Assessment Schedule
Chemistry 2.5 90309
Describe the structural formulae and reactions of compounds containing selected organic
functional groups.
While the writers of this assessment have worked to compile a resource that meets NCEA requirements, it has no
official status and teachers may wish to adjust questions and the assessment schedule as they see fit.
Question
ONE
1
Evidence
(a)
(b)
(c)
Achievement
methanol
2-bromobutane
3-chloropropene
(d)
4 correct answers.
Merit
6 correct
answers.
OH
CH3 C
(e)
H
C
O
C CH3
(f)
O CH2 CH3
CH3 CH2 CH2 C
O
TWO
1
Cl H
Cl Cl
Cl C C H
H C C H
H H
H H
1,1-dichloroethane
2
3
THREE
1,2-dichloroethane
No, there needs to be a double C=C bond to restrict
rotation about the carbon-carbon bond
CH3CH2OH ethanol
1
CH3CH2Br
2
CH3CH=CH2
3
CH3
Both formulae
correct.
Both names
correct.
No – plus mention
of double bond.
Formula and
name correct.
3 correct answers.
CH CH CH3
OH OH
4
FOUR
1
CH3CH2CH2CH2CO2H
Cl H Cl H Cl H
~C C C C C C~
H H H H H H
Correct structure.
Complete
answer
including
non-rotation.
4 correct
answers.
Excellence
2
Q
Five
1&2
Evidence
CH3 CH CH3
OH
propan-2-ol
CH3 CH2 CH2
OH
propan-1-ol
Achievement
1 name and
formula.
Merit
Both names
and formulae
plus correct
identification
of major
product.
Unsymmetrical
recognised
Partial
explanation
Propan-2-ol is the major product
3
SIX
1(a)
The requirements are that the alkene be
unsymmetrical around the double bond with an
unsymmetrical reagent adding on to it.
CH3CH2CH2CH2CH=CH2 + Br2
Excellence
Clear explanation
including both
molecules being
unsymmetrical.
Correct equation
CH3CH2CH2CH2CH(Br)–CH2Br
.
1(b)
(i) Bromine changes from orange to colourless
One observation
Two
observations
One reason
Two reasons
Method and
observation: only
one compound
referred to.
Method and
observation
described for
both
compounds.
One correct.
Both correct.
Ester is insoluble
in water.
Detail of
difference in
solubility.
(ii) two colourless layers remain
1(c)
(i) Bromine reacts with hexene in an addition
reaction and a colourless product is formed.
(ii) Organic product is insoluble in water so two
layers form.
2
Reagent: Cr2O72–/H+
Obs: Reagent turns orange to green with ethanol; no
change with ethanoic acid
Reason: Ethanol can be oxidised ethanoic acid is not
oxidised
Full answer
including
explanation
OR CO32– or active metal
No reaction with ethanol; fizzing with ethanoic acid.
Reason: Ethanoic acid shows typical acid properties;
ethanol does not show acid properties.
OR: moist litmus (or other indicator)
No change in colour with ethanol. Ethanoic acid will
change the colour of acid-base indicators.
Reason: Ethanoic acid shows typical acid properties;
ethanol does not show acid properties.
SEVEN
1
(a) Catalyst/speeds up the reaction
(b) To neutralise any remaining acid.
2
(a) & (b)
3
The ester is insoluble in water while methanol and
propanoic acid are soluble in water. Hence the
reactants dissolve in the water and the ester forms a
separate upper layer because it is less dense than
water.
CH3OH + CH3CH2CO2H
CH3 O C
CH2 CH3 + H2O
Equation without
H2O
O
4
Sweet/fruity smell
correct
Correct
Full discussion on
solubility plus
density.
3
Sufficiency Statement. These exams are not pre-tested. The statements below are a guide and schools should feel
free to adjust them to get a sensible distribution.
Achieved
Merit
Excellence
A total of EIGHT opportunities answered correctly at Achieved or higher.
A total of TEN opportunities answered correctly with 4 at Merit and 6 at Achieved
A total of TWELVE opportunities answered correctly with 2 at Excellence, 4 at Merit and 6 at Achieved.