coupling between nuclei of the same atom is unlikely.  However, they can often couple with any nearby protons, which will lead to a splitting of their NMR signal.  Experimental methods are available to remove this effect, which is called decoupling, to give a very simple NMR spectrum consisting only of single lines. 

In this experiment we will be examining the ¹³C NMR spectrum of the compounds given in Table 1, and using the changes in the chemical shift of one C to probe the effects of electronegativity. All of the compounds in this study contain a grouping of carbon and hydrogen atoms known as the n-butyl group (CH₃-CH₂-CH₂-CH₂-) attached to a single atom or another grouping of atoms, called a substituent. The chemical shift of the carbon (called the alpha carbon) to which the substituent is attached will be shown to depend on the substituent’s electronegativity. This relationship will be explored using more complex substituents and used to predict the chemical shifts of compounds not studied.  

Experimental^4,5,6

Prepare Lewis dot structures for the compounds listed in Table 1.  Click here to review the organic section of the Molecular Models 1 exercise for help in identifying functional groups.  Hints: 1) the n-butyl portion of these molecules in the same in all cases, and 2) for n-butylisocyanide and valeronitrile the given structural formula shows the how the carbons and nitrogen atoms are arranged.  Use VSEPR theory to predict the structure of the atoms in the n-butyl group and in the substituent; identify the hybridization of each carbon atom. You may be want to build models of one or more of these molecules (ask your instructor for assistance). Note that the functional group C₆H₅ (called a phenyl group) is an aromatic ring like benzene except one H is replaced by the n-butyl group and because of the resonance in the ring it can't be adequately modeled using our model kits.

Table 1.  Compounds to be examined in this exercise.

Click here to obtain the ¹³C NMR spectra of the compounds listed in Table 1.  Note that each spectrum shows only the region from 0 to 100 ppm, which is where the carbons in the n-butyl group will occur.  Some compounds (e. g., n-butylbenzene) have additional peaks outside this range due the other carbons in these molecules.

 Assign each line on the spectrum to a specific C atom.  The following information will be useful as you make your assignments.  If your spectrum shows a set of three lines near 78 ppm, ignore it.  This signal arises from the solvent (d-chloroform, CDCl₃). Start assigning peaks by comparing the spectra of 1-chlorobutane, 1-bromobutane and 1-iodobutane.  The carbon furthest from the substituent will show the smallest change when the substituent changes (no more than 1 ppm); assign it first.  The next carbon (reading left to right in Table 1) will be at a slightly larger chemical shift than the first carbon (because it is bonded to two carbons, instead of just one) and it will change a little more (range of about 4 ppm) as the substituent changes.  The third carbon from the left will have a higher chemical shift than the second carbon and surprisingly changes very little with a change in the substituent (about 2 ppm).  Finally, assign the alpha carbon's signal by whichever one is left; it should vary widely as the substituent is changed.  Take care when assigning the spectrum of n-pentane.  Because the molecule is symmetric, the substituent (the CH₃ shown in red in Table 1) is identical to the carbon on the other end of the molecule as are the alpha carbon and the second carbon from the end.  There will only be three lines in n-pentane's ¹³C NMR spectrum (the two end carbons, the two carbons that are one in from the end and the center carbon).  Note that some of the substituents have another carbon atom, which has does not appear in the given spectra because the spectra have been edited to show only the region of interest. The n-butylisocyanide shows an interesting splitting of the alpha-carbon apparently by coupling to ¹⁴N (I = 1).  For this compound use the middle of the three listed peaks for chemical shift of the alpha carbon.

Results and Analysis

Tabulate your results (chemical shift of all the carbons in each compound as a function of χ) in your notebook.  Use the values of χ in Table 2.

Table 2.  Selected electronegativities of the elements.

Prepare a graph in Excel of the alpha carbon's chemical shift as a function of the substituent's electronegativity for 1-bromobutane, 1-chlorobutane and 1-iodobutane.  Determine the equation of the best-fit line through the data.  Print out the graph and attach it to your notebook.

Prepare tables in your notebook like Tables 3 and 4.  Fill in the first column of each table with the alpha carbon's chemical shift in each compound. 

In Table 3 calculate the predicted chemical shift of the alpha carbon using the equation that you just found by substituting χ for the atom in the substituent that is bonded to the alpha carbon into the equation. 

Table 3. Experimental ¹³C chemical shift of the alpha carbon, and the predicted chemical shift for the alpha carbon calculated from the electronegativity of the atom in the substituent directly bonded to the alpha carbon. and the experimental group electronegativity for the substituent calculated from the chemical shift of the alpha carbon.

In Table 4 use the experimental chemical shift and the equation of your best-fit line to calculate the group electronegativity for the substituent.

Table 4. Experimental ¹³C chemical shift of the alpha carbon, and the calculated experimental group electronegativity for the substituent calculated from the chemical shift of the alpha carbon.

With the equation of your best-fit line, predict the chemical shift of the alpha carbon in butane, 1-fluorobutane and in n-butyl lithium from your equation.  In these compounds the substituents are H, F and Li, respectively.

Conclusions

Use the outline for a measurement exercise as the starting point for your conclusions.  Be sure to include a discussion of the following points.

Not all of the predicted chemical shifts in your Table 3 will be that accurate.  Discuss why this is so.

From your data identify which groups are more electron withdrawing (have a higher electronegativity) than the central atom in the group, which are less electron withdrawing and which are about the same.  What do you conclude about the effect that the other atoms in the substituent have on the effective electronegativity of the atom that is directly bonded to the alpha carbon?

A more accurate prediction for the chemical shift of the alpha carbon in butane (substituent is H) is 13.7 ppm, very similar to the carbon furthest from the substituent in the compounds we examined. Is your calculated result for butane consistent with the more accurate prediction?  Why might it not be?  On the other hand, why would we expect the prediction for 1-fluorobutane to be very accurate?  Hint: think about the difference in the C-F and C-H and the compounds that we used to build the calibration curve.

Do you think that the predicted chemical shift for the alpha carbon in n-butyl lithium will be accurate?  Why? What does the electronegativity difference say is a better way to describe the C-Li bond?  Does this help to explain the fact that n-butyl lithium reacts spontaneously and exothermically (often violently) with almost all hydrogen-containing compounds (even the weakest acids) to give butane (C₄H₁₀)? How? FYI: the deuterium in the solvent used (CDCl₃, deuterochloroform) is very weakly acidic and the substance that results when it is deprotonated (CCl₃^-) is extremely reactive.  So, that's why we didn't even attempt to obtain the NMR of n-butyl lithium in CDCl₃.

Huai and coworkers and Bratsch have separately calculated a group electronegativity for the methyl group (-CH₃) of 2.28.^7,8 Does your value match this theoretical value?  Why might your number not agree with their calculation?  Other studies suggest 2.28 is a good estimate for the methyl group's electronegativity.  Why doesn't your result agree with the literature values?  What did we not take into account in our analysis? Hint: there are other compounds that  we could have chosen to do this experiment on, for example CH₃-CH₃.

References

1. Pauling, L. J. Chem. Educ. 1988, 65, 375.

 

2. Pauling, L. The Nature of the Chemical Bond, 3^rd Ed.; Cornell University Press: Ithaca, NY, 1967.

 

3. Sproul, G. J. Chem. Educ. 2001, 78, 387-390. Click here to view this article (J. Chem. Educ. subscribers only).

 

4. Greever, J. C. J. Chem. Educ. 1978, 55, 538. Click here to view this article.

 

5. Boggess, R. K. J. Chem. Educ. 1988, 65, 819-820. Click here to view this article.

 

6. Sawyer, D. S.; Heineman, W. R. and Beebe, J. M. Chemistry Experiments for Instrumental Methods; Wiley: New York, 1984.

 

7. Huai, L.; Qingwei, W. and Lixin, L. J. Chem. Educ. 1992, 69, 783-784. Click here to view this article.

 

8. Bratsch, S. G. J. Chem. Educ. 1985, 62, 101-103. Click here to view this article.