Section A
Answer
Question
1
•
a
Butane because it contains carbon and hydrogen
atoms only
• Whereas the other compounds contain additional
elements (oxygen/chlorine)
Alkanes
b
Notes
Total
2
2
Alkenes
c
d
2
Pent-1-yne
• Correct AND as the number of carbon atoms
increases, the number of electrons/size of the
molecule increases
• This increases the strength of the London (dispersion)
forces;
•
More energy is required to overcome these
intermolecular forces
a
2
b
c
d
3
a
1
3
Functional group (isomerism)
Butan-2-ol
1
1
1
i
1
ii
1
4
b
c
cis-but-2-ene
1
2
a
B = Methylbutane
C = 2,2-dimethylpropane
 Isomer C bcause it is the most branched/spherical
isomer
 This means that it has the smallest surface area for
contact between molecules;
 This results in weaker London (dispersion) forces
which require less energy to overcome
2
b
3
Section B
1
•
•
•
•
•
a
Members have the same functional group
Members have the same general formula
Successive members differ by C𝐻2
Members have similar chemical properties
There is a gradual change in the physical properties
b
3
4
Estimated boiling point in the range: 60-70 C;
c
d
2
a
i
1
ii
1
•
•
Would be expected to be lower than ethane
Ethyne has fewer (atoms and) electrons than ethane
AND the intermolecular forces / Van der Waals forces
/ London dispersion forces would be weaker (so the
boiling point should be lower)
2
2
b
3
•
•
2
a
1
b
2
•
4
•
•
5
1,2-dimethylbenzene
1,3-dimethylbenzene
Warming with acidified potassium dichromate(VI) /
𝐾2 𝐶𝑟2 𝑂7with 𝐻2 S𝑂4 / C𝑟2 𝑂72− with 𝐻 + OR Warming
with acidified potassium manganate(VII) / KMn𝑂4 with
𝐻2 S𝑂4 / Mn𝑂4− with 𝐻 +
The primary alcohol will turn from orange to green
The tertiary alcohol will not react / there will be no
color change
3
a
2
b
2
c
6
a
1
i
1
ii
1
b
c
i
ii
d
7
Compound 2 = 2-chloropropanal;
Compound 3 = 2-chloro-but-2-en-1-ol
They are both primary alcohols that would be oxidised /
give the same result (orange to green)
• Reagent: Add bromine water / B𝑟2 (aq)
• Observation: Compound 3 decolorises the bromine
water, while Compound 1 shows no change
Pair 1
• Reagent: Add a named carbonate (e.g. N𝑎2 C𝑂3 )
• Observation: Compound 4 will effervesce / fizz,
while Compound 2 will show no reaction
Pair 2
• Reagent: Warm with Tollens' reagent;
• Observation: Compound 2 will produce a silver /
black mirror, while Compound 4 will show no
reaction
Pair 3
• Reagent: Warm with Fehling's / Benedict's
solution;
• Observation: Compound 2 will produce a red /
orange precipiate, while Compound 4 will show no
reaction
• Water / steam / 𝐻2 O
• (Concentrated) sulfuric acid / 𝐻2 S𝑂4 OR
(Concentrated) phosphoric acid / 𝐻3 P𝑂4
Section C
2
1
2
2
2
1
a
b
i
ii
Five
•
•
•
•
2-methylpentane
3-methylpentane
2,2-dimethybutane
2,3-dimethylbutane
i
1
ii
1
c
2
a
b
3
1
2
2
•
•
x is a primary amine
y is a secondary amine
2
2
4