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Worksheet-01-Answers
Introduction to Organic Chemistry
Chem-160-Summer-2015
1
New Way Chemistry for Hong Kong A-Level Book 3A
21.1 What is Organic Chemistry (SB p.4)
(a) How was organic chemistry defined before 1800s?
(a) The knowledge of organic and inorganic
compounds was raised during the 1780s.
Scientists defined organic chemistry as the
study of compounds that could be obtained
from living organisms. They believed that the
synthesis of organic compounds took place in
living organisms only.
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Answer
21.1 What is Organic Chemistry (SB p.4)
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(b) How is organic chemistry defined nowadays?
(b) Nowadays, scientists have discovered that
many organic compounds can be synthesized
from inorganic substances. The updated
definition of organic chemistry is the study of
carbon compounds, except for carbon
monoxide, carbon dioxide, carbonates,
hydrogencarbonates, carbides and cyanides.
These compounds have been traditionally
classified under inorganic chemistry.
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21.2 The Unique Nature of Carbon (SB p.5)
Why is carbon able to catenate?
The ability to catenate of carbon is chiefly due to
the high strength of the CC single bond (bond
enthalpy of C  C single bond is 356 kJ mol-1).
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21.2 The Unique Nature of Carbon (SB p.6)
Would you expect silicon, which is just below carbon in the
Periodic Table, to catenate to form diverse molecular
structures? Explain your answer.
Answer
Silicon, unlike carbon, does not catenate to form diverse
molecular structures. Carbon is able to catenate because
carbon atoms have a relatively small atomic size. This
enables a carbon atom to form strong covalent bonds
with other carbon atoms. However, due to the greater
atomic size of silicon, its ability to catenate is much lower
than that of carbon.
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21.2 The Unique Nature of Carbon (SB p.7)
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Would you expect sulphur, which has an
electronegativity value very close to carbon, to
catenate? Why?
Answer
The electronic configuration of sulphur is 1s22s22p63s23p4. It
has only two unpaired electrons. Its atomic size is larger than
that of carbon. So it has a much lower tendency to catenate
than carbon.
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21.3 Classification of Organic Compounds (SB p.14)
Identify the functional group(s) in the following compounds:
(a)
Answer
(a) Carbon-carbon double bond (
chloro group (Cl)
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) and
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21.3 Classification of Organic Compounds (SB p.14)
Identify the functional group(s) in the following compounds:
(b)
Answer
(b) Carbonyl group (
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21.3 Classification of Organic Compounds (SB p.14)
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Identify the functional group(s) in the following compounds:
(c)
Answer
(c) Amino group (
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) and carboxyl group (
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21.3 Classification of Organic Compounds (SB p.15)
To which homologous series does each of the following
compounds belong?
(a)
(a) Ester
Answer
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21.3 Classification of Organic Compounds (SB p.15)
To which homologous series does each of the following
compounds belong?
(b)
(b) Amide
Answer
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21.3 Classification of Organic Compounds (SB p.15)
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To which homologous series does each of the following
compounds belong?
(c)
(c) Acid anhydride
Answer
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21.3 Classification of Organic Compounds (SB p.15)
State whether each of the following pairs of compounds
belongs to the same homologous series. Explain your answer.
(a)
(a) No, the first one is a carboxylic acid and the
second one is an ester.
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21.3 Classification of Organic Compounds (SB p.15)
State whether each of the following pairs of compounds
belongs to the same homologous series. Explain your answer.
(b)
(b) Yes, both of them are alcohols.
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21.3 Classification of Organic Compounds (SB p.15)
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State whether each of the following pairs of compounds
belongs to the same homologous series. Explain your answer.
(c)
(c) No, the first one is an amide and
the second one is an amine.
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21.3 Classification of Organic Compounds (SB p.16)
(a) Name the homologous series of organic compounds
that contain oxygen atoms in their functional groups.
(a)
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Alcohol, ether, aldehyde, ketone,
carboxylic acid, ester, acyl halide,
amide and acid anhydride
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21.3 Classification of Organic Compounds (SB p.16)
(b) Identify and name the functional groups in glucose
which has the following structure.
(b)
OH (hydroxyl group)
and  O  (oxy group)
Answer
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21.3 Classification of Organic Compounds (SB p.16)
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(c) Identify and name the functional groups in the
following compounds:
(c)
Br (bromo),
(aldehyde),
(acyl chloride),
(carbon-carbon
double bond) groups
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.18)
Why is oil immiscible with water?
Oil molecules do not have free OH groups,
so they cannot form hydrogen bonds with
water molecules.
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.20)
The relative molecular mass of glucose is 180.0, but it
is soluble in water. Why?
Answer
Glucose molecules have OH groups, so
they are able to form hydrogen bonds with
water molecules. Therefore, glucose is
soluble in water despite it has a high
molecular mass.
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.20)
Despite the fact that butan-1-ol and ethoxyethane have the
same relative molecular mass, they have very different
boiling points. The boiling points of butan-1-ol and
ethoxyethane are 117oC and 35oC respectively. Explain the
difference.
Answer
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.20)
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There is an OH group in butan-1-ol. Thus, butan-1-ol molecules are
able to form hydrogen bonds with one another and the energy required to
separate butan-1-ol molecules would be much greater. Whereas for
ethoxyethane, the attraction among the molecules is weak van der Waals’
forces only. The amount of energy required to break the forces would not
be great. Therefore, the boiling point of ethoxyethane is lower than that of
butan-1-ol.
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.21)
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Explain why propan-1-ol is soluble in water but
1-chloropropane is insoluble in water.
Answer
The  OH group of propan-1-ol molecules enables it to form
hydrogen bonds with water molecules. Thus it is soluble in water.
Although 1-chloropropane is a polar molecule, it does not form
hydrogen bonds with water molecules. So it is insoluble in water.
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.21)
Which molecule would have a higher boiling point,
1-bromobutane or 2-bromobutane? Why?
Answer
1-bromobutane would have a higher boiling point.
1-bromobutane is a straight-chain molecule while
2-bromobutane is a branched-chain molecule. Straight-chain
molecules have a greater surface area in contact with each other,
so greater intermolecular forces exist among the molecules.
Higher energy is required to break down the intermolecular
forces among the molecules of 1-bromobutane.
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.22)
1-Chlorobutane and 2-chloro-2-methylpropane have the
same molecular mass, yet their melting points differ. The
melting point of 1-chlorobutane is –123oC while that of 2chloro-2-methylpropane is –27.1oC. Explain the difference.
Answer
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.22)
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Melting point is a measure of how efficient the molecules are packed
together in the solid state instead of just comparing the van der Waals’
forces among molecules. Hence melting point is a function of the
efficient packing of molecules but not the contact surface area.
1-Chlorobutane is a straight-chain molecule while
2-chloro-2-methylpropane is a branched-chain molecule. As
2-chloro-2-methylpropane is more spherical and symmetrical, its
molecules are packed more efficiently in the solid state.
1-Chlorobutane is linear in shape and flattened, its packing in the
solid state is not so efficient. Hence, it has a lower melting point than
2-chloro-2-methylpropane.
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.24)
(a) What are the major factors that affect the physical
properties of organic compounds?
Answer
(a)
The physical properties of organic compounds
are mainly affected by the structure of the
functional groups, dipole moment of the
molecule, the formation of hydrogen bonding
between molecules, and the length of carbon
chains of the molecule.
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.24)
(b) The melting point and boiling point of pentane are
–130oC and 36.3oC respectively while the melting point
and boiling point of 2,2-dimethylpropane are –15.9oC
and 9.5oC respectively. Account for the difference in
melting point and boiling point between the two
isomers.
Answer
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.24)
(b)
Pentane is a straight-chain molecule, while 2,2-dimethylpropane is
a branched-chain molecule. Straight-chain molecules have a
greater surface area in contact with each other than branchedchain molecules. Straight-chain molecules are held together by
stronger intermolecular forces. Therefore, pentane has a higher
boiling point than 2,2-dimethylpropane. Molecules of 2,2dimethylpropane are more spherical in shape and are packed
more efficiently in the solid state. Molecules of pentane are linear
in shape and flattened, so their packing in the solid state is not
efficient. Since extra energy is required to break down the efficient
packing of 2,2-dimethylpropane, 2,2-dimethylpropane has a higher
melting point than pentane.
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.24)
(c) Which molecule, hexane or cyclohexane, would have a
higher melting point? Explain your answer.
Answer
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.24)
(c)
Cyclohexane has a higher melting point than hexane. Molecules
of cyclohexane are more spherical in shape and are packed more
eff iciently in the solid state. Molecules of hexane are linear in
shape and flattened, so their packing in the solid state is not
efficient. Since extra energy is required to break down the efficient
packing of cyclohexane, cyclohexane has a higher melting point
than hexane.
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.24)
(d) Arrange the following molecules in increasing order of
boiling points. Explain your answer.
Answer
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21.4 Factors Affecting the Physical Properties of Organic Compounds
(SB p.24)
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(d)
The boiling points increase in the order:
butane < propanal < propan-1-ol
Molecules of butane are non-polar. Their molecules are held
together by weak instantaneous dipole-induced dipole interactions.
A relatively small amount of energy is required to separate the
molecules in the process of boiling. Both propanal and propan-1-ol
are polar molecules. Molecules of propanal are held together by
relatively weak dipole-dipole interactions, while molecules of
propan-1-ol are held together by intermolecular hydrogen bonds.
Since the intermolecular forces present in molecules of propan-1ol are stronger than those present in molecules of propanal, a
larger amount of energy is required to separate the propan-1-ol
molecules in the process of boiling.
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