organic chemistry textbook solution of chapter 13

7/22/2019 Organic chemistry textbook solution of Chapter 13

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258 Solutions Chapter 13: Nuclear Magnetic Resonance Spectroscopy

CHAPTER 13Solutions to the Problems

Problem 13.1 Calculate the ratio of nuclei in the higher spin state to those in the lower spin state, Nh/Nl, for13C at 25C in

an applied field strength of 7.05 T. The difference in energy between the higher and lower nuclear spin states in thisapplied field is approximately 0.030 J (0.00715 cal)/mol.

The important equation relates the change in energy of two spin states to their equilibrium concentrations:

Nl

Nh-RTlnG =

Rearranging this expression in terms of Nh/Nlgives:

=lnNl

NhRT

GSubstituting in the appropriate values for R (8.314 Jdeg -1mol-1), T (298 K) and G (0.030 Jmol-1) gives:

1.0000121

1.0000000

0.9999879 =

NhNl =

= -1.21 x 10-5=lnNl

Nh(8.314 Jdeg-1mol-1)(298 deg)

- 0.030 Jmol-1

Problem 13.2 State the number of sets of equivalent hydrogens in each compound and the number of hydrogens in eachset.

(a) 3-MethylpentaneNumbers have been added to the carbon atoms of the structures to aid in referring to specific hydrogens. Usethe "test atom" approach if you have trouble understanding the answers.

CH

CH 3

CH 2 CH 3CH 2CH 3

1 2 3 4

6

ab

c

d

ba 5

There are four sets of equivalent hydrogens. Set a: 6 hydrogens from the methyl groups of carbon atoms 1 and5. Set b: 4 hydrogens from the -CH2- groups of carbon atoms 2 and 4. Set c: 3 hydrogens from the methylgroup of carbon atom 5. Set d: 1 hydrogen from the -CH- group of carbon atom 3.

(b) 2,2,4-Trimethylpentane

CH

CH 3

CH 3 CH 2 C

CH 3

CH 3

CH 38

12345

6 7b

b

a

a

a

c

d

There are four sets of equivalent hydrogens. Set a: 9 hydrogens from the methyl groups of carbon atoms 1, 7,and 8. Set b: 6 hydrogens from the methyl groups of carbon atoms 5 and 6. Set c: 2 hydrogens from the -CH2-

group of carbon atom 3. Set d: 1 hydrogen from the -CH- group of carbon atom 4.


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Chapter 13: Nuclear Magnetic Resonance Spectroscopy Solutions 259

Problem 13.3 Each compound gives only one signal in its 1H-NMR spectrum. Propose a structural formula for eachcompound.

In order for these molecules to give a single absorption peak, each of the hydrogen nuclei must be in anidentical environment. This will only occur in symmetrical molecules.

(a) C3H6O (b) C5H10 (c) C5H12

CH 3C

CH 3

O

C C

CC

CH

H

H

H

HH

H

H

H H

CH 3 C

CH 3

CH 3

CH 3

(d) C4H6Cl4

CH 3-CCl2-CCl2-C H3

Problem 13.4 The line of integration of the two signals in the 1H-NMR spectrum of a ketone of molecular formulaC7H14O rises 62 and 10 chart divisions, respectively. Calculate the number of hydrogens giving rise to each signal, andpropose a structural formula for this ketone.

The ratio of signals is approximately 6:1, which corresponds to a 12:2 ratio of the 14 hydrogens. Thus, thelarger signal represents 12 hydrogens and the smaller signal represents 2 hydrogens. A structure consistentwith this assignment is 2,4-dimethyl-3-pentanone as shown below:

CC

O

C

CH 3

CH 3

H

H3C

H

H3C

Larger Signal

Larger Signal

Smaller Signal Smaller Signal

Problem 13.5 Following are two constitutional isomers of molecular formula C4H8O2.

CCH3

O

CH3CH2O COCH3

O

CH3CH2

(1) (2)

(a) Predict the number of signals in the 1H-NMR spectrum of each isomer.

Each isomer will have three signals.

(b) Predict the ratio of areas of the signals in each spectrum.In each spectrum, the ratio of areas of the three signals will be 3:2:3.

(c) Show how to distinguish between these isomers on the basis of chemical shift.

The -CH3singlet signals will be diagnostic. Isomer (1) is the only one with a -CH3attached to the carbonylcarbon atom, while isomer (2) is the only one with a -CH3attached to an ester sp3oxygen atom. Therefore, thespectrum of isomer (1) will be the one with a -CH3singlet at 2.1-2.3, and the spectrum of isomer (2) will bethe one with a -CH3singlet at 3.7-3.9.


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Problem 13.6 Following are pairs of constitutional isomers. Predict the number of signals in the 1H-NMR spectrum ofeach isomer and the splitting pattern of each signal.

CCH3

O

CH3OCH2 COCH3

O

CH3CH2and(a)

The molecule on the left will have three signals that are all singlets, and the molecule on the right will havethree signals with splitting patterns as indicated.

CCH 3

O

CH3OCH2 COCH3

O

CH 3CH 2

quartettriplet singletsingletsingletsinglet

aa b c b c

CH3CCH3

Cl

Cl

ClCH2CH2CH2Cland(b)

The molecule on the left will have one signal and the molecule on the right will have two signals with splitting

patterns as indicated.

CH 3CC H3

Cl

Cl

tripletquintet

b a

singlet

aa

ClCH2CH 2CH 2Cl

a

triplet

Problem 13.7 Following is the spectrum of 2-butanol. Explain why the CH2protons appear as a complex multiplet insteadof a simple quintet.

079 8 6 5 4

a ab

b

e

e

c

c

d

d

3 2 1 ppm

Chemical Shift (d)blow-up

Solvent

OHc

()

1234CH 2CH 3 CH C H3

a c

OH

*e

b

d

Carbon atom 2 is a chiral center and the two hydrogens labeled as group c on carbon atom 3 are diastereotopic.Thus, these two protons are in different environments and give rise to different chemical shifts, greatlycomplicating the corresponding signals in the observed spectra.


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Problem 13.8 Explain how to distinguish between the members of each pair of constitutional isomers based on the numberof signals in the proton-decoupled 13C-NMR spectrum of each member.

CH2 CH3

b

(a) and

c

de

c

d

bc

de

f

a

g

a

These molecules can be distinguished because they have different numbers of nonequivalent carbon nuclei andthus will have different numbers of 13C-NMR signals. Different signals are indicated by different letters onthe above structures. The molecule on the left has higher symmetry and will have 5 signals corresponding to thecarbon atoms labeled as a - e, while the molecule on the right has less symmetry and will have 7 signalscorresponding to the carbon atoms labeled as a - g.

C C

H3C

H

CH2CH2CH3

H

C C

H3CH2C

H

CH2CH3

H

bb afed

c

a

cc

a

b(b ) and

These molecules can also be distinguished because they have different numbers of nonequivalent carbon nucleiand thus will have different numbers of 13C-NMR signals. Different signals are indicated by different letters

on the above structures. The molecule on the right has higher symmetry and will only have 3 signalscorresponding to the carbon atoms labeled as a - c, while the molecule on the left has less symmetry and willhave 6 signals corresponding to the carbon atoms labeled as a - f.

Problem 13.9 Assign all signals in the 13C-NMR spectrum of 4-methyl-2-pentanone.

0100150200250 50 ppm

208

52

30

24

22

0100150200250 50 ppm

52

30

24

22

(a)

(b)

CDCl3

CH

CH3

CH2

The appropriate peaks have been assigned using the DEPT spectra to distinguish the CH 3, CH2and CHgroups. The different CH3groups were distinguished by the fact that the CH3group adjacent to the carbonylgroup is less shielded (higher chemical shift) than the other two methyl groups, which are equivalent.

CC H2CH(CH3)2

O

CH 3

4-Methyl-2-pentanone

30 20 8

52 24 22


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Problem 13.10 Calculate the index of hydrogen deficiency of these compounds:(a) Aspirin, C9H8O4 (b) Ascorbic acid (vitamin C), C6H8O6 (c) Pyridine, C5H5N

(20-8)/2 = 6 (14-8)/2 = 3 (13-5)/2 = 4(nitrogen correction)

(d) Urea, CH4N2O (e) Cholesterol, C27H46O (f) Dopamine, C8H11NO2

(6-4)/2 = 1 (56-46)/2 = 5 (19-11)/2 = 4(nitrogen correction) (nitrogen correction)

Interpretation of 1H-NMR and 13C-NMR SpectraProblem 13.11 Complete the following table. Which nucleus requires the least energy to flip its spin at this applied field?Which nucleus requires the most energy?

1H

13C

19F

Energy (J/mol)Radio frequency (MHz)Applied field (T)Nucleus

7.05

7.05

7.05 282

75.5

300 0.120

0.0302

0.113

Based on the entries in the table, the 13C requires the least energy to flip its spin and the 1H requires the most.

Problem 13.12 The natural abundance of 13C is only 1.1%. Furthermore, its sensitivity in NMR spectroscopy (a measureof the energy difference between a spin aligned with or against an external magnetic field) is only 1.6% that of 1H. Whatare the relative signal intensities expected for the 1H-NMR and 13C-NMR spectra of the same sample of Si(CH3)4?

A given 13C signal is (0.011)(0.016) = 0.000176 as strong as a given 1H signal. There are three times as many Hatoms as C atoms in Si(CH3)4, so overall the ratio of H to C signals is 1 : (0.000176/3) = 1 : 0.000059. (Notethat this is the same as 17,000 to 1)

Problem 13.13 Following are structural formulas for three constitutional isomers of molecular formula C7H16O and three

sets of 13C-NMR spectral data. Assign each constitutional isomer its correct spectral data.

OH

CCH3

CH3

CH2CH2CH2CH3

CH3CH2CCH2CH3

OH

CH2CH3

Spectrum 362.9332.79

31.8629.1425.7522.6314.08


29.2126.6023.2714.09


7.73

(c)

(b)

CH3CH2CH2CH2CH2CH2CH2OH(a)

These constitutional isomers are most readily distinguished by the number of sets of nonequivalent carbonatoms and thus different 13C signals. Using the following analysis, it can be seen that compound (a) has 7nonequivalent carbon atoms corresponding to Spectrum 3, compound (b) has 6 sets of nonequivalent carbonatoms corresponding to Spectrum 2, and compound (c) has 3 sets of nonequivalent carbon atomscorresponding to Spectrum 1.

OHCCH 3

CH 3

CH 2CH 2CH 2CH 3

g ff

e

ee dd cc bb aaCH 3CH 2CH 2CH 2CH 2CH 2CH 2O H

CH 3CH 2CC H2CH 3

OH

CH 2CH 3c

c c

b

b ba


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Problem 13.14 Following are structural formulas for the cisisomers of 1,2-, 1,3-, and 1,4-dimethylcyclohexane and threesets of 13C-NMR spectral data. Assign each constitutional isomer its correct spectral data.

CH3

CH3

CH3

CH3

CH3

CH3

(a) (b) (c)

Spectrum 131.35

30.6720.85

Spectrum 234.20

31.3023.5615.97

Spectrum 344.60

35.1432.8826.5423.01

These constitutional isomers are most readily distinguished by the number of sets of nonequivalent carbonatoms and thus different 13C signals. Using the following analysis, it can be seen that compound (a) has 4 setsof nonequivalent carbon atoms corresponding to Spectrum 2, compound (b) has 5 sets of nonequivalent carbonatoms corresponding to Spectrum 3, and compound (c) has 3 sets of nonequivalent carbon atomscorresponding to Spectrum 1. The different sets of equivalent carbon atoms are indicated by the letters.

CH 3

CH 3

CH 3

CH 3

CH 3

CH 3

d

c

c

b

ba

a

e

e

d

c

c b

a

a

c

c

b

b

b

b

a

d

a

Problem 13.15 Following are structural formulas, dipole moments, and 1H-NMR chemical shifts for acetonitrile,fluoromethane, and chloromethane.

CH3 C N CH3 F CH3 Cl

Chloromethane

1.87 D

3.05

Fluoromethane

1.85 D

4.26

Acetonitrile

3.92 D

1.97

(a) How do you account for the fact that the dipole moments of fluoromethane and chloromethane are almost identical eventhough fluorine is considerably more electronegative than chlorine?

Recall that dipole moment is proportional to the partial charge times the distance of charge separation.Fluorine is a much smaller atom than chlorine, so it makes shorter bonds leading to relatively short chargeseparation distances. The differences in bond lengths happens to almost exactly offsets the differences inelectronegativities between fluorine and chlorine and the dipole moments come out almost the same.

(b) How do you account for the fact that the dipole moment of acetonitrile is considerably greater than that of eitherfluoromethane or chloromethane?

Again, the key is distance. The acetonitrile has partial charge distributed over more atoms and thus a largerdistance than fluoromethane or chloromethane.

(c) How do you account for the fact that the chemical shift of the methyl hydrogens in acetonitrile is considerably less thanthat for either fluoromethane or chloromethane? (Hint:Consider the magnetic induction in the pi bonds of acetonitrile.)

A magnetic field is induced in the pi system of the nitrile that is against the applied field, thus decreasing thechemical shift.

Problem 13.16 Following are three compounds of molecular formula C4H8O2, and three1H-NMR spectra. Assign each

compound its correct spectrum and assign all signals to their corresponding hydrogens.

C

O

CH3 OCH2CH3 C

O

H OCH2CH2CH3 CH3OC

O

CH2CH3

(1) (2) (3)

For the spectral interpretations in the rest of this chapter the chemical shift () is shown on the structureadjacent to the appropriate hydrogen atom.


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Compound A:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C4H8O2Compound A

2H 2H

1H

3H

()

The spectrum for Compound A corresponds to compound 2: 1H-NMR 8.1 (1H, singlet, H-C(O)-), 4.2 (2H,triplet, -O-CH2-), 1.7 (2H, multiplet, -CH2-), 1.0 (3H, triplet, -CH3).

C

O

H O CH2CH 2CH 31.01.74.28.1

Compound A

( 2 )

Compound B:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C4H8O2

Compound B

2H

3H

3H

()

The spectrum for Compound B corresponds to compound 3: 1H-NMR 3.7 (3H, singlet, CH3-C(O)-), 2.3 (2H,quartet,-C(O)-CH2-), 1.2 (3H, triplet, -CH3).

CH 3 O C

O

CH2CH33.7 2.3 1.2

Compound B

( 3 )


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Compound C:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C4H8O2Compound C

2H

3H

3H

()

The spectrum for Compound C corresponds to compound 1: 1H-NMR 4.1 (2H, quartet, -O-CH2-), 2.0 (3H,singlet, CH3-C-), 1.2 (3H, triplet, -CH3).

C

O

CH 3 O CH 2CH 31.24.12.0

Compound C

( 1 )

Problem 13.17 Following are 1H-NMR spectra for compounds D, E, and F, each of molecular formula C6H12. Eachreadily decolorizes a solution of Br2in CCl4. Propose structural formulas for compounds D, E, and F, and account for theobserved patterns of signal splitting.

Each of the compounds has an index of hydrogen deficiency of 1, in the form of a double bond as evidenced bythe reaction with Br2. The rest of the detailed structures can be deduced from the spectra.

Compound D:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C6H12Compound D

2H2H

1H 1H

6H

()

C C

HCH

H

CH 2

H

CH 3

H3C

4.95

5.00.9 1.9

5.8

0.9

1.6

1H-NMR 5.8 (1H, multiplet; this is more complex than expected because the adjacent vinylic hydrogens arenot equivalent, -CH=), 4.95-5.0 (2H, multiplet, =CH2; this is asymmetric because these two vinylic hydrogensare not equivalent and the hydrogen transto the hydrogen on the other vinylogous carbon has the larger signal


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splitting so it is the signal at 5.0), 1.9 (2H, multiplet; doublet of doublets, -CH2-), 1.6 (1H, multiplet, -CH-),0.9(6H, one doublet, -CH3).

Compound E:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C6H12Compound E

2H1H

3H

3H

3H

()

CH 3-C H2-CH=C

CH 3

CH 3

1.6,1.7

1.6,1.70.9 2.0 5.1

1H-NMR 5.1 (1H, triplet, -CH=), 2.0 (2H, multiplet, -CH2-), 1.6 and 1.7 (6H, two singlets, =C(CH3)2),0.9 (3H,triplet, -CH3)

Compound F:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C6H12Compound F

1H

3H

1H

2H

3H

2H

()

C C

HCH

H

H3C

H

CH 3

CH 2

4.95

5.01.3

5.7

2.00.8

1.0

1H-NMR 5.7 (1H, multiplet; this is more complex than expected because the adjacent vinylic hydrogens arenot equivalent, -CH=), 4.9-5.0 (2H, multiplet, =CH2; this is asymmetric because these two vinylic hydrogensare not equivalent and the hydrogen transto the hydrogen on the other vinylogous carbon has the larger signalsplitting so it is the signal at 5.0), 2.0 (1H, multiplet, -CH-), 1.3 (2H, multiplet, -CH2-), 1.0 (3H, doublet,-CH-CH3), 0.8 (3H, triplet, -CH2-CH3)


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Problem 13.18 Following are 1H-NMR spectra for compounds G, H, and I, each of molecular formula C5H12O. Each isa liquid at room temperature, slightly soluble in water, and reacts with sodium metal with the evolution of a gas. (a)Propose structural formulas of compounds G, H, and I.

The index of hydrogen deficiency is 0 for these molecules, so there are no rings or double bonds. The fact thatthe compounds are slightly soluble in water and react with sodium metal indicates that each molecule has an-OH group. The chemical shifts associated with each set of hydrogens are indicated on the structures.

Compound G:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

Compound G

3H

6H

1H1H

1H

C5H12O

()

CH 3 CH

CH 3

CH

OH

CH 3

0.9

0.9 1.15

1.6

1.85

3.5*

1H-NMR 3.5 (1H, multiplet, -CH-OH-), 1.85 (1H, doublet, -OH), 1.6 (1H, multiplet, -CH-(CH3)2), 1.15 (3H,doublet, -C(OH)-CH3), 0.9 (6H, overlapping doublets; this is more complex than expected because it is adjacentto a chiral center -CH-(CH3)2).

Compound H:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C5H12O

Compound H

2H

6H

2H 1H1H

()

CH 3

CH3CH2CHCH2OH

0.8-0.9

0.8-0.9

1.1

1.4-1.6

2.2

3.4-3.5*

1H-NMR 3.4-3.5 (2H, multiplet; this is more complex than expected because it is adjacent to a chiral center-CH2-OH), 2.2 (1H, broad triplet, -OH), 1.4-1.6 (2H, multiplet; this is more complex than expected because it isadjacent to a chiral center CH3-CH2-), 1.1 (1H, multiplet, -CH-), 0.8-0.9 (6H, broad multiplet, both -CH3groups).


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Compound I:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C5H12O

Compound I

2H

1H

2H

3H

Hint: This signal isa pair of overlapping

2H multiplets

()

2.9

CH3CH 2CH 2CH 2CH2OH1.4 3.61.551.40.9

1H-NMR 3.6 (2H, broad multiplet, -CH2-OH), 2.9 (1H, broad peak, -OH), 1.55 (2H, multiplet, -CH

2-CH

2-

OH), 1.4 (4H, multiplet, CH3-CH2-CH2- and CH3-CH2-CH2-), 0.9 (3H, triplet, -CH3)

(b) Explain why there are four lines between 0.86 and 0.90 for Compound G.

Carbon atom 2 (with the -OH group) is a chiral center. That makes the two methyl groups diastereotopic, sothey have different chemical shifts. The four lines are actually two doublets.

(c) Explain why the 2-H multiplets at 1.5 and 3.5 ppm for compound H are so complex.

Carbon atom 2 is a chiral center. The chiral center makes the adjacent -CH2protons diastereotopic, so thesignals are complex.

Problem 13.19 Propose a structural formula for compound J, molecular formula C3H6O, consistent with the following1H-

NMR spectrum:

Compound J:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

expansion

C3H 6O

Compound J

1H

3H

1H

1H

3 2

1H

1H1H

()

From the molecular formula, there is an index of hydrogen deficiency of 1 indicating that there is one ring or pibond.

O CH 3

H

H

H

1.3

3.02.4

2.75


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1H-NMR 3.0 (1H, multiplet, Hc), 2.4 and 2.75 (2H, multiplets; these hydrogens are not equivalent because thecarbon-carbon bonds cannot rotate freely), 1.3 (3H, doublet, -CH3)

Problem 13.20 Compound K, molecular formula C6H14O, readily undergoes acid-catalyzed dehydration when warmed

with phosphoric acid to give compound L, molecular formula C6H12, as the major organic product. The1H-NMR

spectrum of compound K shows signals at 0.90 (t, 6H), 1.12 (s, 3H), 1.38 (s, 1H), and 1.48 (q, 4H). The 13C-NMRspectrum of compound K shows signals at 72.98, 33.72, 25.85, and 8.16. Deduce the structural formulas of compounds Kand L.

From the molecular formula, there is a hydrogen deficiency index of 0, so there are no rings or pi bonds incompound K. From the 13C-NMR peak at 72.98, there is a carbon bonded to an -OH group. The rest of thestructure can be deduced from the 1H-NMR spectrum. The chemical shifts associated with each set ofhydrogens are indicated on the structure.

CH 3CH 2CC H2CH 3

OH

CH 3

1.38

1.12

1.48 0.901.480.90

1H-NMR 1.48 (4H, quartet, -CH2), 1.38 (1H, singlet, -OH), 1.12 (3H, singlet, -C(OH)-CH3), 0.90 (6H, triplet,both -CH2-CH3 groups)

Dehydration of compound K gives the following alkene as compound L:

C C

CH 3

CH 2CH 3H

CH 3

Problem 13.21 Compound M, molecular formula C5H10O, readily decolorizes Br2in CCl4, and is converted by H2/Ni into

compound N, molecular formula C5H12O. Following is the1H-NMR spectrum of compound M. The 13C-NMR

spectrum of compound M shows signals at 146.12, 110.75, 71.05, and 29.38. Deduce the structural formulas ofcompounds M and N.

Compound M:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C5H10O

Compound M

1H 1H1H

1H

6H

()

From the reaction with Br2and H2/Ni it is clear the compound M has a carbon-carbon double bond. These

conclusions are supported by the13

C-NMR signals corresponding to the sp2

carbons at

146.12 and 110.75.There is also a carbon bonded to an -OH group judging from the signal at 71.05. The rest of the structure isdeduced from the 1H-NMR spectrum.

C C

C

OH

H

H

H

CH 3

CH 3

5.2

5.0

1.9

1.3

1.3

6.0


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1H-NMR 6.0 (1H, doublet of doublets), 5.0 and 5.2, (2H, doublet of doublets; the vinylic hydrogens are notequivalent and the hydrogen transto the hydrogen on the other vinylogous carbon has the larger signalsplitting so it is the signal at 5.2), 1.9 (1H, singlet, -OH), 1.3 (6H, singlet, C-(CH3)2)

Upon hydrogenation, compound M is reduced to the alcohol shown below as compound N:

OH

C

CH 3

CH 3CH3-C H2

Problem 13.22 Following is the 1H-NMR spectrum of compound O, molecular formula C7H12. Compound O reacts

with bromine in carbon tetrachloride to give a compound of molecular formula C7H12Br2. The13C-NMR spectrum of

compound O shows signals at 150.12, 106.43, 35.44, 28.36, and 26.36. Deduce the structural formula of compound O.

Compound O:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C7H12Compound O

2H

6H

4H

()

The molecular formula indicates that there is an index of hydrogen deficiency of 2, so there are two rings and/orpi bonds. The 13C-NMR indicates there is only one double bond because there are only two resonancescorresponding to sp2carbon atoms (150.12 and 106.43). Therefore, compound O must have one ring and one pibond.

CH 2

2.11.5

1.5

1.5 2.1

4.6

1H-NMR 4.6 (2H, singlet, =CH2), 2.1 (4H, broad peak, two -CH2- groups), 1.6 (6H, broad peak, three -CH2-groups)

Problem 13.23 Treatment of compound P with BH3followed by H2O2/NaOH gives compound Q. Following are1H-

NMR spectra for compounds P and Q along with 13C-NMR spectral data. From this information, deduce structuralformulas for compounds P and Q

22.6828.1337.59

72.71

27.4529.1432.12

132.38

(Q)(P)

13C-NMR

C7H14O

(Q)2) H2O2, NaOH

1) BH3

(P)

C7H12


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Compound P:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C7H12Compound P

4H

2H

4H

2H

()

The molecular formula for P indicates that it has an index of hydrogen deficiency of 2 so that is has two ringsand/or pi bonds. The 13C-NMR spectral data shows that there is an sp2 carbon atom (132.38). Since theremust be two sp2carbon atoms to make a pi bond, then the molecule must be symmetric so that both sp 2carbonatoms are equivalent. This also explains why there are so few other 13C-NMR signals. Since there is

presumably only one pi bond, then there must be one ring in the molecule. The rest of the structure can bededuced from the 1H-NMR spectrum.

2.1

1.5

1.71.5

2.1

5.85.8

1H-NMR 5.8 (2H, triplet, both =CH-), 2.1 (4H, multiplet, two -CH2- groups), 1.7 (2H, quintet, the unique-CH2- group), 1.5 (4H, multiplet, two -CH2- groups).

Compound Q:

10 9 8 7 6 5 4 3 2 1 0 ppm

Chemical Shift (d)

C7

H14

O

Compound Q

1H

13H

()

Given the structural formula for P, it is clear that compound Q would be the hydroboration/oxidation product,

namely the alcohol shown below. This structure is consistent with the 13C-NMR spectral data provided aswell as the 1H-NMR spectrum.

OH1.4-1.9

1.4-1.9

1.4-1.9

1.4-1.91.4-1.9

2.0

1.4-1.9

3.8


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1H-NMR 3.8 (1H, broad peak, -C(OH)H-), 2.0 (1H, sharp singlet, -OH), 1.4-1.9 (12H, broad multiplets, all theremaining hydrogens on the ring. The peaks are so broad and the patterns so complex because this ring doesnot have a double bond to hold it rigid, so it has a great deal of flexibility)

Problem 13.24 The 1H-NMR of Compound R, C6H14O, consists of two signals: 1.1 (doublet) and 3.6 (septet) in theratio 6:1. Propose a structural formula for compound R consistent with this information.

There is a hydrogen deficiency index of 0, so there are no rings or pi bonds in compound R. The simplicity ofthe 1H-NMR spectrum indicates a highly level of symmetry in the molecule, with each methyl group beingbonded to a carbon with a single hydrogen atom. The only structure consistent with all of this information isthe following ether. The chemical shifts associated with each set of hydrogens are indicated on the structure.

OHC CH

CH 3

CH 3

H3C

H3C

3.63.6

1.1

1.11.1

1.1

Compound R

Problem 13.25 Following are eight structural formulas along with the 13C-NMR and DEPT spectral data. Given this data,assign each carbon in each compound its correct 13C chemical shift.

CHCH3

Br

CH3CH2CH213.40 21.00

26.46

43.22

51.55

13C DEPT

51.55 CH

CH243.22

26.46 CH2CH321.00

13.40 CH3

(a) C=CH2

CH3

CH3CH2

-

12.23

22.47

30.56 108.33

147.70

13C DEPT

147.70

CH2108.33

30.56 CH2CH322.47

12.23 CH3

(b) CHCH3

CH3

CH2=CHCH2(c)

CH322.26

28.12 CH

CH243.35

115.26 CH2

137.81

DEPT13C

CH

137.81

115.26 43.35

28.12

22.26

22.26

CCH2Br

CH3

CH3

CH3

(d)

CH328.72

33.15

49.02

DEPT13C

CH2

49.02

33.15

28.72

28.72

28.72

-

CCH2CH3

O

CH3CH2

-

35.1

7.5 7.5

35.1

207.8

13C DEPT

207.8

CH235.1

7.5 CH3

(e) CCH3

O

CH3CH2

-

(f)

9.2 CH3

CH330.1

37.6 CH2

208.7

DEPT13C

208.7

37.6 30.1

9.2

COCH3

O

CH

CH3

CH3

-

177.4819.01

19.01

33.94 51.50

13C DEPT

177.48

CH351.50

33.94 CH

CH319.01

(g) CHCH3

CH3

CCH3

O

OCH2CH2(h)

-13C DEPT

171.17

CH263.12

37.21 CH3CH25.05

24.45 CH2CH321.02

171.17

63.12

24.45 25.05

37.21 21.02

21.02


16/18


Problem 13.26 Write structural formulas for the following compounds:(a) C2H4Br2: 2.5 (doublet, 3H) and 5.9 (quartet, 1H)

5.92.5

CH3-CHBr2

(b) C4H8Cl2: 1.60 (doublet, 3H), 2.15 (multiplet, 2H), 3.72 (triplet, 2H), and 4.27 (multiplet, 1H)

CH 3-CHCl-CH2-C H2C l1.6 4.27 2.15 3.72

(c) C5H8Br4: 3.6 (singlet, 8H)

C

CH2B r

CH2B r CH2B r

CH2B r

3.6

3.6

3.6

3.6

(d) C4H8O: 1.0 (triplet, 3H), 2.1 (singlet, 3H), and 2.4 (quartet, 2H)

CH 3CH 2 C

O

CH 32.12.41.0

(e) C4H8O2: 1.2 (triplet, 3H), 2.1 (singlet, 3H) and 4.1 (quartet, 2H); contains an ester group

CH 3 C

O

O CH 2 CH 3

1.22.1 4.1

(f) C4H8O2:1.2 (triplet, 3H), 2.3 (quartet, 2H) and 3.6 (singlet, 3H); contains an ester group

CH 2 C

O

O CH 3CH 3

1.2 2.3 3.6

(g) C4H9Br: 1.1 (doublet, 6H), 1.9 (multiplet, 1H), and 3.4 (doublet, 2H)

CH 3

CHCH 3 CH 2B r3.4

1.9

1.1

1.1

(h) C6H12O2: 1.5 (singlet, 9H) and 2.0 (singlet, 3H)

CH 3 C

O

O C

CH 3

CH 3

CH 31.5

1.5

1.5

2.0

(i) C7H14O: 0.9 (triplet, 6H), 1.6 (sextet, 4H), and 2.4 (triplet, 4H)

CH3-C H2-C H2 C

O

CH2-C H2-C H3

2.4 2.4 1.61.6 0.90.9

(j) C5H10O2: 1.2 (doublet, 6H), 2.0 (singlet, 3H) and 5.0 (septet, 1H)

CH 3 C

O

O C

CH 3

H

CH 31.2

1.2

5.0

2.0


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(k) C5H11Br: 1.1 (singlet, 9H) and 3.2 (singlet, 2H)

CH 3

CCH 3

CH 3

CH 2B r3.2

1.1

1.1

1.1

(l) C7H15Cl: 1.1 (singlet, 9H) and 1.6 (singlet, 6H)

CH 3

CCH 3

CH 3

C

CH 3

CH 3

Cl1.6

1.6

1.1

1.1

1.1

Problem 13.27 The percent s-character of carbon participating in a C-H bond can be established by measuring the 13C-1Hcoupling constant and using the relationship

Percent s-character = 0.2 J(13C-1H)

The 13C-1H coupling constant observed for methane, for example, is 125 Hz, which gives 25% s-character, the valueexpected for an sp3hybridized carbon atom.

(a) Calculate the expected 13C-1H coupling constant in ethylene and acetylene.

For ethylene and acetylene the carbon atoms are sp2and sp hybridized and thus 33% and 50% s-character,respectively. Using the above equation gives coupling constants of 165 Hz and 250 Hz, respectively.

(b) In cyclopropane, the 13C-1H coupling constant is 160 Hz. What is the hybridization of carbon in cyclopropane?

The carbon atoms in cyclopropane are (0.2)(160) = 32% s-character. This corresponds roughly to an sp2

hybridized carbon atom.

Problem 13.28 Ascaridole is a natural product that has been used to treat intestinal worms. Explain why the two methylson the isopropyl group in ascaridole appear in its 1H-NMR as 4 lines of equal intensity, with two sets of two eachseparated by 7 Hz.

0 0

0 0

0 0

0 0

O

O

Ascaridole

*

* *

*

Ascaridole is a chiral molecule, having two chiral centers as shown. As a result, the NMR spectrum is complexand the two methyl groups of the isopropyl groups are split. The best way to think about it is that the twomethyl groups are themselves diastereotopic, thus giving rise to split signals.

Problem 13.29 The 13C-NMR spectrum of 3-methyl-2-butanol shows signals at 17.88 (CH3), 18.16 (CH3), 20.01(CH3), 35.04 (carbon-3), and 72.75 (carbon-2). Account for the fact that each methyl group in this molecule gives adifferent signal.

Because of the chiral center at carbon-2, the two methyl groups of carbon-4 and carbon-5 are diastereotopic andthus have two different chemical shifts. In addition, carbon-1 is not equivalent to 4 or 5, so it also has a uniquechemical shift.

35.04

72.751234

CHCH 3 CH C H3

OH

*CH 3

20.01

517.88

18.16


18/18


Problem 13.30 Sketch the NMR spectrum you would expect from a partial molecule with the following parameters.

RC

CC

RHa

R

Hb

R

Hc

Hc

Ha= 1.0 ppm

Hb= 3.0 ppm

Hc= 6.0 ppm

Jab= 5.0 Hz

Jb c= 8.0 Hz

Jac= 1.0 Hz

The Jaccoupling of 1 Hz is so small that it can be safely ignored since it will not show up on normal resolutionspectra. Assuming the R groups are not involved with signal splitting, expect the signal for Haat 1.0 ppm,integrating to a single hydrogen atom, to be split into a doublet with coupling constant J ab= 5.0 Hz. Expectthe signal for Hcat 6.0 ppm, integrating to two hydrogen atoms, to be split into a doublet with couplingconstant Jbc= 8.0 Hz. Expect the signal for Hbat 3.0 ppm, integrating to a single hydrogen atom, to be splitinto a doublet of triplets with coupling constants Jab= 5.0 Hz and Jbc= 8.0 Hz. The following is drawn asexpanded views around the signals.

6.0 ppm 3.0 ppm 1.0 ppm

Hc

Hb

Ha

8 Hz

5Hz

8 Hz 8 Hz

5Hz

5Hz

5Hz

Note that if the Jac= 1.0 Hz coupling is seen, the signal for Hawill be split into a doublet of closely spacedtriplets, and the signal for H

cwill be a doublet of closely spaced doublets.

6.0 ppm 3.0 ppm 1.0 ppm

Hc

Hb Ha

8 Hz

5Hz

8 Hz 8 Hz

5Hz

5Hz

5Hz

organic chemistry textbook solution of chapter 13

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