# Determination of Molecular Structure by MOLECULAR SPECTROSCOPY

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2 Figure 1. Infrared spectrum of butane. wavelength, with units of inverse centimeters (cm 1 ). For electromagnetic radiation, wavelength times frequency is equal to the speed of light, = c. Therefore, frequency = speed of light / wavelength, = c /. Because the speed of light is a constant, frequency is proportional to the inverse of the wavelength. An infrared spectrum of a substance shows which frequencies of infrared radiation are absorbed by the substance. It is a plot of the percent of the radiation absorbed versus the frequency of the radiation. The frequencies of the absorbed radiation reveal which bonds the molecules contain. Figure 1 is an infrared spectrum of butane. The spectrum shows three larger absorptions, a very strong one at 3 cm 1, a second weaker one at 14 cm 1, and a third even weaker at 13 cm 1. These three absorptions are characteristic of a compound containing hydrogen atoms bonded to carbon atoms with sp 3 orbitals. The absorption at 3 cm 1 results from stretching of bonds. The other two absorptions result from bending of bonds (the bond angle oscillates Figure 2. Infrared spectrum of 1-butene.

3 3 Table 2. Infrared Absorptions Bond Frequency (cm 1 ) Bond Frequency (cm 1 ) (sp 3 ) / O =O 1725 (sp 2 ) 3 31 O 3 (sp) 33 F 135 = l around the 9E tetrahedral angle). Figure 2 shows the infrared spectrum of a close relative of butane, namely 1-butene. It has the same three absorptions as butane, plus several more. Prominent additions are at 165 cm 1, 9 cm 1, and 9 cm 1. There is also an additional absorption at high frequency, namely 3 cm 1. These new absorptions are all characteristic of hydrocarbons containing a double bond between carbon atoms. The absorption at 3 cm 1 corresponds to the stretching of the bond between hydrogen and sp 2 hybridized carbon atoms. The absorption at 165 cm 1 results from vibration of the = double bond. The absorptions at 9 and 9 cm 1 result from bending of a bond out of the plane defined by the = 2 atoms. The infrared frequencies of many different types of bonds have been identified and are tabulated in numerous reference books such as the R andbook of hemistry and Physics. Table 2 contains a small excerpt from this large collection of information. This table can be used to identify some of the absorptions found in the infrared spectrum of acetone, which is shown in Figure 3. In the infrared spectrum, the absorption at 3 cm 1 results from the 3 groups. The strong absorption at 172 cm 1 is produced by the =O group in the center of the molecule. The 3% O 3 Figure 3. Infrared spectrum of acetone.

4 4 absorption at 14 cm 1 and the % absorption at 13 cm 1 are from the 3 groups. The absorption at 12 cm 1 is not listed in Table 2, but from information from Table 1, we can interpret it as a result of stretching of the bonds. Infrared spectroscopy can provide information about the nature of the bonds in a substance. It can reveal the presence (or absence) of certain types of bonds. owever, it can seldom reveal by itself the location of the bonds within the molecule. NULEAR MAGNETI RESONANE SPETROSOPY Nuclear magnetic resonance spectroscopy is extremely useful in determining the arrangement of atoms in a molecule. As its name implies, this form of spectroscopy relies on properties of the nuclei in a molecule. The particles that compose the nucleus, protons and neutrons, have spin, as do electrons. The total spin of the nucleus is determined by the spin of its component particles. Some nuclei have a range of spin states; some, notably those with even numbers of protons and neutrons (e.g., 12 and 16 O), have zero spin. Like electrons, several nuclei of some common elements have two spin states, +½ and ½. These include 1, 13, 19 F, and 31 P. Under usual conditions, 13 nuclei are randomly distributed between the two spin states, because both states have the same energy. As with electron spin, nuclear spin is associated with magnetic effects. Therefore, when a substance containing carbon atoms is placed in a strong external magnetic field, the 13 spin that lines up with the external magnetic field is lower in energy than the spin in the opposite orientation. Therefore, the tiny magnets of the nuclei line up with the external field. It takes a small amount of energy to flip the aligned nuclear magnets to the opposite orientation. This amount of energy occurs in the microwave region of the electromagnetic spectrum. A carbon-containing sample placed in a magnetic field will absorb microwave radiation of an energy equal to the difference between the two spin states. The energy required depends on the strength of the magnetic field the stronger the field, the more energy it takes to flip the nuclear spin to its opposite state. What makes this nuclear magnetic absorption so revealing about molecular structure involves the electrons in a molecule. The electrons produce their own small magnetic field. This magnetic field shields the nuclei somewhat from the external magnetic field. The greater the concentration of electrons around a nucleus, the more the nucleus is shielded from the external magnetic field. Thus, a nucleus with a low concentration of electrons around it is more exposed to the external field, and so requires more energy to flip its nuclear spins. onversely, a nucleus surrounded by a high concentration of electrons will absorb at lower energy. Therefore, the electrons affect to a small Figure NMR spectrum of 2-chlorobutane (decoupled).

5 5 degree the energy required to change the nuclear spins from one state to another, and the different carbon nuclei in a molecule absorb microwave radiation of slightly different energies. Figure 4 shows a 13 nuclear magnetic resonance spectrum ( 13 -NMR) for 2-chlorobutane. The horizontal axis displays the frequency of the microwave energy absorbed and the vertical axis is proportional to the amount of radiation absorbed. The scale on the horizontal axis shows the difference between the frequency of the radiation absorbed by 2-chlorobutane and a base frequency. The reason for doing this is to make the axis easier to read. For ths spectrum, the base frequency is Mz (,, s ). The frequency of the absorption on the right is,,4 z. The absorption on the left occurs at,2, z. If the axis were labeled with the complete frequency, the numbers would be difficult to read. The four absorptions in Figure 4 show that there are four different carbon atoms in the molecule. These four carbon atoms do not absorb the same energy; the absorption on the left, the high-energy absorption, corresponds to the carbon atom with the lowest concentration of electrons. This would be the carbon atom bonded to chlorine, because chlorine is a very electronegative element, and draws electrons away from the carbon atom to which it is bonded. The absorption on the right occurs at lowest energy, so it corresponds to the carbon atom with the highest concentration of electrons. This would be the one furthest from the chlorine atom. It is harder to determine which of the two central absorptions corresponds to which of the other two carbon atoms in the molecule. There is an experimental technique that allows the other two absorptions to be assigned. This technique relies on the fact that the nuclei of the hydrogen atoms in the molecule have spins of +½ and ½, and so generate tiny magnetic fields of their own. onsider the carbon atom bonded to the chlorine in 2-chlorobutane. It is also bonded to one hydrogen atom. The 1 nucleus of this hydrogen atom can have a spin of +½ or ½. Its spin is either aligned with or opposed to the external field. Therefore, some of the 2-chlorobutane molecules will have a carbon atom bonded to a hydrogen with its spin aligned with the external field, and some will have a carbon atom bonded to a hydrogen with its spin opposed. In these two different cases, the carbon nuclei will be exposed to two slightly different magnetic fields, and they will absorb at two slightly different energies. This generates an effect called coupling; the carbon nuclei are coupled to the hydrogen nuclei. The spectrum shown in Figure 4 was measured in such a way that coupling was suppressed. This kind of spectrum is called a decoupled spectrum. Figure 5 shows the 13 -NMR spectrum of 2-chlorobutane in which coupling is allowed. Each of the four absorptions observed in Figure 4 is Figure coupled 13 -NMR spectrum of 2-chlorobutane.

6 Number of bonded to 6 Table 3. Splitting caused by 1 coupling. 1 spin values Number of 13 absorptions 1 1 +½ ½ (+½ +½ ) ( +½ ½ ) ( ½ ½ ) 1 1 (+½ +½ +½) (+½ +½ ½) (+½ ½ ½) ( ½ ½ ½) +1½ +½ ½ 1½ 3 4 split into two or more absorptions. This splitting is caused by the hydrogen atoms to which the carbon atoms are bonded. When a carbon atom is bonded to one hydrogen, it s adsorption is split into two corresponding to the +½ and ½ spins of the hydrogen atoms. When a carbon atom is bonded to two hydrogen atoms, the spins of the two hydrogen atoms add together to give three values: +1,, and 1. When a carbon atom is bonded to three hydrogen atoms, the spins of the 1 nuclei add to four different values: +1½, +½, ½, 1½. The four cases are summarized in Table 3. As a general rule, the NMR absorption for a 13 nucleus is split into a pattern with one more peak than the total number of hydrogen atoms bonded to it. In the 1 -coupled spectrum in Figure 5, the absorption at 2 z is split into two. This means that the corresponding carbon is bonded to one hydrogen atom. This agrees with the assignment of this absorption to the carbon bonded to chlorine. That carbon is bonded to one hydrogen atom. The absorption at 4 z is split into four. This indicates that the corresponding carbon is bonded to three hydrogen atoms. This confirms the earlier assignment of this absorption to the carbon furthest from the chlorine atom, because that carbon atom is bonded to three hydrogen atoms. The remaining two absorptions can now be easily assigned, as well. The absorption at 135 z is split into three, which means the corresponding carbon is bonded to two hydrogen atoms. This is the carbon in the middle of the molecule, adjacent to the one bonded to chlorine. The absorption at z is split into four, so its corresponding carbon atom is bonded to three hydrogen atoms. This is the carbon atom at the end of the molecule and adjacent to the one bonded to chlorine. NMR spectroscopy is a very powerful tool for determining the structure of molecules. Because many nuclei have spin, the arrangements of these nuclei in molecules can be investigated by NMR spectroscopy. ombined with infrared spectroscopy, NMR spectroscopy can reveal both the arrangements atoms in molecules and how they are bonded together.

7 7 STUDY QUESTIONS 1. What is the frequency of vibration in z (sec 1 ) of the O bond? (See Table 2.) What is the wavelength of the light in nm that is absorbed by this bond? What is the energy, in Joules, of one photon of this light? 2. In each of the following pairs of bonds, select the one that stretches at the highest frequency. A. O or ==O B. O or l. == or // D. or O E. or O 3. For each of the following compounds, indicate how many absorptions will occur in the decoupled 13 -NMR spectrum, and indicate into how these will split in the coupled NMR spectrum. A. B. O O. l D. l E. F.

8 8 4. Below are the infrared spectra for three compounds, ethanol, dimethyl ether, and butanone. Identify which spectrum is which. The structures of the molecules can be found in the hemistry & hemical Reactivity textbook, on pages 423 and

9 9 5. There are four different molecules (isomers) with molecular formula 3 6 l 2. A. Draw Lewis structures for all four isomers. B. Below are the 13 NMR spectra for the four isomers. Identify each spectrum with its structural isomer.

10 6. To identify a certain compound, several tests were performed with it. An elemental analysis revealed its empirical formula to be 5 O. Its molar mass was determined to be 86 grams per mole. Below are its infrared spectrum and its 1 -coupled 13 -nmr spectrum. Draw a structural formula for the compound

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