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2011
Chemisstry GA 1:
1 Written
n examinaation 1
GENERA
AL COM
MMENTS
A total of 9048 students saat the June 2011 Chemistry examination. Overall, studdents performeed well. A scoore of just
w
a score of just over 79
7 per cent waas needed to reeceive an A. The
T mean
over 85 per ccent was needed for an A+, while
score for the examination was
w 60 per ceent, which corrresponded to the
t middle off the C+ grade range. These statistics
t be slightly more
m
challengging than the corresponding
c
g 2010 examinnation.
suggested thaat the examinaation proved to
However, thee vast majorityy of students were
w able to finish
f
the exam
mination, to thhe best of theirr ability, within the allottedd
90 minutes.
There were ssignificant insttances in whicch students did not make efffective use off supplied dataa or informatioon; for
example, in Section
S
A, Qu
uestion 13. Onne of the aims of the study is that studentss are able to apply
a
their undderstanding
of chemistry to both familiar and new siituations.
wered correcttly, and five quuestions that
In Section A, there were 11 questions thhat more than 60 per cent off students answ
less than 50 pper cent of stuudents answerred correctly.
Question 8 proved
p
to be thhe most difficuult question inn this section. The majority of students prroceeded alonng the
carboxyl + am
mino  peptiide path withoout considering the requirem
ment that the product,
p
ibuprrofen lysine, needed
n
to be
more solublee in water than
n ibuprofen. The
T question required
r
students to recogniise that a deprrotonated carbboxyl group
increases solubility, a poinnt associated with
w ‘soluble’ aspirin.
For Questionn 9 it was deciided that two responses
r
couuld be correct since it was not made clearr in the questioon that the
ethanol woulld most likely be in excess. Even with thiis allowance, it was evidentt that the propperties of the reactants,
r
products andd the catalyst inn biodiesel prroduction weree not well undderstood.
While studennts would have performed tiitrations as paart of their praactical work thhroughout the unit, understaanding of the
‘equivalencee point’ was noot strong. Thiss was particularly evident inn Question 11
1.
Question 13 emphasised th
he importancee of not only reading
r
the quuestion carefullly, but also ussing the suppllied data
The term ‘heatted to constannt mass’ may have
h
encouragged many studdents to assum
me that all the water
w
was
effectively. T
driven off thee sample. How
wever, a quickk check of n(H
H2O) removedd showed that this was not thhe case.
The overall performance
p
on
o Question 155 was surprisiing. Many students simply did
d not make the required use
u of the
ratio n(NaN3)/n(N2).
Question 18 challenged stuudents to conssider the key factors
f
that deetermine the IR
R absorption band
b
for a covvalent bond.
These factorss include bond
d strength andd the relative masses
m
of the two
t atoms in the bond. Oveerall question performance
showed that m
most studentss were drawn to
t electronegaativity as the key
k factor. Un
nderstanding of
o the principlees of
spectroscopyy is part of thee study.
Question 20 showed that students’
s
undeerstanding of the
t effect of liight from the light
l
source on
n the metal attoms in the
flame of an aatomic absorpttion spectrom
meter could be improved. It suggested
s
thatt many studennts did not maake the
distinction beetween oxidattion and the trransition of eleectrons to highher energy levvels due to thee absorption of
o energy.
As a part of ttheir examinattion preparatioon, students shhould be encoouraged to crittically review multiple-choiice questions
on which theey struggle. Teeachers are enncouraged to provide
p
criticaal analysis and
d feedback to their
t
students on their
performance on multiple-cchoice questioons.
Section B proovided studennts with a mixtture of questio
on types and a wide varietyy of challengess.
Performancee on Question 1d. reiterated the impressio
on that biodiessel is a topic worth
w
addressiing. Question 1f., also
linked to fattty acids, required students to recognise thhe need to use the ratio n(Brr2)/n(compounnd) to determiine the
number of C=
=C bonds present in each molecule
m
and then
t
identify which
w
of the thhree fatty acid
ds in the list qualified.
q
Responses too Question 2 demonstrated
d
t challengess that descripttive responses pose for som
the
me students. It appears that
the message about the need for a (+) chaarge on speciees causing peaaks on a mass spectrum is not
n yet well knnown.
Chemistry GA 1 Exam
© VICTORIAN CURRICULUM AND
A
ASSESSME
ENT AUTHORIT
TY 2011
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Question 2aii. emphasisedd that interprettation of the suupplied data, iin this case th
he mass spectru
rum, in terms of
o the contextt
l
on the y-of the questioon is a challennge for a signiificant proporttion of studennts. Many misrread or misunderstood the label
axis, while others focused on the relativve abundancess of the atoms rather than thhe molecular ioons. Questionn 2b. was well
handled. Most students weere able to linkk the informattion provided on the 1H NM
MR spectrum to
t molecular structure.
s
N
spectrosccopy.
Students conntinued to resppond well to questions on NMR
In answeringg Question 3cii. most students focused on the truncated peak on the espresso
e
coffee chromatograam and
ignored the ppresence of thee peak area inn the data tablee. Students neeed to be awarre that, when suggesting
s
thaat a solution
should be dilluted, it is apppropriate that parameters
p
bee set. For exam
mple, in Questtion 3c. ‘to briing the peak area
a within
the range of the
t calibrationn graph’.
Question 4a. was a stoichiometry questiion that includded much dataa. The challenge for studentts was to identtify the data
olution most efficiently.
e
Coonfusion resullted for those students who felt the need to
t use all the
relevant to geetting to the so
data. In prepaaring for exam
minations studdents should be
b encouraged to think throuugh the calcullation steps beefore
beginning a ssolution, ratheer than just usiing the data inn the order in which it appeears. In Question 4b. most sttudents
interpreted thhe question ass if the moistuure in the origiinal precipitatee (in 4a.) had been an experrimental errorr. The intent
of the questioon was that booth techniquess, in 4a. and 4b., were validd, deliberate an
nd accurate. Consequently,
C
the
calculated peercentage P2O5 should havee been the sam
me for both tecchniques.
Question 5a. assessed stud
dents’ ability to
t write and coombine half-eequations. This is a skill thaat is regularly assessed
a
and
p
Maany students who
w provided a correctly baalanced oxidattion halfwith which students are exxpected to be proficient.
equation in Question
Q
5ai. did
d not accuraately combine it with a giveen reduction half-equation in
n 5aii.
Question 5b. was a stoichiiometry questiion that was handled
h
much better than Qu
uestion 4a. Stu
udents seemedd more
comfortable with titration--based calculaations. With caalculations rellated to back-ttitration, studeents should usse labels such
as ‘supplied’, ‘excess/unreeacted’ and ‘reeacted/reactin
ng’ appropriateely. In Questiions 5bi. and 5bii.
5
it was neecessary to
m of Cr2O72–– three times.
work out the number of mole
T associatedd questions reevealed some significant
s
Question 7a. provided studdents with an organic reactiion pathway. The
uired students to draw a struucture of an alcohol produceed from a chlooroalkane. Errrors included
issues. Questtion 7aii. requ
the incorrect number of caarbon atoms annd inappropriaate representaation of the boond between thhe hydroxyl grroup and
carbon.
r
to the use of correctt systematic nomenclature. In the chemiccal name
Questions 7aaiii. and 7aiv. related
3-methylbutaan-1-ol (3-metthyl-1-butanol), the numberrs 1 and 3 are an essential part
p of the systematic name and are used
because alterrnative locatio
ons of the methhyl and/or hyddroxyl groupss would repressent a differen
nt compound. On an
ethanol moleecule there is no
n other possiible location but
b C-1 for thee hydroxyl funnctional group
p so there is noo number in
the systematiic name. Manyy students inccorrectly incluuded the numbber.
In Question 77av., while H2SO4 or H2SO
O4(l) were acceeptable as reprresenting ‘conncentrated’ sullfuric acid, H2SO4(aq) was
not.
Responses too Question 7b.. suggested that, on seeing the
t words ‘fraactional distillation’ in the question,
q
manyy students
immediately defaulted to the
t fractional distillation
d
off crude oil andd fractionating
g towers ratherr than addressing the
a
to a range of separattions or, in thiis case in partticular, the
fundamental principles of fractional disttillation as it applies
collection off an ester. Disccussion of the method of coollection of ann ester and sepparation of an ester from waater and
unreacted aciid and alcoholl is a logical adjunct
a
to coverage of ester production. This
T question tested
t
studentts’ ability to
apply their unnderstandingss in different ssituations. Thee key factor deetermining diffferent boiling
g temperature in a mixture
of substancess is intermoleccular attractioon.
Question 7c. proved challeenging for a majority
m
of stu
udents. The reqquirement to ‘explain
‘
how the
t evidence provided
p
by
m the reaction mixture
m
had been achieved’’ was not welll
the spectra inndicated that a complete sepparation of baanana oil from
interpreted. Some
S
studentss were able to draws links between
b
the allcohol IR specctrum and the banana oil IR
R spectrum to
argue for sepparation from the
t alcohol, but few were able
a to effectivvely argue for separation fro
om the acid. A useful
ensuing discuussion point might
m
have beeen how the information on the ester spectrum alone shhows separatioon from the
acid and the alcohol.
In Question 88c. the main errors
e
were noot ‘showing alll bonds’ and not
n showing thhe amino grouup as protonated.
Chemistry GA 1 Exam
Published: 3 November 2011
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SPECIFIC
C INFORM
MATON
Section A – Multiple-choice question
ns
The table beelow indicates the percenttage of studen
nts who chosee each option. The correctt answer is in
ndicated by
shading.
% Noo
Question
%B
%C
%D
Com
mments
%A
Answeer
Ethanol (C
CH3CH2OH), ethylamine (C
CH2CH2NH2),
and ethano
oic acid (CH3COOH)
C
all disssolve in
75
5
13
7
0
1
water via hydrogen
h
bon
nding. Ethane (CH3CH3) is
non-polar and is less soluble in waterr.
2
3
80
12
5
0
3
2
2
95
1
0
4
12
8
73
7
0
5
76
12
9
4
0
6
15
19
8
58
1
7
6
7
79
8
0
8
7
10
77
6
0
9
38
31
17
14
4
0
Chemistry GA 1 Exam
The moleccular formula of 2,2,4-trimeethylpentane
is C8H18.
X is the bo
ond between amino
a
acid ressidues in the
primary sttructure of thee protein. Thiss covalent
bond betw
ween C and N in the linking peptide
group is allso known as an amide bonnd.
Y is the bo
ond formed when
w
–SH on thhe side
groups of cysteine moleecules reacts inn
establishin
ng the tertiary structure of thhe protein.
The covaleent S–S bond formed is alsoo known as a
disulfide bond.
b
All amino acids containn the amino, –N
NH2, group,
which is basic
b
and woulld be expectedd to react
with 1.0 M HCl(aq).
X – 1,1,2-trichloroethen
ne
Step 2 is a substitution reaction,
r
H suubstituted by
Br.
Ibuprofen lysine is moree soluble in water
w
than
ibuprofen.. So when ibup
profen reacts with lysine
the producct, ibuprofen lysine,
l
must haave a
structure thhat enables it to dissolve inn water more
readily thaan ibuprofen.
Option C would
w
result from
f
a condennsation
reaction beetween the –C
COOH group on
o ibuprofen
and the ––NH2 group onn lysine. The product
would disssolve in waterr via hydrogenn bonding.
Option D would
w
result from
f
an acid-bbase reaction
between th
he –COOH grroup on ibuproofen and the
–NH2 gro
oup on lysine. This productt would
dissolve inn water via ion
n-dipole bondding and
hydrogen bonding.
b
The negatiive ion, formeed when ibuprrofen
molecules donate H+, diissolves via ioon-dipole
bonding. The
T positive io
on, formed whhen lysine
molecules gain H+, dissolves via ion--dipole and
hydrogen bonding.
b
Since ion--dipole bondinng is stronger than
hydrogen bonding,
b
optio
on D best shoowed the
structure of
o ibuprofen lyysine.
The biodieesel ethyl steaarate is producced via the
reaction: canola
c
oil + 3 ethanol
3 ethyl stearate + glycerol
in the pressence of potassium hydroxidde catalyst.
Published: 3 November 2011
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Question
%A
%B
%C
%D
% Noo
Answeer
10
8
56
27
10
0
0
11
17
45
32
6
0
12
12
15
61
12
2
0
Chemistry GA 1 Exam
Com
mments
Since the ethyl
e
stearate is collected inn layer A,
layer B muust contain glyycerol, potasssium
hydroxide and unreacted ethanol. Thiis separation
reflects thee non-polar naature of the biiodiesel as
distinct fro
om the polar ethanol
e
and gllycerol, and
ionic KOH
H. On this basis, the correctt answer was
option D. Ideally, the etthanol should be in excess
to ensure complete
c
reacction of the cannola oil.
However, this was not made
m
clear in the question,
and it was possible that students mayy have
assumed th
hat the ethano
ol and canola oil
o were
present in the exact stoichiometric rattio for
complete reaction.
r
In thhis case, layer B would
contain onnly glycerol annd potassium hhydroxide.
Hence, option A was alsso accepted ass correct.
Consider the
t amounts of
o CH4 and CO
O2 produced
starting froom 1 mol C6H12O6.
Step 1: prroducts are
2 mol CH3CH2OH
O + 2 mol C
CO2
Step 2: all the CH3CH2OH from stepp 1 is
coonverted to CH
H3COOH, hennce 2 mol
CH3CH2OH
O  2 mol CH
C 3COOH
Step 3: 2 mol
m CH3COOH
 1 mol Ca(CH
C
3COO)2 + 1 mol CO2
Step 4: 1 mol
m Ca(CH3COO)2
 2 mol CH4 + 1 mol CO2
Total n(CO
O2) produced = 4 mol
n(CH4) prooduced = 2 mol
m
Ratio n(CH
H4) produced : n(CO2) prodduced = 2:4
= 1:2
Since the CH
C 4 and the CO
C 2 may be asssumed to be
collected at
a the same tem
mperature andd pressure,
the ratio V(CH
V
)
produc
ced:
V(CO2) prroduced
4
equals 1:2.
When a strrong base succh as NaOH(aqq) is added to
an acid, weak
w
or strong,, the key factoor influencing
the amounnt of NaOH(aq
q) required to reach the
equivalencce is the n(H+) available froom the acid.
Since therre was 25.0 mL
L of both acidds, and both
were 0.10 Mn(acid) preesent
= 0.10×
×25.0×10–3
= 2.0×10–3 mol
Both acidss are monoprootic, so the n(H
H+) available
for titratio
on is the same for each acid.. Hence, the
same amouunt of 0.20 M NaOH(aq) iss required to
reach the equivalence
e
point for both acids.
a
Representiing the acids by
b the generall formula
HA(aq), thhe equation fo
or the titration reaction in
both casess is
HA(aq) + NaOH(aq)  NaA(aq) + H2O(l)
Since the n(HA)
n
is the same
s
for both acids, the
n(NaOH) required
r
is thee same for botth acids.
This questtion required students
s
to maake effective
use of the information in
n Table 11 off the Data
Book (Aciid-base Indicaators). Accordding to the
informatioon in the table:
Thymol bllue is yellow above
a
pH = 2..8
Published: 3 November 2011
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Question
%A
%B
%C
%D
% Noo
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Com
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Methyl redd is yellow abbove pH = 6.3
Phenolphtthalein is colou
urless below pH
p = 8.3
This suggeests that the pH
H of the soluttion was
above pH = 6.3 and beloow pH = 8.3, i.e. between
pH = 6.3 and
a pH = 8.3.
13
33
38
14
14
4
1
14
6
14
18
61
1
15
28
59
10
3
1
16
Chemistry GA 1 Exam
2
69
25
3
0
Overall peerformance on
n this questionn suggested
that a signnificant numbeer of students were unable
to effectiv
vely interpret the
t informatioon in Table 11
of the Dataa Book. Teachhers should ennsure that
students use the Data Book throughouut the study.
n(CaSO4.2
2H2O) = 172
2.1/172.1
= 1.0
000 mol
The loss in
n mass duringg heating was ddue to the
release of H2O.
n(H2O) relleased = 27.0//18.0
= 1.50 mol
The ratio
n(CaSO4.2
2H2O) reacting
g: n(H2O) releeased = 1:1.5
= 2:3
So the coeefficients of CaSO4.2H2O annd H2O in thee
equation must
m be in the ratio 2:3.
Alternativ
vely, 1 mol CaaSO4.2H2O relleases 1.5
mol H2O, leaving 2 – 1..5 = ½ mol H2O remaining
in the hydrrated compou
und, so the cheemical
formula off the product of
o the reactionn must be
CaSO4.½H
H2O. This is co
onsistent withh the equation
2CaSO4.2H
H2O(s)  2CaSO4.½H2O(ss) + 3H2O(l).
d(gas) = 2.86 grams perr litre at STP
Volume off 1 mol gas att = 22.4 L
Hence the M(gas) = 2.886 × 22.4
= 64.1 g mol–1
M(SO2) = 64.1 g mol–1
The inflatiion of the airb
bag to 62.0 L was
w due to
the releasee of N2(g); i.e.. the V(N2) preesent at 100
kPa and 366.6C = 62.0 L
n(N2) prodduced = P(N2) × V(N2)/RT
= 100 × 62.0/[8.31 × (36.6+273)]
= 2.41 mol
Accordingg to the equation for the reaction
n(NaN3)/nn(N2) = 10/16 = 5/8
Hence n(N
NaN3) = (5/8)) × n(N2)
= (5/8) × 2.41
= 1.51
1 mol
m(NaN3) = n(NaN3) × M(NaN3)
= 1.51 × 65.0
= 97.9 g
If the starcch molecule iss formed from
m 500
C6H12O6 molecules,
m
4999 H2O molecuules will be
produced during
d
the conndensation poolymerisation
reaction.
M(starch) = 500 × M(C
C6H12O6) – 4999 × M(H2O)
= 500 × 1800.0 – 499 × 188.0
= 90000 – 8982
8
= 81018 g mol
m –1
Option C (90
( 000 g moll–1) overlookeed the
Published: 3 November 2011
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Question
%A
%B
%C
%D
% Noo
Answeer
17
71
11
6
12
2
0
18
51
20
17
11
1
19
26
6
5
62
2
0
20
12
40
29
18
0
Chemistry GA 1 Exam
Com
mments
condensatiion nature of the
t formation of starch
from glucoose.
In thin-layyer chromatoggraphy, the stroonger the
adsorptionn to the stationnary phase, thee smaller the
distance trravelled up thee stationary phhase and the
lower the Rf value.
At ordinarry temperaturees covalent boonds vibrate,
and the vibbrational enerrgies of moleccules have
defined ennergy (quantum
m) levels (stattes) in the
same mannner as electron
nic energy levvels.
Transitionns between vibbrational energgy levels
depend onn absorption off infrared radiiation of
energy maatching the diffference betweeen
vibrationaal energy levells. The energyy absorbed is
proportionnal to the frequuency of the IR
I radiation,
which is proportional
p
to
o the wavenum
mber.
Covalent bonds
b
in moleecules are not rigid, but
may be coompared to stifff springs thatt can be
stretched and
a bent.
The exact frequency (w
wavenumber) at
a which a
given vibrration occurs is
i determined by the
strength off the bonds innvolved and thhe masses of
the atoms in the bond.
–
C–H bondds (bond energ
gy 410 kJ mol–1
) are
stronger thhan C–O bondds (bond energgy 326 kJ
mol–1).
The mass effect can be attributed to the
t ‘different’
atom in th
he C–O and C––H bonds. Thee lighter H
atom prod
duces a higher vibration freqquency, so
the IR radiiation required
d to bring aboout the
transitionss to higher vibbrational energgy levels in
C–H bondds will be of hiigher wavenum
mber than forr
C–O bondds.
So, the low
wer IR wavenu
umber for bondd stretching in
a C–O bonnd can be attrib
buted to largerr atomic mass
of oxygen atoms comparred to hydrogeen atoms.
A mixture of hydrocarbo
on molecules would be
quite volattile and most effectively
e
sepparated into
its compon
nent compounnds by gas chroomatography.
Each comp
ponent could then be identiified by mass
spectroscoopy.
Aqueous solutions
s
containing the Cu2+(aq) ion
are blue beecause in the presence
p
of white
w
light
they absorrb wavelengthhs in the red reegion of the
visible speectrum, and so
o transmit wavvelengths in
the blue reegion.
When CuS
SO4(aq) is intrroduced into an
a atomic
absorptionn spectrometerr, the sample is
i atomised
in the high
h temperature flame. As thee light from
the Cu lam
mp passes thro
ough the flamee, electrons inn
the Cu atooms in the flam
me absorb eneergy and are
promoted to higher enerrgy levels, butt are not lost
from the atoms.
a
As exciited electrons return to
lower enerrgy levels, eneergy emitted is
i in the greenn
region of the
t visible speectrum.
Because th
he absorption of energy by the Cu atoms
in the flam
me from the ligght source doees not cause
Published: 3 November 2011
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Question
%A
%B
%C
% Noo
Answeer
%D
Com
mments
loss of electrons from thhe atoms, oxiddation does
not occur..
Section B – Short answer questions
q
For each queestion, an outliine answer (orr answers) is provided.
p
In some cases thee answer given
n is not the onnly answer
that could haave been awarded marks.
Question 1
With the excception of Queestions 1d. andd 1f., this quesstion was welll done.
1a.
Marks
%
A
0
10
1
90
Average
0.9
1b.
Marks
%
H
0
14
1
86
Aveerage
0
0.9
1c.
Marks
%
D and E
0
10
1
28
2
63
1d.
Marks
%
F
0
64
1
36
Aveerage
0
0.4
Averaage
1.66
b
is geenerally a methhyl ester of a fatty acid. Strructure F was the
t only esterr
Most studentts overlooked the fact that biodiesel
in the table oof structures prrovided. Struccture E was chhosen by manyy students, inddicating that they
t
did not reealise that
while glycerool is a productt of the production of biodiesel by transeesterification itt is not a com
mponent of bioodiesel.
1e.
Marks
%
B
0
15
1
85
Aveerage
0
0.9
1f.
Marks
%
F
0
65
1
35
Aveerage
0
0.4
Students needed to realise that to react with
w Br2, the substance
s
had to contain C=
=C bonds; thatt is, carbon-caarbon double
bonds. n(Br2) reacting = 0.320/160.0 = 0.002
0
mol. Sinnce 0.001 mol of the substaance reacts completely withh 0.002 mol
Br2, each moolecule of the substance
s
must contain twoo C=C bonds. A similar calcculation was required
r
on thhe 2009
examination..
Chemistry GA 1 Exam
Published: 3 November 2011
7
201
11
Assess
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Rep
port
There were thhree fatty acidds in the table; options D, F and G (see bbelow) . Saturaated fatty acidds have the genneral formula
CnH2n+1COO
OH or CnH2nO2. Unsaturatedd fatty acids with
w the same number
n
of C atoms
a
as a satuurated fatty accid have two
fewer H atom
ms for each C=
=C present. Thherefore,
D.
F.
G.
C13H27COOH – saturated
s
C17H31COOH – polyunsaturate
p
ed with 2 C=C
C bonds
C17H29COOH – polyunsaturate
p
ed with 3 C=C
C bonds
Question 2
2ai.
Marks
0
1
42
5
58
%
[CH3CH2]+/C
CH3CH2+/C2H5+
Averaage
0.66
a well as expected on this question.
q
Studdents need to be
b aware that the species prroducing
Students did not perform as
mass spectrum carry a positiive charge. Thhis issue has allso arisen on previous
p
exam
minations.
peaks on a m
2aii.
2
Averaage
1
0
Marks
33
25
4
43
1.1
%
One mark waas awarded fo
or a descriptionn of relative abundances
a
(eiither of):
 the rrelative abunddances of the 779Br and 81Br isotopes are approximately
a
y equal
 the 79Br isotope iss slightly morre abundant thhan the 81Br isootope.
One mark forr reference to the molecularr ion peaks (eiither of):
 the ppeaks at m/z = 108, i.e. [C2H579Br]+ and m/z
m = 110, i.e. [C2H581Br]+ are approxim
mately the samee height
+
79
 the ppeak at m/z = 108, i.e. [C2H5 Br] is slig
ghtly higher thhan the peak at
a m/z = 110, i.e. [C2H581Br]]+.
Performancee on this questiion suggestedd that many stuudents strugglled to effectively interpret and/or
a
use the information
given about tthe molecular ion peaks. Fuundamentally, this was abouut the differennt peak heightss with the highher peak
height at m/zz=108 indicatiing a greater abundance
a
of the
t lighter moolecule – the one
o with the lighter Br atom
m.
A number off students focuused on the peeaks for 79Br and
a 81Br ratherr than the molecular ion peaaks. Expressioons such as
the peak at m
m/z = 108 bein
ng 60 per cent abundant sug
ggested that thhe ‘relative abu
undance’ scalle in the conteext of mass
spectroscopyy where the mo
ost abundant species
s
is set at
a relative abuundance 100 was
w not well understood.
u
2bi.
Marks
%
Structure 1
0
4
1
7
2
89
Averaage
1.9
Structu
ure 2
This questionn was well donne.
Chemistry GA 1 Exam
Published: 3 November 2011
8
201
11
Assess
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Rep
port
2bii.
0
1
2
3
Averagee
Marks
14
6
23
58
%
2
One mark eaach was awardded for:
 circling the correct structure inn Question 2bii. (Structure 1)
 apprropriate refereence to the siggnals on the 1H NMR spectrrum
 a deescription of thhe different H environmentts in the two compounds.
Possible respponses includeed:
 the N
NMR spectruum of CH3CHB
Br2 shows twoo sets of peakss/two signals
 the N
NMR spectruum of CH3CHB
Br2 shows a doublet
d
and a quartet
q
 this is consistent with the struccture because there
t
are the two
t different hydrogen
h
enviironments – one for the
a
to CH
H2Br, and onee for the H on CHBr2 attach
hed to CH3
threee hydrogen attoms on CH3 attached
 this is not consisttent with CH2BrCH
B
Br
stru
cture
in
which
h
all
H
atoms
are
in
the
sam
me
environmennt
2
 the splitting patteern is consistennt with the diffferent hydroggen environmeents on the CH
H3CHBr2 struccture, – a
ms on CH3 atttached to CH2Br, and a quaartet for H on CHBr
C
ed to CH3.
douublet for three hydrogen atom
2 attache
Overall perfoormance on thhis question was
w strong, aideed in part by tthe fact that it was possible to access fulll marks
without referrring to the splitting patternn on the spectrrum. Many stuudents who didd refer to the signal
s
splittingg explained itt
very well. Sttudents continuue to do well on NMR-relaated questions..
Question 3a.
0
Marks
1
Average
33
67
%
0.7
128 ppm (126–129 ppm was
w accepted)
on were associated with inaaccurate readinng of the graph.
Most errors oon this questio
3b.
0
Marks
1
Averaage
12
8
88
0.99
%
There is no ppeak at the reteention time off caffeine/no peak
p
at 96 secconds.
3ci.
0
1
Averaage
Marks
71
2
29
0.33
%
The caffeine peak area is beyond
b
the rannge of the caliibration graphh. Extrapolatioon outside the range of the sstandard
solutions maay not be accurrate.
Many incorreect responses referred to thee caffeine peaak on the chrom
matogram rath
her than the calibration grapph and
argued that since the top of
o the peak waas not visible, the peak area could not be measured.
m
Ho
owever, the peeak area of
the largest peeak (caffeine) for espresso coffee
c
was giv
ven in the resuults summary table.
3cii.
0
Marks
1
Average
82
18
0.2
%
Dilute the esppresso coffee sample (eitheer of):
 to bbring its caffeine concentration within thee range of the calibration cuurve
 by a factor > 12.
While many responses refe
ferred to dilution, the term on
o its own wass not sufficiennt. Students were expected to
t either state
c
currve, or suggestt a dilution facctor that woulld bring the
the purpose oof the dilutionn with respect to using the calibration
caffeine conccentration withhin the range of the calibrattion graph. Reesponses to suuch questions should not bee superficial.
Chemistry GA 1 Exam
Published: 3 November 2011
9
201
11
Assess
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Rep
port
Question 4a.
Marks
0
24
%
1
20
2
21
3
22
4
13
Average
1.8
= 4.141 g / 245.3
2
g mol–1
–22
= 1.688×10 mol
n(P2O5) = ½ × n(MgNH
H4PO4.6H2O)
m
= 8.441×10–3 mol
m(P2O5) = 8.441×10–3 mol
m × 142.0 g mol
m –1
= 1.199 g
% P2O5 = (1.199 / 3.2566) × 100
= 36.81 %
n(MgNH4PO
O4.6H2O)
One mark eaach was awardded for:
 corrrectly calculatting n(MgNH4PO4.6H2O)
 accuurately calculaating n(P2O5)
 accuurately calculaating the m(P2O5)
 accuurately calculaating the perceentage P2O5 too four significcant figures.
Many studennts struggled too correctly ideentify the dataa relevant to thhe calculationns and attemptted to use all data.
d
The key
to an efficiennt response waas realising that all the P in 3.256 g of ferrtiliser ends up
p in 4.141 g of
o MgNH4PO4.6H2O.
Correctly callculating the n(P
n 2O5) was paarticularly chaallenging, witth most studennts not linkingg it to
n(MgNH4PO
O4.6H2O). Sincce n(P) = n(M
MgNH4PO4.6H
H2O) and n(P2O5) = ½ × n(P)
then n(P2O5) = ½ × n(MgN
NH4PO4.6H2O).
O
n full marks because
b
their final
f
answer was
w not given to
t four signifiicant figures. This
T was
Some studennts did not gain
expected beccause all data that
t needed too be used in thhe calculationss was given to
o four significaant figures.
4b.
0
1
2
Averaage
Marks
72
6
22
0.5
%
The same, beecause:
 no P is lost on heating the preccipitate
 the mass of the prrecipitate is diivided by the M(MgNH4PO
O4) in calculatiing the percenntage P2O5.
t precipitatee collected hadd been deliberrately heated above
a
100 C to
The key poinnt in this questtion was that the
completely cconvert the preecipitate to MggNH4PO4 beffore weighing.. The implicattion was that it
i was known that
t the
precipitate coollected was now
n MgNH4PO4. So while the
t mass of prrecipitate collected will be lower,
l
the calculated
n(MgNH4PO
O4) will be the same as the n(MgNH
n
. 2O) in 5a. because the mass
m of precip
pitate is divideed by a lower
4PO4.6H
molar mass. This question
n was intendedd to test the mo
ore able studeents.
Question 5ai.
0
1
Average
Marks
34
66
0.7
%
Either of:
 TeO
O2(s) + 2H2O(ll)  H2TeO4(aq)
(
+ 2H+(aq) + 2e–
2–
 TeO
O2(s) + 2H2O(ll)  TeO4 (aaq) + 4H+(aq) + 2e–.
Most errors w
with this equaation were assoociated with th
he location annd number of H+(aq) and e–.
5aii.
Marks
0
1
Average
56
44
0.5
%
2–
+
3TeO2(s) + 1Cr
1 2O7 (aq) + 8H (aq)  3H
3 2TeO4(aq) + 2Cr3+(aq) + 1H2O(l)
Chemistry GA 1 Exam
Published: 3 November 2011
10
201
11
Assess
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Rep
port
Many of the students who had providedd the correct an
nswer for 5ai. did not combbine the two half-equations effectively.
or H2O(l) was quite common. Further atteention should be paid to thee combinationn of halfAn incorrect coefficient fo
equations to give the overaall redox equaation.
5b. and 5bii..
Average
Marks
1
2
3
4
0
14
8
12
16
50
%
2.8
Many studennts were confuused about thee classificationns of dichromaate ions – ‘suppplied’, ‘in ex
xcess’ and ‘reaacting with
the tellurite’.. While most students
s
handlled the mathem
matics of suchh questions effficiently, logiical sequencinng proved
challenging for
f some.
5bi.
n(Fe2+)
= 0.00525 × 19.71 x 10–3
= 1.0035 × 10–3 mol [1.035×10–3]
n(Cr2O72–) in
n excess (unreeacted)
= n(Fe2+)/6
= 1.035×10–3/6
= 1.72×10–4 mool
5bii.
n(Cr2O72–) suppplied
n(Cr2O72–) reactting
= 0.003052 × 50.00
0×10–3
–3
= 1.5526×10 moll
= 1.5526×0–3 – 1.72x10-4
= 1.3354×10–3 moll
5biii.
0
1
2
Averaage
Marks
35
19
4
46
1.1
%
2––
n(TeO2) = 3 × n(Cr2O7 ) reacting
–
= 3 × 1.354×10–3
–3
= 4.062×10 mol
m
m(TeO2) = 4.062×10–3 mol
m × 159.6 g mol
m –1
= 0.6482 g ... (uunits needed too be included))
2
Students needed to multiplly the n(Cr2O72–) reacting as
a calculated inn 5bii. by the mole ratio forr TeO2/Cr2O72–
from the
5 Some stud
dents used an incorrect mollar mass, that of H2TeO4, which
w
was surpprising since M
M(TeO2) was
equation in 5aii.
given at the beginning
b
of the
t question.
Chemistry GA 1 Exam
Published: 3 November 2011
11
201
11
Assess
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Rep
port
Question 6ai.
Marks
0
6
%
1
94
Average
1
G
A
T
A
C
C
T
A
T
G
6aii.
0
Marks
26
%
12 hydrogen bonds
1
74
Average
0.8
ply the fact thhe there are thrree hydrogen bonds in eachh G–C link
Many studennts were eitherr not aware off or did not app
and two hydrrogen bonds inn each A–T liink.
6b.
0
1
Marks
8
4
%
All of:
 phosphoric acid/pphosphate
 deoxxyribose
 adennine.
2
13
3
76
Averagee
2.6
Students whoo did not gain full marks onn this question
n often did nott correctly nam
me all the reacctants. The general groups
‘sugar’ and ‘nitrogen base’ were given instead
i
of the specific comppounds ‘deoxyyribose’ and ‘adenine’.
Question 7ai.
1
0
Marks
20
80
%
Any of:
 OH–
 NaO
OH
 potaassium hydroxxide.
7aii.
Marks
%
0
37
1
63
Average
0.8
Average
0.7
Given that thhe question askked for a struccture rather thhan the structuure showing alll bonds, it waas not necessarry to show alll
bonds. Howeever, it had to be clear that it
i is the O endd of the OH thhat is bonded to C, and that there
t
were tw
wo CH3
groups bondeed to carbon number
n
3.
Since the patthway showed
d that that B was
w an alcoholl produced froom ClCH2CH2CH(CH3)2, annd so had to be
HOCH2CH2C
CH(CH3)2, it was
w surprisingg that many sttudents showeed a structure with
w the incorrrect number of
o carbon
atoms.
Chemistry GA 1 Exam
Published: 3 November 2011
12
201
11
Assess
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Rep
port
7aiii.
0
1
Average
Marks
48
52
0.5
%
3-methylbutaan-1-ol or 3-m
methyl-1-butannol
t part of the question inddicated that many students who
w had the correct
c
structurre in 7aii. did
The performaance data for this
not provide a correct namee for the comppound. Many students referrred to a dimetthyl option ratther than identifying the
longest chainn of C atoms.
7aiv.
Marks
%
Ethanol
0
44
1
56
Average
0.6
Ethan-1-ol orr 1-ethanol waas a relativelyy common resp
ponse for this question; how
wever, this is not
n the system
matic name.
There is no oother option foor the –OH funnctional group
p but to be on C-1, hence th
he number is not
n part of thee systematic
name.
7av.
0
Marks
49
%
Any of:
 H2S
SO4
 sulffuric acid
 H2S
SO4(l).
1
51
Average
0.5
This questionn was poorly done.
d
Sulfuricc acid is used as
a a catalyst inn ester producction. A relativvely commonn error was to
give the answ
wer as H2SO4(aq)
(
or dilute sulfuric
s
acid.
7avi.
Marks
%
Reaction II
0
15
1
85
Average
0.9
Ethanoic acidd is produced from ethanol by oxidation..
7b.
0
1
2
Averaage
Marks
31
42
26
1
%
Compounds in the mixturee are separatedd according too their boiling temperatures/intermolecular attraction and
a (any of):
 the most volatile compound (loowest boiling temperature) will be colleccted first (at thhe lowest tempperature)
c
(hiighest boiling temperature) will be colleccted last (at thhe highest tempperature)
 the least volatile compound
 the compounds arre collected inn order of incrreasing boilingg temperature.
Many responnses to this question focusedd on the fractiional distillatioon of crude oiil with emphasis on the fracctionating
tower and coollection of hy
ydrocarbon fraactions at diffeerent levels in the tower witth minimal apppropriate, in the
t context off
the question, reference to the
t role of boiiling temperatture or intermoolecular attracction. Given thhat the producction of esters
s
is the endpoiint of chemicaal pathways coovered in this unit, awareneess of the technnique by whicch an ester is separated
from other coompounds preesent in the reaaction mixturee is a reasonabble expectatioon. Students shhould be awarre that the keyy
factor that innfluences boiliing temperaturre is the strenggth of intermoolecular attracction, not moleecular mass.
Chemistry GA 1 Exam
Published: 3 November 2011
13
201
11
Assess
sment
Rep
port
7c.
0
1
2
Averaage
Marks
46
50
4
0.6
%
Both of:
 com
mpound B has a distinct O–H
H (alcohol) ab
bsorption bandd around 3300
0 cm–1 (3200––3550 cm–1). T
This is not
pressent on the speectrum of bannana oil
 the IR spectrum of
o banana oil does
d
not show
w an O–H (acid) (2500–3300 cm–1) absorrption band.
Students werre required to explain how the
t evidence provided
p
by thhe spectra suppported the claaim that ‘compplete
separation off banana oil frrom the reactioon mixture’ haad been achieved. Since thee reaction mixxture would haave contained
ethanoic acidd and 3-methyylbutan-1-ol, itt was necessarry to indicate how the specttra showed the lack of the acid
a or the
alcohol in thee final bananaa oil product.
While slightlly more than half
h the studennts were able to
t argue effecctively that theere was no alccohol present, few were
able to explaain how the baanana oil specttrum showed that
t the acid w
was not presennt. Many students gave reassons how the
banana oil sppectrum showeed that bananaa oil was an ester, but that was
w not relevaant to the quesstion.
There was soome confusionn around the absorption
a
ban
nd near 3100 cm
c –1 on the baanana oil specttrum, with a significant
s
number of stuudents suggessting that it waas an O–H (accid) band. Studdents should have
h
been able to recognisee the
characteristicc broad shape of an O–H (aacid) band.
Question 8a.
Marks
0
21
%
1
79
Average
0.8
The side grouups on the braadykinin moleecule section should
s
have ennabled students to identify the
t amino acidds from
Table 8 of thhe Data Book.
8b.
0
1
Marks
18
20
%
Both of:
OH
 carbboxyl or COO
 amino, NH2 or NH
H3+.
2
62
Averaage
1.5
It was decideed that either the
t name or thhe chemical fo
ormula of the functional gro
oups was a reaasonable respoonse to
‘identify’ thee two functionnal groups. NH
H3+ was accepted because inn 6 M HCl it is
i fair to assum
me that the NH
H2 group on
the amino accids would be protonated.
Students should be aware that
t the hydroolysis of proteins converts peptide
p
groupss in amino andd carboxyl grooups.
Chemistry GA 1 Exam
Published: 3 November 2011
14
201
11
Assess
sment
Rep
port
8c.
Marks
%
0
59
1
41
Average
0.4
A surprising number of stuudents did nott show the O–
–H bond, despiite the instrucction in the question to show
w all bonds.
s
it as an H+.
Other commoon errors incluuded missing atoms and eitther not includding the positiive charge or showing
Chemistry GA 1 Exam
Published: 3 November 2011
15