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CHEMISTRY *** SEMESTER 01-2008 FINAL EXAMINATION CONTD.
Answer Key
Section I
EBCDB CDBEE CBEAE DCDBC BEAAE DCADC CACCB BDEAC
General remarks:
Students failed to follow instructions. For example, they answered more than the required number of
alternatives (questions 2 & 4 in section II).
Students failed to address questions. For example, they gave equations or mechanisms where reaction schemes
were asked for, they gave one equation where more than one was required, they gave skeletal or molecular
formulae where structural formulae were required etc. etc.
Students in many cases showed ignorance of the basic principles of chemistry (of which the most fundamental
involve charge). They gave formulae for carbocations without any charge, they omitted the charges on familiar
ions such as Cl- and carboxylate, they represented bromine as Br instead of Br2, and elemental magnesium as
Mg2+ instead of Mg, they wrote formulae showing carbon forming more than 4 bonds, and so on. In many cases
they seemed to be completely ignorant of the Brønsted-Lowry theory of acids and bases. Many students cannot
distinguish between groups, atoms, molecules and ions. The words “mixture” and “compound” are also
confused.
Many students’ English is very poor: many students were unable to distinguish between singular and plural
forms. This applied particularly to question 3b in section II, where clearly more than one equation was
required. Many students were unable to express themselves clearly and succinctly, confusing similar words
with different meanings, and using words with the wrong meaning. Many students’ appeared to have little grasp
of the meanings of technical terms. Many students do not understand the correct application of pronouns such
as “it”.
Section II
1) Briefly describe how you would prepare a pure sample of cyclohexene from a sample of cyclohexanol. (8 4)
Cyclohexanol is mixed with concentrated phosphoric acid(1) in a distillation flask. The mixture is warmed
for some time(1) and then the temperature raised until the cyclohexene distils off(1). The crude distillate is
shaken with saturated sodium chloride solution(1) in a separating funnel(1), and the upper, cyclohexene
layer is collected(1). This is then dried with anhydrous calcium chloride(1) and finally redistilled.(1)
Note: this mark was adjusted to 4. It was felt that the question was not sufficiently clear. Thus any 4 of the
above could earn full marks on this question.
a) Complete the following reaction schemes. Show stereochemistry where appropriate and name the
organic product.
Cl2
(2)
i)
Cl
H
H
Cl
trans-1,2-dichlorocyclohexane
HBr
(2)
ii)
Br
bromocyclohexane
KMnO4
iii)
OH-, cold
(2)
H
OH
H
cis-1,2-cyclohexanediol
OH
b) For the reaction of hydrogen bromide with 1-methylcyclohexene,
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CHEMISTRY *** SEMESTER 01-2008 FINAL EXAMINATION CONTD.
CH3
i)
draw the structures of the two possible carbocation intermediates.
(2)
H
C
C
CH3
CH3
H
ii) Which of your structures is the more stable? Use this to explain why the observed product is 1bromo-1-methylcyclohexane rather than 1-bromo-2-methylcyclohexane.
(3)
The left hand carbocation is the more stable (tertiary carbocations are more stable than secondary) and
so will be formed much more rapidly than the right hand one. Attack of a Br- ion on the left hand
structure leads to the observed product.
2) In THREE of the following cases suggest how you could carry out the conversion indicated by writing a
reaction scheme using structural formulae. (All of them can be accomplished in not more than two steps.)
Mention reagents used and any special conditions.
(12)
NB Many students answered more than the required 3 questions. Since one of the primary purposes of an
examination is to see whether students can follow directions, only the FIRST 3 answers were marked.
a) Bromoethane to 1-propanol.
H H
H H
H H H
H
C
C
Mg
Br
H
C
C
Mg
Br
dry ether
1.CH2O
H
C
C
C
H
H
H
OH
2.H2O
H H
H
b) Benzene to 4-bromonitrobenzene.
H
Br
Br
Br2
conc. HNO3
Fe
Warm
Br
+
NO2
O2N
c) 1-Iodopropane to 1-butanamine.
H H H
H C C C
KCN
I
H H H
H C C C CN
H H H
LiAlH4
H H H
H H H H
H C C C C NH2
H H H H
d) Ethene to ethanoic acid.
e) Propane to 2-propanol.
H H H
Cl2
H C C C H
hν
H H H
H Cl H
H C C C H
NaOH
H OH H
H C C C H
H H H
H H H
f) Bromobenzene to benzoic acid.
Br
Mg
Mg Br
CO2, H2O
COOH
Dry ether
3) Benzene reacts with hydrogen at high temperature and pressure in the presence of a suitable catalyst to form
cyclohexane. Cyclohexene reacts under milder conditions in the presence of the same catalyst to form the
same product.
a) Give the name of one catalyst which would be appropriate for the hydrogenation of cyclohexene.
(1)
Nickel (or platinum etc.)
b) Write balanced equations (using structural formulae) to show the hydrogenation reactions mentioned
above. (2)
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CHEMISTRY *** SEMESTER 01-2008 FINAL EXAMINATION CONTD.
H
H
H
H
C
C
C
H
+ 3H2
C
C
H
C
H
H
H
H
H
H
H
H
H
H
C
H
C
+ H2
C
C
C
H
C
C
H
H
H
H
C
H
C
H
C
H
H
H
H
H
H
H
H
C
H
C
H
H
C
C
C
C
H
C
H
C
H
H
H
(“skeletal” or “line angle” formulae were also allowed.)
Note: some students unaccountably failed to give the equations here, but presented them when required to
calculate the resonance energy – hence throwing away 2 marks!
c) Sketch the carbon skeleton of the benzene molecule and indicate on your diagram the lowest energy π
molecular orbital.
(2)
d) Calculate the resonance energy of benzene, given that the molar heat of hydrogenation of benzene
(∆Hh) is -208 kJ mol-1 and the equivalent value for cyclohexene is -120 kJ mol-1.
(3)
Calculated molar heat of hydrogenation of benzene = 3 x molar heat of hydrogenation of cyclohexene
= 3 x -120 = -360 kJ mol-1
The difference between this calculated value and the observed value is the resonance energy of benzene
= |-360 – (-208)| kJ mol-1 = 152 kJ mol-1
e) Benzene requires more vigorous conditions than cyclohexene to react with hydrogen, as noted above.
How does resonance help to explain this?
(2)
Resonance makes benzene more stable than cyclohexene and so less reactive than cyclohexene.
(8)
4) Explain any TWO of the following facts relating to the properties of alcohols and phenols.
NB Many students answered more than the required 2 questions. Since one of the primary purposes of an
examination is to see whether students can follow directions, only the FIRST 2 answers were marked.
a) Methoxymethane boils at a much lower temperature than ethanol has despite the fact that both
compounds have the same relative molecular mass.
The fact that the two molecules have the same RMM means that they experience similar London
Dispersion forces. However, ethanol (CH3CH2OH) has an OH group, which means that molecules can
hydrogen-bond with one another. Methoxymethane (CH3OCH3) has no OH group so molecules cannot
hydrogen-bond with one another. The stronger intermolecular forces produced by the hydrogen-bonding
account for the higher boiling point of ethanol
b) Ethanol is miscible with water in all proportions, whilst chloroethane is virtually insoluble in water.
Ethanol (CH3CH2OH) is very soluble in water because the OH group can act as both a hydrogen-bond
acceptor and donor with water molecules. Chloroethane (CH3CH2Cl) only contains H bonded to C, and
does not contain O, N or F, and so cannot act as either a hydrogen-bond acceptor nor donor.
c) Carboxylic acids are very much stronger acids than alcohols, despite the fact that both possess the -OH
group.
Carboxylic acids form the carboxylate anion (RCOO-) when they ionise in water. This is relatively
stable because the negative charge on the ion is delocalised over the two oxygen atoms by resonance.
When alcohols ionise they form the alkoxide ion (RO-) with the negative charge localised on only one
oxygen atom. The relative stability of the carboxylate anion means that carboxylic acids dissociate to a
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CHEMISTRY *** SEMESTER 01-2008 FINAL EXAMINATION CONTD.
much greater extent than alcohols.
d) 1-hexanol is only slightly soluble in water whilst on the other hand 2,3-hexanediol
(CH3CH(OH)CH(OH)CH2CH2CH3) is miscible in all proportions with water.
1-hexanol contains a single OH group which can act as both an H-bond acceptor a donor with water
molecules. Its solubility is limited because of the large hydrocarbon chain, which is hydrophobic.
2,3-hexanediol has similar hydrocarbon chain, but has two OH groups which can H-bond with water
molecules. These extra hydrogen-bonds make 2,3-hexanediol much more soluble.
e) Phenol is a much stronger acid than ethanol.
Phenol ionises to form the phenoxide ion whereas ethanol ionises to form the ethoxide ion. In
phenoxide the negative charge is on an oxygen atom adjacent to a benzene ring and is therefore
delocalised around the ring by resonance. In the ethoxide ion no such stabilisation is possible. Thus
phenol ionises to a much greater extent than ethanol.
5) The compound 2-iodobutane may be hydrolysed by boiling it with aqueous sodium hydroxide. Most of the
product is formed by an SN2 reaction.
a) What is the main product of this reaction? Give a perspective formula and a name.
(2)
C 2H7
C
H
OH
H 3C
(S)-2-butanol
b) Give a mechanism for the reaction showing the transition state and explain why it leads to inversion of
configuration.
(5)
(3)
The hydroxide ion attacks from the side of the central carbon opposite to the departing iodine atom.
This flips the other groups to the other side of the molecule, like an umbrella being blown inside out. (2)
Note: many students were not able to unambiguously describe how the OH- ion attacked. Many people
called it a “hydroxyl group” which it is not before it becomes attached to the carbon.
c) Hydrolysis of S-3-methyl-3-iodohexane under appropriate conditions produces a racemic mixture of
2-methyl-2-hexanol.
i) Explain the meaning of the term racemic mixture.
(2)
This is a 50/50 mixture of the enantiomers of an optically active compound which shows no optical
activity.
ii) Suggest a name for the mechanism that is operating here.
(1)
SN1
(or “unimolecular nucleophilic substitution”)
iii) Why does this mechanism lead to racemisation?
(3)
This mechanism leads to the formation of a planar carbocation intermediate which can be attacked from
either side by a nucleophile, leading to a mixture of both enantiomers.
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