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
SPECTROSCOPY 325 13.12 (b) The two methyl carbons of the isopropyl group are equivalent CH3 Four different types of carbons occur in the aromatic ring and two different types are present in the isopropyl group. TheC NMR spectrum of isopropylbenzene contains six signals (c) The methyl substituent at C-2 is different from those at C-I and C-3 The four nonequivalent ring carbons and the two different types of methyl carbons give rise to a"C NMR spectrum that contains six signals (d) The three methyl carbons of 1, 2, 4-trimethylbenzene are different from one another: C H,C Also, all the ring carbons are different from each other The nine different carbons give rise to nine separate signal (e) All three methyl carbons of 1, 3, 5-trimethy benzene are equivalet Because of its high symmetry 1, 3, 5-trimethylbenzene has only three signals in itsCNMR 13 13 sp-Hybridized carbons are more shielded than sp-hybridized ones. Carbon x is the most shielded, and has a chemical shift of 8 20 ppm. The oxygen of the OCH3 group decreased the shielding of carbon z: its chemical shift is 8 55 ppm. The least shielded is carbon y with a chemical shift of 20 H H OCH 13.14 The C NMR spectrum in Figure 13 22 shows nine signals and is the spectrum of benzene from part(d)of Problem 13.12. Six of the signals, in the range 8 127-138 ppm, are due to Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
13.12 (b) The two methyl carbons of the isopropyl group are equivalent. Four different types of carbons occur in the aromatic ring and two different types are present in the isopropyl group. The 13C NMR spectrum of isopropylbenzene contains six signals. (c) The methyl substituent at C-2 is different from those at C-1 and C-3: The four nonequivalent ring carbons and the two different types of methyl carbons give rise to a 13C NMR spectrum that contains six signals. (d) The three methyl carbons of 1,2,4-trimethylbenzene are different from one another: Also, all the ring carbons are different from each other. The nine different carbons give rise to nine separate signals. (e) All three methyl carbons of 1,3,5-trimethylbenzene are equivalent. Because of its high symmetry 1,3,5-trimethylbenzene has only three signals in its 13C NMR spectrum. 13.13 sp3 -Hybridized carbons are more shielded than sp2 -hybridized ones. Carbon x is the most shielded, and has a chemical shift of 20 ppm. The oxygen of the OCH3 group decreased the shielding of carbon z; its chemical shift is 55 ppm. The least shielded is carbon y with a chemical shift of 157 ppm. 13.14 The 13C NMR spectrum in Figure 13.22 shows nine signals and is the spectrum of 1,2,4-trimethylbenzene from part (d) of Problem 13.12. Six of the signals, in the range 127–138 ppm, are due to H3C OCH3 H H H H 20 ppm 55 ppm 157 ppm y z y z y H z 3C x CH3 x CH3 x H3C CH3 H3C CH3 x y x y w z n CH3 m CH3 m CH CH3 CH3 x y x y w z m n n SPECTROSCOPY 325 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
326 PECTROSCOPY the six nonequivalent carbons of the benzene ring. The three signals near 8 20 ppm are due to the three nonequivalent methyl groups HaC H3 1, 2, 4-Trimethylbenzene 13.15 The infrared spectrum of Figure 13.31 has no absorption in the 1600-1800-cm region, and so the unknown compound cannot contain a carbonyl(C=O) group. It cannot therefore be acetophenone or benzoic acid The broad, intense absorption at 3300 cm is attributable to a hydroxyl group. Although both phenol and benzyl alcohol are possibilities, the peaks at 2800-2900 cm reveal the presence of hy- drogen bonded to sp'-hybridized carbon. All carbons are sp-hybridized in phenol. The infrared spectrum is that of benzyl a 13.16 The energy of electromagnetic radiation is inversely proportional to its wavelength. Since excitation of an electron for the T-T*transition of ethylene occurs at a shorter wavelength(max=170 nm) than that of cis, trans-1,3-cyclooctadiene(max= 230 nm), the HOMO-LUMO energy difference in 13.17 Conjugation shifts Amax to longer wavelengths in alkenes. The conjugated diene 2-methyl-1, 3- butadiene has the longest wavelength absorption, Amax= 222 nm. The isolated diene 1, 4-pentadiene and the simple alkene cyclopentene both absorb below 200 nm. 2-Methyl-13-butadiene 13.18(b) The distribution of molecular-ion peaks in o-dichlorobenzene is identical to that in the para isomer. As the sample solution to part(a) in the text describes, peaks at m/z 146, 148, and 150 present for the molecular ion (c) The two isotopes of bromine are"Br and"Br. When both bromines of p-dibromobenzene are 7Br, the molecular ion appears at m/z 234. When one is"Br and the other is Br, m/zfor the molecular ion is 236. When both bromines are Br, m/z for the molecular ion is 238 (d) The combinations ofCl, 3Cl, "Br, andR in p-bromochlorobenzene and the values of m/ z for the corresponding molecular ion are as shown. (CL, Br) m/ (Cl, 7Br)or(Cl, Br) m/z=192 (3C.,8Br)m/z=194 13 19 The base peak in the mass spectrum of alkylbenzenes corresponds to carbon-carbon bond cleavage at the benzylic carbo HC、ACH2 子 CH CH Base peak: CsHg Base peak: CH, Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
the six nonequivalent carbons of the benzene ring. The three signals near 20 ppm are due to the three nonequivalent methyl groups. 13.15 The infrared spectrum of Figure 13.31 has no absorption in the 1600–1800-cm1 region, and so the unknown compound cannot contain a carbonyl (C?O) group. It cannot therefore be acetophenone or benzoic acid. The broad, intense absorption at 3300 cm1 is attributable to a hydroxyl group. Although both phenol and benzyl alcohol are possibilities, the peaks at 2800–2900 cm1 reveal the presence of hydrogen bonded to sp3 -hybridized carbon. All carbons are sp2 -hybridized in phenol. The infrared spectrum is that of benzyl alcohol. 13.16 The energy of electromagnetic radiation is inversely proportional to its wavelength. Since excitation of an electron for the A * transition of ethylene occurs at a shorter wavelength (max 170 nm) than that of cis, trans-1,3-cyclooctadiene (max 230 nm), the HOMO–LUMO energy difference in ethylene is greater. 13.17 Conjugation shifts max to longer wavelengths in alkenes. The conjugated diene 2-methyl-1,3- butadiene has the longest wavelength absorption, max 222 nm. The isolated diene 1,4-pentadiene and the simple alkene cyclopentene both absorb below 200 nm. 13.18 (b) The distribution of molecular-ion peaks in o-dichlorobenzene is identical to that in the para isomer. As the sample solution to part (a) in the text describes, peaks at mz 146, 148, and 150 are present for the molecular ion. (c) The two isotopes of bromine are 79Br and 81Br. When both bromines of p-dibromobenzene are 79Br, the molecular ion appears at mz 234. When one is 79Br and the other is 81Br, mz for the molecular ion is 236. When both bromines are 81Br, mz for the molecular ion is 238. (d) The combinations of 35Cl, 37Cl, 79Br, and 81Br in p-bromochlorobenzene and the values of mz for the corresponding molecular ion are as shown. ( 35Cl, 79Br) mz 190 ( 37Cl, 79Br) or (35Cl, 81Br) mz 192 ( 37Cl, 81Br) mz 194 13.19 The base peak in the mass spectrum of alkylbenzenes corresponds to carbon–carbon bond cleavage at the benzylic carbon. CH H3C CH3 CH3 Base peak: C9H11 m/z 119 CH2 CH2CH3 CH3 Base peak: C8H9 m/z 105 CH2 CH3 H3C CH3 Base peak: C9H11 m/z 119 2-Methyl-1,3-butadiene (max 222 nm) H3C CH3 CH3 1,2,4-Trimethylbenzene 326 SPECTROSCOPY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
SPECTROSCOPY 327 13.20 (b) The index of hydrogen deficiency is given by the following formula Index of hydrogen deficiency ==(CH,n+2-CH) The compound given contains eight carbons(CRH&); therefore, Index of hydrogen deficiency =2(C&Hi8-C8Hg) The proDonds(or one triple bond). Since the index of hydrogen deficiency is equal to 5,there blem specifies that the compound consumes 2 mol of hydrogen, and so it contains two must be three rings. (c) Chlorine substituents are equivalent to hydrogens when calculating the index of hydrogen deficiency. Therefore, consider CRHaCl, as equivalent to CHio. Thus, the index of hydrogen deficiency of this compound is 4. Index of hydrogen deficiency ==(C&Hi& -,o) Since the compound consumes 2 mol of hydrogen on catalytic hydrogenation, it must there- fore contain two rings (d) Oxygen atoms are ignored when calculating the index of hydrogen deficiency. Thus, CHOis treated as if it were CrH& Index of hydrogen deficiency =C& -CsHa Since the problem specifies that 2 mol of hydrogen is consumed on catalytic hydrogenation, this compound contains three rings (e) Ignoring the oxygen atoms in CHoo we treat this compound as if it were CHio- Index of hydrogen deficiency =(C8HI8-CgHio Because 2 mol of hydrogen is consumed on catalytic hydrogenation, there must be two rings (f) Ignore the oxygen, and treat the chlorine as if it were hydrogen. Thus, CHO CIO is treated as if it were C&Hio. Its index of hydrogen deficiency is 4, and it contains two rings. 13.21 Since each compound exhibits only a single peak in its H NMR spectrum, all the hydrogens are equivalent in each one. Structures are assigned on the basis of their molecular formulas and chemi sh (a) This compound has the molecular formula C&His and so must be an alkane. The 18 hydrogens re contributed by six equivalent methyl groups (CH3)CC(CH3)3 .,3, 3-Tetramethylbutane (b) A hydrocarbon with the molecular formula Cshio has an index of hydrogen deficiency of 1 and so is either a cycloalkane or an alkene. Since all ten hydrogens are equivalent, this com- und must be cyclopentane (61.5ppm) Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
13.20 (b) The index of hydrogen deficiency is given by the following formula: Index of hydrogen deficiency 1 2 (CnH2n2 CnHx) The compound given contains eight carbons (C8H8); therefore, Index of hydrogen deficiency 1 2 (C8H18 C8H8) 5 The problem specifies that the compound consumes 2 mol of hydrogen, and so it contains two double bonds (or one triple bond). Since the index of hydrogen deficiency is equal to 5, there must be three rings. (c) Chlorine substituents are equivalent to hydrogens when calculating the index of hydrogen deficiency. Therefore, consider C8H8Cl2 as equivalent to C8H10. Thus, the index of hydrogen deficiency of this compound is 4. Index of hydrogen deficiency 1 2 (C8H18 C8H10) 4 Since the compound consumes 2 mol of hydrogen on catalytic hydrogenation, it must therefore contain two rings. (d) Oxygen atoms are ignored when calculating the index of hydrogen deficiency. Thus, C8H8O is treated as if it were C8H8. Index of hydrogen deficiency 1 2 (C8H18 C8H8) 5 Since the problem specifies that 2 mol of hydrogen is consumed on catalytic hydrogenation, this compound contains three rings. (e) Ignoring the oxygen atoms in C8H10O2, we treat this compound as if it were C8H10. Index of hydrogen deficiency 1 2 (C8H18 C8H10) 4 Because 2 mol of hydrogen is consumed on catalytic hydrogenation, there must be two rings. ( f ) Ignore the oxygen, and treat the chlorine as if it were hydrogen. Thus, C8H9ClO is treated as if it were C8H10. Its index of hydrogen deficiency is 4, and it contains two rings. 13.21 Since each compound exhibits only a single peak in its 1 H NMR spectrum, all the hydrogens are equivalent in each one. Structures are assigned on the basis of their molecular formulas and chemical shifts. (a) This compound has the molecular formula C8H18 and so must be an alkane. The 18 hydrogens are contributed by six equivalent methyl groups. (b) A hydrocarbon with the molecular formula C5H10 has an index of hydrogen deficiency of 1 and so is either a cycloalkane or an alkene. Since all ten hydrogens are equivalent, this compound must be cyclopentane. Cyclopentane ( 1.5 ppm) (CH3)3CC(CH3)3 2,2,3,3-Tetramethylbutane ( 0.9 ppm) SPECTROSCOPY 327 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
328 PECTROSCOPY (c) The chemical shift of the eight equivalent hydrogens in C&Hs is 8 5.8 ppm, which is consistent with protons attached to a carbon-carbon double bond 1,3, 5, 7-Cyclooctatetraene (658ppm) (d) The compound CAHg Br has no rings or double bonds. The nine hydrogens belong to three equivalent methyl groups. (CH),CBI r-Butyl bromide(8 1. 8 ppm) (e) The dichloride has no rings or double bonds (index of hydrogen deficiency =0). The four equivalent hydrogens are present as two-CH, CI groups CICH, CH,CI 1. 2-Dichloroethane(8.7 ppm) (f) All three hydrogens in C2H3Cl3 must be part of the same methyl group in order to be CH3CCl3 I, 1, 1-Trichloroethane(8 2.7 ppm) (g) This compound has no rings or double bonds. To have eight equivalent hydrogens it must have our equ valent methyl ene grou CICHCCHCI 1.3-Dichloro-2, 2-di(chloromethyDpropane (63.7ppm (h) A compound with a molecular formula of CnHi has an index of hydrogen deficiency of 4 A likely candidate for a compound with 18 equivalent hydrogens is one with six equivalent CHs groups. Thus, 6 of the 12 carbons belong to CH3 groups, and the other 6 have no hydr gens. The compound is hexamethy benzene. HC A chemical shift of 8 2.2 ppm is consistent with the fact that all of the protons are benzylic ovens (i) The molecular formula of C3H, Br, tells us that the compound has no double bonds and no gs. All six hydrogens are equivalent, indicating two equivalent methyl groups. The com- pound is 2,2-dibromopropane,(CH3)2,CBr2 Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
(c) The chemical shift of the eight equivalent hydrogens in C8H8 is 5.8 ppm, which is consistent with protons attached to a carbon–carbon double bond. (d) The compound C4H9Br has no rings or double bonds. The nine hydrogens belong to three equivalent methyl groups. (CH3)3CBr tert-Butyl bromide ( 1.8 ppm) (e) The dichloride has no rings or double bonds (index of hydrogen deficiency 0). The four equivalent hydrogens are present as two GCH2Cl groups. ClCH2CH2Cl 1,2-Dichloroethane ( 3.7 ppm) ( f ) All three hydrogens in C2H3Cl3 must be part of the same methyl group in order to be equivalent. CH3CCl3 1,1,1-Trichloroethane ( 2.7 ppm) (g) This compound has no rings or double bonds. To have eight equivalent hydrogens it must have four equivalent methylene groups. (h) A compound with a molecular formula of C12H18 has an index of hydrogen deficiency of 4. A likely candidate for a compound with 18 equivalent hydrogens is one with six equivalent CH3 groups. Thus, 6 of the 12 carbons belong to CH3 groups, and the other 6 have no hydrogens. The compound is hexamethylbenzene. A chemical shift of 2.2 ppm is consistent with the fact that all of the protons are benzylic hydrogens. (i) The molecular formula of C3H6Br2 tells us that the compound has no double bonds and no rings. All six hydrogens are equivalent, indicating two equivalent methyl groups. The compound is 2,2-dibromopropane, (CH3)2CBr2. CH3 CH3 CH3 H3C CH3 H3C ClCH2CCH2Cl CH2Cl CH2Cl 1,3-Dichloro-2,2-di(chloromethyl)propane ( 3.7 ppm) 1,3,5,7-Cyclooctatetraene ( 5.8 ppm) 328 SPECTROSCOPY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
SPECTROSCOPY 329 13.22 In each of the parts to this problem, nonequivalent protons must not be bonded to adjacent carbons, because we are told that the two signals in each case are singlets (a) ach signal corresponds to four protons, and so each must result from two equivalent CH groups. The four CH, groups account for four of the carbons of CBHg, leaving two carbons that bear no hydrogens. A molecular formula of C Hg corresponds to an index of hydrogen defi- ciency of 3. A compound consistent with these requirements is HC The signal at 85.6 ppm is consistent with that expected for the four vinylic protons. The sig- nal at 8 2.7 ppm corresponds to that for the allylic protons of the ring (b) The compound has a molecular formula of C HBr and therefore has no double bonds or rings. A9-proton singlet at8 1. 1 ppm indicates three equivalent methyl groups, and a 2-proton singlet at 8 3.3 ppm indicates a CH, Br group. The correct structure is(CH3)3CCH2 Br. (c) This compound (CH,O)has three equivalent CH3 groups, along with a fourth CH3 group that is somewhat less shielded. Its molecular formula indicates that it can have either one double bond or one ring. This compound is( CH3),CCCH (d) A molecular formula of C6H1oo2 corresponds to an index of hydrogen deficiency of 2. The signal at 8 2.2 ppm(6H)is likely due to two equivalent CH3 groups, and the one at 8 2.7 ppm (4H)to two equivalent CH, groups. The compound is CH,CCH, CH, CCH 13.23 (a) A 5-proton signal at87. 1 ppm indicates a monosubstituted aromatic ring With an index of hydrogen deficiency of 4, CHio contains this monosubstituted aromatic ring and no other rings or multiple bonds. The triplet-quartet pattern at high field suggests an ethyl group. CH3 8 2.6 ppm 8 1.2 ppm Ethylbenzene (b) The index of hydrogen deficiency of 4 and the 5-proton multiplet at 87.0 to 7.5 ppm are accommodated by a monosubstituted aromatic ring. The remaining four carbons and nine hydrogens are most reasonably a tert-butyl group, since all nine hydrogens are equivalent. C(CH3)3 Singlet: 8 1.3 ppm (c) Its molecular formula requires that C6Hia be an alkane. The doublet-septet pattern is consis- ent with an isopropyl group, and the total number of protons requires that two of these groups esel ( CH3)2 CHCH(CH3)2 Doublet Septet 60.8ppm81.4pm Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
13.22 In each of the parts to this problem, nonequivalent protons must not be bonded to adjacent carbons, because we are told that the two signals in each case are singlets. (a) Each signal corresponds to four protons, and so each must result from two equivalent CH2 groups. The four CH2 groups account for four of the carbons of C6H8, leaving two carbons that bear no hydrogens. A molecular formula of C6H8 corresponds to an index of hydrogen defi- ciency of 3. A compound consistent with these requirements is The signal at 5.6 ppm is consistent with that expected for the four vinylic protons. The signal at 2.7 ppm corresponds to that for the allylic protons of the ring. (b) The compound has a molecular formula of C5H11Br and therefore has no double bonds or rings. A 9-proton singlet at 1.1 ppm indicates three equivalent methyl groups, and a 2-proton singlet at 3.3 ppm indicates a CH2Br group. The correct structure is (CH3)3CCH2Br. (c) This compound (C6H12O) has three equivalent CH3 groups, along with a fourth CH3 group that is somewhat less shielded. Its molecular formula indicates that it can have either one double bond or one ring. This compound is . (d) A molecular formula of C6H10O2 corresponds to an index of hydrogen deficiency of 2. The signal at 2.2 ppm (6H) is likely due to two equivalent CH3 groups, and the one at 2.7 ppm (4H) to two equivalent CH2 groups. The compound is . 13.23 (a) A 5-proton signal at 7.1 ppm indicates a monosubstituted aromatic ring. With an index of hydrogen deficiency of 4, C8H10 contains this monosubstituted aromatic ring and no other rings or multiple bonds. The triplet–quartet pattern at high field suggests an ethyl group. (b) The index of hydrogen deficiency of 4 and the 5-proton multiplet at 7.0 to 7.5 ppm are accommodated by a monosubstituted aromatic ring. The remaining four carbons and nine hydrogens are most reasonably a tert-butyl group, since all nine hydrogens are equivalent. (c) Its molecular formula requires that C6H14 be an alkane. The doublet–septet pattern is consistent with an isopropyl group, and the total number of protons requires that two of these groups be present. (CH3)2CHCH(CH3)2 Doublet 0.8 ppm Septet 1.4 ppm 2,3-Dimethylbutane C(CH3)3 Singlet; 1.3 ppm tert-Butylbenzene CH2CH3 Quartet 2.6 ppm (Benzylic) Triplet 1.2 ppm Ethylbenzene CH3CCH2CH2CCH3 O O (CH3)3CCCH3 O H2C CH2 SPECTROSCOPY 329 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
