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