CHAPTER 13 SPECTROSCOPY SOLUTIONS TO TEXT PROBLEMS 13.1 The field strength of an NMR spectrometer magnet and the frequency of electromagnetic radia- tion used to observe an NMR spectrum are directly proportional. Thus, the ratio 4.7 T/200 MHz is the same as 1.41 T/60 MHz. The magnetic field strength of a 60-MHz NMR spectrometer is 141T 13. 2 The ratio of H and C resonance frequencies remains constant. When the H frequency is 200 MHz, C NMR spectra are recorded at 50.4 MHz. Thus, when the H frequency is 100 MHz, C NMR spectra will be observed at 25. 2 MHz 13.3 (a) Chemical shifts reported in parts per million(ppm) are independent of the field strength of the NMR spectrometer. Thus, to compare the"H NMR signal of bromoform(CHBr3) recorded at 300 MHz with that of chloroform( CHCI)recorded at 200 MHz as given in the text, the chem- ical shift of bromoform must be converted from hertz to parts per million. The chemical shift for the proton in bromoform is 2065Hz 300 MHZ 6.88 ppm (b) The chemical shift of the proton in bromoform(8 6.88 ppm)is less than that of chloroform (8 7.28 ppm). The proton signal of bromoform is farther upfield and thus is more shielded 13. 4 In both chloroform(CHCI3) and 1, 1, 1-trichloroethane(CH,,) three chlorines are present. In CH-CCI3, however, the protons are one carbon removed from the chlorines, and thus the deshield- ing effect of the halogens will be less. The H NMR signal of CH,CCI3 appears 4.6 ppm upfield from the proton signal of chloroform. The chemical shift of the protons in CH_CCl3 is 8 2.6 ppm. 13.5 1, 4-Dimethylbenzene has two types of protons: those attached directly to the benzene ring and those of the methyl groups. Aryl protons are significantly less shielded than alkyl protons. As shown in text Table 13. 1 they are expected to give signals in the chemical shift range 8 6.5-8.5 ppm. Thus, the 320 Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
CHAPTER 13 SPECTROSCOPY SOLUTIONS TO TEXT PROBLEMS 13.1 The field strength of an NMR spectrometer magnet and the frequency of electromagnetic radiation used to observe an NMR spectrum are directly proportional. Thus, the ratio 4.7 T200 MHz is the same as 1.41 T60 MHz. The magnetic field strength of a 60-MHz NMR spectrometer is 1.41 T. 13.2 The ratio of 1 H and 13C resonance frequencies remains constant. When the 1 H frequency is 200 MHz, 13C NMR spectra are recorded at 50.4 MHz. Thus, when the 1 H frequency is 100 MHz, 13C NMR spectra will be observed at 25.2 MHz. 13.3 (a) Chemical shifts reported in parts per million (ppm) are independent of the field strength of the NMR spectrometer. Thus, to compare the 1 H NMR signal of bromoform (CHBr3) recorded at 300 MHz with that of chloroform (CHCl3) recorded at 200 MHz as given in the text, the chemical shift of bromoform must be converted from hertz to parts per million. The chemical shift for the proton in bromoform is 6.88 ppm (b) The chemical shift of the proton in bromoform ( 6.88 ppm) is less than that of chloroform ( 7.28 ppm). The proton signal of bromoform is farther upfield and thus is more shielded than the proton in chloroform. 13.4 In both chloroform (CHCl3) and 1,1,1-trichloroethane (CH3CCl3) three chlorines are present. In CH3CCl3, however, the protons are one carbon removed from the chlorines, and thus the deshielding effect of the halogens will be less. The 1 H NMR signal of CH3CCl3 appears 4.6 ppm upfield from the proton signal of chloroform. The chemical shift of the protons in CH3CCl3 is 2.6 ppm. 13.5 1,4-Dimethylbenzene has two types of protons: those attached directly to the benzene ring and those of the methyl groups. Aryl protons are significantly less shielded than alkyl protons. As shown in text Table 13.1 they are expected to give signals in the chemical shift range 6.5–8.5 ppm. Thus, the 2065 Hz 300 MHz 320 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
SPECTROSCOPY 321 signal at 87.0 ppm is due to the protons of the benzene ring. The signal at 8 2.2 ppm is due to the methyl protons C t70ppm 13.6 (b) Four nonequivalent sets of protons are bonded to carbon in 1-butanol as well as a fifth distinct pe of proton, the one bonded to oxygen. There should be five signals in the"H NMR spec trum of 1-butanol CHCH CHCH.OH (c) Apply the"proton replacement "test to butane CHa CH CH, CH3 CICH,CH,CH,CH3 CH; CHCH3CH3 CHICH, CHCH3 CH,CH,CH,,CI 2-Chlorobutane a Butane has two different types of protons; it will exhibit two signals in its'HNMR spectrum Like butane, 1, 4-dibromobutane has two different types of protons. This can be illustrated by using a chlorine atom as a test group BrCH,Ch,Ch, Ch,Br BrCHCH, CH,CH, Br BrCH, CHCH,CH, Br BrCH, CH, CHCH, Br BrCH,CH, CH, CHBr 1. 4-Dibromo-l-chlorobutane 14-Dibromo-2-chlorobutane 14-Dibromo-2-chlorobutane 14-Dibromo-l-chlorobutane The H NMR spectrum of 1, 4-dibromobutane is expected to consist of two sig (e) All the carbons in 2, 2-dibromobutane are different from each other, and so protons attached to carbon are not equivalent to the protons attached to any of the other carbons. This com- pound should have three signals in its H NMR spectrum. CH CCH.CH 2.2-Dibromobutane has three (f) All the protons in 2, 2, 3, 3-tetrabromobutane are equivalent. Its H NMR spectrum will consist of one signal CH2C—CCH3 2.2.3.3-Tetrabromobutane Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
signal at 7.0 ppm is due to the protons of the benzene ring. The signal at 2.2 ppm is due to the methyl protons. 13.6 (b) Four nonequivalent sets of protons are bonded to carbon in 1-butanol as well as a fifth distinct type of proton, the one bonded to oxygen. There should be five signals in the 1 H NMR spectrum of 1-butanol. (c) Apply the “proton replacement” test to butane. Butane has two different types of protons; it will exhibit two signals in its 1 H NMR spectrum. (d) Like butane, 1,4-dibromobutane has two different types of protons. This can be illustrated by using a chlorine atom as a test group. The 1 H NMR spectrum of 1,4-dibromobutane is expected to consist of two signals. (e) All the carbons in 2,2-dibromobutane are different from each other, and so protons attached to one carbon are not equivalent to the protons attached to any of the other carbons. This compound should have three signals in its 1 H NMR spectrum. ( f ) All the protons in 2,2,3,3-tetrabromobutane are equivalent. Its 1 H NMR spectrum will consist of one signal. CH3C Br Br Br Br 2,2,3,3-Tetrabromobutane CCH3 2,2-Dibromobutane has three nonequivalent sets of protons. CH3CCH2CH3 Br Br BrCHCH2CH2CH2Br Cl 1,4-Dibromo-1-chlorobutane BrCH2CH2CH2CH2Br 1,4-Dibromobutane 1,4-Dibromo-2-chlorobutane BrCH2CHCH2CH2Br Cl 1,4-Dibromo-2-chlorobutane BrCH2CH2CHCH2Br Cl 1,4-Dibromo-1-chlorobutane BrCH2CH2CH2CHBr Cl ClCH2CH2CH2CH3 1-Chlorobutane CH3CH2CH2CH3 Butane CH3CHCH3CH3 Cl 2-Chlorobutane CH3CH2CH2CH2Cl 1-Chlorobutane CH3CH2CHCH3 Cl 2-Chlorobutane CH3CH2CH2CH2OH Five different proton environments in 1-butanol; five signals H H H H H3C CH3 2.2 ppm 7.0 ppm SPECTROSCOPY 321 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
322 PECTROSCOPY (g) There are four nonequivalent sets of protons in 1, 1, 4-tribromobutane. It will exhibit four sig- nals in its H NMR spectrum BrCCH,CH,CH, Br (h) The seven protons of 1, 1, l-tribromobutane belong to three nonequivalent sets, and hence the H NMR spectrum will consist of three signals Br CCHCHCH L1. 1-Tribromobutane 13.7(b) Apply the replacement test to each of the protons of 1, l-dibromoethene. Br 1.1-Dibromoethene 11-Dibromo-2-chloroethene 11-Dibromo. 2-chloroethene Replacement of one proton by a test group( CD) gives exactly the same compound as replace ment of the other. The two protons of 1, l-dibromoethene are equivalent, and there is only one signal in the H NMR spectrum of this compound (c) The replacement test reveals that both protons of cis-1, 2-dibromoethene are equivalent Br B Br Br C=C cis-1.2-Dibromoethene Z)-1. 2-Dibromo-1-chloroethene (Zr-1, 2-Dibromo-l-chloroethene Because both protons are equivalent, the HNMR spectrum of cis-1, 2-dibromoethene consists (d) Both protons of trans-1, 2-dibromoethene are equivalent; each is cis to a bromine substituent trans-1.2-Dibromoethene (one signal in the H (e) Four nonequivalent sets of protons occur in allyl bromide H Allyl bromide(four signals the H NMR spectrum) Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
322 SPECTROSCOPY (g) There are four nonequivalent sets of protons in 1,1,4-tribromobutane. It will exhibit four signals in its 1 H NMR spectrum. (h) The seven protons of 1,1,1-tribromobutane belong to three nonequivalent sets, and hence the 1 H NMR spectrum will consist of three signals. 13.7 (b) Apply the replacement test to each of the protons of 1,1-dibromoethene. Replacement of one proton by a test group (Cl) gives exactly the same compound as replacement of the other. The two protons of 1,1-dibromoethene are equivalent, and there is only one signal in the 1 H NMR spectrum of this compound. (c) The replacement test reveals that both protons of cis-1,2-dibromoethene are equivalent. Because both protons are equivalent, the 1 H NMR spectrum of cis-1,2-dibromoethene consists of one signal. (d) Both protons of trans-1,2-dibromoethene are equivalent; each is cis to a bromine substituent. (e) Four nonequivalent sets of protons occur in allyl bromide. Allyl bromide (four signals in the 1 H NMR spectrum) C C H H CH2Br H trans-1,2-Dibromoethene (one signal in the 1 H NMR spectrum) C C H Br Br H cis-1,2-Dibromoethene C C H Br H Br (Z)-1,2-Dibromo-1-chloroethene C C Cl Br H Br (Z)-1,2-Dibromo-1-chloroethene C C H Br Cl Br 1,1-Dibromoethene C C Br Br H H 1,1-Dibromo-2-chloroethene C C Br Br H Cl 1,1-Dibromo-2-chloroethene C C Br Br Cl H 1,1,1-Tribromobutane Br3CCH2CH2CH3 BrCCH2CH2CH2Br Br H 1,1,4-Tribromobutane Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
SPECTROSCOPY 323 (f) The protons of a single methyl group are equivalent to one another, but all three methyl groups of 2-methyl-2-butene are nonequivalent. The vinyl proton is uniqu H,C H-C 2-Methyl-2-butene(four signals in the H NMR spectrum 13.8(b) The three methyl protons of 1, 1, 1-trichloroethane(Cl3CCH3) are equivalent. They have the same chemical shift and do not split each others signals. The H NMR spectrum of Cl,CCH3 consists of a single sharp peak (c) Separate signals will be seen for the methylene( CH,) protons and for the methine( Ch)pro- ton of 1.1. 2-trichloroethane Cl1C—CH,C The methine proton splits the signal for the methylene protons into a doublet. The two methylene protons split the methine protons signal into a triplet. (d) Examine the structure of 1, 2, 2-trichloropropane CICH, CCH3 1, 2, 2-Trichloropropane The"H NMR spectrum exhibits a signal for the two equivalent methylene protons and one for the three equivalent methyl protons. Both these signals are sharp singlets. The protons of the methyl group and the methylene group are separated by more than three bonds and do not split ach other's signals (e) The methine proton of 1, 1, 1, 2-tetrachloropropane splits the signal of the methyl protons into a doublet; its signal is split into a quartet by the three methyl protons CICC--CHaDoublet Quartet 1, 1, 1, 2-Tetrachloropropane 13.9(b) The ethyl group appears as a triplet- quartet pattern and the methyl group as a singlet CHaCH,OCH3 Singlet; not vicinal to any other protons in molecule Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
( f ) The protons of a single methyl group are equivalent to one another, but all three methyl groups of 2-methyl-2-butene are nonequivalent. The vinyl proton is unique. 13.8 (b) The three methyl protons of 1,1,1-trichloroethane (Cl3CCH3) are equivalent. They have the same chemical shift and do not split each other’s signals. The 1 H NMR spectrum of Cl3CCH3 consists of a single sharp peak. (c) Separate signals will be seen for the methylene (CH2) protons and for the methine (CH) proton of 1,1,2-trichloroethane. The methine proton splits the signal for the methylene protons into a doublet. The two methylene protons split the methine proton’s signal into a triplet. (d) Examine the structure of 1,2,2-trichloropropane. The 1 H NMR spectrum exhibits a signal for the two equivalent methylene protons and one for the three equivalent methyl protons. Both these signals are sharp singlets. The protons of the methyl group and the methylene group are separated by more than three bonds and do not split each other’s signals. (e) The methine proton of 1,1,1,2-tetrachloropropane splits the signal of the methyl protons into a doublet; its signal is split into a quartet by the three methyl protons. 13.9 (b) The ethyl group appears as a triplet–quartet pattern and the methyl group as a singlet. CH3CH2OCH3 Quartet Singlet; not vicinal to any other protons in molecule Triplet H Cl Cl3CC CH3 Doublet Quartet 1,1,1,2-Tetrachloropropane ClCH2CCH3 Cl Cl 1,2,2-Trichloropropane H Cl2C CH2Cl Triplet Doublet 1,1,2-Trichloroethane 2-Methyl-2-butene (four signals in the 1 H NMR spectrum) C C H3C CH3 H3C H SPECTROSCOPY 323 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
324 PECTROSCOPY (c) The two ethyl groups of diethyl ether are equivalent to each other. The two methyl groups appear as one triplet and the two methylene groups as one quartet. CHCH,OCH,CH (d) The two ethyl groups of p-diethy benzene are equivalent to each other and give rise to a single triplet-quartet pa CHa CH2- -CHCH CH le) split either each other's signals or any of the signals of the ethyl grule ical shift and do not of the Four nonequivalent sets of protons occur in this compound CICH,CH,OCH,CH Vicinal protons in the CICH, CH,O group split one another's signals, as do those in the CH3CH,O group. 13.10 Both H, and H in m-nitrostyrene appear as doublets of doublets. H is coupled to Ha by a coupling constant of 12 Hz and to H by a coupling constant of 2 Hz. H is coupled to H, by a coupling con- stant of 16 Hz and to H, by a coupling constant of 2 Hz 12 HZ 16 hz 2 HZ 2 HZ 2 HZ (diagrams not to scale) 13.11 (b) The signal of the proton at C-2 is split into a quartet by the methyl protons, and each line of this quartet is split into a doublet by the aldehyde proton. It appears as a doublet of quartets (Note: It does not matter whether the splitting pattern is described as a doublet of quartets or a quartet of doublets. There is no substantive difference in the two descriptions. Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(c) The two ethyl groups of diethyl ether are equivalent to each other. The two methyl groups appear as one triplet and the two methylene groups as one quartet. (d) The two ethyl groups of p-diethylbenzene are equivalent to each other and give rise to a single triplet–quartet pattern. All four protons of the aromatic ring are equivalent, have the same chemical shift, and do not split either each other’s signals or any of the signals of the ethyl group. (e) Four nonequivalent sets of protons occur in this compound: Vicinal protons in the ClCH2CH2O group split one another’s signals, as do those in the CH3CH2O group. 13.10 Both Hb and Hc in m-nitrostyrene appear as doublets of doublets. Hb is coupled to Ha by a coupling constant of 12 Hz and to Hc by a coupling constant of 2 Hz. Hc is coupled to Ha by a coupling constant of 16 Hz and to Hb by a coupling constant of 2 Hz. 13.11 (b) The signal of the proton at C-2 is split into a quartet by the methyl protons, and each line of this quartet is split into a doublet by the aldehyde proton. It appears as a doublet of quartets. (Note: It does not matter whether the splitting pattern is described as a doublet of quartets or a quartet of doublets. There is no substantive difference in the two descriptions.) H3C Br C CH O This proton splits the signal for the proton at C-2 into a doublet. These three protons split the signal for proton at C-2 into a quartet. H O2N C C Ha Hb Hc Hb 2 Hz 2 Hz 12 Hz Hc 2 Hz 2 Hz 16 Hz (diagrams not to scale) ClCH2CH2OCH2CH3 Triplet Triplet Quartet Triplet H H H H CH3CH2 CH2CH3 Three signals: CH3 triplet; CH2 quartet; aromatic H singlet CH3CH2OCH2CH3 Quartet Quartet Triplet Triplet 324 SPECTROSCOPY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website