Mass Spectrum Worksheet #2
1.Themassspectrum of pentanelooks like this:
a) What causesthe peak at m/z = 72?
b) Whatcauses thepeak atm/z=57?
c)Writean equation to show howthe species
(molecule or ion)you quoted in part (b) isformed.
d) Whatcauses thepeak atm/z=43?
e)Writean equation to show howthe species you quoted
in part (d) isformed.
f) Whatcauses thepeak atm/z=29?
g)Write an equation to showhow thespecies you quoted
in part(f) isformed.
2.Themassspectrum of2-methylbutane, an isomer ofpentane, lookslike this:
Thepeak atm/z=57 ismuch tallerin
2-methylbutanethan inits isomer n-pentane.
Explain asfully as you
can why thisis, including an equation for the
fragmentation that leadsto it.
3. Themassspectra below are for pentan-2-one and pentan-3-one, butnot necessarily in that order.
Pentan-2-one is CH3COCH2CH2CH3.Pentan-3-one isCH3CH2COCH2CH3.
Decide which iswhich, explaining your thinking asfully aspossible.
A
B
4. Below is the IR and MS for compound X, identify compound X
5. The mass spectrum for chlorine looks like this:
a) Explain why there are two separate groups of peaks.
b) State what causes each of the 5 lines.
c) Explain the approximate relative heights of the lines at 35 and 37.
d) What are the approximate relative heights of the lines at 70, 72 and 74?
e) Why can't you predict the relative heights of the two clusters of lines (35/37 and 70/72/74)?
6.Below is the IR and MS for compound X, identify compound X
7.TheinfraredandmassspectraofanunknowncompoundDaregivenbelow. Deduce the structure of this compound fromthese spectra.
FRAGMENTATIONPATTERNS
1.a)[CHCHCHCHCH].
b)[CH3CH2CH2CH2]+
c)[CH3CH2CH2CH2CH3].[CH3CH2CH2CH2]++.CH3
d)[CH3CH2CH2]+
e)[CH3CH2CH2CH2CH3].[CH3CH2CH2]++.CH2CH3
f)[CH3CH2]+
g)[CH3CH2CH2CH2CH3].[CH3CH2]++.CH2CH2CH3
2.Thelineat m/z = 57 is caused by ionsformed during fragmentation involving the lossof aCHradical.There arethreeCH3groupswhich could break off the molecular ion:
[CH3CHCH2CH3].
CH3
Two of these (theonesshown in red)would break off to leave a secondary carbocation:
[CH3CHCH2CH3].
CH3
Secondary carbocations are more stable than primary ones(asformed in 1(b) and (c) above), and so are more likelyto form.Thereare two ways thatsecondary carbocations could be formed (by loss of either of thered CH3groups).You don't getthisextra stabilisationwhen thepentanemolecular ion fragmentsto lose a CH3radical, and so you get moreof this particular ion with 2-methylbutane than you do with pentane.You will also, of course, get asmallcontribution from the loss of theblueCH3group giving a primary carbocation
(You don'tneed to colour-code this-in fact, in an exam, you may be forbidden fromusing anything other than black orblueink.It justmakesit clearer as a part of an explanation.)
+CH3CHCH2CH3+
.CH3
3.
A
Ahas a very strong peak atm/z=43 which is missing from B.The most likely ion with this massfrom the two compoundswould be [CH3CO]+which comesfrom CH3COCH2CH2CH3.The otherreasonably stableion which you would get from this compound would be [CH3CH2CH2CO]+with a mass of 71.So Ahas to be pentane-2-one,CH3COCH2CH2CH3.
B
Thestrong peak atm/z=57 would come from [CH3CH2CO]+, and thiswould be formed if pentan-3-one fragmented eitherside of the CO group.Thestrong lineat m/z = 29, would come from [CH3CH2]+formed by thesame break, but with the charge on the ethyl group leaving the unpaired electron on the CO group.So B isconsistent with pentan-3-one, CH3CH2COCH2CH3.
4. From the data, we can get a molecular formula of C4H8O matches the MW of 72 g / mol.The following pieces can be deduced from the spectra C=O, -CH3, -CH3, and -CH2-Given these piecesone way they can fit together:
6. We can get a molecular formula of C3H8O which matches the MW = 60 g / mol. In the MS, the molecular ion occurs at m/z = 60 indicating the MW = 60 g / mol.
The IR shows an -OH (3350 cm-1) and C-O (1075 cm-1) but no C=O (around 1700 cm-1), or C=C (around 1600 cm-1). There is only one way they can all fit together:
7. The only ting the IR gives is that we have a C=O group at 1700 cm-1. This weighs 28 but we
have 29 on the mass spectrograph, thus the fragment is probably [HCO]+ for 29. The Molecular ion is at 58 (M+1) peak not required, as 58-28 (C=O) = 29 for a [C2H5]+ ethyl peak .
Thus it is ethanol, which has a molar mass Mr (C2H5CHO ) = 58 amu’s and for 6.02 x 1023 it is
58 g/mol
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