Structure Determination: Mass Spectrometry, IR, and UV Spectroscopy 219

Solutions to Problems

10.1 If the isotopic masses of the atoms C, H, and ° had integral values of 12amu, 1 amu and16amu, many molecular formulas would correspond to a molecular weight of 288 amu.Because isotopic masses are not integral, however, only one molecular formula isassociatedwith a molecular ion at 288.2089 amu.

To reduce the number of possible formulas,assume that the difference in molecular

weight between 288 and 288.2089 is due mainlf.'to hydrogen. Divide 0.2089 by 0.00783,the amount by which the atomic weight of one H atom differs from 1.The answer,26.67, gives a "ballpark"estimate of the number of hydrogens in testosterone. Then,divide 288 by 12, to determine the maximum number of carbons. Since 288 + 12=24,weknow that testosteronecan have no more than 22 carbons if it also includes hydrogen andoxygen. Make a list of reasonablemolecular formulas containingC, Hand ° whose massis 288 and that contain 20-30 hydrogens.Tabulate these, and calculatetheir exact massesusing the values in the text.

The only possible formula for testosterone is C19H2S02.

9CH3I

CH3CH2CH = C\CH3

2-Methylpent-2-ene Hex -2-ene

Fragmentationsoften occur to give relativelystable carbocations.

+ CH3CH2CH= cl\

m/z = 69 CH3 m/z = 55

Spectrum (a), which has a large peak atm/z= 69,corresponds to 2-methylpent-2-ene,andspectrum (b), which has m/z = 55 as its base peak, corresponds to hex-2-ene. Both ofthesepeaks are due to allylic cations.

~ 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Isotopic mass

Molecular Mass Mass Mass Mass offormula of carbons of hydrogens of oxygens molecular ion

C2oH32O 240.0000 amu 32.2504 amu 15.9949 amu 288.2453 amu

C 19H2S02 228.0000 28.2191 31.9898 288.2089

CIsH2403 216.0000 24.1879 47.9847 288.1726

.220 Chapter 10

A((3) In a mass spectrum,the molecular ion is both a cation and a radical. When it fragments,V two kinds of cleavage can occur. (1) Cleavage can fomi a radical and a cation (the species

observed in themass spectrum).Alpha cleavage shows this type of pattern. (2) Cleavagecan form a neutral moleculeand a different radical cation (the speciesobservedin the massspectrum).Alcohol dehydrationand the McLaffertyrearrangementshow this cleavagepattern. \

For each compound, calculatethe mass of the molecular ion and identifythe functional

1gro~ps present. Draw the fragmentationproducts and calculate their masses.(a)

l

' 0

]

+"

1/ I Alpha/C, .H3C :CH2CH2CH3 cleavage

M+ =86 .

l

0

]

+"

,)1 Alpha _ +

H CyC'

CH CH CH I" H3C" + [O=C-CH2CH2CH3 ]3: 2 2 3 C eavage .M+ =86 mlz =71

In theory, alpha cleavage can take place on either side of the carbonyl group, to producethe cations with mlz =43 and mlz =71. In practice, cleavage occurs on the moresubstituted side of the carbonyl group, and the fIrst cation, with mlz =43, is observed.

+

[ H3C-C=O ] +" CH2CH2CH3

mlz =43

(b) I

J

+"

[ ]

+"

- lUOH DehYdratiOn.-O + H20

M+= 100 mlz = 82Dehydration of cyclohexanolproduces a cation radical with mlz =82.

(c)

[

CH3 0

J

+" McLaffertyI 1/ ..CH3CHCH2CCH3 rearrangement

M+ =100

H H~H,\ / 09~ II

H3C_C~C'CH/ C

H 1\H H

+"

[

H,

]

+"

?'iC,

H2C CH3.

mlz =58~

+

?lH2C

H3C/ 'H

The cation radical fragment resulting from McLaffertyrearrangementhas mlz =58.

(d)

[

I

]

+" .CH2CH3 CH2CH3 +

1 Alpha I, N, .. H3C" + 'iN,

H3CtCH2 CH"cH3 cleavage [ H"c CH2CH3 ]M+ =101 mlz =86

Alpha cleavageof triethylamineyields a cation with mlz =86.

iC 20 II Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Structure Determination: Mass Spectrometry, IR, and UV Spectroscopy 221

9.0.4 The molecule,2-methylpentan-2-ol,produces fragments that result from both dehydrationand from alpha cleavage.Two differentalpha cleavage productscan be detected.

[

CH

]

+.

CH3CH2CH2t~3

M+ =102

!Alphacleavage

[

OH

J

+

CH3CH2CH2< + H3C..CH3

mlz =87

Dehydration.

[HO'?"CH3r

CH3 Jmlz =59

/'"'

0.5') We know that: (1) energy increases as wavelengthdecreases, and (2) the wavelengthof X-J radiation is shorter than the wavelengthof infraredradiation.Thus, we estimate that an X

ray is of higher energy than an infrared ray.

E=hv =he/A; h =6.62 X 10-34 J.s; c =3.00 X 108mls

for.A =1 x 1O-{)m (infrared radiation):

E =(6.62X 10-34J.s) (3.00 X108mls) = 2.0 X 10-19J1.0 X 1O-{)m

for A=3.0 x 10-9 m (X radiation):.( -34 )( 8 )

E = 6.62 x 10 J.s 3.00 x 10 mls = 6.6 X 10-17J3.0 x 10-9m

ConfIrmingour estimate, the calculation shows that an X ray is of higher energy thaninfrared radiation.

When comparing radiationexpressedby differentunits, convert radiation in m to radiationin Hz by the equation:

c 3.00 X 108 mls = 3.3 X 1013 Hzv =A = 9.0 X 1O-{)m

The equation E =hv shows that the greater the value of v, the greater the energy. Thus,radiation with v =3.3 X 1013 Hz (A. =9.0 x 10-6m) is higher in energy than radiation withv =4.0 x 109 Hz.

<02011 Cengage Learning. All Rights Reseryed. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

222 Chapter 10

10.6 Use the expression for energy shown in this section. Convert frequency to wavelengthwhen you need to.

(a) E = 1.20 X 10-4kJ/mol = 1.20 x 10-4kJ/molf...(in m) 5.0 x 10-11

= 2.4 X 106kJ/molfor a gammaray.

(b) E = 4.0X 104kJ/molfor an X ray.

(c) v = .£..; f... = £ = 3.0x 108mIs = 5.0 x 1O-8mf... v 6.0 X 1015Hz

E = 1.20 X 19-4kJ/mol = 2.4 x 103kJ/mol for ultraviolet light5.0.'x 10-8 .

(d) E =2.8 X 102kJ/molforvisible light.

(e) E =6.0 kJ/mol for infrared radiation

(f) E =4.0 X 10-2kJ/mol for microwave radiation.

10.7 (a) A compound with a strong absorptionat 1710cm-1contains a carbonyl group and iseither a ketone, an aldehyde, or an ester.

(b)A compound with a nitro group has a strong absorption at 1540 cm-1.~ . ,

(c) A compound showing both carbonyl (1720 cm-1) and -OH (2500-3000 cm-1broad)absorptions is a carboxylic acid.

I~ To use IR spec~oscopy to distinguish between isomers, find a strong IR absorption that isQ present in one isomer but absent in the other. .

(a)CHsCH20H

Strong hydroxyl bandat 3400 - 3640 cm-1

CHSOCHS

.No band in the region .3400 - 3640 cm-1

(b). CHsCH2CH2CH2CH =CH2

Alkene bands at3020-3100 cm-1 andat 1640-1680-1.

oNo bands in alkene region.

(c)

CHsCH2C02H

Strong, broad bandat 2500-3100 cm-1

HOCH2CH2CHO

Strong band at.3400-3640 cm-1

@ 20 II Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a PUblicly accessible website, in whole or in part.

Structure Determination: Mass Spectrometry, JR, and UV Spectroscopy 223

~ ~ (a) An ester next to a double bond absorbs at 1715cm-I. The alkene double bond absorbsD at 1640-1680 cm-I.

(b)The aldehyde carbonyl group absorbs at 1730cm-I. The alkyne CaC bond absorbs at2100-2260 cm-I, and the alkyne H-G bond absorbs at 3300 cm-I.

(c) The most importantabsorptionsfor this compound are due to the alcoholgroup (abroad, intense band at 3400-3650 cm-I) and to the carboxylic acid group, which has aC=O absorption in the range 1710-1760 cm-I and a broad O-H absorptionin therange 2500-3100 cm-I. Absorptions due to the aromatic ring [3030 cm-I (w) and

EJ1450-1600 cm-I (m)] may also be seen.

0.10. 0

HsC II, C

~ 'CHs

H

The compound contains nitrile and ketone groups, as well as a carbon-carbon doublebond. The nitrile absorption occurs at 2210-2260 cm-I. The ketone shows an absorptionat 1690cm-I, a value lower than the usual value because the ketone is next to the doublebond. The double bond absorption occurs at 1640-1680 cm-I. ~

10.11200 nm = 200 x 1O-9m = 2 x 10-7m

400 nm = 400 x 10-9m = 4 x 10-7 m

forA = 2 x 10-7m:

E = 1.20 X 10-4 kl/molA(in ~)

for A =4 x 10-7m:

E = 1.20 X 10-4 kl/mol =A(in m)

= 1.20 X10-4kl/mol = 6.0 x 102kl/mol2.0 x 10-7

1.20 X 10-4 kl/mol = 3.0 x 102kl/mol4.0 x 10-7

The energy of electromagneticradiationin the region of the spectrumfrom 200 om to 400om is 300 - 600 kl/mol. Comparing this with the energy value for IR transitions(calculatedin Section 10.6):

Energy (in kl/mol)

UV

300 - 600

IR

4.8 - 48

The energy required for UV transitions~sgreater than the energy required for IRtransitions.

C 2011 Cengage I:earning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

224 Chapter 10

B.. A

CxlWhere E = molar absorptivity(in L/mol.cm)

A '= absorbanceIn this problem:

E = 50,100 = 5.01 x 104Llmol .cm

1 = 1.00 cm

A = 0.735

C = -LL = 0.735 = 1.47 x 10-5MEX 1 5.01 X 104 Llmol.cm x 1.00 cmeo .13 11compounds having alternatingsingle and multiple bonds should show ultraviolet .

sorption in the range 200-400 nm. Only compound (a) is not UV-active. All of thecompoundspicturedbelowareUVactive. .

1 = sample path length (in cm)

C =concentration(inmol/L)

ooII

(JC, COH

~ OCCHsIIo

Visualizing Chemistry

10.14 (a) The mass spectrum of this ketone shows fragments resulting from both McLaffertyrearrangementand alpha cleavage.McLaffertyrearrangement:

[

0

j

+.

CHsCH29H~CH2C?HsCHs

M+ =114

McLafferty .rearrangement

H ~ /H~

"9~ ~H_C~C"

CH CHs/ C 2H I \H CHs

tH'9

CHsC" C~ "CH2CHsI

H m/z = 86

II:)2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website: in whole or in part.

L

228 . Chapter 10

(b) CSHIZO is the formula Qf an alcohol with M+ =88. The fragment at m/z =70 is due tothe product of dehydration ofM+. The other two fragments are a result of alpha cleavage.Draw the possible Cs alcohol isomers, and draw their products of alpha cleavage. Thetert~aryalcohol shown fits the data.

[

OH

]

+"

CH3CH2+ b-+CH3 Alpha.I I I cleavage

CH3M+=88

[

H3C CH3

]

+"

, /.C=C

/ .,H CH3m/z =70

CH3I .

CH3CH2CH2CHCH3

2-Methylpentane

k[

CH

]

+"

CH3CH2CH2bH~H3

! \+e

+

CH3CH2CH2CHCH3

m/z =71

CH3+ ICH2CHCH3

m/z =57

CH3+1CHCH3

m/z =43

The molecular ion, at m/z =86, is present in very low abundance. The base peak, at m/z =43, represents a stable secondary carbocation. .

Q Before.doing the hydrogenation,familiarizeyourself with the mass spectra of cyclohexeneV and cyclohexane.Note that M+is different for each compound. After the reaction isunderway, inject a sample from the reaction mixture into the mass spectrometer.If thereaction is finished, the mass spectrum of the reaction mixture should be superimposablewith the mass spectrum of cyclohexane.

iO 2011 Cengage Leaming. All Rights Reserved. May nol be scanned, copied or duplicated, or poSled to a publicly accessible website, in whol~ or in part.

-M+=128 mlz =99 m/z =128

Structure Determination: Mass Spectrometry, IR, and UV Spectroscopy 229

6~5 (a) This ketone shows mass spectrumfragments that are due to alpha cleavage and to theMcLaffertyrearrangement.The molecular ion occurs at W =148,and major fragmentshave m/z =120, 105, and 71. (Note that only charged species are shown.)

(b)The fragments in the mass spectrum of this alcohol (CgH160)result from dehydrationand alpha cleavage. Major fragmentshave m/z values of 128(the same value as themolecular ion), 110 and 99.

M+=128 m/z =110

(c) Amines fragmentby alpha cleavage. In this problem, cleavage occurs in the ring,producing a fragment with the same value of m/z as the molecular ion (99).

HI

~N,u-- CHs

M+=99

- Alpha..cleavage

HI

~N'CH3

m/z =99

C>2011 Cengage Leaming_ All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

° +" +

II

aO l O. r(tcAlpha

I \ I

..+ "'Ccleavage

C1OH120 M+ =148 mlz =105 m/z =71

HYJ+"

H,-, +"

Oth

°II McLafferty IC ..

c'O c'O'0 reammgementH2C I

m/z =120

230 Chapter 10-c-

(lO.26)CH3CH2CaCH shows absorptions at 2100-2260 cm-I (CaC) and at 3300 cm-I (CaC-H)"-J that are due to the terminal alkyne bond.

H2C=CHCH=CH2 has absorptions in the regions 1640-1680 cm-I and 3020-3100 thatare due to the double bonds. It also shows absorptions at 910 cm-I and 990 cm-I that aredue to monosubstituted alkene bonds. No absorptions occur in the alkyne region.

CH3CaCCH3.For reasons we won't discuss, symmetrically substituted alkynes such asbut-2-yne do not show a CaC bond absorptionin the IR. There is also no terminal alkyneabsorption.This alkyne is distinguishedfrom the other isomers in that it shows noabsorptions in either the alkyne or aUceneregions.

Two enantiomershave.identicalphysicalproperties (other than the sign of specificrotation). Thus, their IR spectra are also identical.

Diastereomershave differentphysicalpropertiesand chemicalbehavior, and their IRspectraarealsodifferent. .

~ (a) Absorptions at 3300 cm-I and 2150 cm-I are due to a terminal triple bond. PossibleU structures:

CH3CH2C~2C= CH (CH3)2CHC= CH

(b)AnIR absorptionat 3400cm-I is dueto a hydroxylgroup.Sincenodoublebondabs?rptionis present,thecompoundmustbe a cyclicalcohol.

CH3

[>(OH~OH

J>-CH3HO

(c) An absorption at 1715cm-I is due to a ketone. The only possible structure isCH3CH2COCH3.

(d) Absorptions at 1600cm-I and 1500cm-I are du~ to an aromatic ring. Possiblestructures:

~3\ (a)HCaCCH2NH2~ Alkyne absorptions at3300 cm-I, 2100-2260 cm-IAmine absorptionat3300-3500 cm~I

(b) CH3COCH3Strong ketone absorptionat 1715cm-I

CH3

CH3CH2CaNNitrile absorptionat2210-2260 cm-I

CH3CH2CHO

Strong aldehlde absorptionat 1730cm-

C 20 II Cengage Leaming. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website. in whole or in part.

Structure Determination: Mass Spectrometry, IR, and UV Spectroscopy 231

~ Spectrum (b) differs from spectrum (a) in several respects. Note in particulartheabsorptions at 715 cm-1 (strong), 1140cm-1 (strong), 1650cm-I (medium), and 3000cm-I (medium) in spectrum (b). The absorptions at 1650cm-I (C=C stretch)and 3000cm-I (=C-H stretch) can be found in Table 10.1.They al10w.usto assign spectrum (b) tocyc10hexeneand spectrum (a) to cyc1ohexane.

&my absorptionswithmediumto strougintensityare listed.(a). . C=c (b)

o OH aromatIcnng - 0 OCH~C/ 1450-1600 cm-I ~C/ 3

6 aromatic ring C-H 67"' 3030 cm-I 7"'

I carboxylicacidC=O I

~ 1710-1760 cm-I ~carboxylicacid O-H

2500-3100 cm-Imonosubstitutedaromaticring

690-710 cm-I730-770 cm-I

(c) NIIIC

OH

(e)

aromaticring C=C1450-1600 cm-I

aromatic ring C-H3030 cm-I

alcoholO-H3400-3650 cm-I

nitrile C=N2210-2260 cm-I

p-disubstitutedaromaticring

810-840 cm-I

(d) 0

ketone1715 cm-I

aromaticring C=C1450-1600 cm-I

aromaticring C-H3030 cm-I

aromaticester1715cm-I

monosubstitutedaromaticring

690-710 cm-I730-770 cm-1

alkene C=C1640-1680 cm-I

alkene =C.",H3020-3100 cm-I

ketone1715cm-I

ester1735cm-I

(b) CH3COCH=CHCH3,a ketone conjugated with a double bond, shows a strong ketoneabsorption at 1690 cm-I; CH3COCH2CH=CH2shows a ketone absorption at 1715cm-I and monosubstituted alkene absorp~ionsat 910 cm-I and 990 cm-I.

(c) CH3CH2CHOexhibits an aldehyde band at 1730 cm-I; H2C=CHOCH3showscharacteristic monosubstitutedalkene absorptions at 910 cm-I and 990 cm-I.

Q 20 II Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Hexa-l,3,5-trieneAmax = 258nm

H CH3I I

'iC, 'iC, 'iCH2H2C C C

I IH CH3

2,3-Dimethylhexa-l,3 ,5-trieneAmax "'" 268 nm

Structure Determination: Mass Spectrometry, IR, and UV Spectroscopy 235

10.43

In Problem10:42,weconcludedthatonealkylgroupincreasesAmaxof a conjugateddieneby approximately 5 nm. Since 2,3-~imethylhexa-l,3,5-triene has two niethyl substituents,its UV Amaxshould be about 10nm longer than the Amaxof hexa-l ,3,5-triene.

,

10.44

C = ~ = 0.065 = 6.5 x 10-2 =' 5.5 x lO-6MExl 11,900L/mol'cmx 1.00cm 1.19x 104L/mol~

;i 0.45 )The peak of maximum intensity (base peak) in the mass spectrum occurs at mlz =67. This~~ ~~akdoes not represent the molecular ion, however, because M+of a hydrocarbonmust be

an even number. Careful inspection reveals the molecular ion peak at mlz =68. M+ =68corresponds to a hydrocarbonof molecular formula CsHgwith a degree of unsaturationoftwo.

Fairly intense peaks in the mass spectrum occur at mlz =67, 53, 40,39, and 27. Thepeak at mlz =67 corresponds to loss of one hydrogen atom, and the peak at mlz =53represents loss of "amethyl group. The unknown hydrocarbon thus contains a methylgroup.

Significant IR absorptionsoccur at 2130 cm-1 (-C=C- stretch) and at 3320 cm-1(=C-H stretch). These bands indicate that the unknown hydrocarbon is a terminal alkyne.Possible structures for CsHgare CH3CH2CH2C=CHand (CH3hCHC=CH. [Pent-l-yneis correct.]

~Ii0.46 he molecular ion, M+=70, corresponds to the molecular formula CsH10.This\~compound has one double bond or ring. '

The base peak in the mass spectrum occurs at mlz =55. This peak represents loss of amethyl group from the molecularion and indicatesthe presenceof a methyl group in theunknown hydrocarbon. All otber peaks occur with low intensity:

In the IR spectrum, it is possible to distinguish absorptions at 1660cm-1and at 3000cm-1due to a double bond. (The 2960 cm-1 absorption is rather hard to detect because itoccurs as a shoulder on the alkane C-H stretch at 2850-2960 cm-1.)

Since no absorptions occur in the region 890 cm-1- 990 cm-1, we can excludeterminal alkenes as possible structures.The remaining possibilities for CsH10areCH3CH2CH=CHCH3and (CH3hC=CHCH3. [2-Methylbut-2-ene is correct.]

Q 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a pllblicly accessible website, in whole or in part.