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Topic
8
Chemistry of Carbon
Compounds
Unit 29 An introduction to the chemistry of carbon
compounds
Unit 30 Isomerism
Unit 31 Typical reactions of selected functional
groups
Unit 32 Synthesis of carbon compounds
Unit 33 Important organic substances
Key
C o ncepts
An introduction to the chemistry
of carbon compounds
• Homologous series
• Systematic naming
• Effects of functional groups and chain
length on physical properties
Chemistry of
Carbon Compounds
Isomerism
• Structural isomerism — chain
isomerism, position isomerism and
functional group isomerism
• Stereoisomerism — geometrical
isomerism and enantiomerism
Synthesis of carbon compounds
• Synthetic routes for carbon
compounds
• Preparation of simple carbon
compounds
Typical reactions of selected
functional groups
• Reactions of alkanes, alkenes,
haloalkanes, alcohols, aldehydes,
ketones, carboxylic acids, esters and
amides
Important organic substances
• Aspirin
• Soaps and soapless detergents
• Nylons and polyesters
• Carbohydrates
• Lipids
• Proteins
Topic 8
Unit 29 An introduction to the chemistry of carbon compounds
Chemistry of Carbon Compounds
Unit 29
An introduction to the chemistry of carbon
compounds
29.1 – 29.12 & 29.21
Summary
29.1
The value of medicines: longer and healthier lives
29.2
Functional groups: centre of reactivity
29.3
Naming alkanes and alkenes
29.4
Naming carbon compounds with one type of functional
group
29.5
Naming haloalkanes
29.6
Naming alcohols
29.7
Naming aldehydes and ketones
29.8
Naming carboxylic acids
29.9
Naming esters
29.10 Naming amides
1 The following table summarizes the nomenclature of compounds in various
homologous series.
Homologous
series
General
formula
Functional
group it
contains
Alkanes
CnH2n+2
—
Alkenes
CnH2n
C
add appropriate
prefix to –ane
C
–X
Haloalkanes
RX
(X = F, Cl,
Br or I)
Alcohols
ROH
–OH
Aldehydes
RCHO
Ketones
RCOR1
Carboxylic
acids
RCOOH
29.11 Naming amines
29.12 Naming compounds with more than one type of functional
group
29.13 Intermolecular forces and physical properties of carbon
compounds
29.14 Physical properties of haloalkanes
29.15 Physical properties of alcohols
C
29.17 Physical properties of carboxylic acids
29.18 Physical properties of esters
H
29.21 Common names of carbon compounds
C
O
C
OH
O
Esters
RCOOR1
Amides
RCONH2
(unsubstituted)
Amines
RNH2
(primary)
C
O
O
C
N
N
add the name
of halogeno
functional group
as prefix to the
corresponding
alkane
replace the last
letter ‘e’ of the
corresponding
alkane with –al
replace the last
letter ‘e’ of the
corresponding
alkane with –one
O
29.19 Physical properties of amides
29.20 Physical properties of amines
replace ‘ane’ of
the corresponding
alkane with –ene
Example
Structural formula
replace the last
letter ‘e’ of the
corresponding
alkane with –oic
acid
the name consists
of two separate
words, the first
word comes from
the alcohol, the
second word
comes from the
acid
replace the ‘oic
acid’ ending of
the corresponding
acid by –amide
replace the last
letter ‘e’ of the
corresponding
alkane with
–amine
IUPAC name
CH3
2-methylbutane
CH3CHCH2CH3
CH3CH
propene
CH2
Cl
CH3
replace the last
letter ‘e’ of the
corresponding
alkane with –ol
O
29.16 Physical properties of aldehydes and ketones
Nomenclature
CH
CH3
CH3CH2OH
2-chloropropane
ethanol
O
C
CH3CH2
H
propanal
O
C
CH3
propanone
CH3
O
OH
C
CH3CH2CH2
O
CH3
butanoic acid
methyl
C
O
CH3
ethanoate
O
CH3
C
CH3NH2
NH2
ethanamide
methanamine Topic 8
Unit 29 An introduction to the chemistry of carbon compounds
Chemistry of Carbon Compounds
2 Sometimes there is more than one functional group in one compound. This kind
of compound should be named according to the following order of precedence of
functional groups:
–COOH > –COO– > –CONH2 > –CHO > –CO– > –OH > –NH2 > –C=C–
Example
Give the IUPAC names of the following compounds.
a) CH
2COOH
(1 mark)
Exam tips
♦ The longest continuous chain containing the carbon bearing the –OH
group may NOT appear in a straight line. Questions often ask about
the names of such structures.
b)
CH3
O
C
O
(1 mark)
CH2CH(CH3)2
3,4
C2H5
2
H3C
C
1
CH3
OH
2-methylbutan-2-ol
c) CH3CH=CHCH2OH
(1 mark)
d)CH3CH(NH2)CH2COOH
(1 mark)
Answer
a) phenylethanoic acid
(1)
♦ DO NOT spell ‘amine’ as ‘ammine’.
✔
✘
b)2-methylpropyl ethanoate
(1)
c) but-2-en-1-ol
(1)
♦ DO NOT confuse ‘amine’ and ‘amide’.
d)3-aminobutanoic acid
(1)
The general formula of primary amine is RNH2, while that of amide is
(H or R)CONH2.
♦ Students may need to identify the functional groups present in unfamiliar
compounds.
Remarks*
Remarks
➤for (a),the
e.g.
O
CH2CH3
O
C
group is a phenyl group. It is attached to the ethanoic
acid.
➤ (b)
CH2CH3
O
CH3
CH
CH3CH2
O
O
NH2
C
➤ (d) –COOH is the principle functional group. The –NH2 group is named
as a prefix.
oseltamivir
29.13 – 29.20
– amide functional group;
– amine functional group; and
– ester functional group.
CH2CH(CH3)2
➤ (c) The double bond takes the form -en-.
CH3
– C=C bond;
O
this part comes
this part comes from
from ethanoic acid 2-methylpropan-1-ol
NH
Besides the ether linkage, functional groups present in oseltamivir
include:
C
Summary
1 The physical properties (e.g. the boiling point and water solubility) of a carbon
compound are affected by
a) the functional group it contains;
b)the length of the carbon chain in molecules.
Topic 8
Unit 29 An introduction to the chemistry of carbon compounds
Chemistry of Carbon Compounds
2 The following table summarizes the physical properties of members of some
homologous series.
Homologous
series
Intermolecular forces
Physical properties
•permanent dipole-permanent dipole
attractions between molecules
•boiling points higher than those of
alkanes of similar relative molecular
masses
δ+
δ–
CH3
Cl
δ+
δ+
Haloalkanes
CH3
δ–
CH3
Cl
Homologous
series
Intermolecular forces
Physical properties
•permanent dipole-permanent dipole
attractions between molecules
•boiling points higher than those of
alkanes of similar relative molecular
masses
CH3
CH3
δ+
Cl
δ+
O
CH3
δ–
C
•polar molecules can interact with
water molecules
•slightly soluble in water
•h y d r o g e n
molecules
•boiling points much higher than
those of alkanes of similar relative
molecular masses
bonding
between
H
Aldehydes
and
ketones
O
CH3
•hydrogen bonding between aldehyde
/ ketone molecules and water
molecules
H
H
H
•aldehydes and ketones with less
carbon atoms show appreciable water
solubility
H
O
O
O
C
H3C
CH2CH3
H
CH3
key:
O
hydrogen bond
key:
•h y d r o g e n b o n d i n g b e t w e e n
molecules; more extensive than that
in alcohols
hydrogen bond
•hydrogen bonding between alcohol
molecules and water molecules
H
H
H
•alcohols with less carbon atoms
are miscible with water in all
proportions
O
CH3
•alcohols with a long carbon chain in
their molecules are much less soluble
in water
O
H
C
O
H
•boiling points higher than those of
alcohols of similar relative molecular
masses
O
C
CH3
O
key:
hydrogen bond
O
•hydrogen bonding between acid
molecules and water molecules
H
O
H
H
H
hydrogen
bond
O
CH3
C
H
O
H
O
H
H
H
O
Carboxylic
acids
key:
O
CH3CH2
δ–
C
permanent dipole-permanent
dipole attraction
O
CH3CH2
δ+
O
key:
permanent dipole-permanent
dipole attraction
Alcohols
δ–
CH3
key:
CH3CH2
CH3
δ–
C
O
key:
hydrogen bond
H
•the first four acids are miscible with
water in all proportions
Unit 29 An introduction to the chemistry of carbon compounds
Chemistry of Carbon Compounds
•permanent dipole-permanent dipole
attractions between molecules
•boiling points are about the same as
those of aldehydes and ketones of
similar molecular masses
•hydrogen bonding between ester
molecules and water molecules
•simple esters are very soluble in
water
Homologous
series
Intermolecular forces
Physical properties
•hydrogen bonding between molecules
of primary amines; hydrogen bonding
less strong than that in alcohols
•boiling points of primary amines
higher than those of alkanes but
generally lower than those of alcohols
of similar relative molecular masses
3
Physical properties
C
H
H
H
3
H
C
N
O
H
H
2
Intermolecular forces
H
Homologous
series
C
Topic 8
O
CH3CH2
O
H
H
CH2CH3
key:
bonding
between
key:
Amines
hydrogen bond
•h y d r o g e n
molecules
H
N
C
H
CH3
C
N
H
2
Esters
hydrogen bond
•h y d r o g e n b o n d i n g b e t w e e n
primary amine molecules and water
molecules
•boiling points are high
•primary amines with less carbon
atoms are very soluble in water
H
O
H
C
H
N
H
H
N
H
CH3
O
10
C
H
2
C
O
hydrogen bond
C
N
H
H
Exam tips
key:
Amides
3
key:
O
CH3
H
H
H
e.g.
hydrogen bond
•hydrogen bonding between amide
molecules and water molecules
O
H
♦ Questions may ask students to arrange carbon compounds in order of
increasing boiling point.
•simple amides are very soluble in
water
CH3(CH2)2CH3 < CH3(CH2)3CH3 < CH3(CH2)3Cl < CH3(CH2)3OH
Attractions
between molecules
weak instantaneous dipole-
induced dipole attractions
permanent dipole-
permanent dipole
attractions
hydrogen
bonds
♦ The following carbon compounds are miscible with water:
H
– methanol, ethanol and propan-1-ol;
O
CH3
– ethanal and propanone;
C
– methanoic acid, ethanoic acid, propanoic acid and butanoic acid.
N
H
H
H
O
H
O
H
H
key:
hydrogen bond
11
Topic 8
12
Unit 30 Isomerism
Chemistry of Carbon Compounds
Example
Compounds W, X, Y and Z are all colourless liquids. Suggest how you would distinguish
the four compounds from each other.
CH3CH2OH
W (CH3)3COH
CH3(CH2)2Br
CH3(CH2)7OH
X
Y
Z
(5 marks)
Answer
Distinguishing the liquids by water solubility
Add water to the liquids. Both CH3CH2OH and (CH3)3COH are miscible with
water. (1)
Distinguishing the two liquids which are miscible with water
Warm each of these two liquids with acidified potassium dichromate solution.
(1)
Only CH3CH2OH turns the dichromate solution from orange to green.
(1)
Distinguishing the two liquids which are not miscible with water
Warm each of the two liquids which are not miscible with water with AgNO3(aq) in
ethanol. (1)
Only CH3(CH2)2Br gives a creamy precipitate slowly.
Remarks*
Remarks
➤In questions of this type, it is common to distinguish the compounds by
their water solubility before other chemical tests.
➤Hydrocarbons such as cyclohexane and cyclohexene often appear in this
type of questions. They are insoluble in water.
(1)
Unit 30
Isomerism
30.1
Isomerism
30.2
Structural isomerism
30.3
Geometrical isomerism
30.4
Physical properties of geometrical isomers
30.5
Chirality
30.6
Enantiomers
30.7
Test for chirality — plane of symmetry
30.8
Properties of enantiomers
13
14
Topic 8
Unit 30 Isomerism
Chemistry of Carbon Compounds
30.1 – 30.2
Exam tips
Summary
15
♦ When ask about the type of isomerism, give the precise type, instead
of just stating structural isomerism.
e.g.
The following charts show the classification of isomers.
H3C
isomers
different compounds that have the
same molecular formula
OH
and
OCH3
The type of isomerism involved is functional group isomerism.
♦ Questions often ask about methods for distinguishing between isomeric
compounds.
e.g.
structural isomers
atoms are linked in different orders
stereoisomers
atoms are linked in the same way but
with different spatial arrangements
H3C
OH
and
OCH3
can be distinguished by
– a physical method
comparing their boiling points / melting points; H3C
has a higher boiling point / melting point.
structural isomers
OH
– a spectroscopic method
comparing their IR spectra; H3C
OH has a broad and
–1
chain isomers
isomers with the same
functional group but
different carbon skeletons
e.g.
CH3CH2CH2CH2CH3
position isomers
isomers with the same
carbon skeleton and
functional group, but the
position of the functional
group is different
e.g.
and
CH3CH2CH2CH2OH
CH3
CH3CHCH2CH3
functional group
isomers
isomers with the same
molecular formula but
different functional
groups
e.g.
O
CH3CH2C
and
OH
CH3CHCH2CH3
strong absorption at 3 230 – 3 670 cm .
Example
Consider the isomeric compounds X and Y shown below:
OH
OH
CHO
OH
and
O
CH3C
CHO
O
CH3
compound X
compound Y
a) Name the type of isomerism involved.
b)Which of the above compounds has a higher melting point? Explain.
(1 mark)
(3 marks)
Answer
a) Position isomerism
(1)
b)The melting point of compound Y is higher than that of compound X.
(1)
Only compound X can form intramolecular hydrogen bonds.
(1)
Compound Y forms more intermolecular hydrogen bonds than compound X does.
(1)
16
Topic 8
Unit 30 Isomerism
Chemistry of Carbon Compounds
Remarks*
Remarks
17
Example
➤Questions often ask students to compare the melting point / volatility of
isomeric compounds.
Consider the melting points and boiling points of cis-1,2-dichloroethene and trans1,2-dichloroethene.
➤Compound X forms intramolecular hydrogen bonds due to the close proximity
of the –OH group and –CHO group.
O
H
Compound
Melting point (°C)
Boiling point (°C)
cis-1,2-dichloroethene
–80
60
trans-1,2-dichloroethene
–50
48
hydrogen bond
O
C
Explain why
a) cis-1,2-dichloroethene has a higher boiling point;
(3 marks)
b)trans-1,2-dichloroethene has a higher melting point.
(2 marks)
Answer
H
a) The boiling point of a compound depends on its intermolecular attractions.
In a molecule of the cis isomer, the dipole moments of the two polar C–Cl bonds
reinforce each other. Thus, the molecule has a net dipole moment. Thus, these
molecules are held together by permanent dipole-permanent dipole attractions. (1)
In a molecule of the trans isomer, the dipole moments of the two polar C–Cl bonds
cancel each other. Thus, the molecule has no dipole moment. Thus, these molecules
are held together by weaker instantaneous dipole-induced dipole attractions.
(1)
30.3 – 30.4
Summary
1 Stereoisomers that have a different arrangement of their atoms in space due to the
restricted rotation about a carbon-carbon double bond are geometrical isomers.
2 Compounds with two different groups attached to each carbon of the double bond
have two alternative structures, which are geometrical isomers.
CH3
C
CH3
C
H
CH3
C
and
H
H
cis-isomer
b)In addition to intermolecular attractions, the melting point of a compound depends
also on the degree of compactness of molecules in the solid state.
(1)
The cis isomer has a lower degree of symmetry. It fits into a crystalline lattice
relatively poor and thus has a lower melting point.
(1)
H
Remarks*
Remarks
C
➤Questions often ask students to compare the melting points and boiling
points of geometrical isomers.
CH3
e.g.
trans-isomer
3 Geometrical isomers normally have similar chemical properties. However, their
physical properties are often quite different.
CH3OOC
H
C
H3C
Br
C
♦
H3C
C
H3C
C
and
Cl
Cl
H3C
C
are NOT
Br
geometrical isomers. They are identical molecules.
One of the carbon atoms of the double bond has two methyl groups
attached to it.
H
C
H
H
C
COOCH3
m.p. 102 °C
Exam tips
(1)
C
CH3OOC
COOCH3
m.p. –19 °C
Their intermolecular attractions are van der Waals’ forces of comparable
strength.
The trans isomer has a higher melting point because it is more
symmetrical.
Topic 8
18
Unit 30 Isomerism
Chemistry of Carbon Compounds
30.5 – 30.8
Answer
a)
Summary
1 A chiral molecule is defined as one that is not superposable on its mirror image. A
chiral molecule and its mirror image are called a pair of enantiomers.
2 Most simple chiral molecules contain one carbon atom bonded to four different
atoms or groups of atoms. Such a carbon atom is called a chiral carbon.
3 A pair of enantiomers have the same physical property and chemical property,
except
a) their behaviour towards plane-polarized light; and
b)their reactions with chiral reagents.
Exam tips
e.g.
Tamiflu
Aspartame
CH2CH3
CH
O
*
CH3CH2
O
O
O
CH2CH3
NH
CH
C
O
NH
CH2
*
*
H2N
*
NH2
COOH
H
C
HOOC
(1)
H
(1)
b)They rotate the plane of polarization of a beam of plane-polarized light to opposite
directions.
(1)
Remarks*
Remarks
➤Questions often ask students to suggest a different physical property between
a pair of enantiomers. One of them turns the plane of polarization of a
beam of plane-polarized light clockwise, while the other anticlockwise.
♦ Questions may ask students to mark chiral carbons on the structures
of unfamiliar compounds.
C
C
CH3
➤Students need to give good drawings of three-dimensional structures. Use
the conventions commonly used in the representation of three-dimensional
structures.
♦ Questions often ask about enantiomers.
O
CH3
*
CH
C
OCH3
CH2
CO2H
C
CH3
♦ DO NOT confuse the terms ‘chiral’ and ‘achiral’.
Example
Compound X has the following structural formula:
CH(CH3)COOH
The above structural formula can represent two stereoisomers.
a) Draw three-dimensional structures of the two stereoisomers.
b)State a physical property which is different for the two stereoisomers.
(2 marks)
(1 mark)
19
20
Topic 8
Unit 31 Typical reactions of selected functional groups
Chemistry of Carbon Compounds
Unit 31
Typical reactions of selected functional groups
31.1 – 31.7
Summary
31.1
Introduction
31.2
Important reactions of alkanes
31.3
Addition reactions of alkenes
31.4
Addition of hydrogen to alkenes
31.5
Addition of halogens to alkenes
31.6
Addition of hydrogen halides to alkenes
31.7
Substitution reactions of haloalkanes
31.8
Reactions of alcohols
31.9
Reactions of aldehydes and ketones
1 The figure below summarizes the addition reactions of alkenes.
RCH2CH3
alkane
H2(g)
Pt catalyst
RCHCH2
RCH
OH OH
diol
31.10 Reactions of carboxylic acids
HBr(g)
31.12 Hydrolysis of amides
RCHCH3
Br
major product
+
RCHBrCH2Br
alkene
cold alkaline
dilute potassium
permanganate
solution
31.11 Hydrolysis of esters
CH2
Br2
(in organic
solvent)
RCH2CH2Br
minor product
dibromoalkane
Br2(aq)
RCHCH2Br
OH
bromoalcohol
+
RCHBrCH2Br
dibromoalkane
bromoalkanes
2 Markovnikov’s rule for addition reaction of an asymmetric alkene:
When a molecule HA adds to an asymmetric alkene, the major product is the one
in which the hydrogen atom attaches itself to the carbon atom already carrying the
larger number of hydrogen atoms.
3 Substitution reactions of haloalkanes — alkaline hydrolysis of haloalkanes
R
X
NaOH(aq)
reflux
R
OH
21
Topic 8
Unit 31 Typical reactions of selected functional groups
Chemistry of Carbon Compounds
Exam tips
♦ Questions often ask students to predict the major product of an addition
reaction involving an asymmetrical alkene.
e.g.
CH3
CH3
C
C
CH3
H
C2H5
C2H5
C
C
CH3
HBr
Br
HBr
Br
C
C
3
3
Put about 2 cm of ethanol and 1 cm of silver nitrate solution in each of two test
tubes. (1)
Place the test tube in a water bath at 60 °C.
CH3 CH3
23
Answer
(1)
Add several drops of compounds X and Y separately to each test tube.
H
A yellow precipitate forms rapidly in the test tube containing compound Y.
(1)
CH3 H
A white precipitate forms slowly in the test tube containing compound X.
(1)
C2H5 C2H5
Remarks*
Remarks
C
C
➤Questions often ask about chemical tests for distinguishing haloalkanes.
H
H
CH3 H
31.8 – 31.9
I
HI
Summary
C3H7Cl
CH3CH
• reflux
with NaBr +
conc. H2SO4; or
• reflux with
red P + Br2
Deductions:
C3H7Br
•
re
co flu
re nc x w
flu . H it
x 3P h N
w O a
ith 4 ; I
+
re or
d
P
+
I2
C2H5
propan-1-ol
(1° alcohol)
2
C2H5
C
H2C=CH
H
CH3
H
H3C
CH3COOH +
conc. H2SO4
C
HC=CH2
Example
C3H7I
CH3COOC3H7
CH3CH2COOH
1-iodopropane
propyl ethanoate
propanoic acid
CH3CH(OH)CH3
Describe, by giving reagent(s) and stating observations, how you could distinguish
between the following two compounds using a simple test tube reaction.
CH3CH2CHCH3
Cl
compound X
and
CH3CH2CHCH3
I
compound Y
(4 marks)
K2Cr2O7 /
H3O+, heat
1
– Structures of the compound:
C3H7OH
r
;o
ux
efl ux
+ ,r
fl
e
O + , re
an
3
th
/ H 3O
7
ye
H
ox
r 2O /
C O4
th
K 2 Mn
/e
•
K
H4
•
Al +
Li 3O
H
1-bromopropane
– The compound should have a chiral carbon.
• excess
conc. H2SO4,
180 °C; or
• Al2O3,
300 °C
K
di 2 Cr
st 2 O
il 7
of / H
2 Li
ft 3
he O +
H AlH
3O
pr
+ 4
/e
op
th
an
ox
al
ye
th
an
e
– It has a pair of enantiomers.
propanal
1
– Molecular formula C6H12
– It loses its chiral centre after hydrogenation.
CH3CH2CHO
CH2
propene
1-chloropropane
Given information:
– The compound should be an alkene as it can undergo
hydrogenation.
l+
C
.H
nc r
co t; o r
ith ys ; o
w al l 5 l 2
x cat PC SOC
flu 2 h
re Cl it th
• Zn x w wi
i x
m u
• refl
•
♦ Questions may ask students to deduce the structure of a compound
based on the type of isomerism it can exhibit and its reactions.
•
22
propan-2-ol
(2° alcohol)
K2Cr2O7 / H3O+
reflux
1 LiAlH4 / ethoxyethane
2 H3O+
(CH3)3COH
K2Cr2O7 / H3O+
methylpropan-2-ol
reflux
(3° alcohol)
CH3COCH3
propanone
no reaction
24
Topic 8
Unit 31 Typical reactions of selected functional groups
Chemistry of Carbon Compounds
Exam tips
25
31.10 – 31.12
♦ The formation of hydrogen chloride fumes upon reaction with phosphorus
pentachloride is a test for the presence of a hydroxyl group in a
compound.
Summary
♦ The relative ease with which alcohols undergo dehydration shows the
following order:
CH3CH2COOCH2CH3
CH3CH2COOH
CH3CH2OH + conc. H2SO4
ethyl propanoate
Ease of dehydration: 3° > 2° > 1° alcohol
♦ Questions often ask about the oxidation of alcohols:
H 2O /
H3O+
– the oxidizing agent required;
– the name(s) of the product(s).
♦ Remember that LiAlH4 will NOT act upon C=C bonds.
♦ The transformation of –COOH to –CH2OH involves 2 steps: reduction
+
by LiAlH4, followed by treatment with H3O .
propanoic acid
1 LiAlH4 /
ethoxyethane
2 H3O+
NaOH
solution
aqueous
NH3
CH3CH2COOH
CH3CH2COO– Na+
CH3CH2CH2OH
CH3CH2COO– NH4+
propanoic acid
sodium propanoate
propan-1-ol
ammonium propanoate
+
+
CH3CH2OH
CH3CH2OH
ethanol
ethanol
heat
CH3CH2CONH2
Example
propanamide
Describe, by giving reagent(s) and stating observations, how you could distinguish
between the following two compounds using a simple test tube reaction.
OH
CH3
H2O /
+
H3O
NaOH
solution
CH2OH
and
compound X
(2 marks)
compound Y
CH3CH2COOH
CH3CH2COO– Na+
propanoic acid
sodium propanoate
Answer
Warm each compound with acidified potassium dichromate solution.
(1)
Only compound Y turns the dichromate solution from orange to green.
(1)
➤The oxidation reaction is commonly used to distinguish
e.g.
– a primary / secondary alcohol from a tertiary alcohol;
The active ingredient of a superglue has the following structure:
– an aldehyde from a ketone.
Test
Warm with
K2Cr2O7 / H3O+
Alcohols
1° and 2° alcohols
— clear orange
solution turns
green almost
immediately
3° alcohols — no
observable change
♦ When propan-1-ol and propanoic acid are heated under reflux to
produce an ester, the ester can be separated from the reaction mixture
by fractional distillation or using a separating funnel.
♦ Given the structure of an ester, students should be able to analyze the
ester and deduce the alcohol and carboxylic acid forming the ester.
Remarks*
Remarks
Compound
Exam tips
Aldehydes
Ketones
clear
orange
solution
turns
green
no
observable
change
H
CN
C
C
H
C
OCH3
O
H
It is an ester formed from C
H
CN
C
COOH
and CH3OH.
Topic 8
26
Unit 32 Synthesis of carbon compounds
Chemistry of Carbon Compounds
Unit 32
♦ Questions often give carbon compounds with similar structures (such
O
O
as H
C
O
CH3 and CH3
OH ) and compare them.
C
– Whether they have the same molecular formula / relative molecular
mass.
Planning a synthesis
32.3
Problems in devising a synthesis
32.4
Laboratory preparation of simple carbon compounds
– Whether they have the same boiling point.
32.5
Preparing 1-bromobutane in the laboratory
Example
Oil of wintergreen is a common ester. Salicylic acid can be obtained from it in two
steps.
COOCH3
a sodium salt of
salicylic acid
+
CH3OH
Step 1
OH
COOH
Step 2
OH
oil of wintergreen
salicylic acid
a) Give the reagent and condition used in Step 1.
(2 marks)
b)Suggest a structure for the sodium salt of salicylic acid formed in Step 1. (1 mark)
(1 mark)
c) Suggest a reagent that can be used in Step 2.
Answer
a) Sodium hydroxide solution
(1)
(1)
Heat under reflux
COO
Na
–
+
(1)
OH
c) Dilute sulphuric acid / dilute hydrochloric acid
(1)
Remarks*
Remarks
➤To draw the structure of sodium salt of salicyclic acid in (b), just use the
general equation for the hydrolysis of an ester for deduction.
O
R
Introduction
32.2
– Whether they are soluble in water.
– Whether they have the same chemical properties.
32.1
– Whether they have the same odour.
To tackle these questions, first identify the homologous series to which
each compound belongs. Then answer according to the general properties
of the series concerned.
b)
Synthesis of carbon compounds
C
O
O
R1 + NaOH
R
C
O– Na+ + R1
O
H
➤Students may also work backwards from the structure of salicyclic acid to
deduce the structure of its sodium salt.
27
Topic 8
28
Unit 32 Synthesis of carbon compounds
Chemistry of Carbon Compounds
32.1 – 32.3
c)
Cl
heat
Summary
(1)
Steps for devising a synthesis:
a) identify an immediate precursor to the target molecule;
1st precursor
target molecule
(0.5)
conc. H2SO4
heat
b)continue until the starting molecule is reached.
OH
OH–(aq)
2nd precurso
Br2 (in organic
or conc. H3PO4
heat
starting molecule
(1)
solvent)
(0.5)
Br
Br
(1)
Remarks*
Remarks
♦ The syntheses discussed would NOT involve a change in the length of
the carbon chain.
Exam tips
♦ Students should be familiar with the reactions of organic functional groups
in order to suggest workable synthetic routes for the transformations.
➤In part (a), the immediate precursor of the target molecule (an alcohol) may
be a carboxylic acid or a haloalkane. As a carboxylic acid can be obtained
from the starting molecule (an amide) via hydrolysis, so the synthesis can
be done in the two steps shown.
➤Many conversions involve
– the hydrolysis of haloalkanes to alcohols;
Example
– the oxidation of alcohols to ketones and carboxylic acids;
Outline a syntheticc route, in NOT MORE THAN THREE STEPS, to accomplish each of
the following conversions. For each step, give the reagent(s), the conditions and the
structure of the product.
a) CH3CH2CH2CONH2
CH3CH2CH2CH2OH
b) CH
2CH 3
COCH3
(3 marks)
– the reduction of aldehydes, ketones and carboxylic acids to alcohols.
32.4 – 32.5
Summary
(4 marks)
c) Cl
Stage 1 — planning the preparation;
Br
(4 marks)
Br
Answer
a) CH3CH2CH2CONH2
H2O / H3O+
heat (1)
CH2CH3
Stage 2 — carrying out the reaction to produce the desired product;
Stage 3 — separating the crude product from the reaction mixture;
Stage 4 — purifying and drying the product;
CH3CH2CH2COOH (1)
1 LiAlH4 / ethoxyethane
2 H3O+ (1)
b)
The laboratory preparation of a carbon compound involves five main stages:
Stage 5 — calculating the percentage yield of the product.
CH3CH2CH2CH2OH
Common separation and purification methods for products
Type of product
CHBrCH3
•simple distillation
Br2
Liquid product
•fractional distillation
•liquid-liquid extraction
UV light or heat
(1)
Separation and purification method to employ
Solid product
(0.5)
CH(OH)CH3
COCH3
–
OH (aq)
K2Cr2O7 / H3O
heat
reflux
(1)
+
(0.5)
(1)
•re-crystallization
29
Topic 8
30
Unit 32 Synthesis of carbon compounds
Chemistry of Carbon Compounds
Exam tips
♦ DO NOT confuse an experimental set-up for fractional distillation with
that for simple distillation.
31
Answer
a) ♦ Questions often ask students to describe the re-crystallization procedure
for the purification of a crude solid product.
organic layer
Example
aqueous layer
2-chloro-2-methylpropane can be prepared by reacting methylpropan-2-ol with
concentrated hydrochloric acid.
(CH3)3COH(l) + HCl(aq)
This preparation follows the steps outlined below:
(CH3)3CCl(l) + H2O(l)
Step 1 Shake excess concentrated hydrochloric acid with methylpropan-2-ol in a
separating funnel.
Step 2 Separate the organic layer from the aqueous layer.
(2)
b)Any one of the following:
• Allow better mixing of the reactants by vigorous shaking of the separating
funnel.
(1)
• Allow easy isolation of the product by draining out the lower aqueous layer from
the separating funnel.
(1)
Step 3 Wash the crude product with 10% sodium hydrogencarbonate solution.
Step 4 Dry the product with a suitable reagent.
Step 5 Purify the dried product to remove the remaining methylpropan-2-ol.
The table below lists some information of methylpropan-2-ol and 2-chloro-2methylpropane:
Compound
methylpropan-2-ol
2-chloro-2-methylpropane
–3
Density (g cm )
0.78
0.84
Boiling point (°C)
82
51
Water solubility
miscible
very slightly soluble
c) Add slowly 10% sodium hydrogencarbonate solution. Wait until no effervescence
occurs. (1)
Stopper and shake the separating funnel vigorously.
During shaking, open the tap of the separating funnel regularly to release the pressure
built inside the funnel.
(1)
(1)
Remove the aqueous layer. Repeat the washing procedure until no effervescence
occurs upon the addition of sodium hydrogencarbonate solution.
(1)
d)Anhydrous calcium chloride / sodium sulphate
(1)
a) Draw a diagram of the separating funnel after the reaction has taken place in Step
1, labelling clearly the aqueous layer and organic layer.
(2 marks)
e) Simple distillation / fractional distillation
(1)
b)Suggest ONE advantage of using a separating funnel to carry out Step 1. (1 mark)
Remarks*
Remarks
c) Outline the experimental procedure for washing the crude product in Step 3. Include
the necessary safety precautions.
(4 marks)
➤Questions often ask about the purposes of different steps in the preparations
of carbon compounds.
d)Suggest a suitable reagent for drying in Step 4.
(1 mark)
e) Name a method for purifying the dried product in Step 5.
(1 mark)
➤When giving a reagent for drying, remember to include the word
‘anhydrous’.
32
Topic 8
Unit 33 Important organic substances
Chemistry of Carbon Compounds
Unit 33
Important organic substances
33.1 – 33.9 & 33.13
Summary
33.1
Introduction
33.2
Aspirin — from herbal remedy to modern drug
33.3
Analyzing aspirin tablets by back titration
33.4
Detergents
33.5
How do detergents help water to clean?
33.6
The wetting and emulsifying properties of detergents in relation
to their structures
33.7
The cleaning action of detergents
33.8
Manufacture of soaps and soapless detergents
33.9
The cleaning abilities of soaps and soapless detergents in hard
water
33.10 Nylons
33.11 Polyesters
33.12 Carbohydrates
1 a) The structure of the active ingredient of aspirin tablets, acetylsalicylic acid, is
shown below. It contains two functional groups.
O
carboxyl group
C
OH
O
C
CH3
O
ester group
b)The percentage by mass of acetylsalicylic acid in aspirin tablets can be determined
by back titration.
2 a) There are two types of detergents:
i) soap — made from natural fats and oils; and
ii)soapless detergents — made from chemicals derived from petroleum.
b)The structure of a typical anionic detergent is shown below:
33.13 Lipids
33.14 Proteins and polypeptides
hydrophobic
hydrocarbon ‘tail’
c) A detergent helps water to remove dirt by
i) the ability to act as a wetting agent; and
ii)the emulsifying action.
hydrophilic anionic
‘head’
3 Fats and oils are mixed triglycerides.
4 Soap is made by boiling natural oils or fats with a strong alkali, usually sodium
hydroxide or potassium hydroxide solution.
33
34
Topic 8
Unit 33 Important organic substances
Chemistry of Carbon Compounds
a) Animal fat is one of the raw materials in the production of detergent X.
♦ Students should be able to draw the correct structure of the ester linkage
in a triglyceride (see the structure shown below). The three carboxylic
acids bearing long chains are attached to a glycerol backbone.
Exam tips
♦ When fats and oils are heated with sodium hydroxide solution, they are
hydrolyzed first to form glycerol and carboxylic acids. The acids then
react with the alkali to form sodium carboxylates, which are soaps.
Such reactions are called saponification.
H
H
C
O
O
C
R
H
C
O
H
C
R
+
O
H
C
O
3NaOH
alkali
C
C
H
C
H
C
R
i) Name another raw material that is needed in the production of detergent X from
an animal fat.
(1 mark)
ii)Name the type of reaction that takes place in the production of detergent X.
(1 mark)
iii)Write a chemical equation for the reaction involved in the preparation.
(1 mark)
b)An oil tanker was wrecked and spilt a lot of crude oil in the sea. State whether
detergent X is suitable for treating the oil spill. Explain your answer.
(3 marks)
H
O
OH
OH
OH
H
Answer
–
+
+ 3RCOO Na
sodium
carboxylate
(soap)
a) i) Sodium hydroxide
(1)
ii)Saponification / alkaline hydrolysis
(1)
iii)
H
H
triglyceride in fat or oil
H
C
H
H
C
C
– Whether it is biodegradable.
♦ Hydrogenation of vegetable oils produces margarines.
Example
(CH2)nCH3
O
C
O
C
(CH2)nCH3
O
H
C
H
O
C
C
(CH2)nCH3
+
3NaOH
O
C
(CH2)nCH3
H
H
C
O
H
H
C
O
H
H
C
O
H
+ 3CH3(CH2)nCOO– Na+
b)Not suitable
O
C
O
H
The structure of the main chemical constituent of an animal fat is shown below:
O
(CH2)nCH3
H
– Whether it forms scum in hard water / sea water.
C
C
O
– Whether it is made from natural fats and oil or from hydrocarbons
obtained from petroleum.
H
O
O
♦ Questions often give the structure of a soap / soapless detergent and
ask about its properties.
H
O
glycerol
H
(CH2)nCH3
35
(1)
(1)
Sea water contains a lot of metal ions, such as calcium ions and magnesium ions.
(1)
Detergent X will react with these metal ions to form scum.
(1)
Topic 8
36
Unit 33 Important organic substances
Chemistry of Carbon Compounds
Remarks*
Remarks
c) The repeating unit of poly(ethylene terephthalate) is
➤Questions often ask about the purpose of adding concentrated sodium
chloride solution in soap preparation.
Concentrated sodium chloride solution is added to salt out the soap
produced.
➤Questions often ask students to give the products formed from the complete
hydrolysis of triglycerides existing in natural fats and oils.
e.g.
O
C
(CH2)16CH3
O
C
(CH2)7CH
C
O
CH2
CH2
O
from diol
hydrolysis
O
C
b)The open-chain form of glucose contains an aldehyde group and five hydroxyl
groups.
4 a) Fructose can exist in open-chain and cyclic forms.
CH2
OH
CH
OH
b)The open-chain form of fructose contains a keto group and five hydroxyl
groups.
CH2
OH
5 a) When two amino acid molecules undergo condensation reaction, a water molecule
is eliminated and a peptide link forms. The product is a dipeptide.
complete
CH(CH2)7CH3
O
CH2
C
from dicarboxylic acid
O
CH
O
3 a) Glucose can exist in open-chain and cyclic forms.
O
CH2
O
(CH2)16CH3
+ CH3(CH2)16COOH +
CH3(CH2)7
C
H
C
H
H
(CH2)7COOH
R
O
N
C
C
H
H
OH +
H
amino acid 1
+
CH3(CH2)7
C
H
C
H
O
b)Nylons are formed by condensation polymerization.
c) The repeating unit of nylon-6,6 is
H
N
N
C
(CH2)4
C
H
from diamine
H
O
from dicarboxylic acid
O
C
C
H
H
OH
amino acid 2
C
R
O
N
C
C
H
H
R1
O
N
C
C
H
H
OH +
H2O
a dipeptide
.
b)The peptide links in peptides and proteins can be hydrolyzed to release the
individual amino acids.
Exam tips
O
(CH2)6
N
N
peptide link (or amide group)
H
Summary
O
(CH2)7COOH
33.10 – 33.12 & 33.14
1 a) Nylons are polyamides which contain the amide linkage
R1
♦ A condensation reaction is a reaction in which two or more molecules
react together to form a larger molecule with the elimination of a small
molecule such as water.
♦ Nylon and poly(ethylene terephthalate) are thermoplastics as well as
condensation polymers. NOT all thermoplastics are addition polymers.
♦ Questions often give the repeating unit of a polymer and ask about
information concerning the polymer.
e.g.
2 a) Polyesters contain the ester linkage
C
O
.
b)Polyesters are formed by condensation polymerization.
– whether it is an addition polymer or a condensation polymer;
– whether it is formed from one monomer or two different
monomers.
37
Topic 8
38
Unit 33 Important organic substances
Chemistry of Carbon Compounds
Remarks*
Remarks
Example
Twaron is a heat resistant, high strength fibre used in protective clothing. A short
section of the structure of Twaron is shown below.
O
N
C
C
O
N
N
O
C
H
N
C
N
O
a) Draw the repeating unit of Twaron.
Look at the polymer structure.
O
N
H
C
C
O
(1 mark)
b)Draw the structures of the two monomers that could be used to prepare Twaron.
(2 marks)
c) Explain why Twaron has a great strength.
➤Given the structure of a polymer, students should be able to deduce the
structure(s) of monomer(s) used to produce the polymer.
(1 mark)
– If the polymer backbone contains only carbon atoms, then the polymer
is an addition polymer produced from one monomer.
e.g.
H
C6H5 H
C6H5 H
C6H5
C
C
C
C
C
C
H
H
H
H
H
H
from the following monomer:
Answer
a) N
N
O
N
C
C
H
b) H2N
HOOC
NH2
O
COOH
(1)
(1)
(1)
c) There are strong hydrogen bonds between chains of Twaron.
(1)
is an addition polymer formed
H
C6H5
C
C
H
H
– If the polymer backbone contains other atoms, then the polymer is a
condensation polymer, probably produced from two monomers (see Twaron
shown above).
➤Some condensation polymers are produced from one monomer.
e.g.
the repeating unit of polylactide is shown below:
CH3 O
O
C
C
H
It is a condensation polymer formed from one monomer, lactic acid.
CH3 O
HO
C
C
H
lactic acid
OH
39