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 TH20862200 MHz is the same as 1.41 TH2086260 MHz. The magnetic field strength of a 60-MHz NMR spectrometer is 1.41 T. 13.2 The ratio of 1 H and 13 C resonance frequencies remains constant. When the 1 H frequency is 200 MHz, 13 C NMR spectra are recorded at 50.4 MHz. Thus, when the 1 H frequency is 100 MHz, 13 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 1 H NMR signal of bromoform (CHBr 3 ) recorded at 300 MHz with that of chloroform (CHCl 3 ) 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 H9254 H11005H110056.88 ppm (b) The chemical shift of the proton in bromoform (H9254 6.88 ppm) is less than that of chloroform (H9254 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 (CHCl 3 ) and 1,1,1-trichloroethane (CH 3 CCl 3 ) three chlorines are present. In CH 3 CCl 3 , however, the protons are one carbon removed from the chlorines, and thus the deshield- ing effect of the halogens will be less. The 1 H NMR signal of CH 3 CCl 3 appears 4.6 ppm upfield from the proton signal of chloroform. The chemical shift of the protons in CH 3 CCl 3 is H9254 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 H9254 6.5–8.5 ppm. Thus, the 2065 Hz H5007 300 MHz 320 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website signal at H9254 7.0 ppm is due to the protons of the benzene ring. The signal at H9254 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 spec- trum 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 com- pound 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. CH 3 C Br Br Br Br 2,2,3,3-Tetrabromobutane CCH 3 2,2-Dibromobutane has three nonequivalent sets of protons. CH 3 CCH 2 CH 3 Br Br BrCHCH 2 CH 2 CH 2 Br Cl 1,4-Dibromo-1-chlorobutane BrCH 2 CH 2 CH 2 CH 2 Br 1,4-Dibromobutane 1,4-Dibromo-2-chlorobutane BrCH 2 CHCH 2 CH 2 Br Cl 1,4-Dibromo-2-chlorobutane BrCH 2 CH 2 CHCH 2 Br Cl 1,4-Dibromo-1-chlorobutane BrCH 2 CH 2 CH 2 CHBr Cl ClCH 2 CH 2 CH 2 CH 3 1-Chlorobutane CH 3 CH 2 CH 2 CH 3 Butane CH 3 CHCH 3 CH 3 Cl 2-Chlorobutane CH 3 CH 2 CH 2 CH 2 Cl 1-Chlorobutane CH 3 CH 2 CHCH 3 Cl 2-Chlorobutane CH 3 CH 2 CH 2 CH 2 OH Five different proton environments in 1-butanol; five signals HH HH CH 3 H 3 C 2.2 ppm 7.0 ppm SPECTROSCOPY 321 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 sig- nals 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 replace- ment 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) CC H CH 2 BrH H trans-1,2-Dibromoethene (one signal in the 1 H NMR spectrum) CC H Br Br H cis-1,2-Dibromoethene CC H Br H Br (Z)-1,2-Dibromo-1-chloroethene CC Cl Br H Br (Z)-1,2-Dibromo-1-chloroethene CC H Br Cl Br 1,1-Dibromoethene CC Br Br H H 1,1-Dibromo-2-chloroethene CC Br Br H Cl 1,1-Dibromo-2-chloroethene CC Br Br Cl H 1,1,1-Tribromobutane Br 3 CCH 2 CH 2 CH 3 BrCCH 2 CH 2 CH 2 Br Br H 1,1,4-Tribromobutane 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 (Cl 3 CCH 3 ) are equivalent. They have the same chemical shift and do not split each other’s signals. The 1 H NMR spectrum of Cl 3 CCH 3 consists of a single sharp peak. (c) Separate signals will be seen for the methylene (CH 2 ) protons and for the methine (CH) pro- ton 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. CH 3 CH 2 OCH 3 Quartet Singlet; not vicinal to any other protons in molecule Triplet H Cl Cl 3 CC CH 3 Doublet Quartet 1,1,1,2-Tetrachloropropane ClCH 2 CCH 3 Cl Cl 1,2,2-Trichloropropane H Cl 2 CCH 2 Cl DoubletTriplet 1,1,2-Trichloroethane 2-Methyl-2-butene (four signals in the 1 H NMR spectrum) CC CH 3 H 3 C H 3 C H SPECTROSCOPY 323 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 ClCH 2 CH 2 O group split one another’s signals, as do those in the CH 3 CH 2 O group. 13.10 Both H b and H c in m-nitrostyrene appear as doublets of doublets. H b is coupled to H a by a coupling constant of 12 Hz and to H c by a coupling constant of 2 Hz. H c is coupled to H a by a coupling con- stant of 16 Hz and to H b 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.) H 3 C Br CHC 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 O 2 N C C H a H b H c H b 2 Hz 2 Hz 12 Hz H c 2 Hz 2 Hz 16 Hz (diagrams not to scale) ClCH 2 CH 2 OCH 2 CH 3 Triplet Triplet Quartet Triplet HH H H CH 2 CH 3 CH 3 CH 2 Three signals: CH 3 triplet; CH 2 quartet; aromatic H singlet CH 3 CH 2 OCH 2 CH 3 Quartet Quartet Triplet Triplet 324 SPECTROSCOPY 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 13 C 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 13 C 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 13 C NMR spectrum. 13.13 sp 3 -Hybridized carbons are more shielded than sp 2 -hybridized ones. Carbon x is the most shielded, and has a chemical shift of H9254 20 ppm. The oxygen of the OCH 3 group decreased the shielding of carbon z; its chemical shift is H9254 55 ppm. The least shielded is carbon y with a chemical shift of H9254 157 ppm. 13.14 The 13 C NMR spectrum in Figure 13.22 shows nine signals and is the spectrum of 1,2,4-trimethyl- benzene from part (d) of Problem 13.12. Six of the signals, in the range H9254 127–138 ppm, are due to OCH 3 H 3 C H H H H 20 ppm 55 ppm 157 ppm y z y z y z H 3 C x CH 3 x CH 3 x CH 3 H 3 C H 3 C CH 3 yx yx w z n CH 3 m CH 3 m CH CH 3 CH 3 yx yx w z m n n SPECTROSCOPY 325 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website the six nonequivalent carbons of the benzene ring. The three signals near H9254 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-cm H110021 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 H110021 is attributable to a hydroxyl group. Although both phenol and benzyl alcohol are possibilities, the peaks at 2800–2900 cm H110021 reveal the presence of hy- drogen bonded to sp 3 -hybridized carbon. All carbons are sp 2 -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 H9266 A H9266* transition of ethylene occurs at a shorter wavelength (H9261 max H11005 170 nm) than that of cis, trans-1,3-cyclooctadiene (H9261 max H11005 230 nm), the HOMO–LUMO energy difference in ethylene is greater. 13.17 Conjugation shifts H9261 max to longer wavelengths in alkenes. The conjugated diene 2-methyl-1,3- butadiene has the longest wavelength absorption, H9261 max H11005 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 mH20862z 146, 148, and 150 are present for the molecular ion. (c) The two isotopes of bromine are 79 Br and 81 Br. When both bromines of p-dibromobenzene are 79 Br, the molecular ion appears at mH20862z 234. When one is 79 Br and the other is 81 Br, mH20862z for the molecular ion is 236. When both bromines are 81 Br, mH20862z for the molecular ion is 238. (d) The combinations of 35 Cl, 37 Cl, 79 Br, and 81 Br in p-bromochlorobenzene and the values of mH20862z for the corresponding molecular ion are as shown. ( 35 Cl, 79 Br) mH20862z H11005 190 ( 37 Cl, 79 Br) or ( 35 Cl, 81 Br) mH20862z H11005 192 ( 37 Cl, 81 Br) mH20862z H11005 194 13.19 The base peak in the mass spectrum of alkylbenzenes corresponds to carbon–carbon bond cleavage at the benzylic carbon. CH CH 3 H 3 C CH 3 Base peak: C 9 H 11 H11001 m/z 119 CH 2 CH 2 CH 3 CH 3 Base peak: C 8 H 9 H11001 m/z 105 CH 2 CH 3 CH 3 H 3 C Base peak: C 9 H 11 H11001 m/z 119 2-Methyl-1,3-butadiene (H9261 max H11005 222 nm) H 3 C CH 3 CH 3 1,2,4-Trimethylbenzene 326 SPECTROSCOPY 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 H11005 H5007 1 2 H5007 (C n H 2nH110012 H11002 C n H x ) The compound given contains eight carbons (C 8 H 8 ); therefore, Index of hydrogen deficiency H11005 H5007 1 2 H5007 (C 8 H 18 H11002 C 8 H 8 ) H11005 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 C 8 H 8 Cl 2 as equivalent to C 8 H 10 . Thus, the index of hydrogen deficiency of this compound is 4. Index of hydrogen deficiency H11005 H5007 1 2 H5007 (C 8 H 18 H11002 C 8 H 10 ) H11005 4 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, C 8 H 8 O is treated as if it were C 8 H 8 . Index of hydrogen deficiency H11005 H5007 1 2 H5007 (C 8 H 18 H11002 C 8 H 8 ) H11005 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 C 8 H 10 O 2 , we treat this compound as if it were C 8 H 10 . Index of hydrogen deficiency H11005 H5007 1 2 H5007 (C 8 H 18 H11002 C 8 H 10 ) H11005 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, C 8 H 9 ClO is treated as if it were C 8 H 10 . 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 chemi- cal shifts. (a) This compound has the molecular formula C 8 H 18 and so must be an alkane. The 18 hydrogens are contributed by six equivalent methyl groups. (b) A hydrocarbon with the molecular formula C 5 H 10 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- pound must be cyclopentane. Cyclopentane (H9254 1.5 ppm) (CH 3 ) 3 CC(CH 3 ) 3 2,2,3,3-Tetramethylbutane (H9254 0.9 ppm) SPECTROSCOPY 327 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) The chemical shift of the eight equivalent hydrogens in C 8 H 8 is H9254 5.8 ppm, which is consistent with protons attached to a carbon–carbon double bond. (d) The compound C 4 H 9 Br has no rings or double bonds. The nine hydrogens belong to three equivalent methyl groups. (CH 3 ) 3 CBr tert-Butyl bromide (H9254 1.8 ppm) (e) The dichloride has no rings or double bonds (index of hydrogen deficiency H11005 0). The four equivalent hydrogens are present as two GCH 2 Cl groups. ClCH 2 CH 2 Cl 1,2-Dichloroethane (H9254 3.7 ppm) ( f ) All three hydrogens in C 2 H 3 Cl 3 must be part of the same methyl group in order to be equivalent. CH 3 CCl 3 1,1,1-Trichloroethane (H9254 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 C 12 H 18 has an index of hydrogen deficiency of 4. A likely candidate for a compound with 18 equivalent hydrogens is one with six equivalent CH 3 groups. Thus, 6 of the 12 carbons belong to CH 3 groups, and the other 6 have no hydro- gens. The compound is hexamethylbenzene. A chemical shift of H9254 2.2 ppm is consistent with the fact that all of the protons are benzylic hydrogens. (i) The molecular formula of C 3 H 6 Br 2 tells us that the compound has no double bonds and no rings. All six hydrogens are equivalent, indicating two equivalent methyl groups. The com- pound is 2,2-dibromopropane, (CH 3 ) 2 CBr 2 . CH 3 CH 3 CH 3 CH 3 H 3 C H 3 C ClCH 2 CCH 2 Cl CH 2 Cl CH 2 Cl 1,3-Dichloro-2,2-di(chloromethyl)propane (H9254 3.7 ppm) 1,3,5,7-Cyclooctatetraene (H9254 5.8 ppm) 328 SPECTROSCOPY 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 CH 2 groups. The four CH 2 groups account for four of the carbons of C 6 H 8 , leaving two carbons that bear no hydrogens. A molecular formula of C 6 H 8 corresponds to an index of hydrogen defi- ciency of 3. A compound consistent with these requirements is The signal at H9254 5.6 ppm is consistent with that expected for the four vinylic protons. The sig- nal at H9254 2.7 ppm corresponds to that for the allylic protons of the ring. (b) The compound has a molecular formula of C 5 H 11 Br and therefore has no double bonds or rings. A 9-proton singlet at H9254 1.1 ppm indicates three equivalent methyl groups, and a 2-proton singlet at H9254 3.3 ppm indicates a CH 2 Br group. The correct structure is (CH 3 ) 3 CCH 2 Br. (c) This compound (C 6 H 12 O) has three equivalent CH 3 groups, along with a fourth CH 3 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 C 6 H 10 O 2 corresponds to an index of hydrogen deficiency of 2. The signal at H9254 2.2 ppm (6H) is likely due to two equivalent CH 3 groups, and the one at H9254 2.7 ppm (4H) to two equivalent CH 2 groups. The compound is . 13.23 (a) A 5-proton signal at H9254 7.1 ppm indicates a monosubstituted aromatic ring. With an index of hydrogen deficiency of 4, C 8 H 10 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 H9254 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 C 6 H 14 be an alkane. The doublet–septet pattern is consis- tent with an isopropyl group, and the total number of protons requires that two of these groups be present. (CH 3 ) 2 CHCH(CH 3 ) 2 Doublet H9254 0.8 ppm Septet H9254 1.4 ppm 2,3-Dimethylbutane C(CH 3 ) 3 Singlet; H9254 1.3 ppm tert-Butylbenzene CH 2 CH 3 Quartet H9254 2.6 ppm (Benzylic) Triplet H9254 1.2 ppm Ethylbenzene CH 3 CCH 2 CH 2 CCH 3 O O (CH 3 ) 3 CCCH 3 O CH 2 H 2 C SPECTROSCOPY 329 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Note that the methine (CH) protons do not split each other, because they are equivalent and have the same chemical shift. (d) The molecular formula C 6 H 12 requires the presence of one double bond or ring. A peak at H9254 5.1 ppm is consistent with GC?CH, and so the compound is a noncyclic alkene. The vinyl proton gives a triplet signal, and so the group C?CHCH 2 is present. The 1 H NMR spectrum shows the presence of the following structural units: Putting all these fragments together yields a unique structure. (e) The compound C 4 H 6 Cl 4 contains no double bonds or rings. There are no high-field peaks (H9254 0.5 to 1.5 ppm), and so there are no methyl groups. At least one chlorine substituent must therefore be at each end of the chain. The most likely structure has the four chlorines divided into two groups of two. ( f ) The molecular formula C 4 H 6 Cl 2 indicates the presence of one double bond or ring. A signal at H9254 5.7 ppm is consistent with a proton attached to a doubly bonded carbon. The following struc- tural units are present: CC CH 2 H 5.7 ppm (Triplet) C CCH 3 2.2 ppm (Allylic) Cl 2 CHCH 2 CH 2 CHCl 2 H9254 4.6 ppm (Triplet) H9254 3.9 ppm (Doublet) 1,1,4,4-Tetrachlorobutane CC H 3 C H 3 H 2 CH 3 H Triplet Triplet Singlet Singlet Pentet 2-Methyl-2-pentene CC CH 2 H 5.1 ppm (Triplet) 2.0 ppm (Allylic) CC CH 3 CH 3 1.6 ppm (Singlet; allylic) 1.7 ppm (Singlet; allylic) CH 2 CH 3 0.9 ppm (Triplet) 330 SPECTROSCOPY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website For the methyl group to appear as a singlet and the methylene group to appear as a doublet, the chlorine substituents must be distributed as shown: The stereochemistry of the double bond (E or Z) is not revealed by the 1 H NMR spectrum. (g) A molecular formula of C 3 H 7 ClO is consistent with the absence of rings and multiple bonds (index of hydrogen deficiency H11005 0). None of the signals is equivalent to three protons, and so no methyl groups are present. Three methylene groups occur, all of which are different from each other. The compound is therefore: (h) The compound has a molecular formula of C 14 H 14 and an index of hydrogen deficiency of 8. With a 10-proton signal at H9254 7.1 ppm, a logical conclusion is that there are two monosubsti- tuted benzene rings. The other four protons belong to two equivalent methylene groups. 13.24 The compounds of molecular formula C 4 H 9 Cl are the isomeric chlorides: butyl, isobutyl, sec-butyl, and tert-butyl chloride. (a) All nine methyl protons of tert-butyl chloride (CH 3 ) 3 CCl are equivalent; its 1 H NMR spectrum has only one peak. (b) A doublet at H9254 3.4 ppm indicates a GCH 2 Cl group attached to a carbon that bears a single proton. (c) A triplet at H9254 3.5 ppm means that a methylene group is attached to the carbon that bears the chlorine. Butyl chloride CH 3 CH 2 CH 2 CH 2 Cl H9254 3.5 ppm (Triplet) Isobutyl chloride (CH 3 ) 2 CHCH 2 Cl H9254 3.4 ppm (Doublet) 1,2-Diphenylethane H9254 2.9 ppm (Singlet; benzylic) CH 2 CH 2 ClCH 2 CH 2 CH 2 OH H9254 3.7 or 3.8 ppm (Triplet) H9254 2.0 ppm (Pentet) H9254 2.8 ppm (Singlet) H9254 3.7 or 3.8 ppm (Triplet) CCHCH 2 Cl Cl H 3 C Triplet Doublet (H9254 4.1 ppm) Singlet 1,3-Dichloro-2-butene SPECTROSCOPY 331 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) This compound has two nonequivalent methyl groups. 13.25 Compounds with the molecular formula C 3 H 5 Br have either one ring or one double bond. (a) The two peaks at H9254 5.4 and 5.6 ppm have chemical shifts consistent with the assumption that each peak is due to a vinyl proton (C?CH). The remaining three protons belong to an allylic methyl group (H9254 2.3 ppm). The compound cannot be CH 3 CH?CHBr, because the methyl signal would be split into a doublet. Isomer A can only be (b) Two of the carbons of isomer B have chemical shifts characteristic of sp 2 -hybridized carbon. One of these bears two protons (H9254 118.8 ppm); the other bears one proton (H9254 134.2 ppm). The remaining carbon is sp 3 -hybridized and bears two hydrogens. Isomer B is allyl bromide. (c) All the carbons are sp 3 -hybridized in this isomer. Two of the carbons belong to equivalent CH 2 groups, and the other bears only one hydrogen. Isomer C is cyclopropyl bromide. 13.26 All these compounds have the molecular formula C 4 H 10 O. They have neither multiple bonds nor rings. (a) Two equivalent CH 3 groups occur at H9254 18.9 ppm. One carbon bears a single hydrogen. The least shielded carbon, presumably the one bonded to oxygen, has two hydrogen substituents. Putting all the information together reveals this compound to be isobutyl alcohol. (b) This compound has four distinct peaks, and so none of the four carbons is equivalent to any of the others. The signal for the least shielded carbon represents CH, and so the oxygen is at- tached to a secondary carbon. Only one carbon appears at low field; the compound is an alco- Isobutyl alcohol H9254 69.4 ppmH9254 30.8 ppm H9254 18.9 ppm (CH 3 ) 2 CHCH 2 OH Cyclopropyl bromide H Br H9254 12.0 ppm H9254 16.8 ppm Allyl bromide H 2 C H9254 32.6 ppm H9254 134.2 ppm H9254 118.8 ppm CHCH 2 Br 2-Bromo-1-propene H 2 CC Br CH 3 sec-Butyl chloride CH 3 CHCH 2 CH 3 H9254 1.5 ppm (Doublet) H9254 1.0 ppm (Triplet) Cl 332 SPECTROSCOPY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website hol, not an ether. Therefore; (c) Signals for three equivalent CH 3 carbons indicate that this isomer is tert-butyl alcohol. This assignment is reinforced by the observation that the least shielded carbon has no hydrogens attached to it. 13.27 The molecular formula of C 6 H 14 for each of these isomers requires that all of them be alkanes. (a) This compound contains only CH 3 and CH carbons. (b) This isomer has no CH carbons, and two different kinds of CH 2 groups. (c)CH 3 , CH 2 , and CH carbons are all present in this isomer. There are two different kinds of CH 3 groups. (d) This isomer contains a quaternary carbon in addition to a CH 2 group and two different kinds of CH 3 groups. 2,2-Dimethylbutane C CH 3 CH 3 H 3 CCH 2 CH 3 H9254 30.2 ppm H9254 36.5 ppm H9254 8.5 ppm H9254 28.7 ppm 3-Methylpentane H9254 18.4 ppm H9254 36.4 ppm H9254 29.1 ppm H9254 11.1 ppm CH 3 CH 2 CHCH 2 CH 3 CH 3 Hexane H9254 22.8 ppm H9254 31.9 ppmH9254 13.7 ppm CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 2,3-Dimethylbutane H9254 33.9 ppmH9254 19.1 ppm (CH 3 ) 2 CHCH(CH 3 ) 2 tert-Butyl alcohol (CH 3 ) 3 COH H9254 68.9 ppmH9254 31.2 ppm sec-Butyl alcohol H9254 10.0 ppm H9254 32.0 ppm H9254 69.2 ppm H9254 22.7 ppm CH 3 CHCH 2 CH 3 OH SPECTROSCOPY 333 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) This isomer contains two different kinds of CH 3 groups, two different kinds of CH 2 groups, and a CH group. 13.28 The index of hydrogen deficiency of the compound C 4 H 6 is 2. It can have two double bonds, two rings, one ring and one double bond, or one triple bond. The chemical shift data indicate that two carbons are sp 3 -hybridized and two are sp 2 . The most reasonable structure that is consistent with 13 C NMR data is cyclobutene. The compound cannot be 1- or 2-methylcyclopropene. Neither of the carbon signals represents a methyl group. 13.29 Each of the carbons in the compound gives its 13 C NMR signal at relatively low field; it is likely that each one bears an electron-withdrawing substituent. The compound is The isomeric compound 2-chloro-1,3-propanediol cannot be correct. The C-1 and C-3 positions are equivalent; the 13 C NMR spectrum of this com- pound exhibits only two peaks, not three. 13.30 (a) All the hydrogens are equivalent in p-dichlorobenzene; therefore it has the simplest 1 H NMR spectrum of the three compounds chlorobenzene, o-dichlorobenzene, and p-dichloroben- zene. Chlorobenzene (three different kinds of protons) Cl H HH H H x y x y z p-Dichlorobenzene (all protons are equivalent) Cl H HH Cl H x x x x o-Dichlorobenzene (two different kinds of protons) Cl H HH H Cl x x y y HOCH 2 CHCH 2 OH Cl 3-Chloro-1,2-propanediol ClCH 2 CHCH 2 OH H9254 46.8 ppm H9254 72.0 ppm H9254 63.5 ppm OH Cyclobutene H9254 136 ppm H9254 30.2 ppm 2-Methylpentane CH 3 CHCH 2 CH 2 CH 3 H9254 27.6 ppm H9254 14.0 ppm H9254 20.5 ppm H9254 41.6 ppmH9254 22.4 ppm CH 3 334 SPECTROSCOPY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b–d) In addition to giving the simplest 1 H NMR spectrum, p-dichlorobenzene gives the simplest 13 C NMR spectrum. It has two peaks in its 13 C NMR spectrum, chlorobenzene has four, and o-dichlorobenzene has three. 13.31 Compounds A and B (C 10 H 14 ) have an index of hydrogen deficiency of 4. Both have peaks in the H9254 130–140-ppm range of their 13 C NMR spectra, so that the index of hydrogen deficiency can be accommodated by a benzene ring. The 13 C NMR spectrum of compound A shows only a single peak in the upfield region, at H9254 20 ppm. Thus, the four remaining carbons, after accounting for the benzene ring, are four equiva- lent methyl groups. The benzene ring is symmetrically substituted as there are only two signals in the aromatic region at H9254 132 and 135 ppm. Compound A is 1,2,4,5-tetramethylbenzene. In compound B the four methyl groups are divided into two pairs. Three different carbons occur in the benzene ring, as noted by the appearance of three signals in the aromatic region (H9254 128–135 ppm). Compound B is 1,2,3,4-tetramethylbenzene. 13.32 Since the compound has a 5-proton signal at H9254 7.4 ppm and an index of hydrogen deficiency of 4, we conclude that six of its eight carbons belong to a monosubstituted benzene ring. The infrared spectrum exhibits absorption at 3300 cm H110021 , indicating the presence of a hydroxyl group. The com- pound is an alcohol. A 3-proton doublet at H9254 1.6 ppm, along with a 1-proton quartet at H9254 4.9 ppm, suggests the presence of a CH 3 CH unit. The compound is 1-phenylethanol. 1-Phenylethanol CCH 3 H OH H9254 4.9 ppm (Quartet) H9254 1.6 ppm (Doublet) H9254 4.2 ppm (Singlet) H9254 7.4 ppm 1,2,3,4-Tetramethylbenzene (Compound B) CH 3 H 3 C H 3 C CH 3 H9254 16, 21 ppm H9254 128 ppm H9254 134, 135 ppm H 3 CCH 3 H 3 C CH 3 H9254 132 ppm H9254 135 ppm 1,2,4,5-Tetramethylbenzene (Compound A) Chlorobenzene (four different kinds of carbon) Cl x y x y z w p-Dichlorobenzene (two different kinds of carbon) Cl Cl y x x y yy o-Dichlorobenzene (three different kinds of carbon) Cl Cl y x zy x z SPECTROSCOPY 335 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 13.33 The peak at highest mH20862z in the mass spectrum of the compound is mH20862z H11005 134; this is likely to cor- respond to the molecular ion. Among the possible molecular formulas, C 10 H 14 correlates best with the information from the 1 H NMR spectrum. What is evident is that there is a signal due to aromatic protons, as well as a triplet–quartet pattern of an ethyl group. A molecular formula of C 10 H 14 sug- gests a benzene ring that bears two ethyl groups. Because the signal for the aryl protons is so sharp, they are probably equivalent. The compound is p-diethylbenzene. 13.34 There is a prominent peak in the infrared spectrum of the compound at 1725 cm H110021 , characteristic of C?O stretching vibrations. The 1 H NMR spectrum shows only two sets of signals, a triplet at H9254 1.1 ppm and a quartet at H9254 2.4 ppm. The compound contains a CH 3 CH 2 group as its only protons. Its 13 C NMR spectrum has three peaks, one of which is at very low field. The signal at H9254 211 ppm is in the region characteristic of carbons of C?O groups. If one assumes that the compound contains only carbon, hydrogen, and one oxygen atom and that the peak at highest mH20862z in its mass spectrum (mH20862z 86) corresponds to the molecular ion, then the compound has the molecular formula C 5 H 10 O. All the information points to the conclusion that the compound has the structure shown. 13.35 [18]-Annulene has two different kinds of protons; the 12 protons on the outside periphery of the ring are different from the 6 on the inside. These different environments explain why the 1 H NMR spectrum contains two peaks in a 2:1 ratio.The less intense signal, that for the interior protons, is more shielded than the signal for the outside protons. This results from the magnetic field induced by the circulating H9266 electrons of this aromatic ring, which reinforces the applied field in the region of the outside protons but opposes it in the interior of the ring. H H H H H H H H HH HH HH H H H H H 0 H H HH H HH HH H H HH H H H H H H9254 H110021.9 ppm H9254 8.8 ppm 3-Pentanone CH 3 CH 2 CCH 2 CH 3 O p-Diethylbenzene CH 2 CH 3 CH 3 CH 2 H H9254 7.2 ppm (Singlet) H9254 2.7 ppm (Quartet) H9254 1.3 ppm (Triplet) 336 SPECTROSCOPY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Protons inside the ring are shielded by the induced field to a significant extent—so much so that their signal appears at H9254 H110021.9 ppm. 13.36 (a) The nuclear spin of 19 F is H11006 H5007 1 2 H5007 , that is, the same as that of a proton. The splitting rules for 19 F– 1 H couplings are the same as those for 1 H– 1 H. Thus, the single fluorine atom of CH 3 F splits the signal for the protons of the methyl group into a doublet. (b) The set of three equivalent protons of CH 3 F splits the signal for fluorine into a quartet. (c) The proton signal in CH 3 F is a doublet centered at H9254 4.3 ppm. The separation between the two halves of this doublet is 45 Hz, which is equivalent to 0.225 ppm at 200 MHz (200 Hz H11005 1 ppm). Thus, one line of the doublet appears at H9254(4.3H110010.225) ppm and the other at H9254(4.3H110020.225) ppm. 13.37–13.38 Solutions to molecular modeling exercises are not provided in this Study Guide and Solutions Man- ual. You should use Learning By Modeling for these exercises. 13.39 Because 31 P has a spin of H11006 H5007 1 2 H5007 , it is capable of splitting the 1 H NMR signal of protons in the same molecule. The problem stipulates that the methyl protons are coupled through three bonds to phos- phorus in trimethyl phosphite. (a) The reciprocity of splitting requires that the protons split the 31 P signal of phosphorus. There are 9 equivalent protons, and so the 31 P signal is split into ten peaks. (b) Each peak in the 31 P multiplet is separated from the next by a value equal to the 1 H– 31 P cou- pling constant of 12 Hz. There are nine such intervals in a ten-line multiplet, and so the sepa- ration is 108 Hz between the highest and lowest field peaks in the multiplet. 13.40 The trans and cis isomers of 1-bromo-4-tert-butylcyclohexane can be taken as models to estimate the chemical shift of the proton of the CHBr group when it is axial and equatorial, respectively, in the two chair conformations of bromocyclohexane. An axial proton is more shielded (H9254 3.81 ppm for trans-1-bromo-4-tert-butylcyclohexane) than an equatorial one (H9254 4.62 ppm for cis-1-bromo-4-tert- butylcyclohexane). The difference in chemical shift between these stereoisomers is 0.81 ppm. The corresponding pro- ton in bromocyclohexane is 0.67 ppm more shielded than in the equatorial proton in cis-1-bromo-4- tert-butylcyclohexane. The proportion of bromocyclohexane that has an axial hydrogen is therefore 0.67H208620.81, or 83%. For bromocyclohexane, 83% of the molecules have an equatorial bromine, and 17% have an axial bromine. 13.41 The two staggered conformations of 1,2-dichloroethane are the anti and the gauche: Cl Cl HH HH Anti Anti conformation has center of symmetry. Cl H ClH H H Gauche (CH 3 ) 3 C Br H H9254 4.62 ppm cis-1-Bromo-4-tert-butylcyclohexane; less shielded (CH 3 ) 3 C Br H H9254 3.81 ppm trans-1-Bromo-4-tert-butylcyclohexane; more shielded Br H9254 3.95 ppm H Bromocyclohexane CH 3 OP OCH 3 OCH 3 45 Hz H9254 4.3 ppm H9254 4.075 ppmH9254 4.525 ppm SPECTROSCOPY 337 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The species present at low temperature (crystalline 1,2-dichloroethane) has a center of symmetry and is therefore the anti conformation. Liquid 1,2-dichloroethane is a mixture of the anti and the gauche conformations. 13.42 (a) Energy is proportional to frequency and inversely proportional to wavelength. The longer the wavelength, the lower the energy. Microwave photons have a wavelength in the range of 10 H110022 m, which is longer than that of infrared photons (on the order of 10 H110025 m). Thus, mi- crowave radiation is lower in energy than infrared radiation, and the separation between rota- tional energy levels (measured by microwave) is less than the separation between vibrational energy levels (measured by infrared). (b) Absorption of a photon occurs only when its energy matches the energy difference between two adjacent energy levels in a molecule. Microwave photons have energies that match the differences between the rotational energy levels of water. They are not sufficiently high in en- ergy to excite a water molecule to a higher vibrational or electronic energy state. 13.43 A shift in the UV-Vis spectrum of acetone from 279 nm in hexane to 262 nm in water is a shift to shorter wavelength on going from a less polar solvent to a more polar one. This means that the en- ergy difference between the starting electronic state (the ground state, n) and the excited electronic state (H9266*) is greater in water than in hexane. Hexane as a solvent does not interact appreciably with either the ground or the excited state of acetone. Water is polar and solvates the ground state of ace- tone, lowering its energy. Because the energy gap between the ground state and the excited state in- creases, it must mean that the ground state is more solvated than the excited state and therefore more polar than the excited state. 13.44 The dipole moment of carbon dioxide is zero and does not change during the symmetric stretching vibration. The symmetric stretch is not “infrared-active.” The antisymmetric stretch generates a di- pole moment in carbon dioxide and is infrared-active. 13.45 Solutions to molecular modeling exercises are not provided in this Study Guide and Solutions Man- ual. You should use Learning By Modeling for these exercises. SELF-TEST PART A A-1. Complete the following table relating to 1 H NMR spectra by supplying the missing data for entries 1 through 4. Spectrometer Chemical shift frequency ppm Hz (a) 60 MHz ——— 366 1 (b) 300 MHz 4.35 ——— 2 (c) ——— 3.50 700 3 (d) 100 MHz ——— of TMS 4 OOC Symmetric stretch: no change in dipole moment OOC Antisymmetric stretch: dipole moment present as a result of unequal C O bond distances H9004E H9266* n Ground state more solvated in water H9004EH11032 H9266* n Acetone in hexane Acetone in water 338 SPECTROSCOPY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-2. Indicate the number of signals to be expected and the multiplicity of each in the 1 H NMR spectrum of each of the following substances: (a) BrCH 2 CH 2 CH 2 Br (b) (c) A-3. Two isomeric compounds having the molecular formula C 6 H 12 O 2 both gave 1 H NMR spec- tra consisting of only two singlets. Given the chemical shifts and integrations shown, iden- tify both compounds. Compound A: H9254 1.45 ppm (9H) Compound B: H9254 1.20 ppm (9H) H9254 1.95 ppm (3H) H9254 3.70 ppm (3H) A-4. Identify each of the following compounds on the basis of the IR and 1 H NMR information provided (a)C 10 H 12 O: IR: 1710 cm H110021 NMR: H9254 1.0 ppm (triplet, 3H) H9254 2.4 ppm (quartet, 2H) H9254 3.6 ppm (singlet, 2H) H9254 7.2 ppm (singlet, 5H) (b)C 6 H 14 O 2 : IR: 3400 cm H110021 NMR: H9254 1.2 ppm (singlet, 12H) H9254 2.0 ppm (broad singlet, 2H) (c)C 10 H 16 O 6 : IR: 1740 cm H110021 NMR: H9254 1.3 ppm (triplet, 9H) H9254 4.2 ppm (quartet, 6H) H9254 4.4 ppm (singlet, 1H) (d)C 4 H 7 NO: IR: 2240 cm H110021 3400 cm H110021 (broad) NMR: H9254 1.65 ppm (singlet, 6H) H9254 3.7 ppm (singlet, 1H) A-5. Predict the number of signals and their approximate chemical shifts in the 13 C NMR spec- trum of the compound shown. A-6. How many signals will appear in the 13 C NMR spectrum of each of the three C 5 H 12 isomers? A-7. The 13 C NMR spectrum of an alkane of molecular formula C 6 H 14 exhibits two signals at H9254 23 ppm (4C) and 37 ppm (2C). What is the structure of this alkane? PART B The following three problems refer to the 1 H NMR spectrum of CH 3 CH 2 OCH 2 OCH 2 CH 3 . B-1. How many signals are expected? (a)12 (b)5 (c)4 (d)3 B-2. The signal farthest downfield (relative to TMS) will be a (a) Singlet (c) Doublet (b) Triplet (d) Quartet CCH 2 CH 3 O CH 3 OCH 2 COCH 3 O CH 3 CH 2 CCH 2 CH 3 Cl Cl SPECTROSCOPY 339 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-3. The signal farthest upfield (closest to TMS) will be a (a) Singlet (c) Doublet (b) Triplet (d) Quartet B-4. The relationship between magnetic field strength and the energy difference between nuclear spin states is (a) They are independent of each other. (b) They are directly proportional. (c) They are inversely proportional. (d) The relationship varies from molecule to molecule. B-5. An infrared spectrum exhibits a broad band in the 3000–3500-cm H110021 region and a strong peak at 1710 cm H110021 . Which of the following substances best fits the data? (a)C 6 H 5 CH 2 CH 2 OH (c) (b)(d) B-6. Considering the 1 H NMR spectrum of the following substance, which set of protons appears farthest downfield relative to TMS? B-7. Which of the following substances does not give a 1 H NMR spectrum consisting of only two peaks? (a)(c) (b) (d) None of these (all satisfy the spectrum) B-8. The multiplicity of the a protons in the 1 H NMR spectrum of the following substance is (a) Singlet (b) Doublet (c) Triplet (d) Quartet B-9. An unknown compound C 4 H 8 O gave a strong infrared absorption at 1710 cm H110021 . The 13 C NMR spectrum exhibited four peaks at H9254 9, 29, 37, and 209 ppm. The 1 H NMR spectrum had three signals at H9254 1.1 (triplet), 2.1 (singlet), and 2.3 (quartet) ppm. Which, if any, of the fol- lowing compounds is the unknown? H O (a) OH (b) O (c) O (d) None of these (e) (CH 3 ) 2 CCH 2 Cl OH ab Br H 3 C CH 3 C Br CH 3 CH 3 C CH 3 H 3 C CH 3 CH 3 H 3 C OCH 3 C OCH 2 C(CH 3 ) 3 bc a C 6 H 5 CH 2 COCH 3 O C 6 H 5 CH 2 COH O C 6 H 5 CH 2 CCH 3 O 340 SPECTROSCOPY Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-10. How many signals are expected in the 13 C NMR spectrum of the following substance? (a)5 (b)6 (c)8 (d)10 B-11. Which one of the following has the greatest number of signals in its 13 C NMR spectrum? (The spectrum is run under conditions in which splitting due to 13 C– 1 H coupling is not observed.) (a) Hexane (c) 1-Hexene (e) 1,5-Hexadiene (b) 2-Methylpentane (d) cis-3-Hexene B-12. Which of the following C 9 H 12 isomers has the fewest signals in its 13 C NMR spectrum? (a)(b)(c)(d) B-13. Which of the following compounds would best fit a 13 C NMR spectrum having peaks at H9254 16, 21, 32, 36, 115, and 140 ppm? B-14. Which of the following compounds would have the fewest peaks in its 13 C NMR spectrum? (a)(b)(c)(d)(e) B-15. Which of the compounds in the previous problem would have the most peaks in its 13 C NMR spectrum? Cl Cl Cl Cl Br BrClCl BrCl Cl (a) (b) (c) (d) COCH 3 O COCH 3 O SPECTROSCOPY 341 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website