1
Assessment schedule 2010
AS 90309 [Chemistry 2.5] - Describe the structural formulae and reactions of compounds containing selected organic
functional groups.
Question Evidence
2, 4, 4-trichlorohexane
One
(a)
Achievement
Merit
Excellence
2-methyl- 3-chloro-propanoic acid
O
CH3 CH2 CH2 CH2 O
C
CH2 CH2 CH2 CH3
OR
CH3
CH3
CH2 C
Any FOUR
correct isomers in
(b) (i)
CH3
OH
H
(b)(i)
CH2 CH
CH2 CH2 CH3
CH3
C
CH2
CH3
CH3
CH2
CH3
CH2
CH
CH
CH3
CH2
CH2
CH2
CH2
CH3
(b)(ii)
OR
C
CH3
CH2 C
THREE answers
correct in (a)
Not shown are methylcyclobutane and 1,1-dimethylcyclopropane, nor
1,2-dimethylcyclopropane
Any cyclic isomer
Any THREE
correct isomers in
(b)(i) AND (b)(ii)
correct
OR
(c) correct
As for
Achievement plus
(d)(i) correct
AND
identifies in (d)
(ii) that H2 is
symmetrical so
one product AND
HCl is
unsymmetrical so
two products
As for Merit plus
Includes full
discussion of how
to determine
major and minor
products
2
(c)
CH3
CH2 CH3
C
(d)(i)
H
cis-pent-2-ene
CH2 CH CH2 CH2 CH3
H
Cl
major product
(d)(ii)
C
C
H
H
CH3
C
H
CH2 CH3
trans-pent- 2-ene
CH2 CH CH2 CH2 CH3
Cl
H
minor product
The H2 molecule is symmetrical. One product is formed regardless of
how the H2 molecule approaches the double bond in the alkene.
HCl is an unsymmetrical molecule. Two products can be formed
depending on how the HCl molecule approaches the double bond in the
alkene.
If the H atom from HCl bonds to the C atom at the end of the alkene
molecule, this is considered the major product. This is because the H
atom from HCl is bonding to the C in the double bond with the greater
number of H atoms already bonded to it.
If the Hydrogen atom from HCl bonds to the C atom on the other side
of the double bond, this is considered the minor product. This is because
the H atom from HCl is bonding to the C in the double bond with the
smaller number of H atoms already bonded to it.
3
Two
(a)
Product A = CH3 CH2 CH2 CH3 or named as butane
Type = addition
Reactant B = KMnO4 / H+ or K2Cr2O7 / H+
Type = oxidation
OH
OH
CH
(b)(i)
(b)(ii)
CH
CH
3
3
Reactant C = CH3 CH2 CH2
or named as propan-1-ol or propan-2-ol
Type = elimination
Compound A is ethene
Compound B is ethan-1-ol
Compound C is ethanoic acid
Compound D is ethyl ethanoate
Compound A is an alkene because molecular formula is of general
formula CnH2n. Only two carbons, must be ethene.
Compound B is product of addition reaction between an alkene and
water so is an alcohol, only two carbons, so must be ethanol.
Compound B is oxidised to Compound C which is a carboxylic acid.
Carboxylic acids neutralise Na2CO3 to form CO2 gas(bubbles). Two
carbons must be ethanoic acid.
Carboxylic acids react with primary alcohols in presence of conc H2SO4
to form an ester (Compound D). Ethanol + Ethanoic acid form ethyl
ethanoate which has the molecular formula C4H8O2.
SIX of the ten
achievement
answers ticked
correct from (a)
and/or (b)(i)
SEVEN of the ten
achievement
answers ticked
correct from (a)
and (b)(i)
AND
From (b)(ii)
TWO compounds
identified with
correct reasoning.
EIGHT of the ten
achievement
answers ticked
correct from (a)
and(b)(i)
AND
From (b)(ii)
THREE
compounds
correctly
identified with
correct reasoning
and the type of
reactions.
4
Three
(a)(i)
H
H
C
C
H
(a)(ii)
H
(b)(i)
(b)(ii)
(b)(iii)
H
Cl
H
H
H
C
C
C
C
H
C6H5
H
C6H5
H
Circle ester linkage
Ester
H2C
OH
O
(c)
HC
OH
H2C
OH
HO
C
(CH 2)16-CH 3
In both reactions, a colour change from purple to colourless will be seen
as purple MnO4– /H+ is reduced to Mn2+.
(OR colour change from purple to brown precipitate if non-acidified
MnO4–.)
Propene will react to form a diol,
propan-1,2-diol:
MnO4-/H+
CH3CHCH2
CH3CH(OH)CH2(OH)
Propanol will react to form a carboxylic acid, propanoic acid:
MnO4-/H+
CH3CH2CH2OH
CH3CH2COOH
FOUR from (a) or
(b) ticked correct FOUR from (a) or
(b) ticked correct
OR
AND
From (c)
ONE of
From (c)
- Colour
ONE of
change from
- Propene reacts
purple to
to form diol /
colourless.
propan -1,2-diol.
- Propene
- Propanol reacts
reacts to
to form
form diol /
carboxylic acid /
propan-1,2propanoic acid
diol.
- Propanol
Whichever is
reacts to
chosen, it is
form
linked to
carboxylic
reaction type,
acid /
colour change
propanoic
OR equation.
acid.
FOUR from (a) or
(b) ticked correct
AND
From (c)
BOTH of
- Propene reacts
to form diol /
ethan-1,2-diol.
- Propanol
reacts to form
carboxylic acid
/ propanoic
acid.
Both linked to
colour change,
equations, AND
both reactions are
oxidation
reactions. One
omission
allowed.
5
Four
React all three substances with dilute solution of bromine water.
Bromine in hexene will change from orange to colourless immediately.
1, 2-dibromohexane will be formed.
React remaining two substances with sodium carbonate. Propanoic
acid will form bubbles (fizz) as CO2 will be produced. Sodium
propanoate will be formed.
2CH3CH2COOH + Na2CO3  2(CH3CH2COO)Na + H2O + CO2
By process of elimination, the other colourless liquid is hexane.
Hexene identified
with observation
with bromine.
Hexene
identified with
observation with
bromine.
OR
AND
Propanoic acid
identified with
observation with
sodium carbonate.
Propanoic acid
identified with
observation with
sodium
carbonate.
Both hexane and
propanoic acid
samples correctly
identified with
observations AND
equations.
Process of
elimination may be
implied.
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
3A
OR
1M + 1A
3M
OR
2E + 1A
OR
1E + 1M + 2A
3E
OR
2E + 1M + 1A