F6 Chem UT V (11-12) Suggested answers
Section A (36 marks)
1.
(a)
(b)
(i)
2-aminopropanoic acid (2-氨基丙酸)
1
(ii)
4-bromo-3-methylpentanal (3-甲基-4-溴戊醛)
1
(i)
1
(ii)
1
2.
2
3.
% atom economy = 146 / [146 + 4(18)]
1
= 66.97 %
4.
(a)
1+1
and
cis-3-methylhex-2-enoic acid
5.
trans-3-methylhex-2-enoic acid
(b)
cis-trans / geometrical isomerism (順反 / 幾何異構)
1
(a)
conc. H2SO4
1
(b)
alcohol B
carboxylic acid D
CH3CH2OH
CH3CH2CH2COOH
1+1
1
5.
(c)
water out
condenser
water in
pear-shaped flask
a mixture of ethanol, butanoic
anti-bumping granule
acid and conc. H2SO4
heat
(d)
6.
1
Correct labelling
1
There is strong hydrogen bond between hexanoic acid molecules
1
While weak van der Waals’ forces
1
/ dipole-dipole interaction between ester molecules
Any one of the following test with correct observations
Test
Tollen’s test / silver mirror
K2Cr2O7 / H+
Fehling test
7.
Correct set-up
(a)
1+1
Aldehyde
Ketone
Silver mirror
No observable change
Orange to green
No observable change
Brick red ppt formed
No observable change
Assume the mass of compound be 100 g
C
H
O
Mass / g
81.08
8.11
10.81
No. of moles
6.757
8.11
0.676
10
12
1
Simplest mole ratio
Hence, the empirical formula is C10H12O
Since
(C10H12O)n = 148  n = 1
i.e. molecular formula is C10H12O
1
1
1
2
(b)
Since there is an absorption peak at about 1750 cm1, hence W contains C=O group.
W can react with 2,4-dinitrophenylhydrazine to give orange ppt
W does not react with acidified potassium dichromate
i.e. W is aromatic ketone, but not alcohol nor aldehyde.
1
The structure of W is:
O
C
CH3
C
CH3
H
1
W is reduced by LiAlH4 to give S and S can turn acidified potassium dichromate green
i.e. S is alcohol
1
The structure of S is
OH CH3
C
C
H
H
CH3
1
8.
Any TWO of the followings:
-
aldehyde
-
alcohol / alkanol
-
ester
9.
2
linear shape
V-shape
1+1
1+1
10.
Not correct
The large difference of electronegativity between two atoms cause a dipole bond and dipole moment
1
The molecule may have net resultant dipole moment because of asymmetrical shape / dipole moments do not
cancel out.
Section B
1.
A
1+1
Multiple Choices
2.
A
(4 marks)
3.
B
4.
D
4
3
(e)
The molecules of hexanoic acid are held together by hydrogen bonds as well as van der Waals’
forces while the molecules of ethyl butanoate are held together by weak van der Waals’ forces
only. [2] Since the strength of hydrogen bonds is greater than that of van der Waals’ forces,
more energy is needed to overcome strong hydrogen bonds between hexanoic acid molecules
during boiling process. [2]
5.
(a) Hydrogen gas [1]
(b) The general formula is CnH2nO2. [1]
(c) They belong to carboxylic acids. [1]
(d) The functional group is
. [1]
Test the compounds with sodium carbonate solution. [1] They will give colourless gas. [1]
(e) The third member is propanoic acid. [1]
The fourth member is butanoic acid. [1]
6.
(a) It contains OH group and CHO group. [2] (Accept ‘ROR group’ as answer.)
(b) It belongs to aldehydes. [1]
(c) It is soluble in water [1] because it contains OH group and CHO group which can form
hydrogen bonds with water molecules. [1]
(d) It is used as an ingredient in perfume. [1]
II. Multiple Choice Questions
1. B CH3CH2CONH2 undergoes alkaline hydrolysis to give carboxylate ion, CH3CH2COO.
2. C Propan-2-ol reacts with an oxidizing agent to give a ketone. However, ketones are resistant to
further oxidation.
3. D The acid hydrolysis does not go to completion because it is a reversible reaction.
4. B Butanoic acid is reduced to butan-1-ol.
5. B The synthetic route is
1.
(a)
[1]
4
[1]
(b) In CO2, the carbon atom forms 2 double bond pairs with oxygen atoms and no lone pair is
left [1] so the molecule is linear. [1] In SO2, the sulphur atom forms 2 double bond pairs
with oxygen atoms and 1 lone pair is left [1] so the molecule is V-shaped. [1]
2.
(a) Differences in electronegativity among the constituent atoms only ensure the presence of polar
bond(s). [1] A molecule is polar only when there is a non-zero overall dipole moment, [1]
which is determined by the vector sum of all the dipole moments. [1]
(b) The structure of the substance is so symmetrical [1] that all the dipole moments present cancel
out. [1]
3.
(a) & (b)
[1]
For 3 correct polar bond dipole moments [1] and net dipole moment [1]
Trigonal planar [1]
(c) The statement is incorrect. [1] A molecule may contain many different polar covalent
bonds and each bond has different magnitudes in dipole moment. [1] Only when
symmetrical molecules with only one type of polar bond can be non-polar. [1] Also a
symmetrical molecule may not be symmetrical in a way that all the dipole moments
completely cancel out. [1]
4.
(a) X should be greater than 36. [1] Although both chemicals A and B have about the same
molecular mass, B is polar while A is non-polar. [1] Hence, the van der Waals’ forces between
molecules of B are stronger. [1]
(b) Y should be greater than X. [1] Although both chemicals B and C have the same molecular
mass and are polar, the OH group in molecules of C allows the formation of hydrogen bond,
[1] which needs extra energy to overcome [1] and thus the boiling point of C is greater.
(c) The student was wrong. Although both (CH3)3COH and chemical C have the same molecular
mass and both of them can form hydrogen bonds, [1] the branching of the hydrocarbon chain
reduces the surface area of contact between (CH3)3COH molecules. [1] Therefore, (CH3)3COH
5
should have a lower boiling point.
5.
(a) C7H6O2 [1]
(b) 120 because of the trigonal planar structure. [1]
(c) Compound B would have a higher melting point due to the formation of intermolecular
hydrogen bonds. [1] All the molecules are held more firmly and greater amount of energy is
required to overcome the attraction. [1] Compound A would have a lower melting point
because of the formation of intramolecular hydrogen bonds. [1] It is possible due to the close
proximity of the hydroxyl and aldehyde groups. [1] The formation of intramolecular hydrogen
bonds lowers the possibilities of formation of intermolecular hydrogen bonds. [1] As a result,
lower energy is required for melting, leading to a lower melting point.
(d) Compound A [1] has a greater dipole moment because the dipole moments arising from the
polar groups (OH and CHO) cannot be cancelled in 1,2-position. [1] However, it would be
cancelled in a much greater extent in 1,4-position. [1]
II. Multiple Choice Questions
1. A Non-octet structures cannot exist in non-metal elements like hydrogen, oxygen and nitrogen.
2. A
3. A There are 7 outermost shell electrons in I, shared with 4 electrons from Cl, together with 1
incoming electron, totally 12 electrons, i.e. violates the octet rule.
9.
Possible structures:
W
S
O
C
OH CH3
CH3
C
H
T
CH3
C
C
H
H
H
CH3
C
C
CH3
CH3
6