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© 2011 Pearson Education, Inc. 1 Chapter 13 Mass Spectrometry, Infrared Spectroscopy, and Ultraviolet/Visible Spectroscopy Organic Chemistry 6 th Edition.

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Presentation on theme: "© 2011 Pearson Education, Inc. 1 Chapter 13 Mass Spectrometry, Infrared Spectroscopy, and Ultraviolet/Visible Spectroscopy Organic Chemistry 6 th Edition."— Presentation transcript:


2 © 2011 Pearson Education, Inc. 1 Chapter 13 Mass Spectrometry, Infrared Spectroscopy, and Ultraviolet/Visible Spectroscopy Organic Chemistry 6 th Edition Paula Yurkanis Bruice

3 © 2011 Pearson Education, Inc. 2 Spectrally Identifiable Functional Groups

4 © 2011 Pearson Education, Inc. 3 The Mass Spectrometer A mass spectrum records only positively charged fragments, either cations or radical cations m/z = mass-to-charge ratio of the fragment

5 © 2011 Pearson Education, Inc. 4 Information obtained from a mass spectrum: The molecular ion (M): measured to the nearest whole number or up to four decimal places (high-resolution mass spectrometry). Isotope peaks (M + 1, M + 2 etc.). Typically M and the isotope peaks are the highest masses in the spectrum The high-resolution mass of the molecular ion provides the molecular formula directly. The whole-number mass of the molecular ion and the relative intensities of M + 1, M + 2, etc., can also provide the molecular formula. … M + 1 M Exception: a compound whose molecular ion completely fragments

6 © 2011 Pearson Education, Inc. 5 … Fragment masses and intensities together provide structural information. M M-15 The base peak has the greatest intensity in the spectrum. Intense peaks correspond to relatively stable cationic and/or relatively stable radical species lost. The fragments lost also provide structural information. M-29 Base peak For example: Fragment m/z 57 resulted from the loss of methyl (m/z = 15) from the molecular ion. Given its intensity, m/z 57 must be the sec-butyl carbocation (not the primary butyl carbocation).

7 © 2011 Pearson Education, Inc. 6 The Mass Spectrum of Pentane Note weak m/z = 57 peak, primary butyl carbocation

8 © 2011 Pearson Education, Inc. 7 The base peak of m/z 43 in the mass spectrum of pentane indicates the preference for C-2 to C-3 fragmentation: The mass of the radical species lost in a fragmentation is the difference between the m/z values of the fragment ion and the molecular ion All fragments originate from the molecular ion

9 © 2011 Pearson Education, Inc. 8 Note strong m/z = 57 peak, secondary butyl carbocation The Mass Spectrum of Isopentane

10 © 2011 Pearson Education, Inc. 9 2-Methylbutane is more likely than pentane to lose a methyl radical because a secondary carbocation can be formed:

11 © 2011 Pearson Education, Inc. 10 What are the structures of m/z 42 and 41? These ions arise from loss of the ethyl radical and either hydrogen atom or H 2 from the pentane molecular ion: Note: All fragments originate from the molecular ion. Exception: Tandem Mass spectrometry where fragments of fragments are observed. Two-Fragment Loss from the Molecular Ion

12 © 2011 Pearson Education, Inc. 11 Isotopes in Mass Spectrometry M + 1 peak: a contribution from 2 H or 13 C. M + 2 peak: a contribution from 18 O or from two heavy isotopes ( 2 H or 13 C) in the same molecule. A large M + 2 peak suggests a compound containing either chlorine or bromine: a Cl if M + 2 is one-third the intensity of M; a Br if M + 2 is the same intensity as M. To calculate the molecular masses of molecular ions and fragments, the atom mass of a single isotope of an atom must be used.

13 © 2011 Pearson Education, Inc. 12 Fragmentation Patterns of Alkyl Halides 79 Br 81 Br

14 © 2011 Pearson Education, Inc. 13 The Mass Spectrum of 2-Chloropropane 35 Cl 37 Cl 35 Cl 37 Cl

15 © 2011 Pearson Education, Inc. 14  -Cleavage results from the homolytic cleavage of a C—C bond at the  carbon:

16 © 2011 Pearson Education, Inc. 15  -Cleavage occurs because the C—Cl and C—C bonds have similar strengths, and the species that is formed is a relatively stable cation:  -Cleavage is less likely to occur in alkyl bromide because C—C bond is stronger than C—Br bond

17 © 2011 Pearson Education, Inc. 16 Fragmentation Patterns of Ethers

18 © 2011 Pearson Education, Inc. 17 A C—O bond is cleaved heterolytically, with the electrons going to the more electronegative atom:

19 © 2011 Pearson Education, Inc. 18 A C—C bond is cleaved homolytically at an  -position because it leads to a relatively stable cation:

20 © 2011 Pearson Education, Inc. 19 Fragmentation Patterns of Alcohols Because they fragment, molecular ions obtained from alcohols usually are not observed

21 © 2011 Pearson Education, Inc. 20 Like alkyl halides and ethers, alcohols undergo  -cleavage: In alcohols, loss of water results in a fragmentation peak at m/z = M-18:

22 © 2011 Pearson Education, Inc. 21 Common Fragmentation Behavior in Alkyl Halides, Ethers, and Alcohols 1. A bond between carbon and a more electronegative atom breaks heterolytically 2. A bond between carbon and an atom of similar electronegativity breaks homolytically 3. The bonds most likely to break are the weakest bonds and those that lead to formation of the most stable cation

23 © 2011 Pearson Education, Inc. 22 Fragmentation Patterns of Ketones An intense molecular ion peak:

24 © 2011 Pearson Education, Inc. 23 McLafferty rearrangement may occur:

25 © 2011 Pearson Education, Inc. 24 Spectroscopy and the Electromagnetic Spectrum Spectroscopy is the study of the interaction of matter and electromagnetic radiation

26 © 2011 Pearson Education, Inc. 25 Electromagnetic radiation has wave-like properties High frequencies and short wavelengths are associated with high energy

27 © 2011 Pearson Education, Inc. 26 Vibrational Transitions Observed in IR Spectroscopy Functional groups stretch at different frequencies, and IR spectroscopy is used to identify functional groups

28 © 2011 Pearson Education, Inc. 27 Infrared transitions require a bond dipole to occur: The more polar the bond, the more intense the absorptions: The intensity of an absorption band also depends on the number of bonds responsible for the absorption

29 © 2011 Pearson Education, Inc butene — infrared active 2,3-dimethyl-2-butene — infrared inactive 2,3-dimethyl-2-heptene — infrared active, but very weak absorption band Influence of symmetry on IR activity of the alkene stretch:

30 © 2011 Pearson Education, Inc. 29 The Vibrating Bond as a Quantized Harmonic Oscillator Ball-and-spring model: Quantum levels for a stretching vibration: Fundamental transition: o  1 Overtone: o  2 Overtones are twice the frequency of the fundamental transition and are always weak

31 © 2011 Pearson Education, Inc. 30 The approximate wavenumber of an absorption can be calculated from Hooke’s law:  = wavenumber c = speed of light K = force constant M 1 and M 2 = masses of atoms

32 © 2011 Pearson Education, Inc. 31 Hooke’s law predicts that lighter atoms will vibrate at a higher frequency than heavy atoms: C—H~3000 cm –1 C—D~2200 cm –1 C—O~1100 cm –1 C—Cl~700 cm –1 Increasing the s character of a bond (higher K value) increases the stretching frequency: sp sp 2 sp 3

33 © 2011 Pearson Education, Inc. 32 Note the influence of mass and s character on stretching frequency:

34 © 2011 Pearson Education, Inc. 33 An Infrared Spectrum The functional group region (4000–1400 cm –1 ) The fingerprint region (1400–600 cm –1 ) The functional group, or diagnostic region, is used to determine the functional group present The fingerprint region is used for structure elucidation by spectral comparison High energyLow energy

35 © 2011 Pearson Education, Inc. 34 Functional group regions: Both compounds are alcohols Fingerprint regions: Compounds are different alcohols

36 © 2011 Pearson Education, Inc. 35 The exact position of the absorption band depends on electron delocalization, the electronic effect of neighboring substituents, and hydrogen bonding:

37 © 2011 Pearson Education, Inc. 36 Esters have a carbonyl and a C—O stretch Ketones have only a carbonyl stretch Carbonyl overtone

38 © 2011 Pearson Education, Inc. 37 Putting an atom other than carbon next to the carbonyl group causes the position of the carbonyl absorption band to shift: The predominant effect of the oxygen of an ester is inductive electron withdrawal The predominant effect of the nitrogen of an amide is electron donation by resonance

39 © 2011 Pearson Education, Inc. 38

40 © 2011 Pearson Education, Inc. 39 The position of a C—O absorption varies because of resonance release in acids and esters: ~1050 cm –1 ~1250 cm –1 ~1250 cm –1 and 1050 cm –1

41 © 2011 Pearson Education, Inc. 40 Acids are readily distinguished from alcohols Broad OH stretch C═O stretch Higher-frequency C─O stretch

42 © 2011 Pearson Education, Inc. 41 The position and the breadth of the O—H absorption band depend on the concentration of the solution It is easier to stretch an O—H bond if it is hydrogen bonded

43 © 2011 Pearson Education, Inc. 42 The strength of a C—H bond depends on the hybridization of the carbon

44 © 2011 Pearson Education, Inc. 43 Examine the absorption bands in the vicinity of 3000 cm –1

45 © 2011 Pearson Education, Inc. 44 Sharp absorption bands at ~1600 cm –1 and 1500–1430 cm –1. Overtones at 1700–1900 cm –1 for the in-plane and out-of- plane benzene C—H bends. The benzene overtones in the diagnostic region are readily recognized. Benzene ring: Benzene in-plane and out-of-plane C—H bends

46 © 2011 Pearson Education, Inc. 45 Stretch of C—H Bond in an Aldehyde The stretch of the C—H bond of an aldehyde shows one absorption band at ~2820 cm –1 and another one at ~2720 cm –1

47 © 2011 Pearson Education, Inc. 46 Primary amine: two N—H stretches at 3350 cm –1. Amine: N—H bend. “Isopropyl split” at 1380 cm –1 indicates the presence of an isopropyl group. Identifying a functional group by the bending vibrations:

48 © 2011 Pearson Education, Inc. 47 The absence of absorption bands can be useful in identifying a compound in IR spectroscopy Bonds in molecules lacking dipole moments will not be detected Analyzing Infrared Spectra The position, intensity, and shape of an absorption band are helpful in identifying functional groups

49 © 2011 Pearson Education, Inc. 48 wavenumber (cm –1 ) assignment and 890 absence of bands 1500–1430 and 720 sp 2 CH sp 3 CH a terminal alkene with two substituents has less than four adjacent CH 2 groups

50 © 2011 Pearson Education, Inc. 49 wavenumber (cm –1 ) assignment and and sp 2 CH an aldehyde benzene ring a partial single-bond character carbonyl

51 © 2011 Pearson Education, Inc. 50 wavenumber (cm –1 ) assignment OH group sp 3 CH alkyne

52 © 2011 Pearson Education, Inc. 51 wavenumber (cm –1 ) assignment N—H sp 3 CH amide carbonyl N—H Bend

53 © 2011 Pearson Education, Inc. 52 wavenumber (cm –1 ) assignment >3000 < and sp 2 CH sp 3 CH a benzene ring a ketone carbonyl a methyl group

54 © 2011 Pearson Education, Inc. 53 Ultraviolet and Visible Spectroscopy Spectroscopy is the study of the interaction between matter and electromagnetic radiation UV/Vis spectroscopy provides information about compounds with conjugated double bonds

55 © 2011 Pearson Education, Inc. 54 UV and Vis light cause only two kinds of electronic transition: Only organic compounds with  electrons can produce UV/Vis spectra. A visible spectrum is obtained if visible light is absorbed. A UV spectrum is obtained if UV light is absorbed. Forbidden transition: lone pair orthogonal to  system Symmetry: allowed transition

56 © 2011 Pearson Education, Inc. 55 A chromophore is the part of a molecule that absorbs UV or visible light Only compounds with  electrons can produce UV/Vis spectra Allowed Forbidden

57 © 2011 Pearson Education, Inc. 56 The Beer–Lambert Law The molar absorptivity of a compound is a constant that is characteristic of the compound at a particular wavelength A = log(I 0 /I) c = concentration of substance in solution l = length of the cell in cm  = molar absorptivity, a measure of the probability of the transition A =  c l  = ~10,000 M –1 cm –1, Allowed  = <100 M –1 cm –1, Forbidden

58 © 2011 Pearson Education, Inc. 57 Effect of Conjugation on max The max and  values increase as the number of conjugated double bonds increases

59 © 2011 Pearson Education, Inc. 58

60 © 2011 Pearson Education, Inc. 59

61 © 2011 Pearson Education, Inc. 60 If a compound has enough conjugated double bonds, it will absorb visible light ( max >400 nm), and the compound will be colored

62 © 2011 Pearson Education, Inc. 61 An auxochrome is a substituent in a chromophore that alters the max and the intensity of the absorption:

63 © 2011 Pearson Education, Inc. 62 The Visible Spectrum and Color

64 © 2011 Pearson Education, Inc. 63 Uses of UV/Vis Spectroscopy Measure the rates of a reaction Determine the pK a of a compound Estimate the nucleotide composition of DNA

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