Carey - Organic Chemistry - sgchapt13

Carey - Organic Chemistry - sgchapt13

(Parte 1 de 3)

CHAPTER 13 SPECTROSCOPY

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 T 200 MHz isthe same as 1.41 T 60 MHz. The magnetic field strength of a 60-MHz NMR spectrometer is 1.41 T.

13.2The ratio of 1H and 13C resonance frequencies remains constant. When the 1H frequency is 200 MHz, 13C NMR spectra are recorded at 50.4 MHz. Thus, when the 1H frequency is 100 MHz, 13C NMR spectra will be observed at 25.2 MHz.

NMR spectrometer. Thus, to compare the 1H 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

6.8 ppm

(b)The chemical shift of the proton in bromoform ( 6.8 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.

13.4In 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 1H NMR signal of CH3CCl3appears 4.6 ppm upfield from the proton signal of chloroform. The chemical shift of the protons in CH3CCl3is 2.6 ppm.

13.51,4-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 BackForwardMain MenuTOCStudy Guide TOCStudent OLCMHHE 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 1H NMR spectrum of 1-butanol.

Butane has twodifferent types of protons; it will exhibit twosignals in its 1H 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 1H 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 threesignals in its 1H NMR spectrum.

BrBr Br Br

2,2,3,3-Tetrabromobutane

CCH3

2,2-Dibromobutane has three nonequivalent sets of protons.

CH3CCH2CH3 Br

BrCHCH2CH2CH2Br Cl

1,4-Dibromo-1-chlorobutane

BrCH2CH2CH2CH2Br 1,4-Dibromobutane 1,4-Dibromo-2-chlorobutane BrCH2CHCH2CH2Br

1,4-Dibromo-2-chlorobutane BrCH2CH2CHCH2Br

1,4-Dibromo-1-chlorobutane BrCH2CH2CH2CHBr

ClCH2CH2CH2CH3

1-Chlorobutane CH3CH2CH2CH3Butane

CH3CHCH3CH3

Cl 2-Chlorobutane

CH3CH2CH2CH2Cl

1-Chlorobutane

CH3CH2CHCH3

Cl 2-Chlorobutane

CH3CH2CH2CH2OH

Five different proton environments in 1-butanol; five signals

7.0 ppm

SPECTROSCOPY 321

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322 SPECTROSCOPY

(g)There are fournonequivalent sets of protons in 1,1,4-tribromobutane. It will exhibit four signals in its 1H NMR spectrum.

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 1H 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 1H 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.

Allyl bromide (four signals in the 1H NMR spectrum)

CH2BrH H trans-1,2-Dibromoethene (one signal in the 1H NMR spectrum)

C HBr Br cis-1,2-Dibromoethene

C HBr H

C ClBr H

C HBr Cl

1,1-Dibromoethene

C BrBr H

1,1-Dibromo-2-chloroethene

C BrBr H

1,1-Dibromo-2-chloroethene

C BrBr Cl

1,1,1-Tribromobutane Br3CCH2CH2CH3

BrCCH2CH2CH2Br Br

H 1,1,4-Tribromobutane

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(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.

same chemical shift and do not split each other’s signals. The 1H NMR spectrum of Cl3CCH3 consists of a single sharp peak.

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 1H 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.

CH3CH2OCH3

QuartetSinglet; not vicinal to anyother protons in molecule Triplet

Cl Cl3CC CH3 Doublet

Quartet 1,1,1,2-Tetrachloropropane

ClCH2CCH3 Cl

Cl 1,2,2-Trichloropropane

H Cl2CC H2Cl

DoubletTriplet 1,1,2-Trichloroethane

2-Methyl-2-butene (four signals in the 1H NMR spectrum)

H3C H

SPECTROSCOPY 323

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(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.

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.

13.10Both Hband Hcin m-nitrostyrene appear as doublets of doublets. Hbis coupled to Haby a coupling constant of 12 Hz and to Hcby a coupling constant of 2 Hz. Hcis coupled to Haby a coupling constant of 16 Hz and to Hbby a coupling constant of 2 Hz.

13.1(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

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.

Hb Hc

2 Hz2 Hz 12 Hz

2 Hz2 Hz 16 Hz

(diagrams not to scale)

ClCH2CH2OCH2CH3

Triplet

Triplet Quartet Triplet

Three signals:

CH3 triplet;

CH2 quartet; aromatic H singlet

CH3CH2OCH2CH3 Quartet QuartetTriplet Triplet

324 SPECTROSCOPY

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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 sixsignals. (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 threesignals in its 13C NMR spectrum.

13.13sp3-Hybridized carbons are more shielded than sp2-hybridized ones. Carbon xis the most shielded, and has a chemical shift of 20 ppm. The oxygen of the OCH3group decreased the shielding of carbon z; its chemical shift is 5 ppm. The least shielded is carbon ywith a chemical shift of

157 ppm.

OCH3H3C H H

20 ppm55 ppm 157 ppm y zH3C x

CH3 yx yx w z n

CH CH3

CH3 yx w z m

SPECTROSCOPY 325

BackForwardMain MenuTOCStudy Guide TOCStudent OLCMHHE Website the six nonequivalent carbons of the benzene ring. The three signals near 20 ppm are due to the three nonequivalent methyl groups.

13.15The infrared spectrum of Figure 13.31 has no absorption in the 1600–1800-cm 1region, 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 30 cm 1is attributable to a hydroxyl group. Although both phenol and benzyl alcohol are possibilities, the peaks at 2800–2900 cm 1reveal the presence of hydrogen bonded to sp3-hybridized carbon. All carbons are sp2-hybridized in phenol. The infrared spectrum is that of benzyl alcohol.

13.16The energy of electromagnetic radiation is inversely proportional to its wavelength. Since excitation than that of cis, trans-1,3-cyclooctadiene ( max 230 nm), the HOMO–LUMO energy difference in ethylene is greater.

13.17Conjugation shifts maxto longer wavelengths in alkenes. The conjugated diene 2-methyl-1,3- butadiene has the longest wavelength absorption, max 2 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 m z146, 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 m z234. When one is 79Br and the other is 81Br, m zfor the molecular ion is 236. When both bromines are 81Br, m zfor the molecular ion is 238. (d)The combinations of 35Cl, 37Cl, 79Br, and 81Br in p-bromochlorobenzene and the values of m z for the corresponding molecular ion are as shown.

13.19The base peak in the mass spectrum of alkylbenzenes corresponds to carbon–carbon bond cleavage at the benzylic carbon.

CH3

CH2 CH2CH3

CH3

CH2 CH3

2-Methyl-1,3-butadiene ( max 2 nm)

1,2,4-Trimethylbenzene

326 SPECTROSCOPY

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Index of hydrogen deficiency 1

Index of hydrogen deficiency 1

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 C8H8Cl2as equivalent to C8H10. Thus, the index of hydrogen deficiency of this compound is 4.

Index of hydrogen deficiency 1

Since the compound consumes 2 mol of hydrogen on catalytic hydrogenation, it must therefore contain two rings.

Index of hydrogen deficiency 1

Since the problem specifies that 2 mol of hydrogen is consumed on catalytic hydrogenation, this compound contains three rings.

Index of hydrogen deficiency 1

Because 2 mol of hydrogen is consumed on catalytic hydrogenation, there must be two rings.

13.21Since each compound exhibits only a single peak in its 1H NMR spectrum, all the hydrogens are equivalent in each one. Structures are assigned on the basis of their molecular formulas and chemical shifts.

pound must be cyclopentane.

Cyclopentane ( 1.5 ppm)

2,2,3,3-Tetramethylbutane ( 0.9 ppm)

SPECTROSCOPY 327

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(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)

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. Alikely candidate for a compound with 18 equivalent hydrogens is one with six equivalent

CH3groups. Thus, 6 of the 12 carbons belong to CH3groups, and the other 6 have no hydrogens. The compound is hexamethylbenzene.

Achemical shift of 2.2 ppm is consistent with the fact that all of the protons are benzylic hydrogens.

CH3

CH3

ClCH2CCH2Cl CH2Cl

CH2Cl

1,3-Dichloro-2,2-di(chloromethyl)propane ( 3.7 ppm)

1,3,5,7-Cyclooctatetraene ( 5.8 ppm)

328 SPECTROSCOPY

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13.22In each of the parts to this problem, nonequivalent protons must notbe bonded to adjacent carbons, because we are told that the two signals in each case are singlets.

bear no hydrogens. Amolecular formula of C6H8corresponds to an index of hydrogen deficiency of 3. Acompound 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.

bond or one ring. This compound is .

(d)Amolecular formula of C6H10O2corresponds to an index of hydrogen deficiency of 2. The signal at 2.2 ppm (6H) is likely due to two equivalent CH3groups, and the one at 2.7 ppm

(4H) to two equivalent CH2groups. 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.

be present.

Doublet

0.8 ppmSeptet 1.4 ppm

2,3-Dimethylbutane

Singlet; 1.3 ppm tert-Butylbenzene

CH2CH3

Quartet 2.6 ppm (Benzylic)

Triplet 1.2 ppm

Ethylbenzene

SPECTROSCOPY 329

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Note that the methine (CH) protons do not split each other, because they are equivalent and have the same chemical shift.

proton gives a triplet signal, and so the group C?CHCH2is present. The 1H NMR spectrum shows the presence of the following structural units:

Putting all these fragments together yields a unique structure.

therefore be at each end of the chain. The most likely structure has the four chlorines divided into two groups of two.

tural units are present:

C CH2

H 5.7 ppm (Triplet)

C CCH3 2.2 ppm (Allylic)

Cl2CHCH2CH2CHCl2

4.6 ppm (Triplet)

3.9 ppm (Doublet)

1,1,4,4-Tetrachlorobutane

C H3C

H Triplet

TripletSinglet Singlet

Pentet

2-Methyl-2-pentene

C CH2

H5.1 ppm (Triplet)

2.0 ppm (Allylic)

C CH3

CH3 1.6 ppm (Singlet; allylic)

1.7 ppm (Singlet; allylic)

(Parte 1 de 3)

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