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3.5 Organic Chemistry Assessment Schedule 2007
Question
number
ONE
(a)
Answer
(i) 4-hydroxypent-2-ene
(ii)
1
Achievement
Achievement with Merit
3 correct (including
correct use of numbers
for placement of
functional group).
4 correct
1 correct point
2 correct points
Achievement with
Excellence
(iii) ethanamide
(iv)
ONE
(b)
(i) Optical isomers require a chiral/asymmetric
carbon atom/carbon atom with four different groups
attached to it.
(ii)
(iii) The two isomers will rotate plane-polarised light
in opposite directions.
All correct
2
Geometric isomerism
ONE
(c)
ONE
(d)
ONE
(e)
TWO
(a)
TWO
(a)
TWO
(a)
Due to its shape the cis-isomer is more polar and will
have a higher boiling point due to stronger permanent
dipole-dipole interactions. In the trans-isomer the
bond dipoles largely cancel out therefore it has a
lower boiling point.
(i)
(ii)
Heating under reflux increases the rate of reaction and
the condenser returns evaporated organic molecules to
the reaction flask by condensing the vapour that is
produced on heating.
(iii) Sulfuric acid = catalyst (or dehydrating agent )
Anhydrous sodium sulfate = To remove last traces
of water from organic layer
Correct answer
Two correct diagrams
OR two correct names
All correct
Identifies the cis-isomer
as having the higher
boiling point.
Identifies the cis-isomer
as having the higher
boiling point due to
greater polarity
Correct equation (must
have conc. sulfuric acid
and water)
Diagram showing
Round bottomed flask
(or other suitable flask)
and condenser clearly
labelled.
OR
Correct explanation for
heating under reflux
Both correct
Correctly labelled
diagram and full
explanation (no seal on
condenser).
Full explanation and
comparison of each
isomer in relation to
polarity and boiling
point
3
TWO
(a)
TWO
(b)
(iv) The boiling point of methyl ethanoate is between
55◦C and 59◦C so by keeping the distilling
temperature constant you will only collect the ester.
Identifies ester/ methyl
ethanoate b.pt between
55◦C and 59◦C
(i) not conc HCl/ZnCl2
only
or PCl3
(ii)
Correct reagents for (i)
and (iii)
OR
Correct structure for
ethanoyl chloride
Identifies ester/methyl
ethanoate b.pt between
55◦C and 59◦C and links
to collection of ester
All Correct
(iii)
TWO
(c)
THREE
(a)
Because the reaction of an acid chloride and an
alcohol produces HCl fumes.
Correct answer
3 correct
(i)
(ii)
(iii)
(iv)
4 correct
5 correct
4
(v)
Three
(b)
Three
(c)
FOUR
(a)
Four
(b)
FOUR
(c)
Elimination or dehydration reaction
Correct answer
But-1-ene is an unsymmetrical molecule, therefore,
HBr adds across the double bond in two ways to give
two different products. This addition takes place
according to Markovnikov’s rule (rich get richer).
Identifies the alkene as
unsymmetrical about the
double bond (or
mentions Markovnikov
rule etc)
Identifies the alkene as
unsymmetrical about
the double bond and
links this to the
formation of 2 products
according to
Markovnikov’s rule.
Addition polymerisation and C=C double bond
1 correct answer
Correct structure drawn
Both answers correct
Water molecule eliminated
Polyester formed
1 correct answer
2 correct answers
All correct
5
FIVE
(a)
(i)
(ii)
SIX
(a)
Using Tollens reagent [Ag(NH3)2]+
propanal (an aldehyde) is easily oxidised
to a carboxylic acid, forming a silver
mirror on the side of the test tube.
Propanone is not easily oxidised so no
reaction.
(Plus Fehlings or Benedicts or dichromate
or permanganate)
Using an appropriate acid base indicator
(e.g Litmus, UI). 1 amino propane turns
Litmus blue (base) and propanoic acid
turns litmus red. (acid fizzes with
carbonate, amine doesn’t etc).
Amine group (NH2) and Carboxylic acid group
(COOH)
SIX
(b)
SIX
(c)
Sufficiency statement:
Two correct reagents
identified.
Two correct reagents
identified and one set of
observations correct.
Both correct.
Peptide link circled.
Partially correct.
All correct.
Achievement: 11 out of the 21 achievement opportunities
Merit: Achievement, plus 6 out of the 13 merit opportunities
All correct.
6
Excellence: Merit, plus 3 out of the 5 excellence opportunities