Study guide and students solutions manual for organic chemistry

Identify the compound with molecular formula C8H7O2Br that gives the following IR and 1H NMR spectra... Identify the compound with molecular formula C7H6O that gives the following 1H NMR

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Spectroscopy Problems

1 Determine the structure of the straight-chain pentanol that produces the mass spectrum shown here

0 20

80 100

2 The mass spectrum of an ether is shown here Determine the molecular formula of the ether that would

produce this spectrum and then draw possible structures for it

0 20

80 100

From Spectroscopy Problems of Study Guide and Solutions Manual for Organic Chemistry, Seventh Edition Paula Yurkanis Bruice

Copyright © 2014 by Pearson Education, Inc All rights reserved.



3 The mass spectra of pentane and isopentane are shown here Determine which spectrum belongs to which

compound

0 20

80 100

0 20

80 100



4 Which of the following compounds gives the mass spectrum shown here?

0 20

80 100

2 N O

5 An unknown acid underwent a reaction with 1-butanol The product of the reaction gave the mass spectrum

shown here What is the product of the reaction, and what acid was used?

0 20

80 100



6 Identify the compound with molecular formula C9H10O3 that gives the following IR and 1H NMR spectra.

1600 1800

0.2

0.3

0.4 0.5 0.6 0.7 0.9

2.0

A S O B N E

8

1 2



8 Identify the compound with molecular formula C6H12O that gives the following IR and 1H NMR spectra.

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0 WAVENUMBERS

A B S O R B A N C

9 A methyl-substituted benzene is treated with Cl2 in the presence of AlCl3, and a 1H NMR spectrum is taken

of one of the monochlorinated products that is isolated from the reaction mixture Identify this compound

10 Identify the compound with molecular formula C6H14O that gives the following IR and 1H NMR spectra.

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R B A N C E

12 Identify the compound with molecular formula C6H12 that gives the following IR and 1H NMR spectra.

6

~2.2

1600 1800

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R A C E

1 2 7



14 Identify the compound with molecular formula C4H7ClO that gives the following IR and 1H NMR spectra.

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R A C

16 Identify the compound with molecular formula C4H6O that gives the following IR and 1H NMR spectra.

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R A C

18 The 1H NMR spectra of 1-chloro-3-iodopropane and 1-bromo-3-chloropropane are shown here Which

compound gives which compound spectrum?

20 Identify the compound with molecular formula C5H10O2 that gives the following 1H NMR spectrum.

6 3

1

0 5

21 Identify the compound with molecular formula C8H7O2Br that gives the following IR and 1H NMR spectra

1600 1800

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R B A C

22 Identify the compound with molecular formula C7H6O that gives the following 1H NMR spectrum.

10

1

23 The two 1H NMR spectra shown here are given by constitutional isomers with molecular formula C3H7Br

Identify each isomer

a

6 1

24 An unknown alkene with molecular formula C8H16 undergoes ozonolysis Only one product is formed The

IR and 1H NMR spectra of the product are shown here What is the product? What alkene produced this product?

1600 1800

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R A C

25 Identify the compound with molecular formula C4H7BrO2 that gives the following IR and 1H NMR spectra.

1600 1800

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R A C

27 A compound with the following IR spectrum was formed by a reaction with 1-propyne If the number of

carbons in the reactant and product is the same, what compound is formed and what reaction conditions produced this compound?

1600 1800

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R A C

29 Identify the compound with molecular formula C4H8O2 that gives the following 1H NMR spectrum.

2

30 Identify the compound with molecular formula C9H11NO that gives the following 1H NMR spectrum

7 8

6 2

2 1



32 The 13C NMR and 1H NMR spectra of1, 2-, 1, 3-, and 1, 4-ethylmethylbenzene are shown here Determine

which spectrum belongs to which compound

33 Identify the compound with molecular formula C7H14O that gives the following 1H NMR spectrum.

~2.8 ppm



34 Identify the compound with molecular formula C5H10 that gives the following 1H NMR spectrum.

7

1

3 3 3

35 Identify the compound with molecular formula C3H4O that gives the following IR and 1H NMR spectra

1600 1800

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R B A N C E

36 Identify the compound with molecular formula C4H7Cl that gives the following 1H NMR spectrum.

38 An unknown alcohol gives the following 1H NMR spectrum Identify the alcohol (HINT: Because this is a

spectrum of a pure alcohol, the OH proton is split by adjacent protons.)

2 6



40 Identify the compound with molecular formula C3H7NO that gives the following IR and 1H NMR spectra.

1600 1800

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R B A N C E

2

3



41 The 1H NMR spectra shown here are given by constitutional isomers of propylamine (C3H9N) Identify the

isomer that gives each spectrum

a

1

2 3



42 Identify the compound with molecular formula C3H7N that gives the following IR and 1H NMR spectra.

1600 1800

0.1

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R A C

1

1 1

3



44 Identify the compound with molecular formula C7H8O2 that gives the following IR and NMR spectra.

1600 1800

0.2

0.3 0.4 0.5 0.6 0.7 0.8 1.0 2.0

A B S O R A C

3



45 Identify the alcohol that gives the following 1H NMR spectrum.

4

1 3 6

46 Identify the compound with molecular formula C6H12O2 that gives the following 1H NMR data The

num-ber of hydrogens responsible for each signal is given in parentheses

47 Identify the compound with molecular formula C9H10O that gives the following 1H NMR data The

num-ber of hydrogens responsible for each signal is given in parentheses

Answers to Spectroscopy Problems

1 There are three straight-chain pentanols: 1-pentanol, 2-pentanol, and 3-pentanol Because the most stable

fragment for an alcohol is the one formed by a-cleavage, we can see which of the alcohols will form the

base peak shown in the spectrum (that is, a base peak with m >z = 45) as a result of a-cleavage.

For 1-pentanol, only one a-cleavage is possible It will form a cationic fragment with m >z = 31.

CH3CH2CH2

m/z = 45

For 3-pentanol, only one a-cleavage is possible because of the symmetry of the molecule a-Cleavage will

form a cationic fragment with m >z = 59.

CH3CH2

OH

-cleavage α

CH3CH2

m/z = 59

The base peak of the given mass spectrum has m >z = 45 Thus, the mass spectrum is that of 2-pentanol

We also see a significant fragment at m>z = 73, the m/z value of the other a-cleavage product.



2 First, we must first identify the molecular ion The molecular ion, the peak that represents the intact

start-ing compound, has an m>z = 74 Now we can use the rule of 13 to determine the molecular formula.

74

13 = 5 carbons with 9 left overFrom the rule of 13, we end up with a molecular formula of C5H14 Because the compound is an ether, we know that it has one oxygen, so we must add one O and subtract one C and four Hs from the molecular formula The resulting molecular formula is:

C4H10OThere are three ethers that have this molecular formula: methyl propyl ether, diethyl ether, and isopropyl methyl ether

O

3 First, we need to determine what the most abundant cationic fragments would be for each compound

The possible fragments for pentane are:

Fragmentation pattern 2 is expected to give the base peak (the most stable fragment) The cation formed in

pattern 2 (m>z = 43) will be more stable than the cation formed in pattern 3 (m>z = 29), because the

lon-ger fragment will result in greater inductive stabilization (the lonlon-ger chain will do a better job of stabilizing the carbocation)

Fragments from fragmentation patterns 1 and 4 are expected to be less common They each form a mary species, like fragmentation patterns 2 and 3, but the second species they form is a methyl fragment (either a radical or a carbocation), which is less sable than the second species formed in fragmentation patterns 2 and 3

pri-

Only four fragmentation patterns are shown for isopentane Fragmentations that would result in only a

pri-mary fragment and a methyl fragment have been excluded because these would be less abundant than those shown here

The four major fragments occur at the same m/z values (57, 43, 29, and 15) as those found for pentane, but

their relative intensities are different

Fragmentation patterns 1 and 3 are expected to produce the most abundant fragments because they both form a secondary cation, and the stability of the cation is more important than the stability of the radical in

determining the most abundant fragments Therefore, we expect a base peak with m>z = 43 (because the secondary cation is accompanied by a primary radical) and a less intense peak with m >z = 57 (because

the secondary cation is accompanied by a methyl radical)

Both spectra show a base peak at m >z = 43 The major difference in the two spectra is the intensity of the peak with m >z = 57 The spectrum of isopentane should show a more intense peak because it is due to a secondary cation, whereas the peak with m >z = 57 in the spectrum of pentane is due to a primary cation.

Thus, pentane gives the first mass spectrum and isopentane gives the second.



4 The molecular ion for this compound has an m>z = 73 The nitrogen rule states that if a molecular ion has

an odd value, then the structure must have an odd number of nitrogens Therefore, we can eliminate the alkane, the ketone, and the ether

Now we can determine the molecular formula of the compound using the rule of 13 When we subtract 14 (the mass of nitrogen) from 73, we get 59

in each compound Given the relative stability of this fragment, we can anticipate that it will be the base peak of the mass spectrum

CH2

H3C N

CH3

-cleavage α

CH3+

-cleavage α

a-Cleavage of N,N-dimethylethylamine gives a cation with m >z = 58 a-Cleavage of butylamine gives

a cation with m >z = 30 The spectrum shows a base peak with m>z = 58, indicating that the compound

that gives the spectrum is N,N-dimethylethylamine.

N

5 The mass spectrum has two peaks of identical height with m/z values = 136 and 138, indicating the

pres-ence of bromine in the product (Bromine has two isotopes of equal abundance with weights of 79 and

81 amu.)

Now we need to think about the type of reaction that occurred

Under acidic conditions, the starting material (1-butanol) will be protonated

O H

H H

A−

O+



The protonated alcohol now has a leaving group that can be replaced by a nucleophile Because we know that bromine is present in the product, we can assume that bromide ion is the incoming nucleophile.

H H

We now know that the product of the reaction is 1-bromobutane The acid, which must be the source of the nucleophile, is HBr.

6 The two signals near 7 and 8 ppm are due to the hydrogens of a benzene ring Because these signals

inte-grate to 4 protons, the benzene ring must be disubstituted Since both signals are doublets, the protons that

give each signal must each be coupled to one proton (N + 1 = 1 + 1 = 2) Thus, the substituents must

be at the 1- and 4-positions The proton or set of protons that gives the signal at the lowest frequency is labeled a, the next lowest b, and so on.

(C3H6O3 - C2H5O = CHO2) There is one remaining NMR signal, a singlet (9.8 ppm) that integrates to 1 proton

To help with the identification, we turn to the IR spectrum The broad absorption near 3200 cm-1 indicates the O¬ H stretch of an alcohol; the proton of the OH group would give the broad NMR signal at 9.8 ppm The strong absorption at 1680 cm-1 indicates the presence of a carbonyl C=O group

Now that all the fragments of the compound have been identified, we can put them together The

com-pound is ethyl 4-hydroxybenzoate.

OCH2CH3

O



7 The signals at 1.1 and 1.8 ppm have been magnified and are shown as insets on the spectrum (the 2 and 1

represent the ppm scale) so you can better see the splitting The triplet (1.1 ppm) that integrates to 3 tons and the quartet (1.8 ppm) that integrates to 2 protons are characteristic of an ethyl group (The peak to the right of the quartet is actually the beginning of the adjacent signal that integrates to 6 protons.)

pro-The singlet (1.7 ppm) that integrates to 6 protons indicates that there are two methyl groups in the same environment Because the signal is a singlet, the carbon to which they are attached cannot be bonded to any hydrogens The only atom not accounted for in the molecular formula is Br

C

CH3

CH3Therefore, the ethyl group and the bromine must be the two substituents that are attached to the carbon

Thus, the compound is 2-bromo-2-methylbutane.

CH3

Br

CH3

8 A major clue comes from the IR spectrum The strong absorption at ∼1710 cm-1 indicates the presence

of a carbonyl (C=O) group Because the compound has only one oxygen, we know that it must be an aldehyde or a ketone The absence of absorptions at 2820 and 2720 cm-1 tells us that the compound is not

an aldehyde

The absorptions at 2880 and 2970 cm-1 are due to C¬ H stretches of hydrogens attached to sp3 carbons The 1H NMR spectrum has two unsplit signals One integrates to 9 protons, the other to 3 protons A signal

that integrates to 9 protons suggests a tert-butyl group, and a signal that integrates to 3 protons suggests

a methyl group The fact that they are both singlets indicates that they are on either side of the carbonyl

group Thus, the compound is 3,3-dimethyl-2-butanone.

9 From the reaction conditions provided, we know that the product is a monochlorinated toluene.

Cl

Cl2, AlCl3

The singlet (2.3 ppm) that integrates to 3 protons is due to the methyl group

The signals in the 7–8 ppm region that integrate to 4 protons are due to the protons of a disubstituted zene ring Because both signals are doublets, we know that each proton is coupled to one adjacent proton Thus, the compound has a 1,4-substituted benzene ring

ben-Therefore, the compound is 4-chloromethylbenzene.

10 The strong and broad absorption in the IR spectrum at 3400 cm-1 indicates a hydrogen-bonded O¬ H

group The absorption bands between 2800 and 3000 cm-1 indicate hydrogens bonded to sp3 carbons.There is only one signal in the 1H NMR spectrum that integrates to 1 proton, so it must be due to the hydrogen of the OH group The singlet that integrates to 3 protons can be attributed to a methyl group that

is attached to a carbon that is not attached to any hydrogens

Since the other two signals show splitting, they represent coupled protons (that is, protons on adjacent bons) The quartet and triplet combination indicates an ethyl group Since the quartet and triplet integrate

car-to 6 and 4 procar-tons, respectively, the compound must have two ethyl groups

The identified fragments of the molecule are:

Hb

When these fragments are subtracted from the molecular formula, only one carbon remains Therefore, this

carbon must connect the four identified fragments The compound is 3-methyl-3-pentanol.

OH

11 The 1H NMR spectrum contains only one signal, so only one type of hydrogen is present in the

mol-ecule Because the compound has 4 carbons and 9 identical hydrogens, the compound must be tert-butyl