Spectroscopy, Building Bridges to Knowledge

1

Spectroscopy

Building Bridges to Knowledge

Photo exiting a tunnel on the Freeway in Shanghai, China

Introduction

IR spectroscopy deals with that portion of the electromagnetic spectrum defined as the infrared region, and it is small compared to the broad span of the spectrum (Figure 10.1). The wavelengths, λ, listed in figure 10.1 are given in meters.

2

Figure 10.1 Electromagnetic spectrum

A more detailed picture of the electromagnetic spectrum may be found at http://www.google.com/imgres?imgurl=http://mc2.gulf-pixels.com/wp-content/uploads/2009/07/Electromagnetic-Spectrum2.jpg&imgrefurl=http://mc2.gulf-pixels.com/%3Fp%3D281&h=688&w=962&sz=108&tbnid=e65qHUkp2-GCeM:&tbnh=106&tbnw=148&prev=/images%3Fq%3Delectromagnetic%2Bspectrum%2Bwavelength&hl=en&usg=__WItpzH6eWsvubZ8QtPSXIVzW6rg=&ei=2HJ2Srz3Oo3OsQPPsrH_CA&sa=X&oi=image_result&resnum=4&ct=image.

Table 10.1, corresponds with the electromagnetic spectrum, lists the wavelength, λ, in meters; the frequency in reciprocal seconds and the energy in J for regions within the electromagnetic spectrum.

3

Region of the Spectrum Wavelength, m

Frequency, s-1 Energy, J

X-rays 10-9 - 10-11 3x1017 - 3x1019 2x10-16 - 2x10-14 Ultraviolet 4x10-7 - 10-9 7.5x1014 - 3x1017 5x10-19 - 2x10-16

Visible 7x10-7 - 4x10-

7 4.3x1014 - 7.5x1014 3x10-19 – 5x10-19

Infrared 10-4 - 7x10-7 3x1012 - 4.3x1014 2x10-21-3x10-19 Microwaves 0.1-10-4 3x109 - 3x1012 2x10-24 – 2x10-21 Radio waves >0.1 < 3 x 109 < 2x10-24

Nuclear Magnetic Resonance

1-5 3x108 - 5x107 2x10-25 -3x10-26

Table 10.1 regions within the electromagnetic spectrum

Infrared Spectroscopy

The infrared region has a wavelength between 10-44 m to 7x10-7 m,

and an energy range between 2 x 10-21J – 3 x 10-19 J. The specific region that provides information to chemists about the structural identification of molecules is when the wavelength, λ, is around 2.5 x 10-6 m to 1.5 x 10-5 m. This region is traditionally indicated in reciprocal centimeters, i.e., 1/λ where λ is in cm. Therefore, the most useful information for organic chemists is the infrared region between 4000 cm-1

1 – 666 cm-11. This region is referred to as the

vibrational infrared region where the frequencies of radiation correspond in energy to the natural vibrational frequencies of organic molecules. When a molecule is placed in an infrared spectrometer and subjected to infrared radiation with energies between 8.0x10-20 J -1.3x10-20 J, an infrared spectrum results. The spectrum generally consists of percent transmittance versus the wave number in reciprocal centimeters or wavelength in micrometers and has the general appearance of Figure 10.2, the infrared spectrum of an ester. Many of the bands that correspond to transmittance versus wave numbers cannot be correlated with a stretching or bending vibration;

4

however, those that can be identified provide important information about the structure of the organic compound.

Figure 10.2 Sample IR spectrum

An IR spectrum could be reported as absorbance versus wave numbers or wavelengths as well as transmittance versus wave numbers or wavelengths. The following website allows one to interconvert between the absorbance and transmittance of an infrared spectrum.

http://webbook.nist.gov/cgi/cbook.cgi?ID=C674760&Units=SI&Mask=80#IR-Spec

The relationship between transmittance and absorbance is given by Equation 10.1.

5

Equation 10.1

Where A is the absorbance and T is the transmittance. The percent transmittance is the transmittance multiplied by 100, i.e., %T = 100T.

When the frequency of infrared radiation incident on organic molecules equals the frequency of the vibrating bonds attached to atoms in the molecule, then the infrared energy absorbed is equal to the energy associated with the vibrating molecule. The radiation absorbed can be measured or the light transmitted can be measured. The light absorbed or transmitted is related to the wavelength or wave number corresponding to the light absorbed or transmitted.

Every organic compound has its own characteristic infrared spectrum; therefore, the infrared spectrum can be considered as a fingerprint of the molecule. The position of the infrared absorption is important in identifying certain functional groups. For instance, in Figure 10.2, the carbonyl group in the ester exhibits a characteristic absorption at 1752 cm-1. As indicated earlier, wave numbers are the preferred absorption bands used by organic chemists in identifying functional groups in molecules. Equation 10.2 gives the relationship between the wave number and wavelength (in micrometers).

6

Equation 10.2

νcm−1 = 1

λµm

x 104 where νcm−1 is the wave number

Equation 10.2 is simply derived from taking the reciprocal of the wavelength in centimeters, i.e.,

Imagine that the atoms of organic molecules behave like balls connected to springs. The bonds connecting the atoms of the organic molecules can vibrate in two ways. They can stretch, and they can bend. In stretching, the distance between the atoms increases, but the atoms remain in the same bond axis. In bending, the molecules are deformed and the position of the atoms changes relative to the original bond axis. Unlike a ball on a spring, the various molecular vibrations occur at quantized frequencies, i.e., only at specified frequencies that change by an interval of n, where n is a whole number. The various stretching and bending vibrations are called fundamental vibrations and are illustrated in Figure 10.3

Bending Vibrations

7

Stretching Vibrations

Figure 10.3 Fundamental Absorption Modes for a Tri-atomic Molecule The “+” and “-indicate movement above and behind the plane of the two dimensional image of the tri-atomic molecule.

When a ball attached to a spring is allowed to vibrate, the frequency of vibration (oscillation) is a function of the force constant of the spring and the mass of the ball. The equation for calculating the frequency of the vibrating ball on the spring can also be used to approximate the stretching frequency of vibrating bonds. The equation is derived from Hooke’s Law (Equation 10.3).

Equation 10.3 F = -kx

Where F is the restoring force exerted by the atoms; x is the displacement of the end of the spring from its equilibrium position; and k is the force constant of the bond.

The time for an oscillation to be completed is given by Equation 10.4.

Equation 10.4

The frequency is given by Equation 10.5

8

Equation 10.5

Therefore, the frequency of vibration is represented by Equation 10.6

Equation 10.6

and Equation 10.6 can be converted into Equation 10.7

Since there are two atoms attached to the bond in the stretching frequency, m needs to replaced with the reduced mass, μ

9

Where μ equals

Therefore, the wave numbers of atoms undergoing stretching vibrations in the infrared region of the electromagnetic spectrum is given by Equation 10.7.

Equation 10.7

Where m1 is the mass of atom 1 and m22 is the mass of atom 2; k is

the force constant of the bond holding atoms 1 and 2 together; and c is the velocity of light, 2.998 x 108 m/s.

Table 10.2 lists the force constants for some common bonds.

10

Bond k, N/m C-H 500 N-H 650 O-H 760 S-H 400 P-H 310 Si-H 270 C-C 450 C-N 490 C-O 450 C-F 560 C-Cl 340 C-Br 290 C-I 230

C=O 1230 C=C 980 N=O 910 S=O 1000 C≡C 1560 C≡N 1750

Table 10.2 k, N/m, for some common bonds

The force constant of a bond is directly proportional to the strength of the bond. Table 10.2 indicates that the force constant of a single bond is smaller than the force constant of a double bond. The force constant of a double bond is smaller than the force constant of a triple bond.

When covalent bonds of organic and some inorganic molecules are subjected to infrared radiation, the bonds will absorb certain

11

frequencies of infrared radiation exciting them to a higher energy state. The molecular vibrations are quantized and only frequencies of infrared radiation that matches the natural frequencies of the bond vibrations are absorbed. The absorbed energy increases the amplitude of the vibration motion of the bonds and can be detected and measured by the infrared spectrometer. An infrared spectrum is, therefore, a record of infrared absorptions occurring at certain frequencies and wavelengths.

The stronger the chemical bond, the greater will be the absorption frequency of infrared radiation. For example, the C=O bond is stronger than the C-O bond; therefore, the C=O absorption stretch appears at higher frequencies and wave number than the C-O absorption stretching frequency. Functional groups in organic molecules can be determined from the position of the absorption.

Using the data in Table 10.2, the wave number for the O-H absorption stretching frequency can be calculated in the following manner.

1λ= ν

cm−1 = 12 x 3.1416 x 3.998 x 108 m/s

760 N/m16.0 g/6.022 x 1023 x 1.00 g/6.022 x 1023

16.0 g/6.022 x 1023 + 1.00 g/6.022 x 1023

1λcm

= νcm−1 = 5.309 x 10−10 s

m 760 kg m/ms2

1.56 x 10−10 g x 1 kg1000 g

12

Therefore, the O-H stretching frequency in wave numbers is close to 3700 cm-1 (two significant figures). According to Table 10.3 the calculated value is close to the observed value.

A similar calculation can be done for the C-H bond, and the result is close to the observed value found in Table 10.3.

These calculations are only approximations, because the actual values are determined by the strength of the bonds as well as electrical and steric properties in the vicinity of the bond.

The ranges of absorption frequencies for various stretching vibrations are listed in Table 10.3.

1λcm

= νcm−1 = 5.309 x 10−10 s

m x 6.98 x 1014 s−1

1λcm

= 3.7 x 10−5 m−1 x 1 m100 cm

= 3.7 x 103 cm−1

1λcm

= 5.309 x 10−10 sm

500 kg-m/ms2

1.53 x 10−24g x kg1000 g

1λcm

= νcm−1 = 5.309 x 10−10 s

m 500 kg-m/ms2

1.53 x 10−24g x kg1000 g

1λcm

= νcm−1 = 5.309 x 10−10 s

m x 5.71 x 1014 s−1

1λcm

= 3.03 x 105 m−1 x 1 m100 cm

= 3.03 x 103 cm−1

13

Table 10.3 IR functional group stretching vibrations

http://www.chem.ucla.edu/~webspectra/irtable.html has a more detail table of infrared absorptions.

Following is additional information that may be useful in the structure elucidation of organic molecules.

The wave numbers for C-H bending absorption frequencies are between 1475 cm-1 and 1300 cm-1. The wave numbers for C=C-H bending absorption frequencies are between 1000 cm-1 and

14

650 cm-1.

Overtone absorption frequencies can assume values that are ½, ⅓, or ¼ the wavelength ( 2, 3, or 4 times the wave number).

Combination bands result from the sum of two or more different wave numbers.

Difference bands result from the subtraction of two or more wave numbers.

The number of vibrations anticipated for a linear molecule is given by 3n-5 where n equals the number of atoms in the linear molecule. For example, carbon dioxide, CO22

, would theoretically exhibit four predicted vibration modes since it contains three atoms (one carbon and two oxygen atoms), 3(3)-5= 4. Following is a description of the four possible vibration modes for carbon dioxide.

15

The two bending modes, C and D, are degenerate (the same in energy); therefore, only one bending vibration is observed in the infrared (IR). The symmetrical stretching vibration A does not involve a change in dipole moment, i.e., the dipole moment of A is zero; therefore, this stretching vibration is infrared inactive. Infrared vibrations are accompanied by a change in dipole moment. The illustrations above can be repeated for any of the axes, and the results would be the same. No matter how the molecule moves in space, there are only two vibrations that would be IR active – the asymmetrical stretching vibration (accompanied by a change in dipole moment) and the bending vibration.

Figures 10.3 and 10.4 are IR spectra for carbon dioxide. Figure 10.3 represents transmittance versus wave numbers, and Figure 10.4 represents absorbance versus wave numbers.

16

Figure 10.3 Infrared spectrum of carbon dioxide: percent transmittance versus wave numbers, cm-1

http://www.wag.caltech.edu/home/jang/genchem/infrared.htm

The wavelength at approximately 4.3 μm (4.3 x 10-6 m) has a wave number of 2.33 x 105 m-1 or 2330 cm-1. The predicted value for the unsymmetrical stretch is 2640 cm-1

-1.

The wavelength at approximately 14 μm (1.4 x 10-5 m) has a wave number 7.14 x 104 m-1 or 714 cm-1. The predicted value for the bending vibration is 546 cm-1.

Note that if a vibration leads to a zero dipole moment during vibration, then that vibration is infrared inactive, i.e., in order for a vibration to be observed in the IR spectrum, there must be a net dipole moment during the vibration process.

17

Figure 10.4 Infrared spectrum of CO2 : absorbance versus wave numbers

http://science.widener.edu/svb/ftir/ir_co2.html

There are 3n-6 fundamental vibrations for non-linear molecules where n equals the number of atoms in the molecule. Some of the predicted infrared vibrations may not be observed because they are weak or may overlap with other absorptions. Weak bands may also occur that are not predicted due to overtones and combinations of two or more fundamental absorptions.

The predicted number of absorptions will not be observed if:

v There is no change in the dipole moment of the molecule during vibration. This was observed for carbon dioxide.

v Absorption occurs outside the region of the spectrum under observation.

18

v The vibrations result in absorptions so close that they coalesce.

v The absorptions are too weak to be seen.

Non-Fundamental Absorptions

As previously indicated, the three types of non-fundamental vibrations in IR spectra are combination, overtone and difference bands. Generally, these bands have low intensities.

The absorption bands of different functional groups appear at different frequencies. Consequently, the positions of the absorptions (and their intensities) are the basis for interpreting infrared spectra. For example, a strong transmittance band at 1887 cm-1 would suggest the presence of a carbonyl group in the molecule. A strong transmittance band in the region between 3750 cm-1 – 3000 cm-1 would be attributed to the OH stretching vibration. Depending upon other characteristic transmittance bands, the compound could be an alcohol, acid or phenol.

Infrared spectroscopy is frequently used to obtain structural features of molecules; however, comparing infrared spectra with that of known samples is a viable method for identifying organic compounds. Identification of organic compounds without comparison is often made when other spectrometric methods of analyses, e.g., Ultraviolet (UV) spectroscopy, nuclear magnetic resonance (NMR) spectrometry, and mass spectrometry (MS), are used along with infrared spectrophotometry.

19

The two most important regions of an infrared spectrum are 4000 cm-1 – 1300 cm-1 and 909 cm-1 – 650 cm-1. Some important transmittance or absorbance bands are located in these areas, and the bonds causing their absorbance or transmittance are listed in Table 10.3.

Following are some characteristic transmittance bands for organic molecules in the infrared region of the electromagnetic spectrum.

Hydrocarbons, CnH(2n+2)

Saturated straight chain hydrocarbons are characterized by relatively simple spectra that show only C-H and C-C bending and stretching vibrations.

Figure 10.5, the spectrum of octane, is an example of a straight chain hydrocarbon spectrum.

20

Figure 10.5 Infrared Spectrum of octane

http://orgchem.colorado.edu/Spectroscopy/Spectroscopy.html

Figure 10.5 shows the C-H stretching (asymmetrical and symmetrical) of the CH3 group at 2971 cm-1; the –CH2- bending (asymmetrical and symmetrical) at 1470 cm-1; the C-H bending at 1383 cm-1; and the –CH2- rocking at 728 cm-1.

Branched Hydrocarbons

Branched hydrocarbons can be characterized by the following observations.

The isopropyl group is characterized by a strong doublet of about equal intensity between 1388 cm-1 – 1380 cm-1 -1as a consequence of bending vibrations.

The t-butyl group is characterized by two bending vibrations at 1395 cm-1 – 1385 cm-1 and near 1370 cm-1, and the latter transmittance band is more intense (Figures 10.6).

If the compound has an internal carbon with two methyl groups attached, the spectrum is characterized by a doublet in the same region as the isopropyl and tertiary butyl bending vibrations at 1395 cm-1 – 1380 cm-1.

21

Figure 10.6 Infrared spectrum of 2,2,4-trimethylpentane the spectra was taken from http://www.google.com/search?hl=en&lr=&rlz=1R2GGLL_en&q=%22infrared+spectrum

+of+2%2C2%2C4-trimethylpentane%22&btnG=Search&aq=f&oq=&aqi

Cyclic Hydrocarbons, CnH2n

Cyclic hydrocarbons, unless there is ring strain, have infrared spectra similar to straight chain hydrocarbons. Cyclic hydrocarbons have a slight shift in the -CH

2 – bending vibration to higher

wavelength (lower wave number). For example, the -CH2 – bending in n-octane (Figure 10.5) occurs at 1470 cm-1, but the CH2 – bending in cyclohexane (Figure 10.7) occurs at 1461 cm-1.

22

Figure 10.7 Infrared Spectrum of cyclohexane

Alkenes, CnH2n

Alkenes possessing a terminal double bond, C=C-H, exhibit a C-H transmittance or absorbance band at 3058 cm-1 unless the transmittance of the absorbed infrared radiation obscured by other transmittance bands. The transmittance band at 3058 cm-1 due to the terminal alkene C-H stretch is illustrated in the IR spectrum of 1-pentene (Figure 10.8).

In addition, there is a weak to moderate transmittance band between 1667 cm-1 – 1634 cm-1 due to C=C stretching vibration. The C-H alkene bending vibration exhibits two transmittance bands between 1000 cm-1 – 650 cm-1. These bands are more intense than the other alkene transmittance bands.

23

Figure 10.8 Infrared spectrum of 1-pentene

The C=C stretching vibrations for trans alkenes are generally weaker than the C=C vibrations for the cis alkenes.

Figure 10.9 Infrared Spectrum of cyclohexene taken from

24

http://neon.otago.ac.nz/chemlect/chem111/module4/overlays/lecture4/lecture.pdf

If the alkene is di-substituted, symmetrical, and internally located, then the C=C stretching vibration will be very weak or absent.

Akynes, CnH(2n-2)

The spectra of alkynes exhibit C≡C stretching vibration between 2260 cm-1 – 2100 cm-1

1 (Figure 10.10). This observation does not occur if the triple bond is symmetrical. The infrared spectrum of 1-hexyne shows a transmittance band for the C-H stretching vibration of the terminal triple bond at 3324 cm-1. The C-H stretching vibration between 3333 cm-1 – 3267 cm-1 is generally strong for a terminal alkyne.

Figure 10.10 Infrared spectrum of 1-heptyne taken from http://orgchem.colorado.edu/hndbksupport/irtutor/alkynesir.html

The C≡C stretching vibration at 2126 cm-1 will not occur with a

25

symmetrical triple bond.

Amines, RNH2 and R2NH

The transmittance bands of primary and secondary amines are characterized by N-H stretching vibrations. The IR spectra for dilute solutions of primary amines exhibit asymmetrical and symmetrical transmittance bands at approximately 3496 cm-1 and 3390 cm-1. Figure 10.11 is an example of the infrared spectrum of primary amines.

Figure 10.11 Infrared spectrum of n-butylamine spectra taken from http://books.google.com/books?id=ega5c11VHvkC&pg=PA900&lpg=PA900&dq=%22infrared+spectrum+of+n-butylamine%22&source=bl&ots=tM4rWQhGup&sig=lyLYg13ewFXj3fbc6YaXYmPWCHM&hl=en&ei=Hmp3SpW9LYfatgPIqOjlBA&sa=X&oi=book_result&ct=result&resnum=3#v=onepage&q=%22infrared%20spectrum%20of%20n-butylamine%22&f=true

The IR spectrum of n-butylamine exhibits the characteristic –NH2

stretching vibration at 3496 cm-1 and 3390 cm-1.

26

The N-H stretching vibration appears as a single transmittance band at around 3367 cm-1 – 3322 cm-1.

Ethers, ROR

Strong transmittance bands in the region between 1150 cm-1 – 1085 cm-1 characterize aliphatic ethers.

The infrared spectrum of dibutyl ether shows a broad and strong transmittance band at approximately 1136 cm-1.

Aryl alkyl ethers show symmetrical and asymmetrical stretching vibration between 1275 cm-1 – 1200 cm-1 and 1075 cm-1 – 1020 cm-1 respectively.

Amides

The characteristic transmittance bands of amides result from the N-H and C=O stretching vibrations. The C=O stretching vibration of amides depends on the physical state of the amide and the extent of hydrogen bonding. The C=O stretching vibrations usually occurs at shorter wave numbers than the C=O stretching vibration of ketones and aldehydes. Table 10.4 lists some common wave numbers for

27

various structural classes of amides.

amide solid state, wave number, cm-1 solution, wave number, cm-1

primary 1650 1690 secondary 1640 1680-1700

tertiary 1680 – 1630 1647-1615 Table 10.4

The wave numbers for the stretching vibrations of amides are impacted by other factors including the physical state of the amide and the electrical effects and the nature of the solvent dissolving the amide.

The nature of the N-H stretching vibration of amides also depends upon structural features of the amide, i.e., whether it is a primary, secondary or tertiary amide.

Primary amides in dilute solution show two symmetrical and asymmetrical N-H stretching vibrations near 3520 cm-1 and 3400 cm-1. In the solid state, these transmittance bands near 3350 cm-1 and 3180 cm-1 are due to the ability of these systems to molecularly associate (hydrogen bond).

Amides can also molecularly associate linearly.

28

The N-H stretching vibration for secondary amides in dilute solution occurs near 3500 cm-1 – 3400 cm-1. In the solid state, the N-H stretching vibration appears as multiple transmittance bands near 3330 cm-1 – 3060 cm-1.

Figure 10.13, the infrared spectrum of butyramide, and Figure 10.14 is the infrared spectrum of N-methylformamide.

Figure 10.13 Infrared spectrum of butyramide

29

Figure 10.14 Infrared spectrum of N-methylformamide

Observe the multiple bands in Figure 10.14 around 3330 cm-1 – 3400 cm-1

-1 for the N-H stretching vibration of N-methylformide.

Organic Acids and Acid Halides

Organic acids show characteristic absorption for the O-H and C=O

30

stretching vibration. Organic acids in the solid and liquid states and in solution (unless the solution is very dilute) exist as dimers. Hydrogen bonding causes the dimerization of organic acids.

The O-H stretching vibration is influenced by hydrogen bonding and appears as a transmittance broad and intense band in the region between 3300 cm-1 – 2500 cm-1. The carbonyl stretching frequencies for acids are usually more intense than for ketones and aldehydes. The C=O stretching vibration occurs between 1720 cm-1 – 1695 cm-1 for aliphatic acids and 1710 cm-1 – 1680 cm-1 for aromatic acids in which the carbonyl group of the carboxylic acid is attached to the aromatic ring.

Figure 10.15 is the infrared spectrum of propanoic acid, an aliphatic acid. Observe the sharp absorption at around 1700 cm-1.

Also, observe the broad band due to O-H stretching vibration between 3500 cm-1 – 2500 cm-1.

31

Figure 10.15 Infrared spectrum of propanoic acid

Figure 10.16 is the infrared spectrum of benzoic acid, an aromatic acid. Observe the carbonyl stretching vibration at 1687 cm-1.

32

Figure 10.16 Infrared spectrum of benzoic acid

The characteristic transmittance peaks observed in the infrared spectrum of stearic acid are listed in Table 10.5.

Transmittance Wave number

broad intense OH stretch vibration 2841 cm-1 carbonyl, C=O, stretch vibration 1692 cm-1 O-H out of phase bending vibration 931 cm-1 Table 10.5 primary transmittance signals for stearic acid

33

Acid Halides

Acyl halides exhibit C=O stretching vibration at 1815 cm-1 – 1785 cm-1 in contrasts to the carbonyl stretching vibration of non-conjugated aliphatic acids that occur between 1720 cm-1 – 1695 cm-1.

The predominant transmittance for butyryl chloride is at about 1785 cm-1. The C=O bond is strengthen by the inductive effect of the chloride; therefore, the stronger bond strength of the carbonyl bond will lead to a higher wave number and lower wavelength.

butyryl chloride

Esters

Two strong transmittance bands are characteristics of esters. These transmittance bands occur at higher wave numbers than ketones as a consequence of inductive effects. The C=O stretching vibration occurs at 1750 cm-1 – 1735 cm-1 for aliphatic esters and C-O stretching vibration at 1250 cm-1 and 1051 cm-1. For aromatic

34

esters, the transmittance bands are located in the region between 1730 cm-1 – 1715 cm-1. The C-O transmittance band is generally broad and strong. Depending on the nature of the compound, sometimes only one C-O transmittance is observed.

The characteristic transmittance peaks observed in the infrared spectrum of ethyl acetate are listed in Table 10.6.

Transmittance Wave number carbonyl, C=O, stretching vibration 1751 cm-1 C-O stretching vibration 1250 cm-1 and 1051 cm-1

(two asymmetric coupled vibrations) Table 10.6 primary transmittance signals for ethyl acetate

Nitriles

R-C≡N

Nitriles exhibit a moderate and sharp transmittance between 2260 cm-1 – 2240 cm-1. The transmittance band for aromatic nitriles is between 2200 cm-1 – 2222 cm-1. The transmittance of the C≡N stretching vibration can be distinguished from the C≡C stretching vibration transmittance band and C=C stretching vibration band. Table 10.7 compares the wave numbers for the stretching vibrations for these three systems.

Transmittance bands Wave number C≡N 2260 cm-1 – 2222 cm-1 C≡C 2260 cm-1 – 2100 cm-1 C=C 1667 cm-1 – 1640 cm-1

Table 10.7 primary transmittance bands for C≡N; C≡C and C=C

35

Aldehydes and Ketones

Ketones and aldehydes exhibit characteristic transmittance bands due to C=O stretching vibrations between 1870 cm-1 – 1540 cm-1. Transmittance bands between these wave numbers are high intensity and are seldom obscured by other transmittance bands.

The C-H stretching vibration for aldehydes exhibits a transmittance band between 2830 cm-1 – 2695 cm-1. However, the aldehydic C-H stretching vibration can be obscured by C-H stretching vibrations of aromatic and aliphatic compounds.

The strong C=O stretching vibration for aldehydes and ketones can be clearly observed in the infrared spectra of acetone and propanal. Acetone exhibits a strong transmittance band at 1724 cm-1 due to C=O stretching vibration. Also, propanal exhibits a strong transmittance band at 1724 cm-1 due to C=O stretching vibration.

Aldehydes exhibit two weak aldehydic C-H transmittance bands at about 3390 cm-1 and 2941 cm-1

Alcohols, ROH

The O-H stretching vibration is very characteristic of alcohols. The

36

position of the transmittance band depends upon hydrogen bonding of the OH group. An O-H group that is not involved in hydrogen bonding exhibits a strong transmittance band between 3650 cm-1 – 3584 cm-1. Hydrogen bonding causes the O-H stretching vibration to shift to smaller wave numbers that appear between 3550 cm-1 – 3200 cm-1. Hydrogen bonding causes the force constant to be reduced; therefore, causing a decrease in wave number. Hydrogen bonding also causes the transmittance band to broaden. The transmittance spectrum of 1-butanol exhibits an O-H stretching vibration between 3000 cm-1 – 3200 cm-1 or 3.3 μm – 3.1μm. The neat solutions (pure solutions) of primary, secondary and tertiary alcohols can be distinguished by the position of the C-O stretching vibration. This is indicated in the characteristic transmittance bands listed in Table 10.8.

Transmittance bands Wave number C-O broad stretching vibration for CH3CH2CH2CH2OH, 1-butanol, a primary alcohol

1080 cm-1 – 1020 cm-1

C-O broad stretching vibration for CH3CH2CH(CH3)OH,2-butanol, a secondary alcohol

1160 cm-1 – 1060 cm-1

C-O broad stretching vibration for (CH3)3COH, 2-methyl-2-propanol, a tertiary alcohol

1250 cm-1 – 1120 cm-1

Table 10.8 C-O broad stretching vibrations for primary, secondary and tertiary alcohols

Observe the shift toward higher wave numbers of the C-O stretching vibration as we move from primary to secondary to tertiary alcohols.

37

Anhydrides

Anhydrides are characterized by two strong transmittance bands due to asymmetric and symmetric C=O stretching vibrations at about 1786 cm-1 and 1724 cm-1.

Maleic anhydride, compound I, is an unsaturated five-member cyclic anhydride with a higher wave number than noncyclic anhydrides.

I

Compound I exhibits a transmittance band for the C=O stretching vibration at about 1865 cm-1 and 1782 cm-1.

Table 10.9 is a tabulation of the wave numbers for saturated noncyclic anhydrides, conjugated noncyclic anhydrides, and conjugated cyclic anhydrides.

38

Transmittance bands Wave number

C=O stretching vibration for saturated noncyclic

anhydrides

1818 cm-1 and 1750 cm-1

C=O stretching vibration for conjugated noncyclic

anhydrides

1175 cm-1 – 1720 cm-1

C=O stretching vibration for conjugated cyclic

anhydrides

1865 cm-1 – 1782 cm-1

Table 10.9 C=O stretching vibrations for cyclic and noncyclic anhydrides

Aromatic Hydrocarbons

The C-H stretching vibrations of aromatic hydrocarbons occur in the region between 3100 cm-1 – 3000 cm-1. The C-H in plane bending vibration is more predominant than the C-H aromatic stretching vibration. The bending vibrations are strong transmittance bands between 1300 cm-1 – 1000 cm-1.

The C-C stretching vibrations for aromatic carbons are found in the region between 1600 cm-1 – 1585 cm-1 and 1500 cm-1 – 1400 cm-1.

The nature of ring substitution in mononuclear aromatic compounds can be ascertained in the infrared spectra of neat solutions. This is accomplished by studying the weak overtones and combination transmittance bands in the region between1627 cm-1 – 2000 cm-1. The characteristic appearance of the bands expected for various substitution types are demonstrated in the following manner.

(1) Following is a sketch of the pattern for the weak overtone and combination transmittance bands between 1627 cm-1 – 2000 cm-1

1 for mono-substituted aromatic compounds.

39

(2) Following are sketches of the weak overtone and combination transmittance bands between 1627 cm-1 – 2000 cm-1

-1 for di-substituted aromatic compounds (ortho, meta and para patterns).

40

(3) Following are sketches of the weak overtone and combination transmittance bands between 1627 cm-1 – 2000 cm-1

-1-1 for tri-substituted aromatic compounds.

41

(4) Following are sketches of the weak overtone and combination transmittance bands between 1627 cm-1 – 2000 cm-1

-1 for tetra-substituted aromatic compounds.

42

(5) Following is a sketch of the weak overtone and combination

43

transmittance bands between1627 cm-1 – 2000 cm-1 for penta-substituted aromatic compounds.

(6) Following is a sketch of the weak overtone and combination transmittance bands between 1627 cm-1 – 2000 cm-1 for hexa-substituted aromatic compounds.

For example, the infrared spectrum of toluene, compound I, exhibits the infrared spectrum represented by Figure 10.17.

44

I

Figure 10.17 http://orgchem.colorado.edu/Spectroscopy/Spectroscopy.html

Note the following about Figure 10.17:

The overtone transmittance bands and its pattern between 2000 cm-1 – 1667 cm-1 which is the characteristic pattern for monosubstituted aromatic compounds.

45

C=C aromatic stretching vibration at 1614 cm-1, 1506 cm-1 and 1465 cm-1

The characteristic strong aromatic C=C out-of-plane bending vibration at 738cm -1 and 694 cm-1

46

Ultraviolet-Visible Spectroscopy

Refer to http://www.kayelaby.npl.co.uk/chemistry/3_8/3_8_7.html to obtain additional information about the operational functions of Ultraviolet (UV)-Visible spectroscopy.

UV-Visible Absorption spectroscopy is effective in identifying molecules with a high degree of conjugation. The Beer-Lambert Law (Equation 10.8) is a relationship between the absorbance and the concentration of the absorbing species and the path length of the light.

Equation 10.8

This equation can be used to determine the concentrations of the absorbing species; where ε is the molar absorptivity or extinction coefficient in

Where L is the path length through the sample in cm; c is the concentration of the absorbing species in moles/L.

Many organic compounds exhibit color, e.g., chlorophyll is green; 2,4-dinirophenylhydrozones of aldehydes and ketones exhibits colors between yellow and red depending on the extent of

o

IA = -log = ε c L

I

47

conjugation in the molecules.

These compounds are colored, because they exhibit wavelengths from electronic excitations in the visible region of the electromagnetic spectrum.

There are four kinds of electronic excitations in ethylene. Ethylene contains two types of bonding molecular orbitals, the π molecular orbital and the σ molecular orbital. The following figure illustrates the four types of electronic transitions possible in ethylene (ethene).

48

σ→σ* with an energy difference,

that is greater than 1.2 x 10-18 J and a wavelength that is less than 165 nm. As can be observed from the diagram, the σ→σ* transition has the largest value for ΔE and the smallest λ ,

.

The π→π* transition has an energy smaller than 1.2 x 10-18 J and a wavelength about 190 nm which is greater than 165 nm.

The π→π* transitions are susceptible to signals in ultraviolet-visible spectra only when conjugation results in decreasing the energy difference between the π→π* transition; thereby increasing the value

hcE = λ

Δ

49

for λ.

Functional groups or chromophores absorb light resulting in electrons being excited to antibonding molecular orbitals in a unique pattern. For example, electrons in the sigma frame exhibit σ→σ* transitions, electrons in nonbonding molecular orbitals exhibit n→π* transitions, electrons in unsaturated systems can exhibit π→π* transitions; and electrons in nonbonding molecular orbitals adjacent to the π frame can exhibit n→π* transitions as well as π→π* transitions.

Cyclopentene exhibits σ→σ* transitions; σ→π* transitions; π→σ* transitions; and π→π* transitions. The π→π* transitions has the smallest value for ΔE but the largest value for λ.

In addition to σ→σ* transitions; σ→π* transitions; π→σ* transitions; and π→π* transitions, aldehydes and ketones exhibit n→σ* transitions and n→π* transitions. Considering these transitions, the n→π* transition is the longest wavelength, i.e.., between 200nm-400 nm. The π→π* transitions are 10 to 100 times more intense than the n→π* transitions. Both transitions are useful in identifying unknowns.

Absorptions below 200 nm (the vacuum ultraviolet) are of little use in interpreting UV spectroscopy. Absorptions in this region are σ→σ* and σ→π*.

As indicated previously, conjugated dienes exhibit longer wavelengths than isolated systems.

50

The conjugated diene system exhibits delocalization energy due to resonance; therefore, the energy difference between the highest occupied bonding molecular orbitals and the lowest unoccupied molecular orbital decreases and the wavelength increases. The more conjugated the system, the smaller the energy difference between the highest occupied molecular orbital and the lowest unoccupied molecular orbital.

The π* antibonding molecular orbital, the excited state, is more polar than the π bonding molecular orbital, the ground state; consequently, polar solvents would solvate the excited state to a greater extent than the ground state. This would result in a decrease in the value of ΔE and an increase in λ toward the visible (a bathochromic shift or red shift) region of the electromagnetic spectrum. This effect is demonstrated in the following diagram:

51

ΔE > ΔE’ and λ < λ’

For molecules exhibiting n→π* transitions, the ground state molecularly associates with hydrogen bonding solvents better than the excited state. This is a blue shift or a hypsochromic shift. Following is an illustration of the hypsochromic or blue shift:

52

ΔE < ΔE’ and λ > λ’

Solvents that molecularly associate would interact with the ground state better than the excited state resulting in a blue shift or hypsochromic shift in the ultraviolet-visible spectrum.

Following is an example of a UV-Visible spectrum of 2-methyl-1,3-butadiene (http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/UV-Vis/spectrum.htm.)

53

Woodward Correlation Rules

Dr. Robert Woodward of Harvard developed a correlation table (Table 10.10) that can be used to approximate the λmax of conjugated systems that have certain functional groups associated with the conjugation. For example, the wavelength, λmax, for π→π* transition for 1,3-butadiene, the parent conjugated system, is 217 nm, and if this system has any of the functional groups in table 10.10 attached to diene systems., then the Woodward correlation table would add the designated nanometers in table 10.10 to 217 nm. Functionality Associated with the Conjugated System

Additions to the Parent System, nm

Exocyclic* Double Bond (Conjugated Double Bond)

5

Extention of double bond 30

Alkyl, R, Groups 5

Homoannularity** (double bonds in the same ring)

36

-Cl 5

-Br 5

-NR3 60

-SR 30

-OR 6

-OOCCH3 0

54

Table 10.10 Woodward Correlation Table for approximating λmax of conjugated systems

*double bond directly attached, but external to a ring, e.g.,

**Example of homoannularity

The Woodward rules can be used to approximate the λ max for the

following terpene.

Parent 217 nm

Extension of the double bond system 30 nm

Homoannularity 36 nm

Four alkyl groups at 5nm each 20 nm

Approximate λ max 303 nm

The Woodward rules can be used to approximate the λmax for the

55

following steroid.

Parent 217 nm

Extension of the double bond system 30 nm

Homoannularity 36 nm

Four alkyl groups at 5 nm each 20 nm

Exocyclic double bond 5 nm

Approximate λmax 308 nm

The Woodward rules can be used to approximate the λma max for β-

carotene.

56

Parent 217 nm

9 x Extension of the double bond system

9 x 30 nm= 270 nm

ten alkyl groups at 5nm each 50 nm

Approximate λmax 537 nm

The literature values for the λm max of β-carotene is 450 nm and 478

nm.

Table 10.11 gives the additional nanometers added to 215 nm for n→ π* transition of the parent carbonyl system.

Functionality Associated with the

Conjugated Carbonyl System Additions to the Parent System

Extension to the Conjugated Carbonyl 30 nm

R α β

γ and higher

10 nm 12 nm 18 nm

-OH α β

γ and higher

35 nm 30 nm 50 nm

CH3COO-

α, β, and γ

6 nm

-OCH3

α β

35 nm 30 nm

57

γ

δ

17 nm 31 nm

-Cl

α β

15 nm 12 nm

-Br

α β

25 nm 30 nm

Exocyclic double bond 5 nm Homoannularity 39 nm

Table 10.11 gives approximate nanometers that should be added to 215 nm for n→ π* transitions in carbonyl systems.

Following are applications of the table for n→ π* transitions for two carbonyl systems.

Parent 215 nm

Extension of the double bond system 30 nm

Exocyclic double bond 5 nm

γ- alkyl substituent 18 nm

δ - alkyl substituent 18 nm

Approximate λmax 286 nm

Predicted λm max for

58

Parent 215 nm

Extension of the double bond system 30 nm

Homoannularity 39 nm

γ - alkyl substituent 18 nm

2 x δ - alkyl substituent 2 x 18 nm

Approximate λmax 338 nm

Aromatic compounds exhibit π→π* transitions between 229 nm – 330 nm depending on the type of functional group attached to the aromatic system (Table 10.12).

Compound Wavelength, nm

Benzene, C6H6 255

Phenol, C6H5OH 210

Nitrobenzene, C6H5NO2 252 and 280

Biphenyl, , C6H5 - C6H5, 246 and 330

Styrene, C6H5CH=CH2 244 and 280

Table 10.12 Approximate π→π* transitions for aromatic systems

59

Magnetic Resonance Spectroscopy

Infrared spectra result from bond vibrations, UV-visible spectra result from loosely attached electrons transitioning to higher energy levels, and proton nuclear magnetic resonance (NMR) spectra result from spinning nuclei of protons in different environments. NMR spectrometry gives more information about molecular structure than infrared or ultraviolet -visible spectrometry. When NMR spectrometry is used in conjunction with infrared spectroscopy and mass spectrometry, the structure of many unknown organic compounds can be easily elucidated.

The nuclei of certain atoms are constantly spinning (analogous to a spinning top or gyroscope) and since atomic nuclei have positive charges, the spinning results in a nuclear angular momentum referred to as “I,” the spin angular momentum.

The spin angular momentum is quantized just like in the case of vibrational frequencies. The value of “I” depends on the number of protons and neutrons in atoms. If the number of protons and neutrons is even, then the spin angular momentum of the nucleus is zero, i.e., the nucleus does not spin. If the number of protons is odd and the number of neutrons even or vice versa, the spin angular momentum is ½ or greater. Nuclei having “I” equal to ½ are ideally suited for NMR study while nuclei with an I value of zero or a value greater than ½ are not suited for magnetic resonance studies. Examples of nuclei with a value of ½ are H1; F19; C13; and P31. Nuclear magnetic resonance spectrometry can be used to study these molecules; however, H1 is the most studied nucleus since it is

60

found in large isotopic abundance and is common to organic molecules.

Nuclear magnetic resonance studies on the C13 nuclei have been hampered in the past because of the low abundance of the isotope; however, newer computer techniques and increased instrument sensitivities have made C13 NMR an important and exciting component of organic chemistry. C13 NMR studies can lead to a determination of the carbon skeleton of organic compounds.

A nucleus with a spin angular momentum “I,” according to quantum theory, can have 2(I)+1 orientations when placed in an applied magnetic field. Thus the nucleus of H1 can have two orientations, 2(½)+1; it can align itself with the applied magnetic field (parallel to the field) or against the field (anti-parallel to the field). The situation is analogous to the needle of a compass which aligns itself with the earth’s magnetic field.

Alignment with the field represents a lower energy state, -μHHo, and alignment against the field represents a higher energy level, +μHHo. The difference in energy, ∆E, between these two orientations is determined by the magnitude of the magnetic moment of the proton, μH, as well as the strength of the applied magnetic field, Ho.

∆E = E2 – E

1 = μ

HH

o – (-μ

HH

o) = 2 μ

HH

o

where μ H represents the magnetic moment and Ho represents the

strength of the applied field

61

This relationship can be represented by Figure 10.18

Figure 10.18 representation of the change in energy for the nuclear energy levels generated by an external magnetic field

According to Figure 10.18,

is greater than

therefore, > .

As the applied magnetic field increases, the energy difference, , between orientations increases.

62

In addition to alignment with or against the applied field, the nucleus will also rotate about the field. This nuclear phenomenon is called precession and again is analogous to a spinning gyroscope. Figure 10.19 is a representation of the motion of a nucleus in a magnetic field.

Figure 10.19 the motion of a nucleus in a magnetic field

The principle of NMR spectrometry requires that the precessing proton changes orientation from a lower to higher energy state. This is accomplished by holding the applied magnetic field, Ho, constant and applying electromagnetic radiation (of the radio frequency range) in such a way that its magnetic component H1 is rotating at right angles to the applied field and in the same direction as the precessing proton. By increasing the frequency of the electromagnetic radiation, the rotation of H1 is increased. When the rate of rotation of H1 is equal to the angular velocity of the precessing proton, energy is absorbed and the proton flips to the higher energy orientation (Figure 10.20).

63

Figure 10.20 illustration of proton transition from a low energy level to high energy orientation

At this point, the proton is said to be in resonance, and the absorption of energy is recorded as a peak, and a spectrum is obtained. Holding H1 constant and varying Ho will achieve the same results. This procedure is easier to accomplish and is what NMR instruments do. Protons located in different environments in the molecule will absorb different amounts of energy; consequently, a spectrum of the molecule is obtained. A rough schematic of an NMR spectrometer and an example of an NMR spectrum are displayed in Figures 10.21 and 10.22 respectively.

64

Figure 10.21 schematic diagram of an NMR spectrometer

Figure 10.22 http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/nmr/nmr1.htm, is an example spectrum of the NMR of p-xylene.

Chemical Shift

The environment of the proton determines the extent the nucleus would experience the affect of the applied magnetic field. Protons in

65

different environments experience different magnetic field effects. For instance, a proton on an OH group would experience a different applied field effect than the protons of a methyl, CH3, group. These differences are attributed to the differences in electron densities surrounding the nuclei of protons. For instance, the proton bonded to an oxygen atom has a lower electron density than the protons bonded to the carbon atom of a methyl group. This difference is due to the electron withdrawing capability of the oxygen atom bonded to the proton of the OH group. Protons exhibiting this effect are described as being deshielded. On the other hand, the methyl protons are described as being shielded since they experience little or no electron withdrawing effect; consequently, the higher the electron density around the proton, the greater the shielding effect. As the shielding increases, the applied field necessary for resonance to take place increases. Therefore, the radio frequency radiation necessary for resonance to take place by any proton (shielded or deshielded) is dependent on the spin angular momentum, μH, the shielding constant, σ and the strength of the applied field, Ho, and can be expressed by equation 10.8.

Equation 10.8

where h is Planck’s constant

Equation 10.8 can be rearranged to give equation 10.9 where Ho is the applied field strength necessary to execute resonance of any

66

proton.

Equation 10.9

Equation 10.9 indicates that an increase in the shielding constant would result in an increase in the applied field necessary to cause resonance. It is difficult to measure absolute values for Ho ; therefore, the applied field, Happlied field, necessary to produce resonance for a particular proton is measured against a reference. The difference between the the field, Ho, of the actual proton and the field, Hreference, of the standard divided by the applied field, Happlied field, is called the chemical shift (Equation 10.10).

Equation 10.10

chemical shift = δ and

The most popular reference is tetramethylsilane (TMS), (CH3)4Si. The chemical shift of TMS is assigned a value of zero, and the resonances of other protons are measured relative to the TMS standard since the protons of TMS are highly shielded. TMS protons appear at a higher applied field strength than most organic products because of the electron density about them. Protons of other organic compounds have a field effect that is lower than TMS;

67

therefore, they would have chemical shifts higher than 0, the chemical shift for TMS.

Since the frequency necessary for resonance to take place in the sample and the references are large values with slight differences, it is more convenient to express chemical shifts in parts per million; consequently, the equation for the chemical shift, δ can be written as:

The NMR spectrum of p-xylene (http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/nmr/nmr1.htm) shows the chemical shifts for the different types of protons relevant to the standard, TMS, at 0 ppm or 0 δ. The single line at about 2.25 ppm (δ) is the signal from the six protons of the two methyl groups in the para position. These protons are more shielded than the single line for the four aromatic protons at 7.0 ppm (δ).

The chemical shifts for a number of protons in various environments have been determined and listed in Table 10.12. These values are important in deciding the nature of the environment of the proton, and can lead to the identification of the compound. For example, let’s consider the protons in diethyl ether, compound I, and 2-bromoethanol, compound II.

68

The six protons labeled “a” of compound I are equivalent, i.e., they are in the same environment; consequently, they will experience the same effect of the applied field. The four protons labeled “b” are in the same environment, but are in a different environment than the “a” protons. The “b” protons are more deshielded than the “a” protons since they are adjacent to an electron-withdrawing group, the oxygen atom. Therefore, the “b” protons will experience a higher δ value than the “a” protons.

In compound II, the two a’ protons are equivalent; therefore, they will have a chemical shift value that is different from the b’ protons. Oxygen is more electronegative than Br; consequently, the two b’ hydrogen atoms will be more deshielded than the two a’ protons, and the NMR spectrum of compound II will show two different resonance signals.

Table 10.13 is a tabulation of approximate additional δ values that would add to the standard δ values in the proton magnetic spectra of methyl hydrogen atoms, CH3; methylene hydrogen atoms, CH2; and methine hydrogen atoms, CH, in proximity to a specified functional group. The standard proton magnetic resonance position of the hydrogen atoms of a methyl group is 0.90 δ. The standard

69

proton magnetic resonance position of the hydrogen atoms of a methylene group is 1.20 δ. The standard proton magnetic resonance position of the hydrogen atoms of a methine group is 1.55 δ. The α hydrogen atom (Figure 10.23) is the hydrogen atom adjacent to a specified functional group. The β hydrogen atom (Figure 10.23) is the hydrogen atom that is one carbon atom removed from the specified functional group.

Figure 10.23

Table 10.14 list chemical shift values for methyl protons with tetramethylsilane (TMS) as a standard

Designated Protons

Chemical Shift range, hertz (cps)

Chemical Shift range, ppm (δ): hertz/60.

R-CH2-R

78

1.5

R3CH

120

2.0

R2C=CH2

2.0 x 102

5.0

R2C=CHR

2.7 x 102 – 3.9 x 102

4.5 – 6.5

70

3.9 x 102 – 5.1 x 102

6.5 - 8.5

RC≡C-H

1.08 x 102 – 1.86 x 102

1.8 – 3.1

R2C=CRCH3

90 - 1.56 x 102

1.5 – 2.6

R-CH2Cl

2.22 x 102

3.7

R-CH2Br

1.38 x 102

2.3

R-O-CH3

3.48 x 102

5.8

RCHO

5.4 x 102 – 6.0 x 102

9-10

RO-H

30-3.00 x 102

0.5 – 5

3.6 x 102 – 4.8 x 102

6.0-8.0

RCOOH

6.0 x 102 – 7.8 x 102

10 – 13

Table 10.12 chemical shift values for methylene protons with tetramethylsilane (TMS) as a standard

71

Table 10.13 predicts the rough proximity locations of the chemical shifts (ppm) of methyl, methylene, and methine protons attached to primary (alpha) and secondary (beta) electron withdrawing groups.

The results of adding the α and β values to the standard values give a close approximations to the experimental values.

72

There are four non-equivalent protons in 1,2-dichloropropane, compound III.

73

Protons a and b are in different environments; therefore, they will give rise to separate H1 NMR signals. At first glance, it would appear that protons “a” and “b” are equivalent; however, a careful examination of the molecule (including making a model) in the format shown in III would indicate that protons a and b are in different environments.

There are four different kinds of protons in t-butyl isobutyl ether, compound IV.

The different protons are: nine protons for “a;” two protons for “b;” six protons for “c;” and one proton for d; therefore, the ratio of different types of protons would be 9:2:6:1.

How many H1 NMR signals would cyclobutane exhibit? How many

74

H1 NMR signals would ethanol, CH3CH2OH, exhibit?

Proton magnetic resonance spectroscopy cannot distinguish between mirror images. Compounds IV and V, enantiomers of 2-chlorobutane, would not give separate H1

1 NMR signals.

The geometric isomers cis-2-bromo-2-butene, compound VI, and trans-2-bromo-2-butene, compound VII, have the vinylic proton in different environments. Consequently, the chemical shifts would be slightly different.

75

Spin-spin Interaction

High resolution H1 NMR spectra of organic compounds exhibit peak patterns due to spin interaction of neighboring nuclei. Spin-spin interactions or peak splitting provides information about protons on adjacent carbon atoms. Spitting patterns are helpful in structure elucidation. Figure 10.24 is the low resolution H1

1 NMR spectrum of ethyl alcohol. Figure 10.25 is the high resolution H1 NMR spectrum of ethyl alcohol.

Figure 10.24 the low resolution H11 NMR spectrum of CH3CH2OH

76

Figure 10.25 the high resolution H1 NMR spectrum of CH3CH2OH found at http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/nmr1.htm

Spin-spin interaction results from the tendency of bonding electrons to pair its spin with the spin of the nearest proton nucleus. The spin transfer effect will, in turn, affect the second bonding electron and this affect will be felt by the nearest adjacent protons. One proton has influence on the spin orientation of an adjacent proton through the bonding electrons. The terminology that is used is that the protons couple each other. This coupling causes the splitting of protons on adjacent carbon atoms. For example, the following system would exhibit two types of signals:

77

The magnetic field the “a” protons experience is slightly increased or decreased by the spin of the tertiary proton “b.” The field experienced by proton “a” is increased if the tertiary proton is aligned with the applied field or decreased if the tertiary proton is aligned against the applied field. Therefore, for ½ of the molecules, absorption by the “a” protons is shifted slightly downfield and for the remaining ½ of the molecules, the absorption is slightly upfield. The singlet for protons “a” would be split into a doublet of equal intensity. On the other hand, the field experienced by proton “b” can be increased, decreased or unaffected by the spins of protons “a.” Protons “a” can align with the applied field, align against the field or have no impact on the field in two different ways as demonstrated in the following manner:

Proton “b” would appear as a triplet with peak intensities of 1:2:1 as indicated in Figure 10.26.

78

Figure 10.26 spin-spin interaction of protons in 1,1-dichloro-2-bromoethane

The number of peaks found in the multiplet, resulting from proton coupling, depends on the number of ways the spin of the interacting nuclei can be arranged to give different magnetic moments.

For example, the spin-spin interaction of the methyl and methylene protons in CH3CH2X would give rise to the following spectrum.

If one of the methyl hydrogen atoms is replaced with a deuterium atom to give XCH2CH2D, then the spin-spin interaction would give rise to the following spectrum.

If two of the methyl hydrogen atoms are replaced with deuterium atoms to give XCH2CHD

2, then the spin-spin interaction would give

79

rise to the following spectrum.

If three of the methyl hydrogen atoms are replaced with deuterium atoms to give XCH2CD3, then the spin-spin interaction would give rise to the following spectrum.

The peak ratio of multiplexes can be determined by using the coefficients of the binomial expansion (a+b)n where n is equal to the number of protons on adjacent carbon atoms.

For example, protons on a carbon atom adjacent to carbon atoms

80

containing six protons would exhibit a septet (seven lines) with intensities calculated by expanding the binomial equation:

Therefore, the ratio for the septet would be 1:6:15:20:15:6:1.

The proton magnetic resonance spectrum of ethanol, CH3CH2OH, exhibits a triplet for the methyl hydrogen atoms because of the spin-spin interactions of the hydrogen atoms on the carbon atom adjacent to the methyl group, the methylene group in ethanol. The hydrogen atoms on the methyl group split the signal of the hydrogen atoms on the methylene group, -CH2-, into a quartet. The hydrogen atom of the OH group exchanges rapidly with the hydrogen atoms of other ethanol molecules; therefore, it does not remain in position long enough to see splitting unless conditions are established to slow down the exchange rate or eliminate the exchange process. Figures 10.27 and 10.28 represent the spin-spin interactions of the methylene and methyl protons.

Figure 10.27 possible spin orientations for the methylene, –CH2-, protons

81

Figure 10.28 possible spin orientations for the methyl, -CH3, protons

As indicated in Figure 10.28, the methyl protons can have eight possible spin combinations. Two sets of three combinations are equivalent; therefore, the adjacent protons of the methylene group would experience a quartet with a peak ratio of 1:3:3:1.

The H1 NMR spectrum of 1,1-dichloro-2,2-dibromethane, compound VIII, would exhibit two doublets as showed in Figure 10.29.

82

Figure 10.29 H1 NMR spectrum of compound VIII

The Ha and Hb protons are in different environments; therefore, each proton would be split into a doublet as a consequence of the number of possible spin orientations for each proton as indicated in Figure 10.30.

Figure 10.30 the spin orientations of Ha and Hb protons in compound VIII

Figure 10.30 shows that each proton can have two spin combinations, +½ (↑) and -½(↓). This means that the protons will split each other into doublets of equal intensity. The Hb proton will be located farther downfield from the H

a proton because chlorine has

a greater electronegativity value than bromine; therefore, the Hbb is

more deshielded and would experience a lower field affect.

This information can be used to predict the H1 NMR spectrum for 2-chloropropane compound IX.

83

The six protons of the two methyl groups are equivalent. These six protons will be split into a doublet by the proton on the neighboring methine group. The proton on the methine group will be split into a septet with peak intensities 1:6:15:20:15:6:1. A representation of the H1 NMR spectrum of compound IX is illustrated in Figure 10.31.

Figure 10.31 H1 NMR spectrum of 2-chloroproane

The septet signal for the methine proton would occur at a lower field since it is adjacent to an electronegative chlorine atom. The methine proton septet signal and the methyl protons doublet would exhibit a peak ratio of 1:6 with the intensity of the septet, using the binomial theorem, (a+b)6, of 1:6;15:20:15:6:1, and the intensity of the doublet would be 1:1. These details are reflected in Figure 10.32.

84

Figure 10.32 H1 NMR spectrum of 2-chloroproane showing the spin-spin integration and integration to obtain the ratio of 1:6

The splitting patterns of protons are due to the spin-spin interaction of protons on adjacent carbon atoms. The observed splitting pattern for a particular absorption is always one more than the number of adjacent protons. If the adjacent protons are equivalent, then no splitting pattern will occur, e.g., the H1 NMR spectra for 1,2-dibromoethane, compound X, and dioxane, compound XI, will exhibit singlets (no splitting, because all the hydrogen atoms are equivalent) as shown in Figures 10.33 and 10.34 respectively.

85

Figure 10.33 representation of the H1 NMR spectrum of compound X

Figure 10.34 representation of the H1 NMR spectrum of compound XI

86

The J coupling constant

The distance between the peaks in the multiplex is called the coupling constant, J, and is usually cited in cps. Figure 10.35 illustrates the J coupling constant exhibited by 1,1-dichloro-2-bromoehane.

1, 1-dichloro-2-bromoethane

Figure 10.35 the J coupling constants for the splitting patterns in 1,1-dichloro-2- bromoethane

The J coupling constant is a measure of the effectiveness of spin coupling, and it is independent of the applied field.

87

Diamagnetic Anisotropy

The electron density surrounding the proton, a factor that is influenced by the inductive effect of the bonded groups, as indicated previously, determines the chemical shift. This shielding or deshielding effect is not sufficient to explain the low field resonance of the aldehydic proton (Figure 10.36), and the aromatic protons (Figure 10.36) or the high field resonance of acetylenic protons (Figure 10.36). The unexpected signals for an aldehydic proton, or aromatic protons, or acetylenic protons are influenced by a process referred to as diamagnetic anisotropy.

Figure 10.34 aldehydic proton; aromatic proton, and acetylenic proton

Diamagnetic anisotropic effect is due to the tendency of the π electrons to circulate when placed in an applied magnetic field. The electron circulation produces a small magnetic field which may oppose or augment the applied field. Opposition to the applied magnetic field results in a shielding effect which causes the protons to resonate at a high field. If the anisotropic effect augments the applied field, the protons will experience deshielding, and will

88

resonate at low field.

The magnetic anisotropic effect is common with multiple bonds such as in aromatic compounds, acetylenic compounds, and carbonyl compounds. The diamagnetic anisotropic effect produced by the circulating electrons in benzene is shown in Figure 10.37.

Figure 10.37 anisotropic effect generated by the circulating π electrons in the benzene ring

The spinning nuclei of the aromatic ring generate a magnetic field that augments the external magnetic field; therefore, these protons exhibit a downfield shift in its H1 NMR spectrum.

A close examination of Figure 10.37 suggests that protons in the region above the aromatic ring should be shielded and, therefore,

89

should exhibit an up field shift. This is the case in compound XI where protons b, c, d, and e would exhibit proton resonance as indicated in Table 10.15

Proton Chemical Shift, δ(ppm) a 3.81 b 2.50 c 1.30 d 0.90 e 0.30

Table 10.15 the chemical shifts a proton a, b. c, d and e in compound XI

Figure 10.38, a representation of the spectrum of compound XI, shows that protons d and e reside directly above the aromatic ring and are, therefore, shielded by the magnetic field generated by the circulating π electrons, and exhibit a high field shift (an upfield chemical shift or a low δ value).

90

Figure 10.38 theoretical representation of the H1 NMR spectrum of compound XI

Diamagnetic anisotropy can be used to explain the low field resonance of aldehydic protons. This is illustrated in Figure 10.39.

Figure 10.39 Diamagnetic Anisotropy in Acetylenic Systems

91

Analogous to the aromatic protons in the plane of the ring, the aldehydic protons are positioned so that the field generated by circulating π electrons in the carbonyl group generates a field that augments the applied field; therefore, causing the aldehydic proton to be deshielded and exhibit a downfield δ value around 9 ppm, a much lower field field than would have been predicted without the consideration of diamagnetic anisotropic effect.

13C NMR Spectrometry

Wide-band heteronuclear decoupled 13C spectra indicate the different types of carbon atoms in a compound. The coupled 13C is complex and gives insight into the number of hydrogen atoms attached to the carbon atoms. However, coupled 13C spectra exhibit large J values.

Noise or heteronuclear decoupled spectra are generated by using a noise generator. Following is an example of a 13C NMR decoupled spectrum of diethyl phthalate.

Diethyl phthalate

92

The spectrum exhibits six lines.

The carbon atoms in dethyl phthalate are labeled in the following manner:

The labeled carbon atoms would exhibit the following C-13 downfield signals relevant to TMS as a reference.

93

Clearly 13C NMR like 1H1 NMR gives useful structural information; however, 13C NMR provides direct information about the carbon skeleton of a molecule. C-13 spectra give information about the number and type of carbons and the chemical shifts relevant to the chemical environment of the carbon. The intensity of the signals is weaker than the proton signals; therefore, an increase in the signal-to-noise ratio is required - this is done by pulsed Fourier Transform (FT) – NMR. Fourier Transformation is the mathematical manipulation of the data to plot spectra. 13C NMR chemical shifts are measured relative to the carbons of tetramethylsilane. The following table lists the chemical shifts for some carbon atoms.

1,4-diethoxybenzene exhibits the following decoupled 13C NMR spectrum.

1,4-diethoxybenzene

94

The spectrum exhibits four signals in reference to TMS. This indicates that the molecule has four carbon atoms in different chemical environments.

The labeled carbon atoms give rise to the different signals in the carbon-13 spectrum.

The C-13 decoupled signals would correspond to the following spectrum.

95

13C NMR spectra are not necessarily concerned with electrical integration and spin-spin interaction. The Fourier-Transformed pulse technique distorts the signal intensities for carbons without hydrogen atoms; therefore, the integration can be misleading.

Fundamentally, the different signals in the 13C NMR spectra give rise to the different types of carbon atoms in the molecule, i.e., if the carbon-13 spectrum exhibits five signals, then the molecule has carbon atoms in five different chemical environments.

Table 10.18 lists the chemical shifts for a number of carbon atoms relevant to tetramethylsilane.

96

Type of Carbon Approximate Chemical Shift, PPM, relevant to TMS

R-CH3 8-35 R2CH2 15-50 R3CH 20-60 R4C 30-40 C-O 50-80 C-N 40-60 C-Cl 35-80 C-Br 25-65 C-I 0-40

RCONR2 165-175 RCO2R 165-175 RCO2H 175-185 RCOH 190-200 RCOR 205-220 ≡CR 65-85

alkene carbon 100-150 Aromatic C6Hx 110-170

Table 10.18: C-13 carbon Shifts Relevant to TMS

Other types of Magnetic Resonance Spectrometry

In addition to the tradition proton magnetic spectroscopy students encounter in their first organic chemistry course, their are other types of magnetic resonance spectrometric research tools that chemist have reported in the literature. These include the following:

(a) Distortionless Enhancement of NMR Signals by Polarization Transfer (DEPT), a one Dimensional NMR technique, that enhances the intensities of 13C signals and provides information about the number of attached protons to the carbon atoms

(b) Insensitive Nuclei Enhanced by Polarization Transfer (INEPT), a one dimensional NMR technique, that gives similar results as DEPT, but gives less detail DEPT.

97

(c) Nuclear Overhauser Effect Spectroscopy (NOE), a one Dimensional NMR technique, that involves irradiation of the sample at specific frequencies before a signal is acquired, and enhancing the intensities of nearby nuclei.

(d) Insensitive Nuclei Enhanced by Polarization Transfer (INEPT), a one dimensional NMR technique. This technique is similar to DEPT, but not as detailed.

(e) Nuclear Overhauser Effect Spectroscopy (NOE), a one

Dimensional NMR Technique, where the sample is irradiated at specific frequencies before the signal is acquired, and enhancing the intensities of nearby nuclei.

(f) Nuclear Overhauser Effect Spectroscopy (NOESY), a two

dimentional version of the NOE experiment that yields a display of all atoms that are close in space. The diagonal and the projection on each axis give rise to one dimensional spectra.

(a) Correlated Spectroscopy (COSY), a two dimensional NMR

technique which displays all protons that are coupled. The diagonal and the projection on each axis are the one dimensional spectrum. The off diagonal peaks indicate the presence of coupling between pairs of protons. Following is an example of a COSY spectrum of ethylbenzene copied from http://chem.ch.huji.ac.il/nmr/techniques/2d/cosy/cosy.html

98

(h) Spin Echo Spectroscopy (SECSY), a two dimensional NMR technique, yields the same information as COSY, but with a different format. (i) J-Resolved 2D NMR, a two Dimensional NMR technique, has

normal spectrum on one axis and coupling on the other axis. (j) Heteronuclear Shift Correlation (HETCOR), a two dimensional

NMR technique, correlates the 13C spectrum on one axis and 1H spectrum of the other axis.

(k) Incredible Natural Abundance Double Quantum Transfer

Experiment (INADEQUATE), a two dimensional NMR technique, that directly obtains carbon-carbon connectivity, and, ultimately, the carbon skeleton of a molecule.

(l) Inadequate Sensitivity Improvement by Proton Indirect

99

Detection (INSIPID), a two Dimensional NMR technique, is a reverse detection INADEQUATE experiment that greatly reduces the acquisition time.

DEPT 13C NMR Let’s examine one of these other types of NMR spectra, the DEPT 13C NMR spectrum, in more detail. DEPT uses a second radio frequency transmitter to irritate the sample. The frequency excites the protons affecting the appearance of 13C spectra by transferring polarization to 13C nuclei. The resulting spectra provide information about the number of protons attached to carbon atoms. DEPT spectra are referred to as “edited” spectra since the peak intensities and phases are modified relative to normal 13C spectra. DEPT spectra unambiguously identify the CH3, CH2, and CH carbon peaks. The technique uses three separate spectra: (1) At 45o, DEPT-45 (2) At 90o, DEPT-90 (3) At 135o, DEPT-135.

DEPT does not detect the quaternary carbons or solvents.

Following is the representation of the 13C NMR spectrum of 4-hydroxy-3-methyl-2-butanone (compound I).

CH3 C CH CH2OH

O

CH3

a

bc

d

e

100

I

The three DEPT spectra of 4-hydroxy-3-methyl-2-butanone are:

(1) DEPT-45

DEPT-45 spectra exhibit positive phases (above the horizontal line) for methyl, CH3, carbon atoms; methylene, CH2, carbon atoms; and methane, CH, carbon atoms. The carbon atoms without hydrogen atoms (examples quaternary carbon atoms and carbonyl carbon atoms) and carbon atoms from solvents will not appear in DEPT-45 spectra.

Consequently, the carbonyl carbon, carbon atom “a,” and the carbon atom for the solvent deuterochloroform disappear. Carbon atoms “b”, “c”, “d”, and “e” are in positive phase (above the horizontal line)

(2) DEPT-90

020240 406080100120140160180200220

CDCl3

a

b c d e

020240 406080100120140160180200220

b c d e

101

DEPT -90 provides spectra in which carbon atoms with three hydrogen atoms and two hydrogen atoms attached are in negative phase (below the horizontal line) and carbon atoms with one hydrogen atom attached are in positive phase (above the horizontal line). Consequently, the “c” carbon atom of 4-hydroxy-3-methyl-2-butanone has one hydrogen atom attached and, therefore, is in positive phase (above the horizontal line). Carbon atoms “d”, three hydrogen atoms attached; “e”, three hydrogen atoms attached; and “b”, two hydrogen atoms attached, are in negative phase (below the horizontal line).

(1) DEPT-135

DEPT-135 provides spectra in which carbon atoms with two

020240 406080100120140160180200220

b

c

d e

020406080100120140160180200220

b

c d e

102

hydrogen atoms attached are in negative phase and carbon atoms with one hydrogen atom attached and three hydrogen atoms attached are in positive phase. Consequently, the “b” carbon atom of 4-hydroxy-3-methyl-2-butanone has two hydrogen atoms attached and is in negative phase. Carbon atoms “c”, one hydrogen atom attached; “d”, three hydrogen atoms attached; and “e”, three hydrogen atoms attached, are in positive phase. DEPT spectra assist spectroscopists to identify the types of carbon atoms in molecules.

103

Mass Spectrometry

Mass spectrometry is an analytical instrumental method used to determine structural features of organic molecules. It is based on fragmentation patterns related to

Liquid samples are volatilized in vacuo and passed through an ionization chamber. This is done on the edge of a special insertion probe. The sample is bombarded with 70 electron volts (eV) of energy.

Removing one electron from the bonding or nonbonding molecular orbital ionizes the molecule. The resulting ion is called the parent ion or molecular ion. The molecular ion may fragment into ions of smaller m/e ratios.

A positive potential is used to repel the positively charged fragment from the ionization chamber, and an accelerator plate (2000 V) is used to accelerate the positive ions through a magnetic field.

Depending on the m/e ratio, the fragments are deflected with a certain radius of curvature according to equation 10.11.

104

Equation 10.11

centrifugal force = centripetal force

Where H is the strength of the magnetic field

e is the charge on the ion

v is the velocity of the ion

m is the mass of the ion

r is the radius of curvature of the ion in the magnetic field

Diagram 10.1 is the representation of a working mass spectrometer (http://www.chemguide.co.uk/analysis/masspec/howitworks.html)

105

The output fragmentation data of a mass spectrum is represented by Figure 10.40.

106

Figure 10.40 Representation of the Mass Spectrum of a molecule.

The mass spectrum represented by figure 10.40 shows a molecular ion peak at m/e ratio equal to 132 (this means the molecular mass is 132 g/mol) and fragmentation patterns at e/m at 103, 77, 63, 51 and 39.

The mass spectrum data can be explained by the following mathematical equations. The mathematics begin by recognizing that potential energy = kinetic energy. Equation 10.12 represents this equivalency.

Equation 10.12

107

eV = the potential energy of the ion accelerated by a potential voltage

V is the voltage; e is the charge on the particle; m is the mass of the particle; and v is the velocity of the particle.

The radius of curvature of an accelerating particle in a magnetic field is given by equation 10.13.

Equation 10.13

where H is the strength of the magnetic field; e is the charge on the particle; m is the mass of the particle; r is the radius of curvature of the particle; and v is the velocity of the particle.

Equation 10.13 can be manipulated in the following manner to give the following results of the square of the velocity of the particle.

108

Manipulation of Equation 10.13

Using equation 10.13 and the square of the velocity of the particle accelerating in a magnetic field, equation 10.14 can be derived.

Equation 10.14

Equation 10.14 is a mathematical expression for calculating the mass-to-charge (m/e) ratio of a moving particle of a specified charge in a magnetic field, H, with a voltage V with a designated radius of curvature.

Following is the derivation of Equation 10.14 from Equations 10.12 and the manipulated form of Equation 10.13.

Therefore, every particle that is accelerated in a magnetic field would

H2r2

2V = m

e

109

have a m/e equal to the square of the strength of the magnetic field times the square of its radius of curvature in the magnetic field divided by twice the voltage within the magnetic field. Every fragmentation particle would be exposed to the same voltage and the same magnetic strength; therefore, the difference in their charge-to-mass ratios would be the radius of curvature within the magnetic field. The larger the particle, the larger the radius of curvature, and larger fragments would give rise to greater mass-to-charge values.

Most of the ions that pass through the collector slip have a plus one (+1) charge.

The mass spectrometer generally functions by holding r and H constant and vary V, the accelerating potential. There is an indirect relationship between V and m/e. The result is that the mass of a high m/e ratio is collected first.

The current of the ions entering the collection slit is recorded as a peak on a chart.

The height of the peak is directly proportional to this current.

Compounds containing C, H, O, and S have molecular ion peaks that are even, and odd m/e fragmentation patterns unless rearrangement occurs or there is a loss of a neutral fragment. If rearrangement occurs or if there is loss of a neutral fragment from a radical-carbocation, the molecular ion, then the m/e ratio will be even.

Compounds containing C, H, and N usually have odd m/e molecular

110

ion peaks, and even m/e fragmentation patterns unless rearrangement occurs or there is a loss of a neutral fragment. If rearrangement occurs or if there is loss of a neutral fragment from a radical-carbocation, the molecular ion, then the m/e ratio will be odd.

The base peak in the mass spectrum is the peak with the highest m/e ratio. The parent peak or molecular ion peak is generally the peak of appreciable intensity at the highest m/e ratio. Isotope peaks also exist as the molecular ion peak plus 1 peak or the molecular ion plus 2 peak. Depending on the number of ions present, isotope peaks may exist at mass-to-charge ratios higher than the parent peak plus 2 mass-to-charge ratio. The mass-to-charge ratios for other peaks represent major fragmentation patterns that are characteristic of the molecular structure of the molecule.

Following are some examples of the major fragmentation patterns of molecules that are characteristic of the molecular structures of molecules. These fragmentation patterns occur in a constant magnetic field with slight variation in electrical potentials.

Primary Alcohols

A m/e of 31 would be the base peak for primary alcohols. The

111

molecular ion peak would depend on the size of the primary alcohol, but would follow a pattern of 32, 46, 60, 74, 88, 102, 116, 130, etc.

In addition, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

Secondary Alcohols

The mass-to-charge ratio for secondary alcohols depends on the alkyl groups attached. First, the bulkier alkyl group would be cleaved from the molecular ion leaving a fragment that would exhibit m/e equal to the mass of the smaller alkyl group + carbon atom + hydrogen atom + hydroxyl group. For example, 2-pentanol would lose a propyl radical from the molecular ion to give rise to a base peak at m/e = 45. This base peak would consist of a methyl group + hydrogen atom + carbon atom + oxygen atom + hydrogen atom = 15 + 1 +12 + 16 + 1 = 45.

112

The mass-to-charge ratio for the base peak of secondary alcohols would depend on the alkyl groups attached to the carbon atom that has the -OH group attached.

Tertiary Alcohols

The mass-to-charge ratio for tertiary alcohols depends on the alkyl groups attached. First, the bulkier alkyl group would be cleaved from the molecular ion leaving a fragment that would exhibit m/e equal to the mass of the smaller alkyl groups + carbon atom + hydroxyl group. For example, 2-methyl-2-pentanol would lose a propyl radical from the molecular ion to give rise to a base peak at m/e = 59. This base peak would consist of two methyl groups + carbon atom + oxygen atom + hydrogen atom = 15 + 15 +12 + 16 + 1 = 59.

113

Primary Amines

A m/e of 30 would be the base peak for primary amines. The molecular ion peak would depend on the size of the primary amine, but would follow a pattern of 31, 45, 59, 73, 87, 101, 115, 129, etc.

114

In addition, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

Secondary Amines

The mass-to-charge ratio for secondary amines depends on the alkyl groups attached. First, the bulkier alkyl group minus a methylene group would be cleaved from the molecular ion leaving a fragment that would exhibit m/e equal to the mass of the smaller alkyl group + hydrogen atom + nitrogen atom + carbon atom + two hydrogen atoms. For example, N-methylpropylamine would lose an ethyl radical from the molecular ion to give rise to a base peak at m/e = 44. This base peak would consist of a methyl group + hydrogen atom + nitrogen atom + carbon atom + two hydrogen atoms = 15 + 1 +14 + 12 + 2 = 44.

115

Tertiary Amine

The mass-to-charge ratio for tertiary amines depends on the alkyl groups attached. First, the bulkier alkyl group minus a methylene group would be cleaved from the molecular ion leaving a fragment that would exhibit m/e equal to the mass of the smaller alkyl groups + nitrogen atom + carbon atom + two hydrogen atoms. For example, N, N-dimethylpropylamine would lose an ethyl radical from the molecular ion to give rise to a base peak at m/e = 58. This base peak would consist of two methyl groups + nitrogen atom + carbon atom + two hydrogen atoms = 15 + 15 +14 + 12 + 2 = 58.

116

Straight Chain Alkane

Straight chain alkanes would exhibit a base peak at m/e 43, and the molecular ion peak would depend on the size of the alkane, but

117

would change by 14, the mass of a methylene, CH2, group.

In addition, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

Branched Alkanes

The base peak would be formed at the highest branch as illustrated with isopentane; therefore, the base peak would be at a mass to-charge ratio of 43 for iso-alkanes, and the molecular ion peak would depend on the size of the iso-alkane, but would follow a pattern of 14, the mass of a methylene group.

118

Again, the base peak would be formed at the highest branch; therefore, t-butyl carbocation exhibiting a mass -to-charge ratio of 57 would be the base peak.

The molecular ion peak would depend on the size of the 2,2-dimethylalkane, but would follow a pattern of 14 (the mass of a methylene group), 28, 42, etc. the mass of a methylene group for straight chain alkyl groups.

In addition, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

119

Alkenes

The base peak for propene would be at m/e ratio of 27.

In addition, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

120

The base peak for terminal alkenes would be at m/e = 41, because the allyl carbocation (stabilized by resonance) is more stable than a vinyl carbocation.

In addition, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

Cyclohexene loses a neutral fragment (ethene) as illustrated by the fragmentation patterns above where the ethene is lost from the molecular ion, a radical-carbocation; therefore, the base would have an even m/e ratio. This type of fragmentation pattern was

121

mentioned previously.

In addition, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

Carbonyl Compounds

Methyl ketones give rise to a base peak at m/e 43. The peak at m/e equal to 43 is stabilized by resonance, and is referred to as the methyl acylium ion. The molecular ion peak of methyl ketones depends on the size of the methyl ketone.

122

Phenyl ketones give rise to a base peak at m/e 105. The peak at m/e equal to 105 is stabilized by resonance, and is referred to as the phenyl acylium ion. The molecular ion peak of phenyl ketones depends on the size of the phenyl ketone.

In addition, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

123

Primary Alkyl Bromides

A m/e of 93 would be the base peak for primary alkyl bromides. The molecular ion peak would depend on the size of the primary alkyl bromide, but would follow a pattern of 108, 122, 136, 150, 164, 178, 192, 206, etc. Primary alcohol bromides would also exhibit a characteristic molecular ion + 2 peak that would be about 97% as intense as the molecular ion peak.

In addition, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

124

Esters

125

The base peak of esters depends on the group attached to the carbonyl group. If the group is a methyl group, than the base peak would be at m/e 43. If the group is a propyl group, then base peak would be at m/e 71.

Esters with alkyl groups that have two or more carbon atoms attached to the oxygen of the Ester can undergo the McLafferty rearrangement.

As illustrated in the above fragmentation pattern, the McLafferty

126

rearrangement results in rearrangement of a hydrogen atom followed by the loss of a neutral fragment. Consequently, the base peak has an even mass to charge ratio. The mass to charge ratio of the base peak depends on the size of the group attached to the carbonyl group. For example, once the molecular ion is formed, if a methyl group is attached to the carbonyl group, and an ethyl group is attached to the oxygen atom, then a hydrogen atom on the methyl group of the ethyl group rearranges to the oxygen of the carbonyl group. Movement of electrons would occur as illustrated in the McLafferty fragmentation pattern resulting in the loss of ethene (a neutral fragment), and the formation of an even m/e fragment at m/e 60.

Other neutral fragments can be lost, but the rearrangement would involve the transferred of a hydrogen atom. For example, if a propyl group is attached to the oxygen atom, the neutral fragment lost would be propene. If an ethyl group is attached to the carbonyl group, then the base peak would be at m/e 74. If a propyl group is attached to the carbonyl group, then the base peak would be at m/e 88. If a butyl group is attached to the carbonyl group, then the base peak would be at m/e 102.

The McLafferty rearrangement is not unique to esters. Compounds with the following molecular arrangements can exhibit the McLafferty rearrangement.

127

Where Z and Q are electronegative atoms with electrons in nonbonding molecular orbitals.

In addition, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions.

Arenes

Aromatic nuclei with alkyl side chains, arenes, form a base peak at a mass to charge ratio equal to 91. The m/e at 91 is due to the formation of benzyl carbocation that rearranges to a tropyllium ion. The benzyl carbocation is formed by cleavage at the carbon atom attached to the aromatic nucleus. The mass-to-charge ratio of the molecular ion or parent peak depends on the structure of the R group.

128

The benzyl carbocation then rearranges to the tropyllium ion by the following pathway.

129

Incidentally, the tropyllium ion obeys Hückel’s rule for aromaticity.

Isotope Peaks

Other isotope peaks appear in mass spectra. For example, sulfur has a molecular ion plus 2 isotope peak that is approximately 5% of the molecular ion peak, and chlorine has a molecular ion plus 2 isotope peak that is approximately 33% of the molecular ion peak.

130

131

The statement throughout this discussion “In addition, there would be a variety of other m/e peaks resulting from the fragmentation of the molecular ion or fragmentations from other ions” can be explained by evaluating the mass spectra of a couple of compounds. For example, the evaluation of the spectrum of benzophenone would lead to the following fragmentation patterns.

benzophenone

Mass Spectrum of Benzophenone

132

The molecular ion peak

133

The m/e =77 is the result of the heterolytic cleavage of the bond between the phenyl group and the carbonyl group. This is an example of “other m/e peaks resulting from the fragmentation of the molecular ion.”

134

The m/e of 51 is the result of the loss of a neutral fragment, acetylene, from the phenyl carbocation. The m/e is odd, because the neutral fragment is the result of the fragmentation of a carbocation by homolytic cleavage. This an example of “other m/e peaks resulting from the fragmentations of other ions.”

The peak at mass-to-charge ratio 152 is more difficult to explain. It is a small peak; therefore, the mechanism for its formation must be more involved then the mechanism for the formation of the other fragmentation patterns. A m/e =152 involves rearrangement and the loss of a neutral fragment. The process described is a stepwise solution that attempts to show the various homolytic cleavages required to arrive at the desired fragment. These processes are probably more concerted than the rationalization presented. The rationale for the formation of the m/e = 152 fragment simply represents a pedagogy for allowing visualization of the formation of the fragment at m/e = 152.

135

Following is a representation of resonance stabilization of the fragment at m/e at 152.

136

For example, the evaluation of the spectrum of n-butylamine would lead to the following fragmentation patterns.

n-butylamine

The mass spectrum of n-butylamine

137

The m/e =43 is the result of the heterolytic cleavage of the bond between the propyl group and the carbon attached to the amino group. This is an example of “other m/e peaks resulting from the fragmentation of the molecular ion.”

138

Problems

Spectrometry

1. Convert 3.96 μm to cm-1 2. Give a brief explanation about the theory of infrared

spectroscopy. 3. Calculate the approximate stretching vibration, in

wavenumbers, for the C-F bond if the force constant, k, is approximately 760 N/m.

4. How man fundamental vibrations are predicted for the non-linear water molecule? How many fundamental vibrations are predicted for the linear carbon dioxide molecule?

5. Lists four reasons why all predicted fundamental vibrations of a bond may not appear in the infrared spectrum.

6. Draw the expected fundamental vibrations for CO2 and label the degenerate modes.

7. Using the data in Table 10.2, calculate the wave numbers for the C≡N and C≡C stretching vibrations.

8. Predict which of the following compounds would give little or no C≡C stretching transmittance bands:

(a) H-C≡C-CH3 (b) H-C≡C-CH2CH3

139

(c) CH3C≡CCH3 (d) CH3CH2 CH2C≡CCH3 9. The carbon-oxygen double bond appears at a higher

wavenumber (higher frequency of infrared radiation) than the carbon-oxygen single bond. Give an explanation for this observation.

10. A certain compound gives a carbonyl absorption at

approximately 1775 cm-1. Which of the following compounds would give rise to a transmittance band at 1775 cm-1? Give an explanation for your selection.

11. When a carbonyl group is conjugated with a C-C double bond, the conjugation lowers the observed wave number by approximately 20 cm-1 as compared to a nonconjugated carbonyl. For example, compound A has a strong transmittance band at 1715 cm-1 and compound B has a strong transmittance band at 1685 cm-1.

Based on the transmittance band for cyclohexanone at

1715 cm-1 and 1724 cm-1 for cyclopentanone, predict the absorption for

CH3 C

O

CH3

O O O

a b c d

CH3

CH3

C H CH2 C CH3

O

C

CH3

CH3

CH C

O

CH3

A B

140

12. How could you use infrared spectroscopy to distinguish

between the following compound pairs? Indicate the characteristic absorptions expected for each structure.

(a)

(b)

(c)

O O

and

C

O

C

O

CH3CH2CH2 OH CH3CH2CH2 H

CH3C

O

C

O

CH3CH2CH2 CH3CH2CH2 OCH3

O O

141

(d)

(e)

(f)

(g)

(h)

CH3 CC CH3CH3CH2C N

OO

CH3

CH3

CH3

C OH CH3CH2CH2CH2OH

CH2 CHCH3CH2CH2CH2CH3 CH3CH CH3

CH3 CH3

CH3

O

CH3CH2CH2CH3CH2C OCH3

142

(i)

(j)

13. Sketch the H1

1 NMR spectrum for

O

CCH3CH2CH2OH CH2 OH

CH3

CH3

CH3

CH

CH C

C

O

O

O

H3C

H3C

CHH3C C

O

OH

C C

CH3

O

C

H

H

H

a. b.

C C

CH3

H

H O

H

143

14. The aroma of coffee is, in part, related to compound A, C5H8O2.

Compound A can be represented by the following theoretical H1 NMR spectrum. C5H8O2, exhibits a strong IR transmittance at 1750 cm-1. Suggest a structure for compound A.

.

H1 NMR of Compound A 15. 2’-deoxyuridine, compound B, has the following H1

1 NMR spectrum:

144

Assign as many signals as you can to the appropriate protons in 2’-deoxuridine. 16. Determine structures for the products from the following

pathway:

O

N

N

H

HH

O

H

H

CH2OH

HH

H

HO

1

234

5

1'

2' 3'

4'

5'

6

Compound B

145

Following is the H1

1 NMR of B:

H1 NMR of Compound B

(CH3)3CCH=CHC(CH3)3

B2H6

C10H23B (infrared spectrum at 2500 cm-1)A

(1) 30% H2O2

(2) H2O

C10H22O

B

146

17. Suggest structures for molecules exhibiting the following spectra. Give rationales for your answers by explaining the chemical shifts, spin-spin interactions and, if provided, the electrical integration for each spectrum.

(a) Determine the structural formula for C7H7ClO; the

compound exhibits the following spectra:

H1 NMR for C7H7ClO

13C NMR for C7H7ClO

147

(b) Determine the structural formula for C9H9ClO; the

compound exhibits the following spectra:

H1 NMR for C9H9ClO

13C NMR for C9H9ClO

148

(c) Determine the structural formula for C12H14O4 , the compound exhibits the following spectra:

H1 NMR for C12H14O4

13C NMR for C12H14O4

149

(d) Determine the structural formula for C5H9O4N , the compound exhibits the following spectra:

H1 NMR for C5H9O4N

13C NMR for C5H9O4N

150

(e) Determine the structural formula for C6H14O, the compound exhibits the following spectra:

H1 NMR for C6H14O

13C NMR for C6H14O

(f) Determine the structural formula for C7H12O4, the compound

151

exhibits the following proton NMR and C-13 spectra:

H1 NMR for C7H12O4

13C NMR for C7H12O4

(g) Determine the structural formula for C4H10O, the compound exhibits the following spectra:

152

H1 NMR for C4H10O

13C NMR for C4H10O

153

(h) Determine the structural formula for C11H9NO4, (challenging) the compound exhibits the following spectra:

Characteristic infrared absorptions at about

2222 cm-1

3333 cm-1

1751 cm-1.

18. Determine the structure of C7H8S if the compound exhibits the following spectra data:

TMS

δ δδ δ δ0

4

11

3

1.72.55.78.2

154

H1 NMR for C7H8S

13C NMR for C7H8S

Partial mass spectrum of C7H8S

155

19. Determine the structure of C9H10 if the compound exhibits the following proton magnetic resonance and carbon-13 magnetic resonance spectra.

156

20. Determine the structure of C12H14O4 if the compound exhibits the following the following proton magnetic spectrum:

21. Suggest a reasonable structure for C8H10 from its 13C NMR spectrum, its 13C NMR DEPT-45 spectrum, its 13C NMR DEPT-90 spectrum, and its 13C NMR DEPT-135 spectrum.

157

The 13C NMR spectrum for C8H10

13C NMR DEPT-45 spectrum for C8H10

158

13C NMR DEPT-90 spectrum for C8H10

13C NMR DEPT-135 spectrum for C8H10