Download MS and IR Exam Qs

16
7*
Organic compound C has the following percentage composition by mass:
C, 54.5%; H, 9.1%; O, 36.4%.
The infrared spectrum and mass spectrum of compound C are shown below.
100
transmittance
50
(%)
0
4000
3000
1000
2000
1500
wavenumber / cm–1
500
100
80
relative
intensity
60
40
20
0
10
20
30
40
50
m /z
60
70
80
90
In the mass spectrum, a secondary carbocation is responsible for the peak with the greatest
relative intensity.
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17
Identify compound C.
In your answer you should make clear how your conclusion is linked to all the evidence.
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END OF QUESTION PAPER
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16
7*
Compound F is a trans stereoisomer which is a useful intermediate in organic synthesis.
The results of elemental and spectral analysis of compound F are shown below.
Percentage composition by mass: C, 68.6 %; H, 8.6 %; O, 22.8 %.
Infrared spectrum
100
transmittance
(%)
50
0
4000
3000
2000
1500
wavenumber / cm–1
500
1000
Mass spectrum
100
80
60
relative
intensity
40
20
0
10
20
30
40
50
m /z
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60
70
80
17
In the mass spectrum, the peak with the greatest relative intensity is caused by the loss of a
functional group from the molecular ion of compound F.
Determine the structure of compound F.
Explain your reasoning and show your working.
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END OF QUESTION PAPER
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PMT
21
7
A student was provided with a mixture of two structural isomers. Each isomer has the percentage
composition by mass C, 29.29%; H, 5.70%; Br, 65.01%. The relative molecular mass of each
isomer is less than 150.
(a) Determine the structures of the two structural isomers.
Show your working.
In your answer you should link the evidence with your explanation.
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[5]
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PMT
22
(b) The student heats the mixture of the two structural isomers from (a) under reflux with
aqueous sodium hydroxide to form two compounds, E and F. The student separates the two
compounds.
Compound E is heated under reflux with acidified potassium dichromate(VI) to form
compound G, which gives the infrared spectrum below.
100
transmittance
(%)
50
0
4000
3000
2000
wavenumber /
(i)
1500
1000
500
cm–1
Analyse the information and spectrum to determine the structures of E, F and G.
Include an equation for the formation of G from E.
In your answer you should link the evidence with your explanation.
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PMT
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(ii)
Compound G is heated with compound F in the presence of a small amount of
concentrated sulfuric acid to form organic compound H.
Draw the structure of the organic compound H.
[2]
[Total: 13]
END OF QUESTION PAPER
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11
5
Propanoic acid, CH3CH2COOH, is a member of the homologous series of carboxylic acids.
(a) Suggest the general formula for a carboxylic acid.
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(b) The displayed formula for propanoic acid is shown below.
H
(i)
H
H
C
C
H
H
O
C
O
H
State the shape and bond angle around a carbon atom in the alkyl group of propanoic
acid. Explain the shape.
Shape ................................................................................................................................
Bond angle ........................................................................................................................
Explanation .......................................................................................................................
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(ii)
Suggest a value for the C–O–H bond angle in propanoic acid.
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12
(c) Compound D is a neutral compound which is a structural isomer of propanoic acid,
CH3CH2COOH.
The infrared spectrum of compound D is shown below.
Item removed due to third party copyright restrictions.
peak at
at
peace
1700
3300
cm
1
on
Suggest two possible structures of compound D.
Explain all your reasoning.
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13
(d) 2-Chloropropanoic acid, CH3CHClCOOH, can be made by reacting propanoic acid with
chlorine in a radical substitution reaction.
(i)
State the conditions for the reaction.
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(ii)
Write the overall equation for the reaction.
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(iii)
The first step in the reaction mechanism involves homolytic fission of a chlorine molecule
to form two chlorine radicals.
Why is this step an example of homolytic fission?
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(iv)
Write two equations to show the propagation steps in the mechanism for this reaction.
Use dots,•, to show the unpaired electrons on radicals.
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(v)
Draw the displayed formula of the radical formed in the first propagation step.
Use a dot,•, to show the position of the unpaired electron.
[1]
(vi)
Further substitution forms a mixture of organic products.
Draw the structure of an organic product formed from 2-chloropropanoic acid by further
substitution.
[1]
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