Fragmentation Patterns in the Mass Spectra of Organic Compounds

7/26/2019 Fragmentation Patterns in the Mass Spectra of Organic Compounds

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Fragmentation Patterns in the Mass Spectra of Organic

Compounds

This page looks at how fragmentation patterns are formed when organic molecules

are fed into a mass spectrometer, and how you can get information from the mass

spectrum.

The origin of fragmentation patterns

The formation of molecular ions

When the vaporised organic sample passes into the ionisationchamber of a mass spectrometer, it is bombarded by a stream of

electrons. These electrons have a high enough energy to knock an

electron off an organic molecule to form a positive ion. This ion is

called themolecular ion- or sometimes theparent ion.

Note:If you aren't sure about how a mass spectrum is

produced, it might be worth taking a quick look at the page

describinghow a mass spectrometer works.

The molecular ion is often given the symbolM+or - the dot in

this second version represents the fact that somewhere in the ion

there will be a single unpaired electron. That's one half of what was

originally a pair of electrons - the other half is the electron which

was removed in the ionisation process.

Fragmentation

The molecular ions are energetically unstable, and some of them

will break up into smaller pieces. The simplest case is that a

molecular ion breaks into two parts - one of which is another

positive ion, and the other is an uncharged free radical.

Note:A free radical is an atom or group of atoms which contains
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a single unpaired electron.

More complicated break-ups are beyond the scope of A'level

syllabuses.

The uncharged free radical won't produce a line on the mass

spectrum. Only charged particles will be accelerated, deflected and

detected by the mass spectrometer. These uncharged particles will

simply get lost in the machine - eventually, they get removed by the

vacuum pump.

The ion, X+

, will travel through the mass spectrometer just like anyother positive ion - and will produce a line on the stick diagram.

All sorts of fragmentations of the original molecular ion are possible

- and that means that you will get a whole host of lines in the mass

spectrum. For example, the mass spectrum of pentane looks like

this:

Note:All the mass spectra on this page have been drawn using

data from the Spectral Data Base System for Organic

Compounds(SDBS) at the National Institute of Materials and

Chemical Research in Japan.

They have been simplified by omitting all the minor lines with

peak heights of 2% or less of the base peak (the tallest peak).
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It's important to realise that the pattern of lines in the mass

spectrum of an organic compound tells you something quite

different from the pattern of lines in the mass spectrum of an

element. With an element, each line represents adifferentisotopeof that element. With a compound, each line

represents a differentfragmentproduced when the molecular ion

breaks up.

Note:If you are interested in themass spectra of elements,you

could follow this link.

The molecular ion peak and the base peak

In the stick diagram showing the mass spectrum of pentane, the

line produced by the heaviest ion passing through the machine (at

m/z = 72) is due to the molecular ion.

Note:You have to be a bit careful about this, because in some

cases, the molecular ion is so unstable that every single one ofthem splits up, and none gets through the machine to register in

the mass spectrum. You are very unlikely to come across such a

case at A'level.

The tallest line in the stick diagram (in this case at m/z = 43) is

called thebase peak.This is usually given an arbitrary height of

100, and the height of everything else is measured relative to this.

The base peak is the tallest peak because it represents thecommonest fragment ion to be formed - either because there are

several ways in which it could be produced during fragmentation of

the parent ion, or because it is a particularly stable ion.

Using fragmentation patterns

This section will ignore the information you can get from the

molecular ion (or ions). That is covered in three other pages which
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you can get at via the mass spectrometry menu. You will find a link

at the bottom of the page.

Working out which ion produces which line

This is generally the simplest thing you can be asked to do.

The mass spectrum of pentane

Let's have another look at the mass spectrum for pentane:

What causes the line at m/z = 57?

How many carbon atoms are there in this ion? There can't be 5

because 5 x 12 = 60. What about 4? 4 x 12 = 48. That leaves 9 to

make up a total of 57. How about C4H9+then?

C4H9+would be [CH3CH2CH2CH2]

+, and this would be produced by

the following fragmentation:

The methyl radical produced will simply get lost in the machine.

The line at m/z = 43 can be worked out similarly. If you play around

with the numbers, you will find that this corresponds to a break

producing a 3-carbon ion:


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The line at m/z = 29 is typical of an ethyl ion, [CH3CH2]+:

The other lines in the mass spectrum are more difficult to explain.

For example, lines with m/z values 1 or 2 less than one of the easy

lines are often due to loss of one or more hydrogen atoms during

the fragmentation process. You are very unlikely to have to explain

any but the most obvious cases in an A'level exam.

The mass spectrum of pentan-3-one

This time the base peak (the tallest peak - and so the commonest

fragment ion) is at m/z = 57. But this isn't produced by the same ion

as the same m/z value peak in pentane.

If you remember, the m/z = 57 peak in pentane was produced by

[CH3CH2CH2CH2]+

. If you look at the structure of pentan-3-one, it'simpossible to get that particular fragment from it.

Work along the molecule mentally chopping bits off until you come

up with something that adds up to 57. With a small amount of

patience, you'll eventually find [CH3CH2CO]+- which is produced by

this fragmentation:

You would get exactly the same products whichever side of the CO


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Applying the logic of this to fragmentation patterns, it means that asplit which produces a secondary carbocation is going to be moresuccessful than one producing a primary one. A split producing atertiary carbocation will be more successful still.

Let's look at the mass spectrum of 2-methylbutane. 2-methylbutaneis an isomer of pentane - isomers are molecules with the samemolecular formula, but a different spatial arrangement of the atoms.

Look first at the very strong peak at m/ ! "#. $his is caused by adifferent ion than the corresponding peak in the pentane massspectrum. $his peak in 2-methylbutane is caused by%

$he ion formed is a secondary carbocation - it has two alkyl groupsattached to the carbon with the positive charge. As such, it isrelatively stable.

$he peak at m/ ! & is much taller than the corresponding line in

pentane. Again a secondary carbocation is formed - this time, by%

(ou would get the same ion, of course, if the left-hand )*#groupbroke off instead of the bottom one as we've drawn it.

+n these two spectra, this is probably the most dramatic eample ofthe etra stability of a secondary carbocation.


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Examples involving acylium ions, [RCO]+

+ons with the positive charge on the carbon of a carbonyl group,)!, are also relatively stable. $his is fairly clearly seen in the

mass spectra of ketones like pentan-#-one.

$he base peak, at m/!&, is due to the )*#)*2)0ion. 1e've

already discussed the fragmentation that produces this.

Note:There are lots of other examples of positive ions with extra

stability and which are produced in large numbers in a massspectrometer as a result. Without making this article even longer

than it already is, it's impossible to cover every possible case.

Check past exam papers to find out whether you are likely to

need to know about other possibilities. If you haven't got past

papers, follow the link on thesyllabusespage to find out how to

get hold of them.

Using mass spectra to distinguish between compounds

Suppose you had to suggest a way of distinguishing between

pentan-2-one and pentan-3-one using their mass spectra.

pentan-2-one CH3COCH2CH2CH3

pentan-3-one CH3CH2COCH2CH3

Each of these is likely to split to produce ions with a positive charge
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on the CO group.

In the pentan-2-one case, there are two different ions like this:

[CH3CO]+

[COCH2CH2CH3]+

That would give you strong lines at m/z = 43 and 71.

With pentan-3-one, you would only get one ion of this kind:

[CH3CH2CO]+

In that case, you would get a strong line at 57.

You don't need to worry about the other lines in the spectra - the

43, 57 and 71 lines give you plenty of difference between the two.

The 43 and 71 lines are missing from the pentan-3-one spectrum,

and the 57 line is missing from the pentan-2-one one.

Note:Don't confuse the line at m/z = 58 in the pentan-2-onespectrum. That's due to a complicated rearrangement which you

couldn't possibly predict at A'level.

The two spectra look like this:


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Computer matching of mass spectra

As you've seen, the mass spectrum of even very similar organic

compounds will be quite different because of the different

fragmentations that can occur. Provided you have a computer data

base of mass spectra, any unkown spectrum can be computer

analysed and simply matched against the data base.

Questions to test your understanding

If this is the first set of questions you have done, please read theintroductory

pagebefore you start. You will need to use the BACK BUTTON on your

browser to come back here afterwards.

questions on fragmentation patterns

answers

Where would you like to go now?

To the mass spectrometry menu . . .

To the instrumental analysis menu . . .

To Main Menu . . .
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EIMass!pectra of "ssorted Organic #ompounds

$he mass apectra of three different saturated hydrocarbons are displayed below. $wo

are isomeric heanes and the third is cycloheane. )omments regarding the

fragmentation patterns are presented in the bo to the right of each spectrum. +ons are

sometimes characteried by loss of a specific neutral fragment from the molecular ion.

or eample, a 3-4& ion is identified as loss of a methyl group. dd-electron ions,

including the molecular ion, are colored orange when marked. 5ven-electron ions are

colored magenta. $he 6$oggle 5amples6 button at the bottom will display a different set

of spectra in which the influence of a particular functional group may be eamined.

7epeated clicking of this button will cycle through fifteeen spectra. +n each eample the

molecular ion is designated by M $0.

4. $hese three eamples are hydrocarbons having no functional groups.

---------------------------------------------------------------------------

*eane shows the same fragmentation pattern as other unbranched alkanes. $hus, alkyl

carbocations at m/!4&, 28, "# and & 9a provide the dominant peaks in the spectrum.

$he m/!& butyl cation :3-28; is the base peak, and the m/!"# and 28 ions are alsoabundant.

4

2

#


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2. )hain branching clearly influences the fragmentation of this isomeric heane. $he

molecular ion at m/!y having the si carbons of heane closed to a ring, the fragmentation is profoundly

changed. $o begin with, the molecular ion at m/!


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entanal displays a set of ions associated with the alkyl chain :e.g. m/!&, "#, "4, 28 ?

2;. $he molecular ion at m/!


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----------------------------------------------------------------------------

+n 4-pentanol the hydroyl group is at the end of the five-carbon chain. $here are two

significant odd-electron fragment ions, one at m/!C :loss of water;, and the other at

m/!"2 :loss of water and ethene;. $he fragment ion at m/!&& is probably due to a

methyl radical loss from the m/!C ion. $he m/!#4 ion may be a protonatedformaldehyde ion, formed by alpha-fragmentation.


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-------------------------------------------------------------------------

$he mass spectrum of4-aminopentane is remarkably simple, thanks to the directive

influence of nitrogen. Alpha-fragmentation generates the m/!#C even-electron cation,

which is the only significant fragment ion. $he molecular ion :m/!


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$he molecular ion in the mass spectrum of ethyl acetate is rather weak. $he base peak

results from an alpha-fragmentation of ethoyl radical to give an m/!"# ion. +t is

interesting that the alternative alpha-cleavage to an 3-4& ion is very weak. $he loss of

water to generate the odd-electron ion at m/!C is curious. Bmall methyl and ethyl ions

are found at m/!4& ? 28.

4". $he isomeric ester, methyl propanoate, has a more abundant molecular ion than

ethyl acetate. $he alpha fragmentation of methoyl radical generates the strong m/!&

ion. Alpha-fragmentation of the ethyl group leads to both m/!&8 ? 28. A smaller methyl

peak is also seen.

4&. $he stability of dimethylformamide :93; is evident in the abundance of its

molecular ion :m/!#;, which is also the base peak. A small 3-4& peak is observed, but

the second most abundant ion is loss of *) by an alpha-cleavage :m/!"";. ther

ions at m/!"2, #C ? 2< are probably derived from the m/!"" ion.

End of this supplementary topic


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#ommon %ragment Ions and &eutral %ragments

Common Small Ions

m/z composition 15 Da CH3

17 OH

18 H2O

19 H3O, F

26 C2H2, CN

27 C2H3

28 C2H4, CO, H2CN

29 C2H5, CHO

30 CH2NH2

31 CH3O

33 SH, CH2F

34 H2S

Common Neutral Fragments

mass loss composition 1 Da H

15 CH3

17 OH

18 H2O

19 F

20 HF

27 C2H3, HCN

28 C2H4, CO

30 CH2O

31 CH3O

32 CH4O, S

33 CH3+ H2O, HS


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35(37) Cl

36(38) HCl

39 C3H3

41 C3H5, C2H3N

42 C3H6, C2H2O, C2H4N

43 C3H7, CH3CO 44 C2H4O

46 NO2

56 C4H8

57 C4H9

60 CH4CO2

79(81) Br

80(82) HBr

91 C7H7

127 I

128 HI

33 H2S

35(37) Cl

36(38) HCl

42 C3H6, C2H2O, C2H4N

43 C3H7, CH3CO

44 CO2O, CONH2

45 C2H5O

55 C4H7

57 C4H9

59 C2H3O2

60 C2H4O2

64 SO2

79(81) Br

80(82) HBr

127 I

128 HI

End of this supplementary topic


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'earangement Mechanisms in %ragmentation


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4-nonanone The o!ele"#ro$ %ra&'e$# o$ a#

'* 86 a$ 58 are #he re-l# o% a

."/a%%er# rearra$&e'e$#, $ol$&

#he lar&er all "ha$, a$ a

-e-e$# lo o% e#he$e (#heo-le!."/a%%er# rearra$&e'e$#)

lha!"leaa&e lea #o #he '* 99,

71 a$ 43 o$ The "har&e

aare$#l #r-#e oer o#h

%ra&'e$#

butylpentanoate lha!"leaa&e &e o$ a# '*57

85 Da The ."/a%%er#

rearra$&e'e$# o$ #he a" e

&e$era#e a '*116 o$ S-e-e$#

rearra$&e'e$# o$ #he al"ohol e

&e$era#e '*60 a$ 56 o$ The'*103 o$ roal

C4H9CO2H2(+)


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5-methyl-5-hexen-3-ol

The 'ole"-lar o$ ('*114 Da)

$o# oere -$er ele"#ro$ 'a"#

o$a#o$ "o$#o$ The h&he#

'a o$ ('*85) -e #o a$ alha!"leaa&e o% e#hl: #he o#her alha!

"leaa&e &e$era#e '*59 The

rearra$&e'e$# "leaa&e ho;$ here

&e$era#e #he '*56 o$

4,4-imethylcyclohexene

The lo o% a 'e#hl ra"al &e$era#e

#he ae ea a# '*95 Da The

'*81 67 o$ are 'aller

ho'olo&-e o% #h o$ (14 'a

-$# le) C"lohe


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Section 4.1: Introduction to molecular

spectroscopy

(.)"* The electromagnetic spectrum5lectromagnetic radiation, as you may recall from a previous chemistry or physics class, is composed of

electrical and magnetic waves which oscillate on perpendicular planes. Disible light is electromagnetic radiation.

Bo are the gamma rays that are emitted by spent nuclear fuel, the -rays that a doctor uses to visualie your

bones, the ultraviolet light that causes a painful sunburn when you forget to apply sun block, the infrared light that

the army uses in night-vision goggles, the microwaves that you use to heat up your froen burritos, and the radio-

fre@uency waves that bring music to anybody who is old-fashioned enough to still listen to 3 or A3 radio.

Eust like ocean waves, electromagnetic waves travel in a defined direction. 1hile the speed of ocean waves can

vary, however, the speed of electromagnetic waves F commonly referred to as the speed of light F is essentially a

constant, approimately #CC million meters per second. $his is true whether we are talking about gammaradiation or visible light. bviously, there is a big difference between these two types of waves F we are

surrounded by the latter for more than half of our time on earth, whereas we hopefully never become eposed to

the former to any significant degree. $he different properties of the various types of electromagnetic radiation are

due to differences in their wavelengths, and the corresponding differences in their energies% shorter wavelengths

correspond to higher energy.

*igh-energy radiation :such as gamma- and -rays; is composed of very short waves F as short as 4C -4=meter

from crest to crest. Longer waves are far less energetic, and thus are less dangerous to living things. Disible

light waves are in the range of "CC F CC nm :nanometers, or 4C -8m;, while radio waves can be several hundred

meters in length.

$he notion that electromagnetic radiation contains a @uantifiable amount of energy can perhaps be better

understood if we talk about light as a stream ofparticles, called photons, rather than as a wave. :7ecall the

concept known as Gwave-particle dualityH% at the @uantum level, wave behavior and particle behavior become

indistinguishable, and very small particles have an observable GwavelengthH;. +f we describe light as a stream of

photons, the energy of a particular wavelength can be epressed as%

5 ! hc/!

where 5 is energy in kcal/mol, !:the Ireek letter lambda; is wavelength in meters, cis #.CC 4C


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1hen talking about electromagnetic waves, we can refer either to wavelength or to fre@uency - the two values

are interconverted using the simple epression%

!"! c

where" :the Ireek letter Gnu; is fre@uency in s-4. Disible red light with a wavelength of CC nm, for eample, hasa fre@uency of ".28 4C4"*, and an energy of "C.8 kcal per mole of photons.

$he full range of electromagnetic radiation wavelengths is referred to as the electromagnetic spectrum.

Kotice in the figure above that visible light takes up ust a narrow band of the full spectrum. 1hite light from the

sun or a light bulb is a miture of all of the visible wavelengths. (ou see the visible region of the electromagnetic

spectrum divided into its different wavelengths every time you see a rainbow% violet light has the shortest

wavelength, and red light has the longest.

E+ample

5ercise ".4% Disible light has a wavelength range of about "CC-CC nm. 1hat is the corresponding fre@uency

rangeM 1hat is the corresponding energy range, in kcal/mol of photonsM

Bolution

(.),* Molecular spectroscopy - the basicidea+n a spectroscopy eperiment, electromagnetic radiation of a specified range of wavelengths is allowed to pass

through a sample containing a compound of interest. $he sample molecules absorb energy from some of the

wavelengths, and as a result ump from a low energy Gground stateH to some higher energy Gecited stateH. ther

wavelengths are notabsorbed by the sample molecule, so they pass on through. A detector on the other side of

the sample records which wavelengths were absorbed, and to what etent they were absorbed.
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*ere is the key to molecular spectroscopy% a given molecule will specifically absorb only those wavelengths

which have energies that correspond to the energy difference of the transition that is occurring. $hus, if the

transition involves the molecule umping from ground state A to ecited state >, with an energy difference of #5,

the molecule will specifically absorb radiation with wavelength that corresponds to #5, while allowing other

wavelengths to pass through unabsorbed.

>y observing which wavelengths a molecule absorbs, and to what etent it absorbs them, we can gain

information about the nature of the energetic transitions that a molecule is able to undergo, and thus informationabout its structure.

$hese generalied ideas may all sound @uite confusing at this point, but things will become much clearer as we

begin to discuss specific eamples.

Section 4.2: Infrared spectroscopy

)ovalent bonds in organic molecules are not rigid sticks F rather, they behave more like springs. At room

temperature, organic molecules are always in motion, as their bonds stretch, bend, and twist. $hese comple

vibrations can be broken down mathematically into individualibrational modes, a few of which are illustrated

below.


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$he energy of molecular vibration is quantizedrather than continuous, meaning that a molecule can only stretch

and bend at certain 'allowed' fre@uencies. +f a molecule is eposed to electromagnetic radiation that matches the

fre@uency of one of its vibrational modes, it will in most cases absorb energy from the radiation and ump to a

higher vibrational energy state - what this means is that the amplitudeof the vibration will increase, but the

vibrational frequencywill remain the same. $he difference in energy between the two vibrational states is e@ual

to the energy associated with the wavelength of radiation that was absorbed. +t turns out that it is

the infraredregion of the electromagnetic spectrum which contains fre@uencies corresponding to the vibrational

fre@uencies of organic bonds.

Let's take 2-heanone as an eample. icture the carbonyl bond of the ketone group as a spring. $his spring is

constantly bouncing back and forth, stretching and compressing, pushing the carbon and oygen atoms further

apart and then pulling them together. $his is thestretching modeof the carbonyl bond. +n the space of one

second, the spring 'bounces' back and forth &.4& 4C4#times - in other words, the ground-state fre@uency of

carbonyl stretching for a the ketone group is about &.4& 4C4#*.

+f our ketone sample is irradiated with infrared light, the carbonyl bond will specifically absorb light with this same

fre@uency, which by e@uations ".4 and ".2 corresponds to a wavelength of &.


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$he value of #5 - the energy difference between the low energy :ground; and high energy :ecited; vibrational

states - is e@ual to ".84 kcal/mol, the same as the energy associated with the absorbed light fre@uency. $he

molecule does not remain in its ecited vibrational state for very long, but @uickly releases energy to the

surrounding environment in form of heat, and returns to the ground state.

1ith an instrument called an infrared spectrophotometer, we can 'see' this vibrational transition. +n the

spectrophotometer, infrared light with fre@uencies ranging from about 4C 4#to 4C4"* is passed though our

sample of cycloheane. 3ost fre@uencies pass right through the sample and are recorded by a detector on the

other side.

ur &.4& 4C4#* carbonyl stretching fre@uency, however, is absorbed by the 2-heanone sample, and so the

detector records that the intensity of this fre@uency, after having passed through the sample, is something less

than 4CCN of its initial intensity.

$he vibrations of a 2-heanone molecule are not, of course, limited to the simple stretching of the carbonyl bond.

$he various carbon-carbon bonds also stretch and bend, as do the carbon-hydrogen bonds, and all of these

vibrational modes also absorb different fre@uencies of infrared light.


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$he power of infrared spectroscopy arises from the observation that different functional groups have different

characteristic absorption frequencies. $he carbonyl bond in a ketone, as we saw with our 2-heanone eample,

typically absorbs in the range of &.44 - &.4< 4C4#*, depending on the molecule. $he carbon-carbon triple

bond of an alkyne, on the other hand, absorbs in the range =.#C - =.


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$here are a number of things that need to be eplained in order for you to understand what it is that we are

looking at. n the horiontal ais we see +7 wavelengths epressed in terms of a unit called waenumber:cm-

4;, which tells us how many waves fit into one centimeter. n the vertical ais we see G/ transmittanceH, which

tells us how strongly light was absorbed at each fre@uency :4CCN transmittance means no absorption occurred

at that fre@uency;. $he solid line traces the values of N transmittance for every wavelength F the GpeaksH :which

are actually pointing down; show regions of strong absorption. or some reason, it is typical in +7 spectroscopy

to report wavenumber values rather than wavelength :in meters; or fre@uency :in *;. $he Gupside downH vertical

ais, with absorbance peaks pointing down rather than up, is also a curious convention in +7 spectroscopy. 1e

wouldnHt want to make things too easy for youO

E+ample

5ercise ".2% 5press the wavenumber value of #CCC cm-4in terms of wavelength :in meter units;.

Bolution


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$he key absorption peak in this spectrum is that from the carbonyl double bond, at 44= cm -4:corresponding to a

wavelength of &.


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As you can see, the carbonyl peak is gone, and in its place is a very broad GmountainH centered at about #"CC cm-

4. $his signal is characteristic of the -* stretching mode of alcohols, and is a dead giveaway for the presence of

an alcohol group. $he breadth of this signal is a conse@uence of hydrogen bonding between molecules.

+n the spectrum of octanoic acid we see, as epected, the characteristic carbonyl peak, this time at 4C8 cm-4.

1e also see a low, broad absorbance band that looks like an alcohol, ecept that it is displaced slightly to the

right :long-wavelength; side of the spectrum, causing it to overlap to some degree with the )-* region. $his is

the characteristic carboylic acid -* single bond stretching absorbance.

$he spectrum for 4-octene shows two peaks that are characteristic of alkenes% the one at 4="2 cm -4is due to

stretching of the carbon-carbon double bond, and the one at #C8 cm-4 is due to stretching of the s bond

between the alkene carbons and their attached hydrogens.


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Alkynes have characteristic +7 absorbance peaks in the range of 24CC-22&C cm -4due to stretching of the carbon-

carbon triple bond, and terminal alkenes can be identified by their absorbance at about ##CC cm-4, due tostretching of the bond between the sp-hybridied carbon and the terminal hydrogen.

+t is possible to identify other functional groups such as amines and ethers, but the characteristic peaks for these

groups are considerably more subtle and/or variable, and often are overlapped with peaks from the fingerprint

region. or this reason, we will limit our discussion here to the most easily recognied functional groups, which

are summaried in table 4 in the tables section at the end of the tet.

As you can imagine, obtaining an +7 spectrum for a compound will not allow us to figure out the complete

structure of even a simple molecule, unless we happen to have a reference spectrum for comparison. +n

conunction with other analytical methods, however, +7 spectroscopy can prove to be a very valuable tool, given

the information it provides about the presence or absence of key functional groups. +7 can also be a @uick andconvenient way for a chemist to check to see if a reaction has proceeded as planned. +f we were to run a

reaction in which we wished to convert cycloheanone to cycloheanol, for eample, a @uick comparison of the

+7 spectra of starting compound and product would tell us if we had successfully converted the ketone group to

an alcohol :this type of reaction is discussed in detail in chapter 4=;.

+nternal Links

Dirtual $etbook of )hem%+7 spectroscopy
http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_16%3A_Oxidation_and_reduction_reactionshttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_16%3A_Oxidation_and_reduction_reactionshttp://chemwiki.ucdavis.edu/Organic_Chemistry/Spectroscopy/Infrared_Spectroscopyhttp://chemwiki.ucdavis.edu/Organic_Chemistry/Spectroscopy/Infrared_Spectroscopyhttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_16%3A_Oxidation_and_reduction_reactionshttp://chemwiki.ucdavis.edu/Organic_Chemistry/Spectroscopy/Infrared_Spectroscopy


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Section 4.3: Ultraviolet and visible spectroscopy

1hile interaction with infrared light causes molecules to undergo vibrational transitions, the shorter wavelength,

higher energy radiation in the PD :2CC-"CC nm; and visible :"CC-CC nm; range of the electromagnetic spectrum

causes many organic molecules to undergo electronic transitions. 1hat this means is that when the energy

from PD or visible light is absorbed by a molecule, one of its electrons umps from a lower energy to a higherenergy molecular orbital.

(.0"* Electronic transitionsLetHs take as our first eample the simple case of molecular hydrogen, * 2. As you may recall from section 2.4A,

the molecular orbital picture for the hydrogen molecule consists of one bonding =3, and a higher energy

antibonding =Q 3. 1hen the molecule is in the ground state, both electrons are paired in the lower-energy

bonding orbital F this is the *ighest ccupied 3olecular rbital :*3;. $he antibonding =Q orbital, in turn, is

the Lowest Pnoccupied 3olecular rbital :LP3;.

+f the molecule is eposed to light of a wavelength with energy e@ual to #5, the *3-LP3 energy gap, this

wavelength will be absorbed and the energy used to bump one of the electrons from the *3 to the LP3 F in

other words, from the =to the =Q orbital. $his is referred to as a $ $1 transition. #5 for this electronic transition

is 2&< kcal/mol, corresponding to light with a wavelength of 444 nm.

1hen a double-bonded molecule such as ethene :common name ethylene; absorbs light, it undergoes a % %1

transition. >ecause >- >Q energy gaps are narrower than = =1 gaps, ethene absorbs light at 4=& nm - a longer

wavelength than molecular hydrogen.

$he electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded bystandard PD spectrophotometers, which generally have a range of 22C F CC nm. 1here PD-vis spectroscopy


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becomes useful to most organic and biological chemists is in the study of molecules with conugated pi systems.

+n these groups, the energy gap for >->Q transitions is smaller than for isolated double bonds, and thus the

wavelength absorbed is longer. 3olecules or parts of molecules that absorb light strongly in the PD-vis region

are called chromophores.

LetHs revisit the 3 picture for 4,#-butadiene, the simplest conugated system :seesection 2.4>;. 7ecall that wecan draw a diagram showing the four pi 3Hs that result from combining the four 2p atomic orbitals. $he lower

two orbitals are bonding, while the upper two are antibonding.

)omparing this 3 picture to that of ethene, our isolated pi-bond eample, we see that the *3-LP3 energy

gap is indeed smaller for the conugated system. 4,#-butadiene absorbs PD light with a wavelength of 24 nm.

As conugated pi systems become larger, the energy gap for a >- >Q transition becomes increasingly narrow, and

the wavelength of light absorbed correspondingly becomes longer. $he absorbance due to the >- >Q transition in

4,#,&-heatriene, for eample, occurs at 2&< nm, corresponding to a #5 of 444 kcal/mol.

+n molecules with etended pi systems, the *3-LP3 energy gap becomes so small that absorption occurs

in the visible rather then the PD region of the electromagnetic spectrum. >eta-carotene, with its system of 44

conugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while

allowing other visible wavelengths F mainly those in the red-yellow region - to be transmitted. $his is why carrots

are orange.
http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter__2%3A_Introduction_to_organic_structure_and_bonding_II/Section_2.1%3A_Molecular_orbital_theory%3A_conjugation_and_aromaticityhttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter__2%3A_Introduction_to_organic_structure_and_bonding_II/Section_2.1%3A_Molecular_orbital_theory%3A_conjugation_and_aromaticityhttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter__2%3A_Introduction_to_organic_structure_and_bonding_II/Section_2.1%3A_Molecular_orbital_theory%3A_conjugation_and_aromaticityhttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter__2%3A_Introduction_to_organic_structure_and_bonding_II/Section_2.1%3A_Molecular_orbital_theory%3A_conjugation_and_aromaticity


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$he conugated pi system in "-methyl-#-penten-2-one gives rise to a strong PD absorbance at 2#= nm due to

a >- >Q transition. *owever, this molecule also absorbs at #4" nm. $his second absorbance is due to the

transition of a non-bonding :lone pair; electron on the oygen up to a >Q antibonding 3%

$his is referred to as an n %1 transition. $he nonbonding :n; 3Hs are higher in energy than the highest

bonding p orbitals, so the energy gap for an n - >Q transition is smaller that that of a >- >Q transition F and thus the

n - >Q peak is at a longer wavelength. +n general, n - >Q transitions are weaker :less light absorbed; than thosedue to > ! >Q transitions.

&xample

5ercise ".#% *ow large is the > ! >Q transition in "-methyl-#-penten-2-oneM

5ercise "."% 1hich of the following molecules would you epect absorb at a longer wavelength in the PD region

of the electromagnetic spectrumM 5plain your answer.

Bolution


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(.0,* 2ooking at 34is spectra

1e have been talking in general terms about how molecules absorb PD and visible light F now let's look at someactual eamples of data from a PD-vis absorbance spectrophotometer. $he basic setup is the same as for +7

spectroscopy% radiation with a range of wavelengths is directed through a sample of interest, and a detector

records which wavelengths were absorbed and to what etent the absorption occurred. >elow is the absorbance

spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated KA9 0 :we'll

learn what it does insection 4=."; $his compound absorbs light in the PD range due to the presence of

conugated pi-bonding systems.

(ouHll notice that this PD spectrum is much simpler than the +7 spectra we saw earlier% this one has only one

peak, although many molecules have more than one. Kotice also that the convention in PD-vis spectroscopy is

to show the baseline at the bottom of the graph with the peaks pointing up. 1avelength values on the -ais are

generally measured in nanometers :nm; rather than in cm-4as is the convention in +7 spectroscopy.

eaks in PD spectra tend to be @uite broad, often spanning well over 2C nm at half-maimal height. $ypically,

there are two things that we look for and record from a PD-Dis spectrum.. $he first is !ma,which is the

wavelength at maimal light absorbance. As you can see, KA90has !ma,! 2=C nm. 1e also want to record how

much light is absorbed at !ma. *ere we use a unitless number called absorbance, abbreviated 'A'. $his contains

the same information as the 'percent transmittance' number used in +7 spectroscopy, ust epressed in slightly

different terms. $o calculate absorbance at a given wavelength, the computer in the spectrophotometer simply

takes the intensity of light at that wavelength beforeit passes through the sample :+C;, divides this value by the

intensity of the same wavelength afterit passes through the sample :+;, then takes the log4Cof that number%

A ! log +C/+
http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_16%3A_Oxidation_and_reduction_reactions/Section_16.4%3A_Hydrogenation%2F%2Fdehydrogenation_reactions_of_carbonyls%2C_imines%2C_and_alcoholshttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_16%3A_Oxidation_and_reduction_reactions/Section_16.4%3A_Hydrogenation%2F%2Fdehydrogenation_reactions_of_carbonyls%2C_imines%2C_and_alcoholshttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_16%3A_Oxidation_and_reduction_reactions/Section_16.4%3A_Hydrogenation%2F%2Fdehydrogenation_reactions_of_carbonyls%2C_imines%2C_and_alcoholshttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_16%3A_Oxidation_and_reduction_reactions/Section_16.4%3A_Hydrogenation%2F%2Fdehydrogenation_reactions_of_carbonyls%2C_imines%2C_and_alcohols


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(ou can see that the absorbance value at 2=C nm :A2=C; is about 4.C in this spectrum.

&xample

5ercise ".&% 5press A ! 4.C in terms of percent transmittance :N$, the unit usually used in +7 spectroscopy

:and sometimes in PD-vis as well;.

Bolution

*ere is the absorbance spectrum of the common food coloring 7ed R#%

*ere, we see that the etended system of conugated pi bonds causes the molecule to absorb light in the visible

range. >ecause the !ma of &2" nm falls within the green region of the spectrum, the compound appears red to

our eyes.

Kow, take a look at the spectrum of another food coloring, >lue R4%


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*ere, maimum absorbance is at =#C nm, in the orange range of the visible spectrum, and the compound

appears blue.

(.0#* "pplications of 34 spectroscopy inorganic and biological chemistryPD-vis spectroscopy has many different applications in organic and biological chemistry. ne of the most basic

of these applications is the use of the ,eer 2ambert 2awto determine the concentration of a chromophore.

(ou most likely have performed a >eer F Lambert eperiment in a previous chemistry lab. $he law is simply an

application of the observation that, within certain ranges, the absorbance of a chromophore at a givenwavelength varies in a linear fashion with its concentration% the higher the concentration of the molecule, the

greater its absorbance. +f we divide the observed value of A at ?maby the concentration of the sample :c, in

mol/L;, we obtain the molar absorptiity, or e+tinction coefficient:5;, which is a characteristic value for a given

compound.

@! A/c

$he absorbance will also depend, of course, on the path length- in other words, the distance that the beam of

light travels though the sample. +n most cases, sample holders are designed so that the path length is e@ual to 4

cm, so the units for molar absorptivity are mol Q L-4cm-4. +f we look up the value of e for our compound at ?ma, and

we measure absorbance at this wavelength, we can easily calculate the concentration of our sample. As an


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eample, for KA90the literature value of @at 2=C nm is 4ecause the etinction coefficient of double stranded 9KA is slightly lower than that of single stranded 9KA, we

can use PD spectroscopy to monitor a process known as 9KA melting. +f a short stretch of double stranded

9KA is gradually heated up, it will begin to GmeltH, or break apart, as the temperature increases :recall that two

strands of 9KA are held together by a specific pattern of hydrogen bonds formed by Gbase-pairingH;.
http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Solution_Manual/Chapter__4_Solutionshttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Solution_Manual/Chapter__4_Solutionshttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Solution_Manual/Chapter__4_Solutionshttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Solution_Manual/Chapter__4_Solutions


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As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of

the double-stranded 9KA breaks apart, or GmeltsH. $he mid-point of this process, called the Gmelting temperatureH,

provides a good indication of how tightly the two strands of 9KA are able to bind to each other.

+n section 4=.eer - Lambert Law and PD spectroscopy provides us with a convenient wayto follow the progress of many different enymatic redo :oidation-reduction; reactions. +n biochemistry,

oidation of an organic molecule often occurs concurrently with reduction of nicotinamide adenine dinucleotide

:KA90, the compound whose spectrum we saw earlier in this section; to KA9*%

>oth KA90and KA9* absorb at 2=C nm. *owever KA9*, unlike KA90, has a second absorbance band

with ?ma! #"C nm and @! =28C mol L-4cm-4. $he figure below shows the spectra of both compounds

superimposed, with the KA9* spectrum offset slightly on the y-ais%
http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_16%3A_Oxidation_and_reduction_reactions/Section_16.8%3A_Observing_the_progress_of_hydrogenation_and_dehydrogenation_reactions_by_UV_spectroscopyhttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_16%3A_Oxidation_and_reduction_reactions/Section_16.8%3A_Observing_the_progress_of_hydrogenation_and_dehydrogenation_reactions_by_UV_spectroscopy


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>y monitoring the absorbance of a reaction miture at #"C nm, we can 'watch' KA9* being formed as the

reaction proceeds, and calculate the rate of the reaction.

PD spectroscopy is also very useful in the study of proteins. roteins absorb light in the PD range due to the

presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores.

>iochemists fre@uently use PD spectroscopy to study conformational changes in proteins - how they change

shape in response to different conditions. 1hen a protein undergoes a conformational shift :partial unfolding, for

eample;, the resulting change in the environment around an aromatic amino acid chromophore can cause its

PD spectrum to be altered.


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!ection (.(* Mass !pectrometry

ur third and final analytical techni@ue for discussion in this chapter does notfall under the definition of

spectroscopy, as it does not involve the absorbance of light by a molecule. +n mass spectrometry :3B;, we are

interested in the mass - and therefore the molecular weight - of our compound of interest, and often the mass of

fragments that are produced when the molecule is caused to break apart.

"."A% $he basics of a mass spectrometry

$here are many different types of 3B instruments, but they all have the same three essential components. irst,

there is an ioniation source, where the molecule is given a positive electrical charge, either by removing an

electron or by adding a proton. 9epending on the ioniation method used, the ionied molecule may or may not

break apart into a population of smaller fragments. +n the figure below, some of the sample molecules remain

whole, while others fragment into smaller pieces.

Ket in line there is a mass analyer, where the cationic fragments are separated according to their mass.

inally, there is a detector, which detects and @uantifies the separated ions.

ne of the more common types of 3B techni@ues used in the organic laboratory is electron ioni6ation. +n the

ioniation source, the sample molecule is bombarded by a high-energy electron beam, which has the effect of

knocking a valence electron off of the molecule to form aradical cation. >ecause a great deal of energy istransferred by this bombardment process, the radical cation @uickly begins to break up into smaller fragments,

some of which are positively charged and some of which are neutral. $he neutral fragments are either adsorbed

onto the walls of the chamber or are removed by a vacuum source. +n the mass analyer component, the

positively charged fragments and any remaining unfragmented molecular ionsare accelerated down a tube by

an electric field.


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$his tube is curved, and the ions are deflected by a strong magnetic field. +ons of different mass to charge :m/;

ratios are deflected to a different etent, resulting in a GsortingH of ions by mass :virtually all ions have charges of

! 04, so sorting by the mass to charge ratio is the same thing as sorting by mass;. A detector at the end of the

curved flight tube records and @uantifies the sorted ions.

".">% Looking at mass spectra

>elow is typical output for an electron-ioniation 3B eperiment :3B data in the section is derived from

the Bpectral 9atabase for rganic )ompounds, a free, web-based service provided by A+B$ in Eapan.
http://riodb01.ibase.aist.go.jp/sdbs/http://riodb01.ibase.aist.go.jp/sdbs/


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$he sample is acetone. n the horiontal ais is the value for m/ :as we stated above, the charge is almost

always 04, so in practice this is the same as mass;. n the vertical ais is the relative abundance of each ion

detected. n this scale, the most abundant ion, called thebase peak, is set to 4CCN, and all other peaks are

recorded relative to this value. or acetone, the base peak is at m/ ! "# - we will discuss the formation of this

fragment a bit later. $he molecular weight of acetone is &r and &CN r;, chlorinated and brominated compounds have very large and recogniable 302

peaks. ragments containing both isotopes of >r can be seen in the mass spectrum of ethyl bromide%


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3uch of the utility in electron-ioniation 3B comes from the fact that the radical cations generated in the electron-

bombardment process tend to fragment in predictable ways. 9etailed analysis of the typical fragmentation

patterns of different functional groups is beyond the scope of this tet, but it is worthwhile to see a few

representative eamples, even if we donHt attempt to understand the eact process by which the fragmentation

occurs. 1e saw, for eample, that the base peak in the mass spectrum of acetone is m/ ! "#. $his is the result

of cleavage at the GalphaH position - in other words, at the carbon-carbon bond adacent to the carbonyl. Alpha

cleavage results in the formation of an acylium ion :which accounts for the base peak at m/ ! "#; and a methyl

radical, which is neutral and therefore not detected.

After the parent peak and the base peak, the net largest peak, at a relative abundance of 2#N, is at m/ ! 4&.

$his, as you might epect, is the result of formation of a methyl cation, in addition to an acyl radical :which is

neutral and not detected;.

A common fragmentation pattern for larger carbonyl compounds is called the Mc2afferty rearrangement%


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$he mass spectrum of 2-heanone shows a '3cLafferty fragment' at m/ ! &


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ther functional groups have predictable fragmentation patterns as well. >y carefully analying the fragmentation

information that a mass spectrum provides, a knowledgeable spectrometrist can often Gput the pule togetherH

and make some very confident predictions about the structure of the starting sample.

Example

5ercise ".


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sample. >ecause a compound's 3B spectrum is a very reliable and reproducible 'fingerprint', we can instruct the

instrument to search an 3B database and identify each compound in the sample.

$he etremely high sensitivity of modern I)-3B instrumentation makes it possible to detect and identify very

small trace amounts of organic compounds. I)-3B is being used increasingly by environmental chemists to

detect the presence of harmful organic contaminants in food and water samples. Airport security screeners alsouse high-speed I)-3B instruments to look for residue from bomb-making chemicals on checked luggage.

"."9% 3ass spectrometry of proteins -applications in proteomics

3ass spectrometry has become in recent years an increasingly important tool in the field of proteomics.

$raditionally, protein biochemists tend to study the structure and function of individual proteins. roteomicsresearchers, in contrast, want to learn more about how large numbers of proteins in a living system interact with

each other, and how they respond to changes in the state of the organism. ne very important subfield of

proteomics is the search for protein biomarkersfor human disease. $hese can be proteins which are present in

greater @uantities in a sick person than in a healthy person, and their detection and identification can provide

medical researchers with valuable information about possible causes or treatments. 9etection in a healthy

person of a known biomarker for a disease such as diabetes or cancer could also provide doctors with an early

warning that the patient may be especially susceptible, so that preventive measures could be taken to prevent or

delay onset of the disease.

Kew developments in 3B technology have made it easier to detect and identify proteins that are present in very

small @uantities in biological samples. 3ass spectrometrists who study proteins often use instrumentation that issomewhat different from the electron-ioniation, magnetic deflection system described earlier. 1hen proteins are

being analyed, the obect is often to ionie the proteins without causing fragmentation, so 'softer' ioniation

methods are re@uired. +n one such method, called electrospray ioni6ation, the protein sample, in solution, is

sprayed into a tube and the molecules are induced by an electric field to pick up etra protons from the solvent.

Another common 'soft ioniation' method is 'matri-assisted laser desorption ioniation' :M"29I;. *ere, the

protein sample is adsorbed onto a solid matri, and protonation is achieved with a laser.

$ypically, both electrospray ioniation and 3AL9+ are used in conunction with a time-of-flight :$; mass

analyer component.


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$he ionied proteins are accelerated by an electrode through a column, and separation is achieved because

lighter ions travel at greater velocity than heavier ions with the same overall charge. +n this way, the many

proteins in a comple biological sample :such as blood plasma, urine, etc.; can be separated and their individual

masses determined very accurately. 3odern protein 3B is etremely sensitive F very recently, scientists were

even able to obtain a mass spectrum of %yrannosaurus re&protein from fossilied boneO :Bcience 8::;, #4=,

2;.

+n one recent study, 3AL9+-$ mass spectrometry was used to compare fluid samples from lung transplant

recipients who had suffered from tissue reection to control samples from recipients who had not suffered

reection. $hree peptides :short proteins; were found to be present at elevated levels specifically in the tissue

reection samples. +t is hoped that these peptides might serve as biomarkers to identify patients who are at

increased risk of reecting their transplanted lungs. :roteomics 8::, &, 4C&;.

)ontributors Organic #hemistry With a ,iological Emphasis by $im Boderberg:Pniversity of 3innesota, 3orris;

)hapter " Bolutions

+n-chapter eercises

E(.)*

Psing ?A! c, we first rearrange toA! c/? to solve for fre@uency.

or light with a wavelength of "CC nm, the fre@uency is .&C 4C4"*%

+n the same way, we calculate that light with a wavelength of CC nm has a fre@uency of ".28 4C4"*.

$o calculate corresponding energies%

Psing hc/?, we find for light at "CC nm%

Psing the same e@uation, we find that light at CC nm corresponds to "C.8 kcal/mol.
http://www.sciencemag.org/content/316/5822/277.abstracthttp://www.sciencemag.org/content/316/5822/277.abstracthttp://www.sciencemag.org/content/316/5822/277.abstracthttp://www.sciencemag.org/content/316/5822/277.abstracthttp://www.sciencemag.org/content/316/5822/277.abstracthttp://onlinelibrary.wiley.com/doi/10.1002/pmic.200401036/abstracthttp://onlinelibrary.wiley.com/doi/10.1002/pmic.200401036/abstracthttp://onlinelibrary.wiley.com/doi/10.1002/pmic.200401036/abstracthttp://onlinelibrary.wiley.com/doi/10.1002/pmic.200401036/abstracthttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasishttp://facultypages.morris.umn.edu/~soderbt/http://facultypages.morris.umn.edu/~soderbt/http://www.sciencemag.org/content/316/5822/277.abstracthttp://www.sciencemag.org/content/316/5822/277.abstracthttp://onlinelibrary.wiley.com/doi/10.1002/pmic.200401036/abstracthttp://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasishttp://facultypages.morris.umn.edu/~soderbt/


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E(.8*

A wavenumber of #CCC cm-4means that #CCC waves fit in one cm :C.C4m;%

1e want to find the length of 4 wave, so we divide numerator and denominator by #CCC%

Bo #CCC cm-4is e@uivalent to a wavelength of #.## mm.

E(.0*

$he >U >Q transition in "-methyl-#-penten-2-one is at 2#= nm, which corresponds to 424 kcal/mol%

E(.(%

3olecule > has a longer system of conugated pi bonds, and thus will absorb at a longer wavelength. Kotice that

there is an sp#-hybridied carbon in molecule > which isolates two of the pi bonds from the other three.

E(.*

1e use the formula%

recall from your high school algebra that if y ! log:;, then ! 4Cy, so%


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. . . so the intensity of light entering the sample :+C; is 4C times the intensity of the l ight :at a particular wavelength;

that emerges from the sample and is detected :+;. >ecause N$ is simply the reciprocal of this ratio multiplied by4CC, we find that A ! 4.C corresponds to 4CN transmission.

E(.=* Psing @! A/c, we plug in our values for @and A and find that c ! #.2 4C-&3, or #2. m3.

E(.;*

Psing @! A/c and plugging in the given values for @and A, we find that c ! 4=.# ng/mL. *owever, this is theconcentration of the diluted solution - the original solution was 2C times higher :&C mL were removed and diluted

to 4CCC mL to take the PD measurement;. $hus the concentration of the original solution is :2C;:4=.#; ! #2=

ng/mL.

E(.>*

$he sample is propanal. $he peak at m/ ! &< is the molecular ion :parent peak;, and the peak at m/ ! 28 is the

formyl acylium ion. $he 304 peak at m/ ! &8 is a molecular ion in which one of the three carbons is a 4#).

5nd-of-chapter problems

(.)*

1ithout doing any calculation, we can answer the first part of the calculation% electromagnetic waves at #"CC cm-

4are shorter than those at 4=8C cm-4:more waves fit into one centimeter; and thus correspond to a higher

fre@uency.


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#"CC cm-4 ! 2.8" mm ! 4.C2 4C4"*

4=8C cm-4! &.82 mm ! &.C 4C4#*

(.8*

42C cm-4corresponds to a wavelength of .C4/42C ! &.


53/54

(.;*

$he change in A#"Cis 9A ! C.22C. Psing the epression @! A/c, we can calculate that this represents a change in

the KA9* concentration of #.&C 4C -&3. $his is in a 4 mL solution, so #.&C 4C-*

>oth starting compounds contain systems of conugated pi bonds which absorb in the PD range. $he

condensation reaction brings these two conugated systems together to create a single, longer conugated pi

system, which absorbs in the blue part of the visible spectrum.

(.@*

(.):*


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$he molecular ion and the fragments from all three of the typical ketone fragmentation patterns would all have

m/ ! 8oth molecules contain alkene and ketone functional groups, however the degree of pi bond conugation is

different. $herefore, PD would be the more useful techni@ue to distinguish the two.

(.)8*

>oth molecules are straight-chain alkanes with a single ketone group, so their +7 spectra are epected to be very

similar and neither will absorb strongly in the PD range. *owever, the different positions of the ketone :at the)" vs )&position; will result in the formation of fragments of different masses in an 3B eperiment.

Documents

Fragmentation Patterns in the Mass Spectra of Organic Compounds