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Organic Chemistry 6th Edition Paula Yurkanis Bruice Chapter 13 Mass Spectrometry, Infrared Spectroscopy, and Ultraviolet/Visible Spectroscopy 1 © 2011 Pearson Education, Inc. Spectrally Identifiable Functional Groups 2 © 2011 Pearson Education, Inc. 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 3 © 2011 Pearson Education, Inc. 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.). M … M+1 Typically M and the isotope peaks are the highest masses in the spectrum Exception: a compound whose molecular ion completely fragments • 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. 4 © 2011 Pearson Education, Inc. • Fragment masses and intensities together provide structural information. • 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. For example: M-29 Base peak M-15 M … • 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). 5 © 2011 Pearson Education, Inc. The Mass Spectrum of Pentane Note weak m/z = 57 peak, primary butyl carbocation 6 © 2011 Pearson Education, Inc. The base peak of m/z 43 in the mass spectrum of pentane indicates the preference for C-2 to C-3 fragmentation: All fragments originate from the molecular ion 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 7 © 2011 Pearson Education, Inc. The Mass Spectrum of Isopentane Note strong m/z = 57 peak, secondary butyl carbocation 8 © 2011 Pearson Education, Inc. 2-Methylbutane is more likely than pentane to lose a methyl radical because a secondary carbocation can be formed: 9 © 2011 Pearson Education, Inc. Two-Fragment Loss from the Molecular Ion What are the structures of m/z 42 and 41? These ions arise from loss of the ethyl radical and either hydrogen atom or H2 from the pentane molecular ion: Note: All fragments originate from the molecular ion. Exception: Tandem Mass spectrometry where fragments of fragments are observed. 10 © 2011 Pearson Education, Inc. Isotopes in Mass Spectrometry • M + 1 peak: a contribution from 2H or 13C. • M + 2 peak: a contribution from 18O or from two heavy isotopes (2H or 13C) 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. 11 © 2011 Pearson Education, Inc. Fragmentation Patterns of Alkyl Halides 79Br 81Br 12 © 2011 Pearson Education, Inc. The Mass Spectrum of 2-Chloropropane 35Cl 37Cl 35Cl 37Cl 13 © 2011 Pearson Education, Inc. a-Cleavage results from the homolytic cleavage of a C—C bond at the a carbon: 14 © 2011 Pearson Education, Inc. a-Cleavage occurs because the C—Cl and C—C bonds have similar strengths, and the species that is formed is a relatively stable cation: a-Cleavage is less likely to occur in alkyl bromide because C—C bond is stronger than C—Br bond 15 © 2011 Pearson Education, Inc. Fragmentation Patterns of Ethers 16 © 2011 Pearson Education, Inc. A C—O bond is cleaved heterolytically, with the electrons going to the more electronegative atom: 17 © 2011 Pearson Education, Inc. A C—C bond is cleaved homolytically at an a-position because it leads to a relatively stable cation: 18 © 2011 Pearson Education, Inc. Fragmentation Patterns of Alcohols Because they fragment, molecular ions obtained from alcohols usually are not observed © 2011 Pearson Education, Inc. 19 Like alkyl halides and ethers, alcohols undergo a-cleavage: In alcohols, loss of water results in a fragmentation peak at m/z = M-18: 20 © 2011 Pearson Education, Inc. 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 21 © 2011 Pearson Education, Inc. Fragmentation Patterns of Ketones An intense molecular ion peak: 22 © 2011 Pearson Education, Inc. McLafferty rearrangement may occur: 23 © 2011 Pearson Education, Inc. Spectroscopy and the Electromagnetic Spectrum Spectroscopy is the study of the interaction of matter and electromagnetic radiation 24 © 2011 Pearson Education, Inc. Electromagnetic radiation has wave-like properties High frequencies and short wavelengths are associated with high energy © 2011 Pearson Education, Inc. 25 Vibrational Transitions Observed in IR Spectroscopy Functional groups stretch at different frequencies, and IR spectroscopy is used to identify functional groups 26 © 2011 Pearson Education, Inc. Infrared transitions require a bond dipole to occur: C H h C H Higher Energy Vibrational State 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 27 © 2011 Pearson Education, Inc. Influence of symmetry on IR activity of the alkene stretch: 1-butene — infrared active 2,3-dimethyl-2-butene — infrared inactive 2,3-dimethyl-2-heptene — infrared active, but very weak absorption band 28 © 2011 Pearson Education, Inc. The Vibrating Bond as a Quantized Harmonic Oscillator Quantum levels for a stretching vibration: Ball-and-spring model: Fundamental transition: o 1 Overtone: o 2 Overtones are twice the frequency of the fundamental transition and are always weak 29 © 2011 Pearson Education, Inc. The approximate wavenumber of an absorption can be calculated from Hooke’s law: 1 v 2c K M 1M 2 Reduced Mass M1 M 2 = wavenumber c = speed of light K = force constant M1 and M2 = masses of atoms 30 © 2011 Pearson Education, Inc. 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 sp2 sp3 sp2 31 © 2011 Pearson Education, Inc. Note the influence of mass and s character on stretching frequency: 32 © 2011 Pearson Education, Inc. An Infrared Spectrum The functional group region (4000–1400 cm–1) High energy The fingerprint region (1400–600 cm–1) Low energy 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 33 © 2011 Pearson Education, Inc. Functional group regions: Both compounds are alcohols Fingerprint regions: Compounds are different alcohols 34 © 2011 Pearson Education, Inc. The exact position of the absorption band depends on electron delocalization, the electronic effect of neighboring substituents, and hydrogen bonding: 35 © 2011 Pearson Education, Inc. Carbonyl overtone Esters have a carbonyl and a C—O stretch Ketones have only a carbonyl stretch 36 © 2011 Pearson Education, Inc. 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 nitrogen of an amide is electron donation by resonance The predominant effect of the oxygen of an ester is inductive electron withdrawal © 2011 Pearson Education, Inc. 37 38 © 2011 Pearson Education, Inc. The position of a C—O absorption varies because of resonance release in acids and esters: ~1050 cm–1 ~1050 cm–1 ~1250 cm–1 ~1250 cm–1 and 1050 cm–1 39 © 2011 Pearson Education, Inc. Acids are readily distinguished from alcohols Higher-frequency C─O stretch Broad OH stretch C═O stretch 40 © 2011 Pearson Education, Inc. 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 41 © 2011 Pearson Education, Inc. The strength of a C—H bond depends on the hybridization of the carbon 42 © 2011 Pearson Education, Inc. Examine the absorption bands in the vicinity of 3000 cm–1 43 © 2011 Pearson Education, Inc. Benzene in-plane and out-of-plane C—H bends Benzene ring: • Sharp absorption bands at ~1600 cm–1 and 1500–1430 cm–1. • Overtones at 1700–1900 cm–1 for the in-plane and out-ofplane benzene C—H bends. • The benzene overtones in the diagnostic region are readily recognized. 44 © 2011 Pearson Education, Inc. 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 45 © 2011 Pearson Education, Inc. Identifying a functional group by the bending vibrations: • 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. 46 © 2011 Pearson Education, Inc. Analyzing Infrared Spectra The position, intensity, and shape of an absorption band are helpful in identifying functional groups 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 47 © 2011 Pearson Education, Inc. wavenumber (cm–1) assignment 3075 2950 1650 and 890 absence of bands 1500–1430 and 720 sp2 CH sp3 CH a terminal alkene with two substituents has less than four adjacent CH2 groups 48 © 2011 Pearson Education, Inc. wavenumber (cm–1) 3050 2810 and 2730 1600 and 1460 1700 assignment sp2 CH an aldehyde benzene ring a partial single-bond character carbonyl 49 © 2011 Pearson Education, Inc. wavenumber (cm–1) 3300 2950 2100 assignment OH group sp3 CH alkyne 50 © 2011 Pearson Education, Inc. wavenumber (cm–1) 3300 2950 1660 1560 assignment N—H sp3 CH amide carbonyl N—H Bend 51 © 2011 Pearson Education, Inc. wavenumber (cm–1) assignment >3000 <3000 1605 and 1500 1720 1380 sp2 CH sp3 CH a benzene ring a ketone carbonyl a methyl group 52 © 2011 Pearson Education, Inc. 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 53 © 2011 Pearson Education, Inc. UV and Vis light cause only two kinds of electronic transition: Symmetry: allowed transition Forbidden transition: lone pair orthogonal to system • 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. © 2011 Pearson Education, Inc. 54 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 55 © 2011 Pearson Education, Inc. The Beer–Lambert Law e = ~10,000 M–1cm–1, A = ec l Allowed e = <100 M–1cm–1, Forbidden A = log(I0/I) c = concentration of substance in solution l = length of the cell in cm e = molar absorptivity, a measure of the probability of the transition The molar absorptivity of a compound is a constant that is characteristic of the compound at a particular wavelength 56 © 2011 Pearson Education, Inc. Effect of Conjugation on lmax The lmax and e values increase as the number of conjugated double bonds increases 57 © 2011 Pearson Education, Inc. 58 © 2011 Pearson Education, Inc. 59 © 2011 Pearson Education, Inc. If a compound has enough conjugated double bonds, it will absorb visible light (lmax >400 nm), and the compound will be colored 60 © 2011 Pearson Education, Inc. An auxochrome is a substituent in a chromophore that alters the lmax and the intensity of the absorption: 61 © 2011 Pearson Education, Inc. The Visible Spectrum and Color 62 © 2011 Pearson Education, Inc. Uses of UV/Vis Spectroscopy • Measure the rates of a reaction • Determine the pKa of a compound • Estimate the nucleotide composition of DNA 63 © 2011 Pearson Education, Inc.