ORGANIC CHEMISTRY
And Inorganic chemistry
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J. MASAITI
m.p/℃
Variation/trend in melting point
across the third period
0
11
12
13 14 15
16
17
18
Proton number
Na, Mg and Al have giant metallic structures with strong metallic bonds hence
have generally higher m.p and b.p. M.p increases from Na to Al because of
increase in strength of metallic bonds. The metallic bonds become stronger
from Na to Al due to:
-decrease in atomic size/radius
-increase in nuclear charge
-increase in number of delocalised electrons
Silicon has the highest m.p because it has a giant molecular structure (like
diamond) with very strong Si-Si covalent bonds. M.p Si < m.p diamond because
the C-C bonds in diamond are stronger such that the C atom is much smaller
and more electronegative than the Si atom therefore there is a greater level of
overlapping between C atoms in diamond. P, S, Cl and Ar have simple
molecular structures with weak intermolecular forces i.e. VDW forces of
attraction.
M.p S > than that of P because S exists as a simple octa-atomic substance i.e. S8
hence larger molecular size and stronger VDW forces. M.p decreases from S to
Ar due to decrease in molecular size, hence decrease in VDW forces. Ar exists
as discrete atomic species hence it has a small atomic size therefore weak
VDW forces.
Variation in Electrical conductivity across
the third period from sodium to Argon
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Na, Mg and Al are good conductors of electricity because they have giant
metallic structure (are metals) with delocalised valence electrons which act as
charge carriers. Electrical conductivity increases from Na to Al due to increase
in the number of delocalised valence electrons per atom. Si is a semiconductor. Semiconductors which are pure elements or compounds are called
intrinsic semiconductors e.g. Si. Si is a semiconductor because the energy gap
between the valence band (contains valence electrons) and the conduction
band is smaller and therefore valence electrons can jump from the valence
band to the conduction band and if a potential difference is applied across,
these delocalised electrons can carry charge through the material.
Uses of Si and some semiconductors: it is used in the electronics industry to
make radio transistors, chips of calculators (Si chips)
Electrical conductivity/Ω-1mol-1
P, S, Cl and Ar are non-metal with no delocalised electrons therefore are poor
conductors of electricity.
Al
Mg
Na
Si
P
11
12
13
14
15
S
Cl
16
Ar
17
18
Proton number
Variation in first ionisation energy across the third period Na to Ar page 265
Briggs.
Ar
Cl
ΔHi /kJmol-1
P
S
Mg
Si
Al
Na
11
12
13
14
15
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Trend: there is a general increase in first ionisation energy across the third
period from Na to Ar.
This increase is due to:
1) Decrease in atomic size/radius. As atomic size decreases, electrons
become closer to the nucleus hence highly attracted to the nuclear
charge therefore not easily lost.
2) Increase in the magnitude/size of nuclear charge. Across the period the
number of protons increases hence the nuclear charge increases thus
increasing the force of attraction. Requires a lot of energy to remove the
outermost electrons.
Discontinuations/anomalies
1) Decrease from Mg to Al [Mg: 1s22s22p63s2 & Al: 1s22s22p63s23p1]. The
outermost electron in Al is removed from a slightly higher energy level
(3p orbital) further from the nucleus than the 3s electrons in the 3s
orbital in Mg. [Al-lower ionisation energy due to inter-electronic
repulsion].
2) A decrease from P to S [P: 1s22s22p63s23p3 & S: 1s22s22p63s23p4].
∆Hi (S) < ∆Hi (P) because of increased repulsion between the two
electrons occupying the same orbital (3p) in S. Also in P, the 3p subshell
is half filled i.e. each of the 3 p orbitals has a single electron hence its
stable (i.e. no force of repulsion) therefore more energy is needed to
remove such electrons.
Periodicity in chemical properties across the
third period from sodium to Argon
Reaction with O2
Na reacts with atmospheric oxygen to give Na2O-a basic oxide. It burns with a
yellow flame to give Na2O, a basic oxide:
2Na(s) + 12O2 (g) → Na2O(s)
Note: Na2O is also formed when there is excess oxygen.
Mg burns with a bright dazzling flame giving MgO-a basic oxide.
Al reacts with O2 to form Al2O3-an amphoteric oxide.
P4 burns with a bright ghest flame to give two oxides. With limited O2, P4O6 is
formed-an acidic oxide:
P4 + 2O3 → P4O6
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With excess oxygen P4O10 is formed- an acidic oxide:
P4 + 5O2 → P4O10
The flame is quite bright hence phosphorus is used to make materials
(combustible material) with a bright blue flame.
S burns in air (O2) to produce two acidic oxides SO2 and SO3.
In limited O2, SO2 is produced:
S + O2 → SO2
In excess O2 (air) SO3 is produced.
Variation in oxidation number
The oxidation number of the element in its oxide corresponds to the number
of electrons used for bonding.
Oxide
Na2O MgO Al2O3 SiO2
P4O6
P4O10 SO2
SO3
Oxidation +1
+2
+3
+4
+3
+6
+4
+6
number
The maximum oxidation number increases across the period from +1 for Na to
+6 for S due to increase in the number of valence electrons used for bonding,
which is the same as the group number. All oxidation numbers are positive
indicating that O2 is the most electronegative element. For P and S, the
oxidation number is determined by the number of electrons used for bonding
e.g. in SO2 S uses 4 valence electrons and for SO3 it uses 6 electrons.
Trend in m.p of the oxides
MgO
m.p/℃
Al2O3
SiO2
Na2O
P4O10
SO3
Proton number
11
12
13
14
15
16
There is a change in structure and bonding across the period due to the
different electronegativities between the element and oxygen.
From giant ionic structure in Na2O, MgO and Al2O3 with very strong ionic bonds
hence have very large m.p. MgO has very strong ionic bonds because it has a
high ∆Hlatt with small Mg2+ ion with a very large charge density.
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The m.p (Al2O3) < m.p (MgO) because the Al3+ ion is very small with a large
charge density hence it slightly polarises the larger O2- ion thereby weakening
the ionic bond.
SiO2-high m.p- has giant molecular structure with very strong Si-O covalent
bond, due to this SiO2 is used:
a) As an important ceramic
b) Tooth filling
c) Bricks for building the blast furnace
P4O10 and SO3 have simple molecular structures with weak VDW forces and
therefore have low m.p.
Acid-Base nature
Na2O and MgO- basic oxides
Al2O3- amphoteric oxide
SiO2, P4O10, P4O6, SO2 and SO3- acidic oxides
Reactions of the oxides with water
Na2O (a basic oxide) reacts with water to form a strong alkaline solution:
Na2O + H2O → 2Na+ + 2OHpH = 13
MgO slightly dissolves to give a weak alkaline solution because it has strong
ionic bonds [ ∆Hsol = -∆Hlatt + ∆Hhyd]:
MgO + H2O → Mg (OH) 2
Mg (OH) 2 ⇌ Mg2+ + 2OHMg(OH)2 is weakly alkaline therefore it is used as an anti-acid and in tablets to
neutralise stomach acids and also in tooth paste.
Al2O3- an amphoteric oxide is insoluble in water because it has strong ionic
bonds since it has a small Al3+ with a large charge density therefore ∆Hlatt is
numerically larger than ∆Hhyd
Uses: used in spark plugs because of strong ionic bonds hence high m.p, as a
conductor
Amphoterism of Al2O3
Amphoterism is the behaviour of a substance having both acidic and basic
properties. Al2O3 is basic because it consists of the oxide ion (O2-) which is a
proton acceptor. It is acidic because it has Al3+ which is small and has a large
charge density hence it is an electron loving specie.
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As a base
Al2O3 + 6HCl → 2AlCl3 + 3H2O
Al2O3 + 6H+ → 2Al3+ + 3H2O
As an acid
Al2O3 + 2NaOH + 3H2O → 2NaAl(OH) 4
Al2O3 + 2OH- + 3H2O → 2[Al(OH)4]SiO2 is insoluble in water because it is giant molecular with strong Si-O covalent
bonds which are difficult to break.
Being an acidic substance it reacts with hot concentrated NaOH or molten
NaOH giving a salt and water i.e.
SiO2 (s) + 2NaOH (l) → Na2SiO3 + H2O
SiO2 (s) + 2OH- → SiO32- + H2O
CO2 (g) + 2OH- → CO32- + H2O
The oxides of P and S react with water/are hydrolysed to give their respective
acidic solutions. They are hydrolysed because of the presence of low lying dorbitals which can form dative bonds with H2O molecules.
P4O6 + 6H2O → 12H+ + 4PO33-
pH= 2 phosphoric (III) acid
P4O10 + 6H2O → 12H+ + 4PO43- phosphoric (V) acid
SO2 + H2O → 2H+ + SO32- sulphurous acid
SO3 + H2O → 2H+ + SO42- sulphuric acid, pH= 1
Chlorides of period 3 elements
Variation in oxidation number
Formula
NaCl
MgCl2
AlCl3
SiCl4
PCl3/PCl5
Oxidation
+1
+2
+3
+4
+3/+5
state
The oxidation number of the element in the chlorides corresponds to the
number of electrons used for bonding. All the oxidation states are positive
showing that Cl is the most electronegative element. The maximum oxidation
state increases across the period from +1 in sodium to +5 in Phosphorus due to
an increases in the total number of valence electrons used for bonding.
Structure and bonding of the chlorides
NaCl and MgCl2 have giant ionic structures with strong ionic bonds hence have
high m.p.
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AlCl3/Al2Cl6 is simple molecular. It exists as a dimer, Al2Cl6 i.e. 2AlCl3 joined by
covalent dative bonds hence low m.p.
Al2Cl6 sublimes at low temperature because dative bonds are weak and easily
break to produce AlCl3 molecules.
The chlorides of Si and P are simple molecular with weak VDW forces hence
have lower m.p.
NaCl
Mp/℃
MgCl2
Al2Cl6 (sublimation point)
PCl5
SiCl4
PCl3
11
12
13
14
15
Proton number
Reactions of the chlorides with water: Acid-base
nature of the chlorides
The chlorides become more acidic across the period with increasing proton
number. NaCl does not react with water but dissolves in water to give a neutral
solution i.e.
NaCl (aq) + (aq) → Na+ (aq) + Cl- (aq)
pH= 7
It is not hydrolysed because it’s a salt of both a strong acid and a strong base.
MgCl2 dissolves and reacts with water to give a weakly acidic solution:
MgCl2 (aq) + 6H2O → [Mg (H2O) 6]2+ + 2ClThe small Mg2+ with a large charge density polarises the water molecules
releasing H+ ions:
[Mg (H2O) 6]2+ → [Mg (OH) (H2O) 5] + + H+
pH = 6.5
AlCl3 reacts with water/undergoes hydrolysis because of the presence of low
lying d-orbitals therefore Al can form dative bonds with water ligands. The
solution is acidic.
AlCl3 + 6H2O → [Al (H2O) 5(OH)] 2+ + H+ + 3ClThe Al3+ is small with a large charge density hence it polarises the water
molecules giving protons which makes the solution acidic. SiCl4, PCl3 and PCl5
react with water through hydrolysis to give strong acidic solutions due to the
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presence of low lying empty d-orbitals which are used in dative covalency with
water ligands leading to hydrolysis.
SiO2 (s) + 4H+ + 4Cl- pH = 1
SiCl4 + 2H2O →
Note: CCl4 is not hydrolysed/does not react with water because:
a) The d-orbitals are high-lying hence are not accessible
b) Steric hindrance- the C atom is surrounded by large Cl atoms hence not
open to attack by water ligands/nucleophiles. The Cl atom offers steric
hindrance.
PCl3 + 3H2O → H3PO3 + 3HCl pH = 2
PCl5 + 4H2O → H3PO4 + 5HCl pH = 1
Group TWO
Be
1s22s2
(He) 2s2
Mg
Ca
1s22s22p63s2
1s22s22p63s23p64s2
(Ne) 3s2
(Ar) 4s2
Sr
(Ar) 3d104s24p65s2
(Kr) 5s2
Ba
(Ar) 3d104s24p65s24d105p66s2
(Xe) 6s2
Be forms covalent compounds whereas the other elements form ionic
compounds. Explain.
Reactions of the elements with Oxygen.
They all react with oxygen to give basic oxides except Be which gives BeO, an
amphoteric oxide.
M(s) + 12O2 (g) → MO (s)
1
2
2
1
2
2
1
2
2
1
2
2
1
2
2
Be +
O → BeO } amphoteric oxide
Mg + O → MgO
Ca + O → CaO
basic oxides
Sr + O → SrO
Ba + O → BaO
Mg burns in air with a bright dazzling flame giving MgO- a basic oxide.
Mg + 12O2 → MgO
Ca burns in oxygen with a bright dark-red flame giving CaO- a basic oxide.
Uses of CaO: to neutralise acid soils, soil acidity is caused by acid rain and
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excessive use of nitrogen fertilizers such as NH4.NO3 which undergoes
hydrolysis to give H+ ions i.e.
CaO + H2O → Ca (OH) 2 in water
Ca (OH) 2 + 2H+ → Ca2+ + 2H2O
NH4+ + H2O → NH3 + H3O+
Sr reacts with atmospheric oxygen/burns in air with a crimson flame giving
SrO- a basic oxide.
Ba reacts with oxygen/burns in oxygen with an apple-green flame to give BaO.
Reactions of the elements with water
They all react with water to give metal hydroxides and H2 (g) i.e.
M + 2H2O → M (OH) 2 + H2
Observation: bubbles of colourless gas, H2
Type of reaction: redox
Reactivity increases down the group because of:
a) Increase in atomic size/radius- outermost electrons loosely held by the
nucleus i.e. easily lost
b) Increase in screening/shielding effect i.e. increased number of inner
electrons shield outer electrons from the attractive power of the nuclear
charge.
Evidence for increased reactivity down the group is shown by decrease in
ionisation energy down the group and also increase in EØ values- showing
greater increase in reducing power (due to the two reasons above)
Mg → Mg2+ + 2e2+
-
EØ = +2.38
Ø
Ca → Ca + 2e
E = +2.87
Ba → Ba2+ + 2e-
EØ = 2.90
Increasing
reducing
power
Explain the trend in EØ values down the group.
Mg reacts slowly with cold water, giving a weakly alkaline solution of Mg (OH)2,
pH = 11
Mg + 2H2O → Mg (OH) 2 + H2
It reacts vigorously with steam giving MgO + H2
Mg + H2O (g) → MgO + H2
Ca reacts vigorously with cold water to give a strongly alkaline solution of
Ca(OH) 2.
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The vigour of reaction increases down the group. Sr and Ba react even more
vigorously giving very strong alkaline solutions.
Sr + 2H2O → Sr (OH) 2 + H2
Ba + 2H2O → Ba(OH) 2 + H2
Behaviour/reactions of the oxides with water
MgO is slightly soluble in water (why) to give a weakly alkaline solution of Mg
(OH) 2 i.e.
Mg + 2H2O → Mg (OH) 2 + H2
Mg (OH) 2 is a weakly alkaline solution, partially soluble because of:
a) Strong ionic bonds- small Mg2+ ion with large charge density therefore
∆Hlatt is numerically large. For this reason Mg(OH)2 is used as an anti-acid
to neutralise stomach acids and in tooth pastes- it partially dissociates to
give a low concentration of OH- sufficient only to neutralise acids.
CaO, SrO and BaO dissolve or react in water giving very strong alkaline
solutions.
CaO + H2O → Ca2+ + 2OH- pH = 13
SrO + H2O → Sr2+ + 2OH- pH = 13
BaO + H2O → Ba2+ + 2OH- pH = 14
The alkalinity/basicity of the hydroxides increases down the group due to
increased extend of dissociation as Hlatt decreases due to increase in size of
metallic ion and decrease in surface charge density.
Note: CaO and Ca(OH)2- used as agricultural lime because:
a) They are basic hence can neutralise acids in soils
b) Are soluble in water
Summary: the solubilities of the oxides increase down the group because as
∆Hlatt decreases down the group, ionic bonds become weaker due to increase
in cationic size coupled with a decrease in charge density.
Thermal stability (decomposition) of
group II carbonates
GII carbonates decompose on heating to give their respective metal oxides
together with CO2 (g).
MCO3 ∆ MO + CO2
Observation: colourless gas evolved which turns limewater milky. White
powdered solid formed.
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Trend: GII carbonates become thermally stable down the group i.e. thermal
stability increases down the group due to:
a) Decrease in polarising power of the metallic ions on the large CO32- ion
b) Polarising power of the cations decreases due to increase in cationic size
𝑐ℎ𝑎𝑟𝑔𝑒
and decrease in their charge densities (charge density ∝ 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
)
c) The decrease in polarising power of the metallic ions leads to increase in
strength of the C-O covalent bond in the CO32-.
Thermal stability also increases due to decrease in stability of the oxide
product.
MgCO3 → MgO + CO2
CaCO3 → CaO + CO2
SrCO3 → SrO + CO2
BaCO3 → BaO + CO2
MgCO3 easily decomposes i.e. thermally unstable whereas BaCO3 is thermally
stable. It decomposes at a higher temperature (why).
Ans: the Mg2+ is small and has a large surface charge density hence it polarises
the large CO32- ion weakening the C-O covalent bond hence MgCO3 easily
decomposes. Contrary, the Ba2+ is large with the smallest charge density hence
no polarising effect on the CO32- causing stability of the BaCO3.
Also, MgO is more stable than BaO therefore MgCO3 easily decomposes since it
gives more stable oxide.
Uses of GII carbonates
CaCO3- is used in cement
Test for CO32Reagent: any dilute acid
Observations: effervescence- colourless gas evolved which turns limewater
milky.
Thermal decomposition of GII Nitrates
GII nitrates decompose on heating to give metal oxide together with NO2 and
O2.
M(NO3)2 → MO (s) + NO2 (g) + 12𝑂2 (g)
Observation: reddish brown gas (gas turns damp litmus paper red). Gas relights
a glowing splint-confirm Oxygen. White powdery solid.
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Trend: GII nitrates become thermally stable/increase in thermal stability down
the group because of:
a) Decrease in polarising power of the metallic ions on the nitrate ion
b) Polarising power of the cations decreases down the group due to
increase in cationic size and a decrease in charge density.
The decrease in polarising power of the metallic ions down the group leads to
increase in the strength of the covalent bonds between N and O in the nitrate
ion resulting in increased thermal stability. Thermal stability also increases
down the group due to decrease in stability of the metal oxide produced.
Mg(NO3)2 → MgO + NO2 (g) + 12𝑂2 (g)
Ca(NO3)2 → CaO + NO2 (g) + 12𝑂2 (g)
Sr(NO3)2 → SrO + NO2 (g) + 12𝑂2 (g)
Ba(NO3)2 → BaO + NO2 (g) + 12𝑂2 (g)
Qn: Mg(NO3)2 decomposes at a lower temperature than Ca (NO3)2. Explain this
observation.
Ans: the Mg2+ ion is smaller than the Ca2+ and it has a large charge density
hence it polarises the large NO3- weakening the N-O covalent bonds thus it
easily decomposes.
The Ca2+ being larger with a smaller charge density has no polarising effect on
the nitrate ion hence Ca(NO3)2 is more stable.
Test for NO3- and NO2Common test: add NaOH + Al foil/powder and heat (NH3 evolved)
Observation: effervescence/bubbles of colourless gas with a pungent irritating
smell.
A gas turns damp litmus paper blue.
Gas is ammonia.
Distinguishing test: add any dilute acid
-No effect on NO3- (stable base)
-A NO2- produces reddish-brown fumes of NO2
i.e. NO2- + 2H+ → NO + H2O
NO + 12𝑂2 (atm) → NO2
(reddish-brown fumes)
Note: all nitrates and nitrites are soluble in water hence cannot be identified
by precipitation.
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Solubility of GII sulphate
General equation for dissociation:
MSO4 (s) + aq → M2+ (aq) + SO42- (aq)
The solubility of an ionic substance is a measure of the relative magnitudes of
the lattice dissociation enthalpy i.e.
(-∆Hlatt)/reverse enthalpy change of lattice and the lattice enthalpy change of
hydration of its ions i.e.
∆HØsoln = -∆Hlatt + ∆Hhyd
Trend: the solubilities of group II metal sulphates decreases down the group
i.e.
MgSO4 - soluble
CaSO4 - sparingly soluble
SrSO4 - insoluble
BaSO4 - insoluble
Explanation: ∆Hlatt is fairly constant due to the large SO42- ion i.e. ∆Hlatt ∝
𝑞+ × 𝑞−
𝑟+ × 𝑟−
Therefore the hydration enthalpy becomes less exothermic, hence energy
evolved is not sufficient to break the ionic bonds since ∆Hlatt is almost constant.
Enthalpy change of hydration decreases numerically due to increase in size of
the metallic ion together with a decrease in their charge density.
Common uses of GII compounds
Compound
MgO
CaO/Ca(OH)2/CaCO3
CaSO4.12H2O
Use
Refractory (heatresistant) lining of
furnaces
Neutralise acidic soils.
To make cement for
concrete in buildings
Plaster casts for broken
limps because on
addition of water
plaster of Paris expands
slightly and sets to form
CaSO4.2H2O
ORGANIC AND
Reason for use
MgO has high m.p due
to strong ionic bonds
They are basic
Absorbs water and sets
to a hard solid
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MgSO4
Laxative- to loosen the
digestive systemcleaning
‘barium meal’- to
enable X-ray analysis of
patients with digestive
problems
BaSO4
Its soluble
Creates an opaque
environment suitable
for X-ray analysis
-less soluble/insoluble
Tests for GII metal ions
Mg2+ tests
1. With NaOH/NH3 (aq)
Observation: white gelatinous precipitate, insoluble in excess.
Ca2+ tests
1. With NaOH
Observation: white ppt, insoluble in excess. (Only in the presence of
high c(Ca2+)…why?)
Precipitation occurs only when Ki > Ksp i.e. for Ki > Ksp
Ca(OH)2 (s) → Ca2+ + 2OH-With any soluble CO32- or SO42Observation: white ppt e.g.
Ca2+ + CO32- → CaCO3
Ca2+ + SO42- → CaSO4
Test for Ba2+
1. With any SO42Observation: a white ppt, insoluble in acid.
Ba2+ + SO42- → BaSO4 (s)
(II) With any soluble CO32Observation: white ppt, soluble in acid.
Ba2+ + CO32- → BaCO3 (s)
(III) With any sulphite
Observation: a white ppt, soluble in dilute acid.
Ba2+ + SO32- → BaSO3 (s)
(IV) With any CrO42Observation: a yellow ppt, soluble in HCl or HNO3 acids
Ba2+ + CrO42- → BaCrO4 (l)
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(V) With Cr2O72Observation: an orange ppt observed
Group IV
Electronic configuration
Classification
C: 1s22s22p2
non-metal
Si: 1s22s22p63s23p2
Ge: [Ar] 3d104s24p2
metalloid
metalloid
Sn: [Kr] 4d105s25p2
metal
Pb: [Xe] 5d106s26p2
metal
Trend: there is a change from non-metal to metalloid and finally metal. There
is an increase in atomic size/radius down the group and outermost electrons
become loosely attracted hence they become mobile.
Increase in shielding effect as number of inner electrons increases down the
group causing outermost electrons to be loosely attracted by the nucleus,
hence they become delocalised.
Variation in m.p
C (graphite)
Mp/℃
Si
Ge
Pb
Sn
Trend: there is a general decrease in m.p due to:
i)
Increase in atomic size/radius- causing an increase in E-E bond length
and hence bond becomes weaker i.e. poor overlapping (except for
decrease from C –Si-Ge)
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Carbon, Silicon and Germanium have giant molecular structures with strong
covalent bonds bonding their respective atoms hence have high m.p and b.p
Sn and Pb have giant metallic structures with strong metallic bonds between
the metallic ions and their mobile electrons.
The metallic bonds in Sn and Pb are weaker than the covalent bonds in the
members with giant molecular structures hence Pb and Sn have lower m.p
than Ge.
Uses of the elements
Diamond is used in drills and saws for cutting through rock (drill heads)
because it is hard. Graphite is used as a lubricant in lead pencils because of its
softness due to the existence of weak VDW forces between its layers. It is also
used as electrodes for electrolysis because it has mobile electrons (sp2
hybridised). Very pure Si is a semi-conductor hence it is used in the electronics
industry to make silicon chips for calculators and computers. Very pure Ge is a
semiconductor hence it is also used in the electronics industry to manufacture
electronic components e.g. radio transmitters. Sn is used in tin plating of steel
e.g. tin cans to prevent rusting and low m.p in electrical solder.
Bonding, molecular shape and volatility
of the tetrachlorides
CCl4
SiCl4
GeCl4
SnCl4
PbCl2
Molecular shape
Have the general formula XCl4. It has four electron pairs which are all bond
pairs hence they have a tetrahedral shape, bond angle ≈ 109.5°
Bonding
They are simple molecular with weak intermolecular forces called Van Der
Waal forces but strong intramolecular covalent bonds. Since they have weak
VDW forces they have low m.p and b.p
Volatility
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They are volatile liquids with simple molecular structures. Have weak
intermolecular VDW forces hence have low m.p and b.p i.e. are volatile.
Trend: volatility of the tetrachlorides decreases down the group due to:
a) Increase in molecular size/electron cloud leading to increase in strength
of the VDW forces.
b) Increase in Mr value
Thermal stability of the tetrachlorides
Trend: thermal stability decreases down the group because:
i)
The X-Cl covalent bond becomes longer and weaker as the size of GIV
atom becomes larger (poor overlapping as atomic size increases)
ii)
The inert pair effect makes the +4 oxidation state less stable whilst
the +2 oxidation state becomes stable.
The +2 oxidation state becomes more stable than the +4 oxidation state
because of the inert pair effect e.g. in Pb [Xe] 4f145d106s26p2, the inner 6s2
electrons are highly attracted by the nuclear charge due to the poor shielding
effect of the inner 5d orbital hence the 6s pair of electrons becomes
inert/unreactive leaving only the outer 6p electrons to participate in bonding
e.g. CCl4 is very stable to heat.
PCl4 is a yellow liquid which is thermally unstable, it slowly decomposes at
room temperature to form the more stable PCl2.
PCl4 → PCl2 (s) + Cl2 (g)
Hydrolysis of the tetrachlorides
CCl4 does not dissolve nor react with water i.e. is not hydrolysed because it has
no low lying d-orbitals i.e. the d-orbitals in C are high lying and therefore are
not accessible to dative covalency with water ligands hence the process has
high Ea. It is also due to steric hindrance i.e. the C atom is surrounded by large
Cl atom hence not open to attack by water molecules.
SiCl4 is hydrolysed giving SiO4 together with HCl.
SiCl4 + 2H2O → SiO4 (l) + 4HCl (aq) pH = ½
Si is hydrolysed because it has low lying d-orbitals accessible to dative bonding
with water ligands.
GeCl4 is also hydrolysed due to the presence of low lying d-orbitals giving a
strong acidic solution:
GeCl4 + 2H2O → GeO2 (s) + 4HCl (g)
pH= ½
SnCl4 and PbCl4 are partially hydrolysed by water:
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SnCl4 + 2H2O → SnO2 + 4HCl
pH= 1
PbCl4 + 2H2O → PbO2 + 4HCl pH= 1
Note: tetrachlorides (except CCl4) are hydrolysed because of:
1. Presence of low lying d-orbitals
2. Polarity of the X-Cl bond, the partial positive Group 4 element is
therefore attacked by the electronegative halogen species.
Acid-base nature of the oxides
1. CO-is a neutral oxide. It only reacts with molten/fused NaOH giving
sodium methanoate:
CO + NaOH → HCOO-Na+
2. CO2-is an acidic oxide. It gives a weakly solution in water:
CO2 + H2O ⇌ 2H+ + CO32- (carbonic acid)
Reacts with NaOH giving Na2CO3
CO2 + 2NaOH → Na2CO3 + H2O
SiO- unstable- does not exist
SiO2- an acidic oxide. It reacts with fused alkaline NaOH, giving a salt and
water:
SiO2 + 2NaOH → Na2SiO3 + H2O
SiO2 + OH- → SiO32- + H2O
SnO and SnO2 are both amphoteric oxides.
SnO + 2H+ → Sn2+ + H2O/SnO2 + 4H+ → Sn4+ + 2H2O
SnO + 2OH- → SnO22- + H2O/SnO2 + 2OH- → SnO32- + 2H2O
GeO and GeO2 are both amphoteric oxides.
GeO + 2H+ → Ge2+ + H2O/GeO2 + 4H+ → Ge4+ + 2H2O
GeO + 2OH- → GeO22- + H2O/GeO2 + 2OH- → GeO32- + H2O
PbO and PbO2 are both amphoteric oxides.
PbO + 2H+ → Pb2+ + H2O
PbO + 2OH- → PbO22- + H2O
PbO2 + 4H+ → Pb4+ + 2H2O
PbO2 + 2OH- → PbO32- + H2O
PbO- yellow oxide
PbO2- chocolate brown
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Pb3O4- red oxide- it’s a mixture oxide, mixture of:
2PbO + PbO2 (2:1 ratio)
With nitric acid, PbO dissolves but PbO2 does not.
Pb3O4 + 4HNO3 → 2Pb(NO3)2 + 2H2O + PbO
Note: PbO2 is thermally unstable hence it decomposes at room temperature to
PbO i.e.
PbO2 ⟶ PbO + 12O2
Reason: inert pair effect i.e. +2 oxidation state (in Pb) becomes more stable
than +4 oxidation state in Pb.
CO2 is thermally stable because +4 oxidation state is more stable than +2
oxidation state.
SiO2 is stable and cannot decompose to SiO which is highly unstable.
GeO2 decrease in stability of the +4 oxidation
SnO2 state. Increase in stability of the +2
PbO2 oxidation state.
CO
SiO
increasing stability of the monoxides i.e.
the +2 oxidation state increases in stability
GeO
due to inert pair effect
SnO
PbO
Variation in stability of the +2 and
+4 oxidation states
C
stability of the
Si
+4 oxidation state
Ge
Sr
decreases
Ba
The above trend is due to inert pair effect i.e. the inner 6s pair of electrons
become inert/unreactive and do not participate in bonding due to the poor
shielding effect of the inner 5d orbital. The 6s pair is highly attracted by the
nuclear charge leaving the outer 6p electrons to participate in bonding, thus
the +2 oxidation state is more stable than the +4 oxidation state e.g. in Pb.
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The following EØ values give evidence to the above trend.
Ge4+ + 2e- ⇌ Ge2+
Sn4+ + 2e- ⇌ Sn2+ +0.15V
Pb4+ + 2e- ⇌ Pb2+ +1.69V
The EØ values become more positive down the group showing increasing
oxidising power of the elements in the +4 oxidation state or it shows decrease
in reducing power of the elements in the +2 oxidation state e.g.
Sn4+ + 2e- ⇌ Sn2+
Fe3+ + 2e- ⇌ Fe2+
+0.15
+0.77
Pb4+ + 2e- ⇌ Pb2+ +1.69
Fe3+ can oxidise Sn2+ to Sn4+:
Sn2+ + 2Fe3+ → Sn4+ + 2Fe2+
EØ = 0.6V
But Fe3+ cannot oxidise Pb2+ to Pb4+ showing increased stability of the +2
oxidation state of Pb. Pb2+ is a very poor reducing agent hence it is very
unreactive. Pb4+ and PbO2 (PbO2 + 4H+ + 2e- ⇌ Pb2+ + 2H2O, EØ = 1.47) are very
strong oxidising agents capable of oxidising Fe2+, I- even H2O2:
PbO2 + 2I- + 4H+ → Pb2+ + I2 + 2H2O
Sn4+ cannot oxidise Fe2+, I- or H2O2 showing reduced oxidising power of Sn4+ to
Pb4+/PbO2. Considering the oxides:
CO
decrease in reducing
GeO
power of the oxides
SnO
PbO
in the +2 oxidation
state.
CO is a very strong reducing agent i.e. C is quite unstable in the +2 oxidation
state. PbO is a very poor reducing agent because Pb is very stable (unreactive)
in the +2 oxidation state.
CO2
increase in oxidising
SiO2
GeO2
power of the
dioxides on descending
SnO2
the group, showing decrease in
PbO2
stability of the +4 oxidation state.
e.g. PbO2 is a very strong oxidising agent whereas CO2 is a very poor oxidising
agent, evidence to the above trend is also given by the ∆Hf of the oxides e.g.
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C (s) + O2 (g) → CO2 (g)
Pb (s) + O2 (g) → PbO2
∆Hf (PbO2) < ∆Hf (CO2) because CO2 is more stable than
effect).
PbO2 (inert pair
Copy table 13.2 Briggs page 297
Test for Group IV cations: test for Pb2+
Pb2+- is a colourless aqueous solution
KI (aq) ⟶ PbI2 (s) yellow ppt
HCl(aq) ⟶ PbCl2 (s) white ppt
Pb2+ (aq)
H2SO4 (aq) ⟶ PbSO4 (s) white ppt
NaOH (aq) ⟶ Pb(OH)2 ⟶ [Pb(OH)4]2- ⟶ PbO2 brown ppt
NH3 (aq) ⟶ Pb(OH)2 insoluble in excess, white ppt
K2CrO4/Na2CrO4 ⟶ PbCrO4 (s) yellow ppt
All lead (II) compounds are insoluble in water except Pb(NO3)2 and (CH3COO)2Pb lead (II) ethanoate therefore most reagents give precipitates with Pb2+.
Pb2+ can be oxidised to PbO2 (brown ppt) by alkaline H2O2 or OCl- i.e.
Pb2+ (aq) + H2O2 (aq) + 2OH- (aq) ⟶ PbO2 (s) + 2H2O
Ceramics
Are materials made from clay and non-metal materials that have been
permanently hardened by using high temperature. They are strong, brittle and
resistant to heat and attack by chemicals. Ceramics like glass are made up of
silicates. Examples of ceramic materials are silicon nitride (Si3N4) used as an
abrasive and as a refractory material. Silicon carbide (SiC) is used in microwave
furnaces, abrasives and refractory materials. Ferrite (Fe3O4) is used in the core
of electrical transformers and magnetic core. Silicates are ceramics based on
SiO2 and are used as insulators and to make bricks for the blast furnace.
Note: ceramics are usually ionic, covalently bonded materials or both.
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Properties of ceramics
Have high m.p because they are compounds of giant molecular substances e.g.
SiO2 together with giant ionic substances like MgO and Al2O3. They are good
thermal and electrical insulators. They have greater rigidity, hardness and
temperature stability than organic polymers. They are resistant to heat and
chemical attack.
Uses of Ceramics (based on the above properties)
1. Manufacture of glass, bricks and tiles
2. Manufacture of dinnerware
3. As a refractory material in the lining of furnaces
4. As power line insulators
5. Making glasses for solar panels
Group VII
F: 1s22s22p5
Cl: 1s22s22p63s23p5
Br: 1s22s22p63s23p64s23d104p5
I: 1s22s22p63s23p64s23d104p65s24d105p5
Physical properties
Element
Molecular
formula
Colour
m.p/℃
b.p/℃
Solubility in
water
Fluorine
F2
Paleyellow
-220
-188
Soluble
Chlorine
Cl2
Yellowgreen
-101
-35
Moderatelysoluble
Bromine
Br2
Dark red
-7
58
Slightly
soluble
Iodine
I2
Black
113
183
Insoluble
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Volatility
All the molecules are simple diatomic substances with weak intermolecular
forces between their molecules in liquid and solid state called VDW forces
(induced dipole type) or London dispersion forces.
Trend: they become less volatile i.e. volatility decreases down the group due
to:
i)
ii)
Increase in strength of VDW forces caused by an increase in
molecular size or electron cloud.
Increase in Mr value
The decrease in volatility is evidenced by:
a) Change in physical state down the group. Fluorine and chlorine are
gases, bromine is a liquid, and iodine is a solid.
b) Increase in b.p and m.p
These two factors are due to an increase in the strength of the VDW forces on
descending the group.
Change in colour (from table)
The elements become darker down the group from F2 to I2 due to absorption
of part of the wave length of light to bring about electron excitation. Due to
increase in atomic size/number, the last occupied quantum has higher energy
therefore the wavelength absorbed from white light gets longer and the
shorter the wavelength of more intense colour is transmitted.
Reactivity of the elements: oxidising power
Oxidising power (reactivity) decreases down the group due to:
a) An increase in atomic size/radius causing a decrease in electron affinity
b) As atomic size increases the effective nuclear charge decreases
c) Increasing shielding effect nullifies the attractive power of the nuclear
charge thereby decreases the tendency to gain electrons
The decrease in oxidising power is evidenced by the decrease in EØ value for
the reduction of the halogens:
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1
2
Cl2 + e- ⇌ Cl-
EØ = +1.36
decrease in
1
2
Br2 + e- ⇌ Br-
EØ = +1.07
oxidising
1
2
2
EØ = +0.54
power
I + e- ⇌ I-
The decrease in oxidising power is shown by the reaction of the elements with
S2O32-. Cl2 and Br2 are stronger oxidising agents hence they oxidise S2O32- to
SO42- in which the S atom is oxidised from +2 to +6 oxidation state:
S2O32- + 4Cl2 + 5H2O → 2SO42- + 10H+ + 8ClS2O32- + 4Cl2 + 5H2O → 2SO42- + 10H+ + 8ClTo obtain equation e.g. with chlorine:
4Cl2 + 8e- → 8Cl-
Adding
S2O32- + 5H2O → 2SO42- +10Cl- + 10H+
I2 is a weaker oxidising agent. It oxidises the S in S2O32- from +2 to +2.5
oxidation state, an increase of only 0.5 showing that it is a weak oxidising
agent.
The decrease in oxidising power is also shown by their reaction with Fe2+. Cl2
and Br2 oxidise Fe2+ to Fe3+ whereas I2 cannot i.e.
Cl2 + 2Fe2+ → 2Fe3+ + 2Cl- EØ = +0.59
Cl2 can displace/oxidise Br- to Br2 and Br2 can oxidise I- to I2 showing a decrease
in oxidising power:
Cl2 + 2Br- → Br2 + 2ClBr2 + 2I- → I2 + 2Br-
Reactions of the halogens with hydrogen
All react with hydrogen gas under different conditions forming hydrogen
halides HX. F2 explodes with hydrogen under all conditions i.e. it reacts
explosively even in darkness at -200℃. Reaction is complete due to H-F bond in
HF which is strong due to maximum overlapping due to small size of the F
atom and electronegativity.
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Chlorine reacts with hydrogen explosively under sunlight but slowly in
darkness below 200℃. Bromine reacts with hydrogen when heated to 200℃ in
the presence of platinum catalyst.
Iodine reacts with hydrogen at 450℃ and in the presence of a platinum
catalyst to give an equilibrium mixture because the H-I bond is very weak, poor
overlapping since the iodine is very large. Reaction does not reach completion.
Note: the extent to which the reaction occurs is determined by the H-X bond
strength i.e. extend of reaction decreases down the group because:
a) Ea increase
b) The H-X bond becomes weaker i.e. it gets longer and weaker due to
increased atomic size of the halogen together with a decrease in
electronegativity
The conditions under which the reaction occurs becomes increasingly strange
e.g. temperature increase, chlorine does not need a catalyst whilst other
members below it need a catalyst. This shows decrease in reactivity as atomic
size increases and electronegativity decreases i.e. there is increase in Ea.
Thermal stability of the halogen hydrides
Trend: thermal stability decreases down the group i.e. HX becomes less stable
to heat due to:
-Decrease in the H-X bond strength i.e.
Bond
Bond energy/kJmol-1
H-Cl
431
H-Br
H-I
364
299
The H-X bond strength decreases as shown by the decrease in bond energy due
to increase in atomic size of the halogen coupled with a decrease in
electronegativity. Hence H-X bond becomes longer and weaker due to poor
overlapping as atomic size increases e.g. HI easily decomposes into H2 + I2 by
putting a red hot platinum wire into the gas.
Acidity of the hydrogen halides
The hydrogen halides react with water forming strong acidic solutions i.e.
HCl + H2O → H3O+ + Clincreasing
HBr + H2O → H3O+ + Br-
acidity
HI + H2O → H3O+ + I-
why?
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The strength of the acids increases down the group from HCl to HI due to
increase in extend of dissociation caused by the decrease in strength of the H-X
bond as atomic size increases down the group hence H-I is the strongest acid.
Reaction of halide ions: test for halide ions
1. With the reagent AgNO3 followed by aqueous NH3 Solutions containing
halide ions i.e. Cl-, Br-, I- are precipitated by AgNO3.
Chloride ions react with AgNO3 giving a white ppt soluble in aqueous
ammonia i.e.
Ag+ + Cl- → AgCl
AgCl + 2NH3 → [Ag(NH3)2]+ + Cl2. Br- reacts with AgNO3 giving a cream ppt of AgBr which is insoluble in
aqueous ammonia but soluble in concentrated ammonia
The solubilities of the silver halides decreases down the group
Silver halide
Solubility product/ mol2dm-6
AgCl
1.8 x 10-10
AgBr
7.7 x 10-13
AgI
8.3 x 10-17
The solubility of the HX in ammonia is determined by the c(Ag+) available for
complexing with the NH3 ligands i.e. solubility of the AgX decreases due to
decrease in c(Ag+) available for complexing as shown by the decrease in Ksp i.e.
c(Ag+) = √𝐾𝑠𝑝. AgCl is soluble in aqueous ammonia because the c(Ag+)
available is a saturated solution of AgCl which is high enough to allow
complexing i.e.
AgCl ⇌ Ag+ + ClThe high c(Ag+) in equilibrium makes Ki < Ksp hence the above equilibrium
shifts to the right to replace Ag+ removed by complexing causing the
dissolution of AgCl.
AgBr is slightly soluble in aqueous ammonia because Ki is almost equal to Ksp
or rather Ki is slightly greater than Ksp. The c(Ag+) available is not high enough
to upset the equilibrium causing dissolution i.e.
AgI ⇌ Ag+ + IThe removal of silver by complexing cannot upset the above equilibrium
because c(Ag+) is very low i.e. Ki < Ksp
Reaction of halide ions with conc H2SO4 acid
H2SO4 is also a reagent used to identify halide ions in solution.
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With Cl- ions
Observation: white fumes of HCl gas
Cl- + H2SO4 (conc) → HSO4- + HCl
E.g. NaCl + H2SO4 (conc) → NaHSO4 + HCl
With BrObservation: white fumes of HBr gas together with traces of Br2 gas
Br- + H2SO4 (conc) → HSO4- + HBr
2HBr + H2SO4 → Br2 + SO2 + 2H2O
E.g. NaBr + + H2SO4 → NaHSO4 + HBr
With I- e.g. NaI
Observation: solid black iodine or purple vapour of iodine i.e.
2I- + H2SO4 → 2NaHSO4 + I2 + 2SO2 + 2H2O
OR
2Cl- → Cl2 + 2e-
-1.36 reducing power
2Br- → Br2 + 2e-
-1.07 of the halide ions
2I- → I2 + 2e-
-0.54 increases.
The extent to which S (in H2SO4) is reduced increases down the group showing
an increase in reducing power of the halide ions. The reducing power of the Xions increases down the group due to increase in their ionic size i.e. electrons
in outermost shell are loosely attracted hence easily lost. The other reason for
increase in reducing power is due to an increase in shielding effect as the
number of inner electrons increases causing the outermost electrons to be
shielded from the attractive force of the nuclear charge hence they are easily
lost. The increase in reducing power is evidenced by the increase in EØ values
as indicated above.
*Other tests are done using Pb(NO3)2, H2O2/H+
Electrolysis of brine page 312 Briggs (copy notes)
Reaction of chlorine with sodium hydroxide (Briggs page 313: copy notes)
Uses of halogen and halogen compounds
NaClO- as bleach for clothes, as a disinfectant due to its oxidising properties.
NaClO3- used as a weed killer.
Cl2 is used to purify water for drinking and swimming (it kills bacteria).
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Explanation for the use of Cl2 to purify water.
Cl2 slightly dissolves in water to form HCl acid and Chlorine (I) acid.
Cl2 + H2O ⇌ HClO + HCl
The HClO is an oxidising agent, it oxidises the living material in bacteria,
microbes and germs thus it kills pathogens.
The HClO is a weak acid and it partially dissociates giving chlorate (I) ions which
are also oxidising. It oxidises pathogens.
HClO ⇌ H+ + ClOThe HClO is 80 times more effective than OCl- in killing bacteria. Considering
the above equilibrium, if pH (water) is high, c(H+) is low, equilibrium shifts
hence c(ClO-) > c(HClO). If pH is low, c(H+) is high for this reason c(HClO) >
c(ClO-). Due to this, the pH of water must be carefully monitored to ensure a
high concentration of HClO in the water. Chlorination must be carefully
monitored to avoid the formation of toxic chloroalkanes formed by the
reaction of Cl2 with traces of organic compounds in water.
Other uses of Cl2
-to make organic solvents e.g. CCl4 and CHCl3
-to make polymers e.g. PVC, refrigerator fluids and aerosol propellants
-AgCl, AgBr and AgI are used in photographic films and paper (salts are
decomposed by light)
-iodine dissolved in KI and alcohol is used as a disinfectant for cleaning
wounds.
-as insecticides e.g. DDT
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Nitrogen and Sulphur
Unreactivity of Nitrogen
Compared to other elements, nitrogen is unreactive due to the very strong
N≡N bond. The N to N triple bond is very strong because 3 pairs of electrons
shared gives a large charge density/electron density shared between the two N
atoms resulting in very strong force of attraction between shared electrons
and the nuclei of the two N atoms.
This leads to maximum overlapping such that the triple bond is short and
therefore very strong.
BE (N≡N) = 994kJmol-1 very strong indeed that is why reactions involving
nitrogen have very high Ea e.g. Haber process.
Formation and structure of ammonium ion
NH3 is a base i.e. a proton acceptor. It accepts a proton hence it is a base. It has
a lone pair of electrons used to form dative bonds with H+ ions from the acid:
NH3 + H+ ⇌ 𝑁𝐻4+
There is a change in structure from pyramidal, bond angle 107° in ammonia to
tetrahedral bond angle 109.5° in the NH4+ ion i.e.
N
H
H + H+
H
H
H N H
H
I.e. the NH4+ ion has four electron pairs around the central N atom of which all
are bond pairs giving a tetrahedral structure bond angle 109.5°.
Production of ammonia
Ammonia is produced by displacement reaction where any ammonium
sulphite reacts with a strong alkali or basic oxide. In the lab, ammonia is
produced by reacting NH4Cl with NaOH on warming.
NH4Cl + NaOH → NH3 + H2O + NaCl
OR
NH4+ + OH- → NH3 (g) + H2O (g)
MgO + 2NH4Cl → MgCl2 + NH3 + H2O
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Ammonia is also produced when a NO3- is heated with a mixture of Al powder
and excess (aq) NaOH i.e. the NO3- ions are reduced to NH3. This is used as a
test for NO3- and NO2-.
Manufacture of ammonia by the Haber process
Briggs page 320
Consider principles of kinetics and equilibrium
Uses of ammonia and other N compounds
Ammonia is used in:
a) manufacture of fertilisers
b) degreasing agent in laundries and oven cleaners
c) manufacture of HNO3 acid
Nitric acid is used to:
a) manufacture nitrate fertilisers e.g. NH4NO3
b) manufacture of explosives e.g. TNT
c) manufacture of organic dyes (azo-dyes)
Environmental effects of uncontrolled use of nitrate
fertilisers
Eutrophication
Excessive nitrate fertilisers are washed off the land by rain water and get
deposited in water bodies such as rivers and lakes. The fertiliser causes
excessive growth of weeds and algae in the fresh water. This clogs the water
ways with vegetation and limits the amount of sunlight that reaches the
aquatic plants. Some of the dissolved oxygen is used by bacteria during decay
causing further decrease in dissolved oxygen. This consumption is measured as
the biological oxygen demand (B.O.D), the B.O.D therefore becomes large. This
process is called eutrophication. It causes death of fish and aquatic animals.
Environmental effects of nitrates
A high concentration of nitrates in the water for consumption is an
environmental problem because nitrates are poisonous. They oxidise iron (II) in
haemoglobin thus reduces the ability/efficiency of haemoglobin to transport
oxygen to the lungs.
Remedy
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Using less fertiliser and other methods to increase yield of food crops e.g.
natural fertilisers from animal wastes, manure from decomposing vegetation
and selecting more productive strains of wheat and rice.
Environmental effects of oxides of nitrogen, NOx
Sources of oxides of nitrogen
1. Internal combustion process in car engines. The high temperature due to
the combustion of fuel in car engines provides energy for the two
atmospheric gases N2 and O2 to react forming NO and NO2.
N2 + O2 → 2NO
NO + 12O2 ⇌ NO2
2. From electricity generating stations and industries and from combustion
of coal and oil.
3. During lightning (flashes) i.e. lightning provides energy for N2 and O2 to
react and form NOx
Oxides of nitrogen
Causes acid rain i.e. NO2 reacts with water to form a mixture of nitrous acid
(HNO2) and nitric acid (HNO3) i.e.
2NO2 + H2O → HNO2 + HNO3
The nitrous acid is then oxidised by atmospheric oxygen to nitric acid i.e.
HNO2 + 12O2 → HNO3
Overall equation: 2NO2 + H2O + 12O2 → 2HNO3
Nitric acid together with sulphuric acid are the major components of acid rain.
NOx also contribute to acid rain formation as they catalyse the oxidation of SO2
to SO3 which then reacts with H2O forming acid rain:
NO2 + SO2 → NO + SO3
Acid rain has the following environmental problems:
a) Soils become acidic such that they cannot support plant life
b) Destruction of buildings
c) Water in lakes and rivers becomes acidic killing fish
d) Destruction of vegetables
Other than acid rain, NOx have the following environmental effects:
a) Reacts with other pollutants to form ozone which irritates the eye
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b) Catalyses oxidation of SO2 to SO3 which reacts with rain water producing
acid rain- major component of acid rain
c) Nitrogen monoxide destroys the ozone layer since it reacts with O3 i.e.
O3 + NO → O2 + NO2
NO2 + O → NO + O2
Ozone depletion occurs which leads to skin cancer. Ozone is destroyed due to
the presence of chemical substances in the atmosphere e.g. NO, CH4, CFCl3,
etc.
Prevention of NOx pollution
The emission of NOx from car engines can be reduced by using catalytic
converters in the exhaust. The mixture of NOx, CO and unburnt hydrocarbons is
passed over platinum (rhodium, palladium also used) catalyst and is converted
into harmless N2, CO2 and steam (the catalytic converter is composed of a
honey-comp coated with Pt metal)
Lead-free petrol must be used because lead poisons the catalyst making it
ineffective. Typical reactions in the converter are:
2NO + 2CO → N2 + 2CO2
2NO2 + 4CO → N2 + 4CO2
Hydrocarbons + NO → CO2 + N2 + H2O
Disadvantages of use of catalytic converters
They use Pt which is very expensive. Engines that use lead-free petrol are less
efficient so more fuel is used. Catalytic converters are only effective at
temperatures above 400℃ which means they are ineffective on short journeys.
Due to this: the amount of NOx produced from burning fuel ca be reduced by
lowering the temperature i.e. spraying water on the burning fuel.
Sulphur dioxide
Other than oxides of nitrogen, SO2 is also a pollutant.
Sources of SO2 in the atmosphere
a) From sulphur contaminated carbon fuels e.g. burning of coal and oil
(fossil fuel).
b) From industries such as burning metal sulphides ores in extraction of
metals.
c) From natural sources such as volcanoes, sea spray, rotting vegetation
and plankton.
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Role of SO2 in acid-rain formation
SO2 reacts with oxygen and water in the air to form sulphuric acid causing
acid rain. The reaction is catalysed by oxides of nitrogen i.e.
Stage 1: SO2 reacts with NO2 i.e.
SO2 + NO2 → SO3 + NO
Stage 2: SO3 reacts with rain water forming acid rain i.e.
SO3 + H2O → 2H+ + SO42- (acid rain)
Stage 3: NO from stage 1 reacts with more atmospheric O2 to produce
more NO2 i.e.
2NO + O2 → 2NO2
This repeats stage 1.
Also, the NO2 reacts with O2 and H2O in air forming a mixture which becomes
acid rain i.e.
2NO2 + 12O2 + H2O → 2HNO3 (acid rain)
Environmental consequences of acid-rain
1. Destruction and corrosion of buildings and statues
CaCO3 + H2SO4 → CaSO4 + CO2 + H2O
2. Corrosion of steel i.e. H2SO4 reacts with iron in vehicles:
Fe + H2SO4 → FeSO4 + H2
3. Acidification of fresh water lakes causing death of fish
4. Acidification of agricultural soil, lowers the pH of soil causing a reduction
in crop yield
Remedy: application of agricultural lime i.e. CaO/Ca (OH) 2
CaO + 2H+ → Ca2+ + H2O
Ca (OH) 2 + 2H+ → Ca2+ + 2H2O
The presence of acids in soils causes heavy metal ions as well as Al3+ to
dissolve. The dissolved ions may enter water bodies e.g. lakes and rivers or
may be absorbed by plants. The Al3+ and heavy metal ions are poisonous
(toxic). Aquatic plants and trees get destroyed.
Prevention of acid rain
a) Burning less sulphur containing fuels particularly coal.
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b) Treating the exhaust gases from industries and power stations with SO2absorbing chemicals such as heated limestone i.e. CaO from the
decomposition of CaCO3 together with air reacts with SO2 to produce
solid particles of CaSO4 which is trapped by filters- this process is called
flue gas desulphurisation (Ramsden page 438)
Uses of SO2
Is an oxidant hence it is used in food preservation. It kills bacteria. It is also
used to bleach wood pulps.
Contact process and use of H2SO4 acid (Briggs page 324-325)
Introduction to transition
metals
Transition metal: is an element which forms one or more stable ions with an
incomplete or partially filled d-subshell of electrons e.g. Fe3+:
1s22s22p63s23p63d5
Electronic configuration in terms of
increasing n value
Element
Electronic configuration
Sc
Ti
1s22s22p63s23p63d14s2 or [Ar] 3d14s2
1s22s22p63s23p63d24s2 or [Ar] 3d24s2
V
1s22s22p63s23p63d34s2 or [Ar] 3d34s2
Cr
1s22s22p63s23p63d54s1 or [Ar] 3d54s1
Mn
Fe
1s22s22p63s23p63d54s2 or [Ar] 3d54s2
1s22s22p63s23p63d64s2 or [Ar] 3d64s2
Co
1s22s22p63s23p63d74s2 or [Ar] 3d74s2
Ni
1s22s22p63s23p63d84s2 or [Ar] 3d84s2
Cu
Zn
1s22s22p63s23p63d104s1 or [Ar] 3d104s1
1s22s22p63s23p63d104s2 or [Ar] 3d104s2
NB: Zn is not a trans-metal
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Ion
Electronic configuration
V2+
1s22s22p63s23p63d3
V3+
Cr2+
1s22s22p63s23p63d2
1s22s22p63s23p63d4
Cr3+
1s22s22p63s23p63d3
Mn2+
1s22s22p63s23p63d5
Mn3+
Fe2+
1s22s22p63s23p63d4
1s22s22p63s23p63d6
Fe3+
1s22s22p63s23p63d5
Co2+
1s22s22p63s23p63d7
Co3+
Ni2+
1s22s22p63s23p63d6
1s22s22p63s23p63d8
Physical properties
Copy table 16.1 Briggs page 333
Trend in m.p: transition elements are all metals. They have high m.p and b.p
because they have very strong metallic bonds because:
i)
ii)
iii)
They have very small atomic sizes
Have a large number of delocalised electrons in both the 3d and 4s
subshells
Due to this transition metals have higher m.p and b.p than nontransition metals like Mg and Na
The smaller the atom the greater the level of close packing. Compared with Ca,
transition metals have higher m.p because of small atomic sizes and a greater
number of delocalised electrons hence stronger metallic bonds.
Trend in atomic radius: transition metals show little variation in atomic radii
i.e. atomic radius is almost invariant.
As the proton number increases across the period Na to Cl, electrons are
added to the outer shell and as the nuclear charge increases i.e. the force of
attraction increases thereby attracting outermost electrons closer to the
nucleus resulting in decrease in atomic radius. But in the case of transition
metals from Sc to Zn the nuclear charge/force of attraction increases, number
of electrons also increases but the added electrons enter the inner 3d subshell
hence the electrons shield the outer electrons from the nuclear charge thus
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cancelling out the increasing force of attraction of the nuclear charge, that’s
why atomic radii remains almost constant across the period from V to Cu.
-Comparison with Cu:
K
0.2
0.18
0.16
0.14
0.12
0.10
0.08
Ca
Na Si
Zn
Mg Ti V Cr Mn Fe Co Ni Cu
Al Si
P Si Cl
Atomic number
Variation in ionic radii
The variation in atomic radius is the same as the variation in ionic radius and is
almost constant. The increase in nuclear charge is nullified by increase in
shielding effect.
Comparison with Ca2+
The Ca2+ ion is longer than its corresponding dipositive transition metals
because transition metals have a larger nuclear charge.
2000
1500
1000
20
Sc Ti V
Cr Mn Fe Co Ni Cu
Zn
21
24
30
22
23
25
26
27
28
29
Proton number
First ionisation energy is almost invariant because:
a) The atomic size remains almost constant from Ti to cu because the
added electrons enter the 3d subshell thereby increasing the shielding
effect which cancels out the increase in nuclear charge.
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b) The attractive force of the outer electrons remains almost constant.
Compared with Ca, ∆Hi of transition metals is greater than that of Ca
because transition metals have smaller atomic sizes and greater nuclear
charge.
Density
Transition metals have greater densities compared to Ca because transition
metals have smaller atomic sizes hence greater level of close packing. The
atomic size is smaller because of a greater nuclear charge.
Electrical conductivity
Transition metals are good conductors of electricity but poorer conductors
than Ca. they are generally good conductors of electricity because of increased
number of delocalised electrons. Ca has a greater conductivity because it has a
larger atomic size and smaller nuclear charge hence greater extent of
delocalisation.
Variable oxidation state
Trans metals have variable oxidation states in their compounds e.g. for Fe +2
and +3. This is so because of a small energy difference between the 4s and 3d
subshells such that not only 4s electrons participate in bonding but also the 3d
electrons take part hence there is an increase in the number of bonding
electrons causing an increase in oxidation state. But, for non-trans elements
like Ca there is only one oxidation state i.e. +2 because only the 4s electrons
participate in bonding.
Oxidation states: copy notes Briggs page 339
Redox systems
Common oxidising agents are H2O2/H+, MnO4-/H+, Cr2O72-/H+ and Fe3+ and even
atmospheric oxygen.
MnO4- + 8H+ + 5e- ⇌ Mn2+ + 4H2O
EØ = +1.52
Cr2O72- +14H+ + 6e- ⇌ 2Cr3+ + 7H2O
EØ = +1.33
H2O2 + 2H+ + 2e- ⇌ 2H2O
EØ = +1.77
Fe3+ + e- ⇌ Fe2+
O2 + 2H+ + 2e- ⇌ H2O
EØ = +0.77
EØ = +1.23
Solution of Fe2+ ions can be oxidised by acidified solutions of MnO4-, Cr2O72-,
H2O2 and even atmospheric oxygen i.e.
MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
ORGANIC AND
EØ = +0.75
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Observation: pale/joint pink colour of Mn2+
With atmospheric oxygen
O2 + 2H+ + 2Fe2+ → H2O + 2Fe3+
EØ = +0.46
Observation: reddish-brown colour of Fe3+
With Cr2O72-/H+ and H2O2 /H+ (write notes and equations)
Solutions of Fe3+ ions are readily reduced by many reducing agents e.g. I- ions
Fe3+ + e- ⇌ Fe2+
EØ = +0.77
1
2
2
EØ = +0.54
I + e- ⇌ I-
Total equation: Fe3+ + I- ⇌ Fe2+ + 12I2
EØ = +0.23V
Observation: reddish-brown colour of iodine.
Transition element complexes
Transition metals form complex ions due to the presence of partially filled dorbitals. The d-subshell electrons pair up creating empty d-orbitals in addition
to 4s and 4p orbitals which are then used for dative bonding with ligands
forming complex ions. A transition complex ion is composed of a central
transition metal ion datively bonded to ligands. A ligand is a molecule or anion
with lone pairs of electrons used in dative bonding with transition metal ions.
The number of dative bonds formed by ligands with transition metal in a
complex ion is called the coordination number. The coordination number is
always 4 in C (i.e. tetrahedral, square planar or octahedral). The coordination
sphere is composed of the central transition metal ion surrounded by its
ligands e.g.
[FeCl4]2- tetrachloro-iron (II) complex ion:
Cl
Cl
Fe
Cl
2Cl
[Fe(CN)6]4- hexacyno-iron (II) complex
CN
CN
4CN
Fe
CN
CN
CN
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[Cu(H2O)6]2+- hexa-aqua-copper (II) complexion ion:
H2O
H2O
2H2O
Cu
H2O
H2O
H2O
Examples of ligands
Monodentate ligands: CN-, H2O, NH3, CO, CO2, X-, OH- etc
Polydentate ligands: EDTA, C2O42-, H2N-CH2-CH2-NH2.
Ligand-ligand exchange
Cu2+ ions in aq solution are each surrounded by 6 water molecules (ligands) as
given above but addition of NH3 aq displaces 4 water ligands forming
[Cu(H2O)2(NH3)4]2+. Water ligands are displaced by ammonia ligands because
NH3 forms stronger dative bonds hence a more stable complex ion i.e.
[Cu(H2O)6]2+ + 6H2O ⇌ [Cu(H2O)2(NH3)4]2+ + 4H2O
Blue
dark blue
The above equilibrium is displaced to the right because [Cu(H2O)2(NH3)4]2+ is
more stable than [Cu(H2O)6]2+. The two Kstab shows that the Kstab of the
following reaction is greater than that of the reverse reaction.
Note: the stability constant is a measure of the stability of the complex ion.
The larger the Kstab the more stable the complex ion is. Examples of other
transition metal ions Briggs page 340 table 16.5. The MnO4- ion is formed in
the following way:
[Mn(H2O)6]7+ → MnO4- + 8H+ + 2H2O
Reference Briggs page 340 table 16.5 and 343 for more examples
The small Mn7+ has a large charge density hence it polarises the water
molecules until it produces MnO4- which is stable.
In the same way CrO42- is formed:
[Cr(H2O)6]6+ → CrO42- + 8H+ + 2H2O
Haemoglobin
Ligand exchange also occurs in haemoglobin between H2O and O2.
Haemoglobin is made up of haem units attached to a polypeptide chain.
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H2O
N
O2
N haemoglobin
Fe2+
N
N
Fe
oxyhaemoglobin
N
N
N
N
N
N
Polypeptide chain
polypeptide chain
Haemoglobin molecules consist of an iron atom and has an oxidation state of
+2, is attached to 5 N atoms and one O atom in the water molecule. It has a
coordination number of 6. The haemoglobin absorbs oxygen from the lungs by
forming oxyhaemoglobin, the oxygen molecule displaces the water molecule
and forms a dative bond with the iron (II) ion. The oxyhaemoglobin carries
oxygen through the blood i.e.
Oxygen + haemoglobin ⇌ oxyhaemoglobin
At high altitude, c(CO2) in atmosphere is lower than at sea level (due to
decrease in atmospheric pressure. By Le Chateliers principle, the
c(oxyhaemoglobin) is decreased resulting in less oxygen being transported in
the blood for respiration. That’s why people living at high altitudes become
acclimatised by increasing the c(oxyhaemoglobin) in the blood whilst those not
acclimatised face respiratory problems.
Poisonous nature of CO and CN-: Fe2+ can form dative bonds with CO and CNions. These bonds are irreversible because they are stronger than F-O2 dative
bonds. This reduces the oxygen carrying ability of haemoglobin causing death.
This is a ligand exchange process.
Formation and colour of complexes
The d subshell contains 5d orbitals i.e. dxy, dyz, dxz, dx2, dx2-y2. The dxy, dyz and dxz
orbitals are oriented between the corresponding axis e.g. in the dxy the orbitals
are oriented between the x and y axis. The dx2-y2 and dz2 are oriented along the
axis.
𝑧
𝑧
𝑥
dxy
𝑦
ORGANIC AND
𝑥
dxz
𝑦
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In the gaseous state all the five d-orbitals are
degenerate i.e. they possess the same energy.
dx2-y2
During the formation of octahedral complexes
the ligands approach the d-subshell in the
direction of the axis, as a result the electrons in
2 2
2
the dz -y and the dz orbitals are under a greater repulsive force than the
electrons in the dxy, dyz, and dx2 orbitals.
Due to this, the d-subshell splits into two
parts having different energies i.e. d-orbitals
become degenerate i.e. dz2-y2 and dz2 orbitals
dz2
will be more energetic than the dxy, dyz,
and dxz orbitals. The difference in energy
between the splitted d-orbitals corresponds
to a given energy found in the region of visible
dyz
light in the electromagnetic spectrum i.e.
∆H = hf = h 𝜆𝑐 where
HE= energy difference
h = planks constant
f = frequency of light
c = velocity of light
𝜆 = wavelength of light
When white light comes in contact with transition
metal ions, electrons in the lower energy 3d subshell absorb photons which
have exactly the same energy as the difference in energy between splitted
subshells and get promoted to the higher d-subshell (d-d electron transition
occurs). The energy absorbed by the electron in the lower d-orbitals
corresponds to a colour wavelength or frequency of visible light.
The colour of the solution will be the difference between the white light and
the colour absorbed by the metal ion. It is called the complementary colour of
the solution.
In tetrahedral complexes, the ligands approach the d-subshell in the direction
between the axis and as a result the electrons in the dxy, dyz and dxz are under
greater repulsive forces than the electrons in the dx2-y2 and dz2.
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For examples, [Cu(H2O)6] is blue. When light passes through the solution, the
lower energy 3 electrons absorb red light and move up into the higher 3d
subshell. The solution appears blue as this white light-red light.
A few transition metal complex ions are colourless. Usually formed in
complexes where the 3d subshell is:
a) Completely full e.g. in Cu+ i.e. no space for electron transit
b) Completely empty of electrons e.g. Ti4+ i.e. no d electrons to absorb
light.
The colour of complex ions depends on the energy gap between the split 3d
orbitals. This energy gap depends on:
a) The oxidation state of the transition metal e.g. Cu+ colourless, Cu2+ blue
b) The ligands in the complex ion i.e. if the ligands are exchanged in a
complex ion the colour also changes e.g. in Cu2+ when the water is the
ligand in [Cu(H2O)6]2+ it is light blue but when the water ligands are
replaced by Cl- the colour changes to yellow.
Organic
Chemistry
Isomerism: is the existence of different compounds with the same molecular
formula but different structural formulae.
Structural formula: shows the sequence in which the atoms in a molecule are
bonded. A structural formula gives the minimum detail using conventional
groups e.g.
CH3CH2CH2OH propan-1-ol
CH3CHCH3 or CH3CH(CH3)2
CH3
2-methyl propane
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Displayed formula- shows the relative placing of atoms and the number of
bonds between them.
Structural isomerism
There are 3 types:
1. Chain (skeletal/nuclear) isomerism
2. Positional isomerism
3. Functional group isomerism
Chain isomerism
The monomers have the same molecular formula but different carbon chains.
They also possess the same functional group and belong to the same
homologous series. It is mostly found in alkanes e.g. isomers of butane C4H10:
CH3CH2CH2CH3
straight chain isomer
CH3CHCH3
branched chain isomer
CH3
Chain isomers of pentane: C5H10
1. CH3CH2CH2CH2CH3: N-pentane straight chain isomer
2. CH3CH(CH3)CH2CH3: 2-methylbutane branched chain isomer
CH3
3.
CH3
3HC-C-CH3
2,2-dimethyl propane branched chain isomer
CH3
Chain isomers can also be show in other homologous series e.g. in alkenes e.g.
C6H12 hexene
1. CH3CH=CHCH2CH2CH3
hex-2-ene/2-hexene
2. CH3CH=CCH2CH3
3-methyl-3-pentene
CH3
CH3 CH3
3. CH3CH=C-CH3
2,3 dimethyl but-2-ene
For example in aldehydes e.g. butanal C4H8O
CH3CH2CH2CHO N-butanal
CH3CHCHO
2-methylpropanal branched chain isomer
CH3
Positional isomers
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Isomers have a substituent group in different positions in the same C skeleton.
The isomers are chemically similar because they have the same functional
group.
In alkenes e.g. hexene C6H12
CH3CHCH2CH2CH3 2-methyl pentane
CH3
positional isomers
CH3CH2CHCH2CH3
3-methyl pentane
CH3
The methyl group changes position whilst the C-chain is maintained. In
alkenes, the C=C double bond changes position in the same (straight) C chain
e.g. butene, C4H8: two positional isomers
CH2=CHCH2CH3 – but-1-ene or 1-butene
CH3CH=CHCH3 – but-2-ene or 2-butene
Note: butene in the form of but-2-ene also shows another type of isomerism
called Cis-trans (geometrical) isomerism i.e.
H
H
H
C=C
CH3
CH3
C=C
CH3
H3C
H
In alcohols whereby the hydroxyl group (functional group) changes position in
the same carbon skeleton e.g. propanol C3H8O, 2 positional isomers are
produced i.e.
CH3CH2CH2OH
propan-1-ol
CH3CH(OH)CH3
propan-2-ol
Benzene derivatives
Cl
Cl
Cl
Cl
1,4-dichloro benzene
Cl
1,2-dichloro
benzene
Cl
1,3-dichloro
benzene
NO2
O-H
NO2
O-H
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-O-H
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Positional isomerism can be exhibited in ketones (by changing the position of
the ketone functional group) in the same carbon chain e.g. pentanone C5H10O
CH3COCH2CH2CH3 pentan-2-one
CH3CH2COCH2CH3 pentan-3-one
In amines (by changing position of the amine group)
Functional group isomerism
A group of substances having the same molecular formula but different
functional groups and belong to different homologous series. Isomers have
therefore different functional groups and belong to different homologous
series e.g.
 Between alcohols and esters
e.g. C2H6O
CH3CH2OH ethanol (functional group- hydroxyl)
CH3-O-CH3 dimethyl ether
 Between aldehydes and ketones
Functional group in aldehydes is the aldehyde group and functional group in
ketones is the ketone group.
C3H6O- two functional group isomers
CH3CH2CHO propanal
CH3COCH3 propanone
 Between carboxylic acids (carbonyl group functional group) and esters
(ester functional group)
e.g. C3H6O2
CH3CH2COOH
CH3CO2CH3
propanoic acid
methylethanoate
Other examples include: butanoic acid (CH3CH2CH2COOH), methylpropanoic
acid (CH3CH2CO2CH3), ethyl ethanoate (CH3CO2CH2CH3) and methylethyl
methanoate (HCO2CH(CH3)2.
Stereo isomerism: whereby 2 or more compounds have the same
functional group but different in arrangement of the groups.
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Cis-trans (geometrical) isomerism
Stereo isomerism has two types:
a) Cis-trans isomerism
b) Optical isomerism
Cis-trans isomerism (also called geometrical isomerism) occurs in compounds
with C to C double bonds i.e. in alkenes. Cis-trans isomerism is caused by
restriction in rotation about the C to C double bond due to the presence of the
𝜋 bond. Rotation about the double bond is not possible (as this would prevent
the overlap of the p-orbitals forming a 𝜋 bond i.e. it would break the 𝜋 bond.
So Cis-trans isomers are distinct compounds and not easily interconvertible.
There are 3 cases:
Case 1: two similar groups each attached to one of the unsaturated C atoms
e.g. but-2-ene
H3C
CH3
C=C
(b.p 3.7℃)
H
H
H3C
H
C=C
H
Cis isomer
CH3
Trans isomer
2,3-dichloro but-2-ene
Cl
Cl
Cl
C=C
H3C
(b.p 0.8℃)
CH3
C=C
CH3
H3C
Cl
*m.p/b.p (Cis 2,3-dichloro but-2-ene > the trans isomer because the Cis isomer
has a resultant dipole hence stronger intermolecular forces).
Butene-dioc acid HOOCCH=CHCOOH
HOOC
COOH
HOOC
C=C
H
H
C=C
H
H
Cis-butenedioc
COOH
trans-butenedioc
The physical and chemical properties of cis-trans isomers differ e.g. butenedioc
acid (m.p of cis isomer is 135℃ and that of the trans isomer is 287℃). In
solubility the cis-isomer is 100 times more soluble than the trans isomer due to
the same reason. In terms of chemical behaviour the cis isomer forms an acid
anhydride on gentle heating but the trans isomer does not. On strong heating
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trans butenedioc acid forms the anhydride of cis-butenedioc acid showing that
rotation about the double bond is possible at higher temperatures.
Case 2: one similar group attached to each of the unsaturated C atoms e.g. 2pentene/pent-2-ene i.e. CH3CH=CHCH2CH3
H
H
H
CH2CH3
C=C
H3C
C=C
CH2CH3
Cis-isomer
H3C
H
trans-isomer
Case 3: the two unsaturated C atoms are bonded to four different groups
e.g. CH3CCl=CBrCH2CH3: 3-bromo-2-chloro pent-2-ene
3HC
CH2CH3
3HC
C=C
Br
C=C
Cl
Br
Cis-isomer
Cl
CH2CH3
trans-isomer
(CH3)2C=CHCH3 does not show cis-trans isomerism because of the presence of
similar groups attached to the same unsaturated C atoms.
Optical isomerism
Optical isomers are two compounds with the same structural formula but one
isomer is the mirror image of the other.
Optical isomers occur when four different groups of atoms are attached to a
Carbon atom (saturated by four covalent bonds). It occurs in substances
possessing a tetrahedral bonding around one of the C atoms. Such a C atom is
said to be asymmetric and is the chiral centre/chiral C atom. Substances that
exist as two optical isomers are said to be optically active and possess a chiral
centre i.e. an asymmetric C atom.
Chiral centres are labelled with an asterisk. Such isomers are mirror images
and non-superimposable images. Optical activity is the ability to rotate the
plane of polarised light and it is measured by a polarimeter. Chiral compounds
have two different types of molecules called enantiomers which are mirror
images of each other. The mirror images form an equimolar/50:50 mixture of
(+) and (-) enantiomers which are optically inactive and is called a ralemic
mixture or ralemate.
Note: equal comp makes the mixture optically inactive. Examples of optically
active substances are: Alanine H2N-CH(CH3)COOH i.e. 2-aminopropanoic acid
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CH3
C
H
CH3
C
COOH
HOOC
NH2
H
NH2
Lactic acid i.e. 2-hydroxypropanoic acid
CH3
C
HO
Mirror plane
H
COOH
CH3
C
H
OH
COOH
There are 3 forms of lactic acid i.e.
1. The racemic mixture which is found in sour milk
2. The (+) enantiomer which is present in muscle
3. The (-) enantiomer does not occur naturally and has to be obtained from
the racemic mixture
All three forms have the same chemical properties and the physical properties
of (-) and (+). The enantiomers are the same.
Identifying chiral centres
A chiral centre is an asymmetric C atom i.e. a C atom bonded to four different
groups e.g. CH3C H(OH)C HBrC (CH3)CH2OH: three chiral centres
Cl O-H
1-chloro-2-hydroxy cyclo-hexane 2-chiral centres
CH(OH)CBr2 CH(CH3)CH2OH (2-chiral centres)
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3HC
CH3
CH3
Short hand
Structure of lomosterol-it has
CH3
6 Chiral centres (encircled)
CH3
HO
3HC
CH3
Qn: does lomosterol show cis-trans isomerism? Give a reason.
Ans: it does not show cis-trans isomerism because it has a similar group (-CH3)
attached to the same unsaturated Carbon atom.
Alkanes
Straight/branched chain alkanes have the general molecular formula CnH2n+2.
Nomenclature rules: Choose the longest C chain, use the smallest locant to
define side chain groups, use alphabetical order to define position.
H3C
CH3
CHCH3
H3C-C-CH3 2,2-dimethyl propane
H3C
CH3
Cyclopropne Cyclobutane Cyclopentane
C3H6
CH3
C4H8
C5H10
CH2CH3
Cyclohexane
C6H12
CH3
(C8H10)
1-ethyl-3-methyl cyclohexane
1,2-dimethylcyclohexane
CH3
(C9H12)
Physical properties
For the hydrocarbons with general molecular formula CnH2n+2 the m.p and b.p
increases as the number of C atoms increases due to increase in strength of
VDW forces as molecular size increases.
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200b.p/℃
C7 C8
C9
C10
C6
100-
C5
C4
0-
C3
-100-
C2
C1
50
100
150
Solubility
Alkanes are insoluble in water because C-H bond is non-polar hence no
interaction occurs with the polar water molecules. The H bonds within the
water molecules are stronger than any interaction which occur between water
molecules and the non-polar alkane molecules. Dissolution is therefore
favoured by energy considerations.
General Unreactivity of alkanes
Alkanes are generally unreactive even polar molecules such as water and aq
NaOH because:
1. All the electrons and orbitals of carbon are used in bonding in alkanes
thus there are no empty d-orbitals available to form dative bonds with
molecules and ions like water and OH- which have lone pair of electrons
2. The empty d-orbitals are high lying (in the upper 3rd subshell) hence no
accessible for dative covalency with ligands.
3. The C-H bond is almost non-polar because of a very small
electronegativity difference between C and H. the slightly negative C
atom in C-H bond repels the negative oxygen in water and OH-
Combustion
Complete combustion of alkanes in excess oxygen produces carbon dioxide
and water e.g.
C2H6 + 72O2 → 2CO2 + 3H2O
Incomplete combustion gives CO, C soot and H2O e.g.
C3H8 + 72O2 → 3CO + 4H2O
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Note: CO is poisonous because it combines with haemoglobin denying the
uptake of oxygen causing suffocation and death. C soot causes extensive
defoliation causing destruction of vegetation and crops. CO2 causes global
warming, an excessive rise in environmental temperature causing poor
agricultural yields.
Reaction with halogens
Alkanes react with Cl2 in the presence of sunlight/UV light. The reaction is
called radical substitution. Hydrogen atoms are substituted by Cl atoms. A
chain reaction occurs. The stages are initiation, propagation and termination
e.g.
CH4 + Cl2 u.v light CH3Cl, CH2Cl2, CHCl3, CCl4
Homolytic fission: is the breaking up of a covalent bond in a molecule so that
each of the combining atoms receives one electron forming free radicals.
Free radical: is an atom or groups of atoms with unpaired electrons
Heterolytic fission: the breaking up of a covalent bond in which one of the
atoms takes both of the bonding electrons to form an anion. This produces
positive and negative ions.
HCl → H+ + Cl-
Mechanism: free radical substitution
Chain initiation
Reaction is initiated by Cl• free radicals or even methyl free radicals. It involves
the homolytic fission of the Cl-Cl bond in the presence of sunlight to produce
Cl• free radicals. This process does not take place in darkness.
Cl2 u.v light or heat above 300℃ 2Cl•
Propagation
In the second stage, the following two reactions are possible:
Cl• + CH4 → CH3Cl + H•
Cl• + CH4 → CH3• + HCl
The second is more likely because the formation of HCl is more exothermic
than the formation of C-Cl bond i.e.
CH4 + Cl•→ CH3• + HCl
CH3• + Cl2 → CH3Cl + Cl•
CH3Cl + Cl•→ •CH2Cl + HCl
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•CH2Cl + Cl2 → CH2Cl2 + Cl•
CH2Cl2 + Cl•→ •CHCl2 + HCl
•CHCl2 + Cl2 → CHCl3 + Cl•
CHCl3 + Cl•→ •CCl3 + HCl
Chain termination (product formation)
•CH3 + Cl• → CH3Cl chloromethane
•CH2Cl + Cl• → CH2Cl2 dichloromethane
•CHCl2 + •Cl → CHCl3 trichloromethane
•CCl3 + •Cl → CCl4 tetrachloromethane
Organic solvents
Cl• + Cl• → Cl2 radicals reacting
•CH3 + •CH3 → CH3CH3
•CH3 + •CH2Cl → CH3CH2Cl
•CH2Cl + •CH2Cl → CH2ClCH2Cl
•CCl3 + •CHCl2 → CHCl2CCl3
Chlorinated ethanes
Note: if a trace of tetramethyl lead Pb(CH3)4 is added (catalyst) the reaction can
take place in darkness or at room temperature. Pb(CH3)4 also dissociates to
produce •CH3 free radicals.
Alkenes
Nomenclature
CH2=CH2 ethene
CH2=CHCH2CH3 butene
CH3CH=CHCH3 but-2-ene
CH2=CHCH=CH2 but-1,3-diene
Cyclopropene
C3H3
cyclobutene
C4H6
cyclopentene cyclohexene cyclohex-1,3C5H8
C6H10
diene C6H8
Functional group in alkenes is the C to C double bond because it contains the π
electrons. Electrophiles i.e. electron loving species will attack the C to C double
bond because it is a region of high electron density.
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Reaction of alkenes
Reaction with water
Takes place in the presence of Ni (cheaper)/palladium (expensive) catalyst at
100℃ and 4atm. The reaction is called an addition reaction or hydrogenation
i.e.
CnH2n + H2 Ni/Pd cat 100/150℃. 4atm CnH2n+2
The reaction is called an addition reaction or hydrogenation.
CH=CH Ni powder, 150℃. 4atm CH3CH3
CH2=CH2 Ni cat 150℃ CH3CH3
Catalytic hydrogenation using Ni powder is important in the food industry
because plant oil such as sun flower seed oil and peanut oil are ‘poly
saturated’. They are esters of carboxylic acids which contain more than one C
to C double bonds. Animal fats are esters of saturated carboxylic acids.
Hydrogenation is used to convert the unsaturated edible oils into edible fats
e.g. margarine.
Reactions with steam
Alkenes react with steam in the presence of phosphoric (V) acid/SiO2 catalyst
at 300℃ to produce alcohols. Type of reaction is called electrophilic addition.
Observation: alcohol smell
CH2=CH2 + H2O (g) H3PO4/SiO2 cat 300℃
CH3CH2OH
The process can produce positional isomers e.g.
CH2(OH)CH2CH2CH3
Butan-1-ol
CH2=CHCH2CH3 + H2O (g) ) H3PO4/SiO2 cat 300℃
CH3CH(OH)CHCH3
Butan-2-ol
+ H2O (g) )
H3PO4/SiO2 cat 300℃
OH
Cyclohexanol
It can also produce diols from dienes.
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CH2(OH)CH2CH2CH2OH
CH2=CH-CH=CH2 + 2H2O H3PO4/SiO2 cat 300℃
Butan-1,4-diol
CH3CH(OH)CH2CH2OH
Butan-1,3-diol
CH3CH(OH)CH(OH)CH3
Butan-2,3-diol
CH(OH)CH3
CH2CH2OH
CH=CH2
+ H2O
H3PO4/SiO2 cat 300℃
Reaction with hydrogen halides
Hydrogen halides: HCl, HBr and HI add across the C to C double bond to
produce halogenoalkanes. The type of reaction is called electrophilic addition.
CH3CH=CH2 + HBr
CH3CHBrCH3
2-bromopropane
CH3CH2CH2Br
1-bromopropane
Positional
isomers
Observation: white fumes of HCl and HBr disappear. The addition of HX to an
asymmetrical alkene gives two possible products e.g. propene
The major product is 2-bromopropane in accordance with Markonkov’s rule
which states: the addition of a hydrogen halide to an alkene (unsaturated
compound) is such that the H atom from the HX becomes attached to the
unsaturated C atom bearing a greater number of H atoms. The major product
(isomer) is determined by the stability of the carbocation (intermediate) i.e.
H
CH3CH2
C
Less stable H
H
CH3
+
+
+
+
produces CH3CH2CH2Cl
C
+ CH3
produces CH3CHBrCH3
+
H +
+
More stable because
of a greater number of electron releasing groups.
Alkyl groups are electron releasing. They tend to decrease the positive charge
on the carbocation and this stabilises it. The greater the number of electron
releasing alkyl groups the more stable the carbocation. The more stable the
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carbocation the lower is the energy of the transition state that precedes it. The
rate of reaction of alkenes increases in the order:
CH2=CH2 < RCH=CH2 < R2C=CH2 where R is an alkyl group
According to the carbocation theory, alkyl groups confer stability on the
carbocation in the following order of decrease in stability/decrease in
reactivity:
R2C+CH3 > RC+HCH3 > C+H2CH3
CH2ClCH2CH2CH3
CH2=CHCH2CH3 + HCl
positional isomers
CH3CHClCH2CH3
Note: 2-butene/but-2-ene has only one product, 2-bromobutane.
The rate of addition increases in the order HF < HCl < HBr < HI in order of
increasing acid strength.
Addition of halogens
Alkenes react with Cl2 gas or with solutions of Br2 and I2 in an organic solvent
(e.g. CCl4) to give dihalogenoalkanes. The reaction takes place in darkness at
room temperature i.e.
CH2=CH2 + X2 → CH2XCH2Br
Type of reaction is called electrophilic addition.
With Cl2 gas
CH2=CH2 + Cl2 (g)
r.t in darkness
CH2ClCH2Cl
1,2-dichloroethane
With Br2 liquid-test for C to C double bond
CH2=CH2 + Br2 (l)
CCl4 r.t in darkness
CH2BrCH2Br 1,2-dibromoethane
Observation: reddish-brown colour of Br2 (l) disappears i.e. decolourisation
With bromine water i.e. bromine aq
Bromine water is also a reagent used to test for the C to C double bond.
Observation: reddish-brown colour of Br2 disappears. Reaction also produces
1,2-dibromoethane (traces) together with 1-bromo-2-hydroxy-ethane.
CH2BrCH2OH + HBr (major product)
CH2=CH2(g) + Br2(aq) + H2O(l)
CH3CH=CH2 + Br2(aq) + H2O
CH2BrCH2Br
CH3CH(OH)CH2Br
positional isomers
CH3CHBrCH2OH
CH3CHBrCH2Br
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1-bromo-2-hydroxy-propane and 2-bromo-1-hydroxy-propane (the
predominant products) are positional isomers and both are optically active.
Mechanism of electrophilic addition
At any instant a bromine molecule is likely to have an instantaneous dipole
because the electrons are in constant motion and at any moment are unlikely
to be distributed exactly symmetrically. The partial positive end of this dipole is
attracted to the electron-rich π bond in the ethene molecule causing
polarisation of the π electron cloud of the double bond i.e.
H
H
Brƍ+-Brƍ-
C=C
H
H
The π electrons become more gradually attracted to the Brƍ+ and the electrons
of the Br-Br bond become more gradually polarised until the association of
ethene and Br2 is transformed into a positive cation called a bromomium ion
(carbocation) and a Br- ion i.e.
H
H
H
Brƍ+-Brƍ-
C=C
H
H
C
+ Br + Br-
H
C
H
H
The positive charge on the carbocation is stabilised by delocalisation of the
negative charge residing on the Br- ion. The positive bromomium carbocation is
attacked by the Br- to form the product i.e. 1,2-dibromoethane.
H
H
C
+ Br + Br-
CH2BrCH2Br
C
H
H
With bromine water i.e. Br2 (aq) the main product is CH2BrCH2OH. Polarisation of
the Br-Br covalent bond by the water molecule/π electron cloud of the double
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bond. The subsequent attraction between the Brƍ+ and the π electrons of the
double bond form the carbocation.
H H
H-C-C
H H
H
O
H
H H
H-C-C-OH + H+
H H
H+ + Br- → HBr
Ethene and bromine water containing sodium chloride give three products: 1bromoethanol, 1-bromo-2-chloroethane and 1,2-dibromoethane.
CH2=CH2 + Br2 + H2O + NaCl → HOCH2CH2Br (major product)
+ ClCH2CH2Br (smaller amount)
+ BrCH2CH2Br (a trace)
Reactions of alkenes with cold MnO4Cold dilute MnO4- is a weak oxidising agent hence it reacts with ethene
(alkenes) to form diols. Type of reaction is called oxidation.
Observation: the purple colour of MnO4- turns green or a brown suspension of
MnO2 i.e.
CH2=CH2 cold dil MnO4- reflux CH2OHCH2OH (ethan-1,2-diol)
OH
optically active
Cold dil MnO4-
OH
This reaction of alkenes with cold dilute MnO4- is a test for the presence of a C
to C double bond (i.e. unsaturation).
Reactions of alkenes with hot concentrated MnO4Hot concentrated MnO4- is a strong oxidising agent which oxidises alkenes
breaking up the C to C double bond (oxidative rupture) producing carboxylic
acids, ketones and CO2 as products. This reagent is used to determine the
position of the C to C double bond (alkene linkages) in larger molecules. Type
of reaction is oxidation.
Observation: the purple colour of MnO4- disappears i.e. decolourisation occurs.
Case 1: formation of carboxylic acids
Carboxylic acids are formed if the C to C double bond is internal and each of
the unsaturated C atoms is bonded to H atoms e.g.
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CH3CH=CHCH3 hot conc MnO4- 2CH3COOH
COOH
COOH hexan-1,0-dioc acid
Hot conc MnO4-
CH=CHCH3
COOH
+ CH3COOH
Hot conc MnO4-
Case 2: formation of ketones
Ketones are formed when the unsaturated C atoms are internal and each
bonded to alkyl or aryl groups e.g.
CH3
O
O
CH3-C=C-CH2CH3 hot conc MnO4- CH3-C-CH3 + CH3-C-CH2CH3
CH3
propanone butanone
H3C
CH3
C=C-CH2OH
COCH3
Hot conc MnO4-
CH3
+ O=C-COOH
Phenylethanone
Note: the hydroxyl group (alcohol group) is oxidised to a carboxylic group.
Case 1 and case 2 combined
CH3CH=C(CH3)2OH hot conc MnO4- CH3COOH + (CH3)2CO
Case 3
Note: if the C to C double bond is terminal i.e. one in which the last/first
unsaturated C atom is bonded to H atom, one of the products is CO2 together
with either a ketone or a carboxylic acid e.g.
CH2=CHCH2CH3 hot conc MnO4- CH3CH2COOH + CO2
CH=CH2
COOH
Hot conc MnO4-
+ CO2
Note: the evolution of CO2 shows that the C to C double bond is at the end and
this process is called decarboxylation i.e. a method of reducing the number of
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C atoms in a chain (opposite of the step-up process of using KCN/NaCN in little
ethanol).
Note: test for CO2- turns limewater milky.
Test for ketones (product of oxidative rupture)
1. Ketones give a negative test with both Fehling’s and Tollens reagent i.e.
they are not oxidised because of the absence of H atoms bonded to the
C atom of the carbonyl group.
2. Ketones (and aldehydes) are carbonyl compounds. They contain the
carbonyl group hence they react with 2,4-DNPH giving an orange/yellow
ppt.
H3C
H
H3C
C=O
H3C
N-N-H
H
C=N-N-H
NO2
H3C
NO2 + H2O
Orange/yellow ppt
H2O
NO2
NO2
H3C
3. All methyl ketones (ketones containing –C=O group) react with alkaline
iodine (i.e. NaOH + I2) to give a characteristic yellow ppt of
triidomethane commonly called iodoform CHI3.
CH3CH2COCH3 + 3I2 + 4NaOH → CH3CH2COO-Na+ + CHI3 + 3NaI + 3H2O
Identification of position of C to C double bond.
A compound X reacted with hot concentrated MnO4- and gave the following
products. Identify X.
CH3COOH
CH3COCH2COCH3 HOOC-CH2-COOH (CH3)2CO
CH3
CH3CH=
CH3
=C-CH2-C=
CH3
=CH-CH2-CH=
=C
CH3
Combining the above fragments gives X i.e.
CH3
CH3
CH3
X: CH3CH=C-CH2-C=CH-CH2-CH=C
CH3
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Qn: give the structural formula of the organic products when the following
compound is reacted with hot MnO4CH3CH=CCH2C(CH3)=CHCH2C(CH3)CHCH=CH2
Products: CH3COOH, O=C-CH2-C=O, HOOC-CH2-C=O, HOOC-COOH
CH3
CH3
Note: CO2 is also produced but it is not an organic product.
Crude oil
Crude oil is a mixture of 150 compounds. Crude oil is fractionally distilled to
obtain fractions of the volatile substances. Each fraction is a mixture of
hydrocarbons which boil over a limited range of temperature.
It contains alkanes, both branched and unbranched and even aromatic
compounds and also oxygen, nitrogen and sulphur containing compounds.
Cracking
Cracking or pyrolysis (splitting by heat) is the breakdown of larger hydrocarbon
molecules to obtain smaller ones (alkanes and alkenes) of lower Mr for use.
Cracking has therefore two useful results:
i)
ii)
Short chains are produced which are more useful as fuels for motor
cars
Some of the products are alkenes which are more reactive than
alkanes and therefore more useful to use as chemicals i.e. feedstock
for conversion to other compounds e.g. ethene to produce ethyl
ethanoate and polyethene.
Different cracking processes are used:
1. In steam cracking the gas oil is mixed with steam and passed through a
furnace at 1100K (827℃) for 0.2sec obtaining H2, CH4, C2H6, C3H8, C4H10,
butan-1,3-diene, petrol and fuel oil.
2. In catalytic cracking, a surface catalyst of Al2O3/SiO2 at 450℃ is used e.g.
Large alkane
𝑣𝑎𝑝𝑜𝑢𝑟 𝑝𝑎𝑠𝑠𝑒𝑑 𝑎𝑡 450℃
𝑜𝑣𝑒𝑟 𝐴𝑙2𝑂3 𝑜𝑟 𝑆𝑖𝑂2 𝑐𝑎𝑡
alkane with smaller molecules + alkene + H2
2CH3CH2CH3 (g) 450℃ Al2O3/SiO2 cat CH4 (g) + CH3CH=CH2 (g) + CH2=CH2 (g) + H2 (g)
E.g. C22H46 450℃ Al2O3 cat C10H22 + C12H24
C22H46 450℃ Al2O3/SiO2 cat C12H26 + 5C2H4
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Reforming
Straight chain alkanes are usually poor motor fuel. They cause knocking in the
engine by detonating rapidly rather than burning steadily. Branched chain
alkanes are much better fuels.
Reforming: is a process where straight chain alkanes are heated under
pressure with Pt catalyst. The chain break up and reform as branched chain
molecules e.g.
Heat, Pt catalyst
2,2,3-trimethylpentane
Reforming also involves converting straight chain alkanes into aromatic
compounds. If platinum is used as a catalyst it is called platforming.
Arenes
Arenes are hydrocarbons containing benzene ring and the aryl group i.e.
phenyl group C6H5/
CH3
CH3
CH3
Benzene
toluene
2,3-dimethyl benzene
naphthalene
1
*draw structure of benzene:
Position 1: position of reference
6
2
5
3
4
Position 2 & 6: ortho positioners
Position 3 & 5: para positioners
Position 4: para positioner
Benzene
Reactions with halogens
Reactions involve the attack of an electron on the cloud of π electrons.
Chlorine and bromine react with benzene at room temperature in the
presence of a Friedel-Crafts catalyst or hydrogen carrier in which the H atoms
in benzene are substituted by Br/Cl atoms. Type of reaction is called
electrophilic substitution. Aluminium and iron act as catalysts because they are
converted into their chlorides and bromides which act as halogen carriers.
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+ Cl2 (g) Al/AlCl3 at r.t
Cl
+ HCl (g)
Br
+ Br2 (l) Fe/FeBr3 at r.t
+ HBr (g)
Observation: white fumes of HCl/HBr.
Iodine cannot react with benzene.
Mechanism of electrophilic substitution using Br2
The bromine molecule approaches a benzene ring (due to the presence of an
instantaneous dipole in bromine). Delocalised 𝜋 electrons in the benzene ring
interact with the Br2 molecule and polarises the Br-Br bond.
Brƍ+−Brƍ-
+ Br-Br
Benzene is less reactive compared to alkenes hence the reaction can only
proceed in the presence of a Lewis acid (halogen carrier) which accepts a pair
of electrons from the partial negative end of the polarised Br2 molecule and
enables the Br-Br bond to split. A bromomium ion (intermediate) and FeBr4complex are formed.
H
Brƍ+----Brƍ- ----FeBr3
Br
+ FeBr4-
Br
+ HBr + FeBr3
Nitration of benzene
Benzene reacts with a mixture of conc HNO3 and conc H2SO4 at 60℃ to
produce nitrobenzene. The temperature is maintained at 60℃ for
mononitration. If HNO3 conc is used, nitration is slow.
NO2
Conc HNO3 + conc H2SO4 60℃
Nitrobenzene
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Type of reaction is called electrophilic substitution reaction.
H atoms are substituted by nitrile cations i.e. NO2+
Observation: pale yellow liquid i.e. nitrobenzene.
If the temperature is raised to 95℃ and fuming conc HNO3 acid is used 1,3dinitrobenzene is produced as well as its positional isomer 1,4-dinitrobenzene.
NO2
NO2
NO2
NO2
NO2
Conc HNO3 + conc H2SO4 95℃
(Major product)
NO2
NO2
Mechanism of nitration of benzene
HNO3 + 2H2SO4 ⇌ NO2+ + H3O+ + 2HSO4H
NO2+ slow
NO2
+ HSO4- fast
NO2
+ H2SO4
Intermediate carbocation
Toluene or methyl benzene
Reactions of toluene: summary
CH2Cl
CHCl2
&
&
CH3
CH3
Cl &
Free radical substitution
Electrophilic substitution
Cl
CH3
CH3
NO2
&
CHO
Electrophilic substitution
NO2
COOH
Further oxid
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Positional isomers
C
CH3
CCl3
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In toluene, the substituent –CH3 group has a positive inductive effect (i.e. + I)
because the CH3 and other alkyl groups are electron releasing.
There will be high electron density on the ortho (2) and para (4) positions
hence electrophiles will attack the three positions giving major product
isomers.
Classification of substituent groups: activating
and deactivating groups.
Activating groups: groups that release electrons activating the benzene ring;
releases electrons hence stabilises the carbocation.
Deactivating groups: withdraw electrons, deactivating the benzene ring i.e.
destabilises the carbocation.
Activating groups: ortho-para
Deactivating groups: meta directors
director
Strongly activating groups
Nitro group:- NO2
NH2 (-NHR, NR2)
N(CH3)3+ trimethyl ammonium
OHCN: nitrile group
Moderately activating
COOH: carboxyl group
-OCH3 (-OCH2CH3 etc)
SO3H
NHCOCH3
CHO: aldehyde, -COR: ketone group
Weakly activating
Deactivating: ortho-para
-C6H5: phenyl group
Halogen group: -F, -Cl, -Br, -I
Alkyl groups e.g. –CH2CH3
Note: when the directive effect on one group opposes that of the other, in
such cases, complicated products are often obtained. Even so, production is
made in accordance with the following generalisations: strongly activating
groups generally win over deactivating groups.
Sequence: NH2, OH- > OCH3, NHCOCH3 > C6H5, CH3 directors
E.g.
CH3
SO3H
NHCOCH3
CN
NO2
OH
OH
HNO3 conc + H2SO4
CH3
NO2
Sole product because –OH is a
stronger activating group than
the –CH3 group
CH3
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NHCOCH3
NHCOCH3
Chief product, amide group is
more activating than the –CH3
group
Br
Br2, FeBr3 cat
CH3
CH3
CHO
CHO
Chief product because strongly
activating groups win over
deactivating groups
Br2, FeBr3 cat
OH
OH
Reactions of methyl benzene
Toluene has two sets of reactions:
1. Substitution reaction of the benzene ring
2. Reaction of the CH3 group- side chain reactions
Toluene is more reactive than benzene because of the presence of electron
releasing –CH3 group which activates the benzene ring. The –CH3 group is
ortho-para directing i.e. substitution takes place at (2) and (4) positions giving
positional isomers of equal quantity.
Reactions of Toluene with halogens
Reactions involving the –CH3 group
Cl2 reacts with Toluene in 2 ways depending on the condition i.e. in the
presence of u.v light or through boiling. Substitution of the –CH3 side group
occurs, H atoms are replaced by Cl atoms. Type of reaction is free radical
substation.
Cl2 + CH3
CH2Cl
CHCl2
CCl3
u.v light
,
&
I.e. chlorine is bubbled through boiling toluene in u.v light. Substitution occurs
in the side chain. Reaction proceeds in the same manner as in alkanes.
Mechanism
Cl2 u.v light, boil 2Cl•
chain initiation
Chain propagation
CH3
+ Cl•
CH2•
+ HCl
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CH2•
CH2Cl
+ Cl2
+ Cl•
CH2Cl
•CHCl
+Cl•
+ HCl
•CHCl
CHCl2
+ Cl2
+ Cl•
CHCl2
•CCl2
+ Cl•
+ HCl
Chain termination
Cl• + Cl• → Cl2
•CH2
CH2Cl
+ Cl•
•CHCl
CHCl2
+ Cl•
•CCl2
CCl3
+Cl•
•CH2
•CH2
+
•CHCl
-CH2-CH2•CH2
-CHCl-CH2-
Oxidation of the CH3 group
Another reaction involving the –CH3 group i.e. side chain reaction is oxidation.
Powerful oxidising agents e.g. MnO4-/H+ or Cr2O72-/H+ will oxidise the –CH3
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group to a carboxylic acid group via the formation of an intermediate,
Benzaldehyde. The reaction mixture must be refluxed for several hours.
Observation: purple colour of MnO4- disappears/green solution of Cr3+ appears.
Milder oxidising agents like MnO2, CrO2Cl2 oxidises the –CH3 group to an
aldehyde group.
CH3
Further
oxidation
CHO
COOH
benzoic acid
MnO4-/H+ or Cr2O72-/H+
Intermediate
CH2CH3
COOH
MnO4-/H+ or Cr2O72-/H+ reflux
CH3
CHO
COOH
CHO
COOH
MnO4-/H+ or Cr2O72-/H+
CH3
Reactions involving the aromatic benzene ring:
reactions with halogens.
Toluene reacts with halogens in the presence of Friedel-Crafts catalyst or
halogen carriers i.e. Fe/FeCl3 or Al/AlCl3 catalyst at room temperature. The
reaction is called electrophilic substitution. The –CH3 is ortho-para directing
hence substitution occurs on the ortho (2) and para (4) positions giving equal
quantities of positional isomers.
Observation: white fumes of HBr/HCl.
CH3
CH3
+ Cl2
CH3
Cl
AlCl3 cat, r.t
positional isomers
Cl
CH3
CH3
+ Br2
CH3
Br
FeBr3 cat, r.t
Br
Mechanism: electrophilic substitution using Br2.
1. Polarisation of the Br2 molecule by the pi electrons of the benzene ring:
CH3
Brƍ+-BrƍORGANIC AND
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2. Acceptance of electrons by Friedel-Crafts catalyst from the partial
negative end of the polarised bromine molecule and the subsequent
interaction between the pi electrons of the benzene ring and the partial
positive end of the intermediate brominated halogen carrier forming the
intermediate:
CH3
CH3
Brƍ+----FeBr4
H
CH3
+ FeBr4-
+ FeBr3 + HBr
Br
Br
Intermediate
Nitration
Toluene/methyl benzene reacts with conc HNO3 + conc H2SO4 acid at 30℃ to
produce 2 positional isomers i.e. 2-nitrobenzene and 4-nitrobenzene.
Type of reaction is called electrophilic substitution reaction. If the temperature
is raised, 2 or 3 nitro groups are introduced forming 2,4-dinitrotoluene and
2,4,6-trinitrotoluene which is an important explosive that when detonated
produces large amounts of heat energy.
CH3
CH3
Conc H2SO4 + conc HNO3
2ON
NO2
Higher temperature
TNT
Mechanism of mononitration of toluene
NO2
2H2SO4 (conc) + HNO3 (conc) → NO2+ + H3O+ + 2HSO4The electrophile i.e. nitrile cation is attracted to the benzene by the delocalised
pi electrons. It displaces H atoms of the benzene ring.
CH3
CH3
NO2+
CH3
+ HSO4-
H
NO2
+ H2SO4
NO2
Note: benzene and other Arenes are stabilised by the delocalised pi electrons.
Halogen derivatives
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General structure RX where R is the alkyl group e.g. -CH2CH3 or aryl group
(phenyl group) aryl chlorides e.g. C6H5Cl chlorobenzene. Examples of
haloalkanes are CH3Cl, CH3CHClCH3,
-CH2Cl etc. Examples of aryl chlorides
are C6H5Cl/
-Cl. Functional group: polarise carbon-halogen bond i.e. *Cƍ+-Xƍis the functional group. C-X bond polarity is determined by the
electronegativity of the halogen.
The partial positive C atom is open to attack by nucleophiles e.g. OH-, CN-, H2O,
etc which attack it substituting the halogen hence the reaction is classified as
nucleophilic substitution.
Classification of halogenoalkanes
They are classified according to the number of C atoms attached to the C atom
bearing the halogen into:
Primary halogenoalkanes- halogenoalkanes in which the C atom bearing the
electronegative element is bonded to only one other C atom e.g. CH3CH2CH2Cl,
-CH2CH2Cl, etc.
Secondary halogenoalkanes- halogenoalkanes in which the C atom bearing the
electronegative element is bonded to two other C atoms e.g. CH3CHClCH3,
-Cl,
-CHClCH3, etc.
Tertiary halogenoalkanes- halogenoalkanes in which the C atom bearing the
electronegative element is bonded to three other C atoms e.g. (CH3)3CCl,
CH3
CH3
C CH2CH3
Cl
CH3
C
CH3
Br
2-bromo-2-methyl butane
2-chloro-2-phenyl propane
CH3
CH3
C
Cl
CH3
2-chloro-2-methyl propane
Reactivity of halogenoalkanes
The Cƍ+-Xƍ- bond is the functional group hence the reactivity is determined by
the C-X bond strength. The C-X bond strength is determined by the nature of
the halogen i.e. smaller and more electronegative halogens like F form a strong
C-F bond (maximum overlapping) which is shorter therefore stronger hence it
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is not easily broken/hydrolysed. The C-I bond is the weakest because I is large
and less electronegative hence bond is larger and weaker i.e. there is poor
overlapping thus the order of decreasing reactivity is BE (C-I) < BE (C-Br) < BE (
C-Cl) < BE (C-F) i.e.
C-I
>
C-Br >
C-Cl
The ease of reaction also depends on the stability of the leaving group thus the
two factors are opposing. The C-X bond strength is also dependant on the
electron releasing alkyl groups i.e. the greater the number of electron releasing
alkyl groups the weaker the C-X bond and the more reactive the haloalkane
(where the halogen is the same) e.g. comparing primary, secondary and
tertiary haloalkanes, the order of decreasing reactivity/ease of hydrolysis is:
Tertiary haloalkane > secondary haloalkane > primary haloalkane
CH3
CH3-C-Cl
>
CH3CHClCH3
>
CH3CH2Cl
CH3
Hydrolysis/nucleophilic substitution
It occurs when the haloalkane is boiled with NaOH aq. The X atom/group is
replaced by the nucleophilic OH- ions to give alcohols hence the reaction is
called hydrolysis or nucleophilic substitution.
Note: substitution reactions are favoured by weakly basic nucleophiles e.g. CNwhilst elimination (of HX forming alkenes) reactions are favoured by strong
bases i.e. hot ethanolic KOH/NaOH giving alcohols:
CH3CH2Br + OH- boiling/reflux CH3CH2OH + BrCH2Cl
CH2OH
+ OH- boiling/reflux
CH2CH2Cl
+ ClCH2CH2OH
+ OH-
boiling, reflux
+ Cl-
The relative extends to which substitution (nucleophilic) and elimination occur
depends on the structure of the haloalkane and the basic strength of the
nucleophile. When tertiary RX reacts with a base the main product is an alkene
i.e. (CH3)3CCl + OH- → (CH3)2CH=CH2 + HCl
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Tertiary RX are very reactive hence no base is mostly required for substitution
to occur. For a given aqueous base, the reactions which take place are:
Primary RX- substitution
Secondary RX- substitution/elimination
Tertiary RX- elimination
Mechanism of hydrolysis/nucleophilic substitution
Reaction: CH3CH2Br + OH- → CH3CH2OH + Br-
H+
H3C
Cƍ+
H
Br-
OH-
slow
CH3
HO------C--------Br
H
H
rds
OH- approaches from the opposite side of X so as to minimise intermediate
repulsion.
CH3
HO---C---Br fast
H
CH3CH2OH + Br-
H
Unstable intermediate
The mechanism is SN2- nucleophilic substitution where SN- nucleophilic
substitution and 2 is the order.
Energy profile diagram
Energy/kJmol-1
Unstable intermediate
Reactants
Products
Progress of reaction
Mechanism of hydrolysis of tertiary halogenoalkanes
Reaction:
CH3
H3C-C-Cl + OH- → H3C-C-OH + ClCH3
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Stage 1:
CH3
CH3
H3C Cƍ+ Brƍslow
H3C-C +
+ BrCH3 intermediate stabilised by
CH3
Presence of many electron releasing –CH3 groups
Stage 2: CH3
H3C C+
CH3
CH3
H3C-C-OH
CH3
OH-
Mechanism is called SN1 because it is monomolecular.
Energy/kJmol-1
Energy profile diagram
Unstable intermediate
Reactants
Products
Progress of reaction
Elimination reactions
CH2=CHCH2CH3
CH3CHClCH2CH3 hot ethanolic KOH reflux
Positional isomers
Halogenoalkanes react with strong bases i.e. hot ethanolic KOH/NaOH to form
alkenes in which HX is eliminated in the process.
Observation: white fumes of HCl/HBr.
CH3CH=CHCH3
CH3CHBrCH3 hot ethanolic KOH reflux CH3CH=CH2 + HBr
Note: but-2-ene exhibits cis-trans isomerism thus the products of elimination
give 3 isomers.
CH2CH2Cl
CH2=CH2
(phenylethene/styrene)
hot ethanolic KOH reflux
+ HCl
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Br
+ HBr
Hot ethanolic KOH reflux
A dihalogenoalkanes produces a diene e.g.
CH3CHClCHClCH3 hot ethanolic KOH CH2=CH-CH=CH2 + 2HCl
Note: the elimination reaction is the reverse of electrophilic addition of HX to
alkenes e.g. CH2=CH2 + HCl → CH3CH2Cl.
Chloroethane/vinyl chloride a monomer in the manufacture of PVC is produced
by elimination i.e.
CH2ClCH2Cl hot ethanolic KOH CH2=CHCl + HCl
Formation of nitriles
Halogenoalkanes react with KCN/NaCN in little ethanol to produce nitriles.
Type of reaction is called nucleophilic substitution i.e. CN- are nucleophiles
which substitute the halogen.
CH3CHClCH3 KCN/NaCN in little ethanol CH3CH(CN)CH3 + HCl
CH3CH2Cl KCN/NaCN in little ethanol CH3CH2CN + NaOH
CH2Cl
CH2CN
KCN/NaCN in little ethanol
The reaction of RX with KCN/NaCN is used as a step up process of increasing
the number of C atoms in a chain (the reverse of decarboxylation).
Note: the nitrile produced can be converted into a carboxylic acid by boiling
with dilute H2SO4/HCl. The reaction is called acid hydrolysis e.g.
CH3CH2CN + H2O H2SO4/HCl, boil CH3CH2COOH + NH4+ (NH3)
CH2Cl
CH2CN
CH3COOH
+ NH4+
+ H2O
KCN in little ethanol
Chloromethyl benzene
add H2SO4/HCl, boil
Phenylethanonitrile
Phenylethanoic acid
Alkaline hydrolysis of the nitrile gives a Na salt of the acid i.e.
CH3CH2CN + H2O NaOH (aq), boil CH3CH2COO-Na+ + NH3 (g)
CH3CH2CN + H2O NaOH (aq), boil CH3CH2COO- + NH3 (g)
Reduction of nitriles using NaBH4 in methanol or LiAlH4 in dry ether gives
amines e.g.
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CH3CN
CH3CH2NH2
LiAlH4 in dry ether
NaBH4 in methanol
Qn: describe the reaction scheme indicating in each step the reagents,
conditions, observations (if any) and type of reaction in the conversion of CH4
to CH3COOH.
ANSWER
CH4 u.v light, limited Cl2 CH3Cl KCN in little ethanol CH3CN H+, boil CH3COOH
Free radical substitution
nucleophilic substitution
acid hydrolysis
Formation of primary amines
Halogenoalkanes react with excess concentrated ammonia on heating to
produce amines. The halogen atom is replaced with the amine group (-NH2).
The reaction is called nucleophilic substitution and ammonia is the
nucleophile. Primary amines are formed from primary haloalkanes.
Observation: white fumes of HCl/HBr.
CH3CH2Cl + NH3 (conc) sealed tube/bomb heat CH3CH2NH2 + HCl
excess
CH3CH2CH2Br + NH3 (conc)
sealed tube or bomb
CH3CH2CH2NH2 + HBr
1-amino propane
Excess
CH2Cl
CH2NH2
+ NH3 (conc)
+ HCl
excess
Reactivity of halogenoalkanes: ease of hydrolysis
Reactivity or ease of hydrolysis is determined by the C-X bond strength. The CX bond strength is determined by the nature of the halogen i.e. its atomic size
and electronegativity. C-F bond is very strong because F is small and very
electronegative, there is maximum overlapping of orbitals hence the bond is
short and very strong and for this reason flouroalkanes are very unreactive/
cannot undergo hydrolysis. C-I is the weakest because I has a large atomic size
and low electronegativity hence poor overlapping thus the bond is long and
therefore weaker hence it is easily hydrolysed.
Reactivity/ease of hydrolysis decreases in the following order:
-C-I > -C-Br
> -C-Cl
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It is also determined by the number of electron releasing alkyl groups. The
more the number of electron releasing groups the weaker the C-X bond (hence
it becomes easier to break) the more reactive the R-X e.g. comparing primary,
secondary and tertiary haloalkanes, tertiary haloalkanes are the most reactive
because of the greater number of electron releasing groups with a positive
inductive effect hence the C-X bond in tertiary haloalkanes is very weak. The
order of decreasing ease of hydrolysis/order of decreasing reactivity is:
Tertiary haloalkanes > secondary haloalkanes > primary haloalkanes
CH3
H3C-C-Cl
CH3
H
>
H3C-C-CH3
Cl
>
CH3CH2Cl
The ease of hydrolysis can be determined by using the reagent AgNO3 and
noting the time taken for the appearance of AgX ppt (white ppt of AgCl, cream
ppt of AgBr and yellow ppt of AgI).
Tertiary haloalkanes- the ppt AgX appears immediately
Secondary haloalkanes- ppt appears after e.g. 1 minute
Primary haloalkanes- ppt appears after e.g. 2 minutes
The shorter the time taken for the appearance of the AgX ppt the greater is the
ease of hydrolysis.
Hydrolysis of acyl chlorides e.g. chlorobenzene
Halogenoarenes have a halogen atom directly attached to the benzene ring
e.g. chlorobenzene.
When the halogen is bonded to the benzene ring, the C-X bond is very strong
e.g. in C6H5Cl the C-Cl bond is very strong. It is caused by one of the chlorine
lone pairs being drawn into the benzene delocalised 𝜋 electrons i.e. the lone
pair of chlorine interacts with the π electrons of the benzene ring forming a π
bond which adds to the C-Cl bond such that it behaves as a pseudo double
bond hence it becomes very strong and not easily broken that’s why C6H5Cl
cannot be hydrolysed. Even if aq AgNO3 is added no ppt is observed. Under
strangest conditions of a temperature of 350℃ and a pressure of 150atm
chlorobenzene is hydrolysed by NaOH i.e.
Cl
O-Na+
+ 2NaOH(aq)
150atm, 350℃
ORGANIC AND
(aq) + NaCl (aq) + H2O
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Note: to get phenol the sodium phenoxide is reacted with HCl (aq).
O-Na+
+ HCl(aq)
OH
(aq) + NaCl (aq)
Hydrolysis of acid/acyl chlorides
Comparing with all halo-substituted organic compounds, acid/acyl chlorides
are the most easily hydrolysed. Acid chlorides react vigorously with cold water
forming carboxylic acid together with white fumes of HBr/HCl (observation).
They also fume in humid/moist air due to hydrolysis (only aliphatic acid
chlorides). They are easily hydrolysed because the C atom is bonded to 2 very
electronegative elements making it a very reactive nucleophilic site towards
nucleophiles.
COCl
COOH
+ H2O
+ HCl
CH3COCl + H2O → CH3COOH + HCl
Note: acyl chlorides are insoluble in water but are hydrolysed giving carboxylic
acids and HCl. Part of HCl dissolves in water to form HCl acid- a strong acid, pH
=1/2. Comparing tertiary, secondary and primary haloalkanes, chlorobenzene
and acyl chlorides the order of decrease in ease of hydrolysis/reaction is:
Acyl/acid chloride > tertiary haloalkane > secondary haloalkane > primary
haloalkane > aryl halides
CH3COCl > (CH3)3CCl > CH3CHClCH3 > CH3CH2Cl > Cl
Ethanoyl
Chloride
2-chloro-22-chloropropane chloroethane
methylpropane
chlorobenzene
Note: if AgNO3 (aq) is added to each of the above: a white ppt is instantly
observed for the acyl chlorides showing its greatest ease of hydrolysis followed
by tertiary, secondary and primary haloalkanes. No ppt is observed for aryl
halides e.g. chlorobenzene.
Uses of flouroalkanes
and chloro-flouro-carbons (CFC’s)
Flouroalkanes and CFCs are unreactive and do not burn. They are unreactive
because the C-F bond is very strong. Because of their unreactive nature these
compounds are used:
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1. As coolants in air conditioners and refrigerators
2. Fire extinguishers
3. Solvents
4. Making foam plastic (such as expandable polystyrene packaging)
5. As propellants in aerosol spray cans
6. Insecticides e.g. DDT
Environmental effects of CFCs
and flouroalkanes
CFCs are chemically inert/unreactive, they do not react with air and water and
are non-biodegradable. The molecules are volatile i.e. they rise up into the
stratosphere where they are decomposed by u.v light (i.e. homolytic fission)
producing chlorine free radicals e.g.
F
Cl
C Cl
u.v. light
F
Cl
C• + Cl•
F
F
The Cl• reacts with O3 in the ozone layer causing its depletion i.e.
Cl• + O3 → ClO• + O2
Ozone absorbs harmful u.v radiation from the sun. Due to ozone depletion, u.v
light reaches the earth’s surface causing:
i)
Skin cancer
ii)
Destruction of vegetation and crops
Corrective measures towards ozone depletion
a) To greatly reduce the use of CFCs
b) To find alternatives to CFCs which do not damage the ozone layer (i.e.
chlorine and fluorine free radicals that can be destroyed by atmospheric
oxygen and water).
Hydroxyl
compounds
Physical properties
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The m.p and b.p of alcohols increases as the number of C atoms increases due
to increase in the strength of VDW forces as molecular size/electron cloud and
Mr value increases (the reason is not H-bonds because they are common in
alcohols). The m.p and b.p of a straight chain isomer of an alcohol is greater
than that of its branched chain isomer e.g. m.p of CH3CHCH2CH3 butan-2-ol is
greater than
CH3
that of 2-hydroxy-2-methyl propane i.e. because
the straight 3HC C OH
chain isomer has a larger molecular size and
stronger
CH3
VDW forces. The m.p and b.p of alcohols is
greater than those of alkanes of similar mass and size because the hydrogen
bonds in alcohols are stronger than the VDW forces in alkanes. Alcohols are
soluble in water because they form H bonds with the water molecules i.e.
Hƍ+‖‖‖‖‖‖OƍCH3-Oƍ-
H
H
Fermentation
Fermentation is the breakdown of simple sugars and starch by enzyme zymase
in yeast at ≈35℃. Yeast is a living plant containing the enzyme zymase which
breaks down the carbohydrates. If it is starch it is first hydrolysed by enzymes
giving glucose which then undergoes fermentation. The ethanol-water solution
is then fractionally distilled i.e. ethanol-water solution is concentrated by
fractional distillation. The component of the lowest b.p is not ethanol but a
binary azeotrope (constant boiling point mixture) which gives a vapour of
constant composition hence cannot be separated further no matter how
efficient the fractionating column is.
C6H12O6 (aq) yeast (zymase) 2CH3CH2OH (aq) + CO2 (g)
Glucose
At higher concentrations, ethanol kills the enzyme zymase i.e. at
concentrations greater than 15% by volume.
Combustion of alcohols
Alcohols burn in air/O2 giving CO2 + H2O. The reaction is highly exothermic
hence the use of alcohols as fuels e.g. ethanol.
CH3CH2OH + 3O2 → 2CO2 + 3H2O
Substitution to give halogenoalkanes
Alcohols react with the nucleophilic reagents HX, PX3, PX5 and SOCl2 giving
halogenoalkanes. The reaction is called nucleophilic substitution.
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Halogenation using hydrogen halides
1. Chlorination
Dry hydrogen chloride is bubbled through the anhydrous alcohol (primary
alcohol) in the presence of a ZnCl2 catalyst. When the solution is saturated, it is
refluxed on a waterbath producing the primary halogenoalkanes.
CH3CH2OH (l) + HCl (g) ZnCl2 cat, waterbath CH3CH2Cl (l) + H2O (l)
For secondary and tertiary alcohols no ZnCl2 catalyst is used because they are
reactive. The order of decreasing reactivity is: tertiary alcohol > secondary
alcohol > primary alcohol.
2. Bromination
HBr is used for bromination, type of reaction is nucleophilic substitution. (HBr
and HCl are produced by reaction of conc H2SO4 and NaX).
3. Iodination
Red phosphorus + I2 are reacted first to give HI (conc H2SO4 is not used because
it oxidises the HI giving iodine vapour).
CH3CH2OH + HI (g) → CH3CH2I (l) + H2O (l)
Type of reaction is nucleophilic substitution.
Halogenation using phosphorus halides (PX3 and PX5)
PI3 (produced by reacting I2 + red phosphorus) reacts with alcohols producing
iodoalkanes. Also, red phosphorus + bromine are used to generate PBr3 in the
reaction mixture (containing P + Br2 + alcohol). Distillation is done to separate
the reaction mixture.
3CH3CH2OH + PI3 → 3CH3CH2I + H3PO3
3CH3CH2OH + Br3 → 3CH3CH2Br + H3PO3
Chlorination usually uses PCl5. PCl5 reacts in the cold with alcohols (PCl3 is not
used) giving chloroalkanes.
CH3CH2OH + PCl5 cold, distil, reflux CH3CH2Cl + POCl3 + HCl
Type of reaction is nucleophilic substitution.
Observation: white fumes of HCl.
Note: the formation of HCl (observed as white fumes) when alcohols react with
PCl5 under cold conditions is used as a test for alcohols.
Chlorination using sulphur dichloride oxide SOCl2.
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SOCl2 reacts with alcohols in the presence of pyridine to form halogenoalkanes.
Pyridine is an organic base which absorbs the acidic fumes of HCl as it is
formed.
CH3CH2OH + SOCl2 reflux with pyridine CH3CH2Cl + HCl + SO2
The use of SOCl2 is an economical method because the by-products are gases
which pass away and hence no distillation is required like in other methods.
Observation: white fumes of HCl.
Reaction with Na metal
Alcohols react with Na metal forming sodium alkoxides in which the H atom of
–OH group is displaced by Na. H2 is also evolved. Type of reaction is redox.
CH3CH2OH + Na → CH3CH2O-Na+ + 12H2
Observation: bubbles of a colourless gas i.e. H2
Note: phenol also reacts in the same manner as alcohols i.e.
OH
O-Na+
Sodium phenoxide
CH3CH2OH + Na → CH3CH2O-Na+ + 12H2
COMPARE
+ 12H2
+ Na
Primary alcohols and aldehydes are oxidised because of the presence of H
atoms attached to the C atom bonded to the OH group in alcohols or to the C
of the C=O group. Controlled oxidation of primary alcohols using Na2Cr2O72- at
room temperature or cool conditions gives aldehydes. Cooler temperatures are
used to avoid further oxidation to carboxylic acids.
CH3CH2OH Cr2O72-/H+ cool/r.t CH3CHO (ethanal)
CH2OH
CHO
Benzaldehyde
Cr2O72-/H+ cool/r.t
Observations: orange colour of Cr2O72- changes to green.
Type of reaction: oxidation.
Note: acidified MnO4- is too powerful an oxidising agent to stop at the
aldehyde. It oxidises primary alcohols to carboxylic acids i.e. aldehydes are
further oxidised to carboxylic acids. Half-equation for oxidation of primary
alcohols to aldehydes:
CH3CH2OH + H2O → CH3COOH + 4H+ + 4e-
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Half equation of complete oxidation to aldehydes:
CH3CH2OH → CH3CHO + 2H+ + 2eCr2O72- + 14H+ + 6e- → 3Cr3+ + 7H2O
3CH3CH2OH + Cr2O72- + 8H+ → 3CH3CHO + 3Cr3+ + 7H2O
If MnO4- is used, brown ppt of MnO2 is observed. Complete oxidation gives
carboxylic acids.
CH3CH2OH Cr2O72-/H+, MnO4-/H+ CH3COOH
Oxidation of secondary alcohols gives ketones as the final products. Ketones
cannot be further oxidised because of the absence of free H atoms bonded to
the C atom of the carbonyl group. Acidified Cr2O72- on warming or MnO4-/H+ is
used:
CH3CH(OH)CH3 Cr2O72-/H+ on warming, MnO4-/H+ (CH3)2CO propanone
OH
O
Cr2O72- on warming/H+, MnO4-/H+
CH(OH)CH3
H3C
Cr2O72- on warming/H+, MnO4-/H+
C=O
Phenylethanone
Observation: the orange colour of Cr2O72- turns green
: The purple colour of MnO4- is decolourised or turns pale pink.
Type of reaction: oxidation.
Tertiary alcohols cannot be oxidised because of the absence of free H atoms
bonded to the C atom bearing the –OH group e.g.
CH3
no hydrogen
to oxidise hence cannot be oxidised.
OH
3HC C
CH3
CH2OH
C=O
COOH
H
CH2OH
CH2OH
Cr2O72-/H+ warming
C=O
C=O
further oxidation
C=O
COOH
H
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Dehydration
Alcohols are dehydrated to alkenes by the elimination of a molecule of H2O.
There are 2 ways of doing this:
1. By heating the alcohol with a dehydrating agent such as H3PO4 acid or
excess conc H2SO4 on a water bath.
2. By passing the alcohol vapour over hot Al2O3 which catalyses the
reaction.
CH3CH2OH heat in excess conc H2SO4 at 170℃ CH2=CH2 + H2O
CH3CH(OH)CH3 hot Al2O3 cat CH3CH=CH2 + H2O
Note: H2SO4 is not a very good dehydrating agent as it can also oxidise
alcohols.
CH2CH2OH
CH=CH2
+ H2O
heat in excess conc H2SO4 at 170℃
OH
+ H2O
heat in excess conc H2SO4 at 170℃
CH3CH(OH)CH2CH3 on dehydration gives three isomers:
CH3CH2CH=CH2
CH3CH(OH)CH2CH3 heat in excess conc H2SO4 at 170℃
CH3CH=CHCH3
But-2-ene exhibits geometrical (cis-trans) isomerism.
H
H
H
C=C
H3C
CH3
C=C
CH3
H3C
Cis isomer
H
trans isomer
2 positional isomers + 1 isomer (cis-trans isomer).
Note: dehydration reaction is also called elimination reaction.
Formation of esters/esterification
1. Between alcohols and carboxylic acids
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Alcohols react with carboxylic acids on heating with a little concentrated H2SO4
acid to give esters. Type of reaction is called condensation since water is
removed during the process.
Observation: sweet/fruity smell of the ester
The H2SO4 has two functions:
It acts as a catalyst i.e. it supplies H+ ions to catalyse the reaction.
It absorbs water produced in the reaction thereby increasing the
yield of ester i.e. the forward reaction is promoted in the
equilibrium system.
i)
ii)
CH3COOH + CH3CH2OH little conc H2SO4, heat, reflux CH3CO2CH2CH3 + H2O
COOH
CO2CH3
+ CH3OH
little conc H2SO4, heat, reflux
Benzoic acid
methyl benzoate
CH2OH
CH2OCOCH3
+ CH3COOH
+ H2O
little conc H2SO4, heat, reflux
2. Between alcohols and acid chlorides
Alcohols also react with acid/acyl chlorides to produce esters. The reaction is
an example of acylation i.e. the substitution of RCO-group into another
compound.
Observation: sweet/fruity smell of ester/white fumes of HCl
CH3COCl + CH3CH2OH pyridine CH3CO2CH2CH3 + HCl
COCl
CO2CH3
+ CH3OH
pyridine
+ HCl
The reaction takes place in the presence of pyridine (a base) to absorb acidic
fumes of HCl.
Characteristic distinguishing reactions
between primary, secondary
and tertiary alcohols.
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Controlled oxidation/mild oxidation of primary alcohols using Cr2O72-/H+ at
room temperature or cool conditions gives aldehydes whilst secondary
alcohols are oxidised to ketones.
Test for aldehydes
1. Fehling’s solution: alkaline CuSO4
Aldehydes give a brick-red ppt (of Cu2O) e.g.
RCHO + 2Cu2+ + 5OH- ⟶ RCOO- + Cu2O + 3H2O
Ketones do not react with Fehling’s solution i.e. cannot be oxidised because of
the absence of H atoms attached to the C atom of the C=O group.
Note: ketones and aromatic aldehydes do not react with Fehling’s solution
2. With Tollens reagent: silver mirror test
Aldehydes react with ammoniacal silver nitrate (a mixture of NH3 + AgNO3) to
give a shiny layer of silver/silver mirror.
Ketones give a negative test i.e. do not react with Tollens reagent.
CH3CHO + 2[Ag(NH3)2]+ + 3OH- ⟶ CH3COO- + 2Ag + 4NH3 + 2H2
3. Using 2,4-DNPH: common test
Aldehydes and ketones are carbonyl compounds because they contain the
carbonyl group as the functional group. The common test for ketones and
aldehydes i.e. test for the C=O group is using the reagent 2,4-DNPH which gives
a characteristic orange/yellow ppt. type of reaction is called condensation e.g.
H
3HC-C=O
H
N-NH NO2
H
H2O
H3C-C=N-NH NO2
NO2
+ H2O
NO2
Yellow ppt
Reactions with alkaline iodine: iodoform reaction
Alcohols which contain the group CH3CH(OH) react with alkaline iodine i.e. a
solution of NaOH + I2 (react to give the oxidant IO-) to give a characteristic
yellow ppt of Triiodomethane commonly called iodoform, CH3I. This reaction is
used as a test for the CH3CH(OH) group.
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The alcohols which give a positive iodoform reaction are:
i)
ii)
iii)
For primary alcohols: only ethanol
For secondary alcohols: all secondary methyl alcohols e.g. propan-2ol i.e. CH3CH(OH)CH3, butan-2-ol i.e. CH3CH(OH)CH2CH3, etc
Tertiary alcohols do not have a H atom attached to the C atom
bearing the OH group hence give a negative test with alkaline iodine.
OH
3HC-C-CH3 + 4I2 + 6NaOH
H
OH
3HC-C-R + 4I2 + 6NaOH
H
CHI3 + CH3COO-Na+ + 5NaI + 5H2O
yellow ppt
CHI3 + RCOO-Na+ + 5NaI + 5H2O
yellow ppt
Test for alcohols
1. With PCl5: alcohols give white fumes of HCl (carboxylic acid also give HCl)
under cool conditions.
2. With ethanoic acid + little H2SO4 (conc) on boiling gives esters which
have a characteristic sweet fruity smell.
3. Boil the alcohol with a little K2Cr2O7. Primary and secondary alcohols are
oxidised, orange Cr2O7- turns green.
Phenol
A compound in which the OH group is attached directly to the benzene ring.
Alcohols and phenols share the same functional group i.e. OH group. To
differentiate them, the hydroxyl group in phenol is called the phenolic group.
OH
OH
OH
OH
Br2
CH3
Phenol
4-methyl
2-bromo
phenol
phenol
naphthalene-1-ol
Physical properties
Phenols have generally high m.p and b.p because they have strong H-bonds
between their molecules. They are moderately soluble in water because they
form H-bonds with water molecules. They are colourless solids.
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Reaction with bases
Phenol is a weak acid, pKa= 10. It partially dissociates in water, being a weak
acid it reacts with bases.
O-
OH
+ H3O+
+ H2O ⇌
Phenols therefore dissolve in bases such as NaOH due to acid-base
reaction/neutralisation.
OH
O-Na+
+ NaOH ⟶
+ H2O
Phenols do not react with CO32- because they are very weak bases.
Note: alcohols are weaker acids than phenol hence alcohols do not react with
bases such as NaOH.
Reactions with sodium
Phenol like alcohols reacts in the same manner with metal Na giving sodium
phenoxide together with H2 gas.
Observation: bubbles of colourless gas, H2
Type of reaction: redox
O-Na+
OH
+ Na
+ 12H2 (g)
⟶
redox reaction
Nitration of phenol
Phenol is more reactive than benzene and toluene because the phenolic group
is a stronger activating group than the –CH3 group in toluene which is orthopara directing. Phenol reacts with dilute HNO3 to produce a mixture of
positional isomers i.e. 2-nitrophenol and 4-nitrophenol (trace of 3nitrophenol).
OH
OH
+HNO3
OH
NO2 &
r.t
NO2
Type of reaction: electrophilic substitution.
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Note: m.p and b.p (4-nitrophenol) > m.p and b.p (2-nitrophenol) because 2nitrophenol forms intramolecular H-bonds hence weak intermolecular VDW
forces operate between its molecules whilst 4-nitrophenol has stronger
intermolecular H-bonds. When conc HNO3 is used 2,4,6-trinitrophenol which is
an important explosive. Alternatively, conc H2SO4 + conc HNO3 is used to form
2,4,6-trinitrophenol.
OH
OH
+ HNO3 (conc)
rt
2ON
NO2
NO2
Halogenation of phenol
An aqueous solution of Br2/bromine water reacts with phenol to produce a
characteristic white ppt of 2,4,6-tribromophenol, no iron catalyst is used.
Observation: white ppt of 2,4,6-tribromophenol and reddish-brown colour of
bromine disappears.
Type of reaction: electrophilic substitution
Br2 (aq) is used as a test for phenol.
Cl2 (aq) reacts in the same manner.
With Cl2(g)/Br2(l) no halogen carrier at room temperature. Phenol gives 2
positional isomers i.e. 2-chlorophenol and 4-chlorophenol. Br2 (l) is a non-polar
solvent. CCl4 or carbon disulphide CS2 gives 2-bromophenol and 4bromophenol, positional isomers.
OH
OH
+ 3Br2 (aq)
r.t
OH
+ Br2 (l)
r.t, CCl4
Br
Br + 3HBr
Br
OH
OH
Br &
+ HBr
Br
Observation: reddish-brown colour of Br2 disappears, white fumes of HBr.
Test for phenol (common test)
1. Add aq Br2
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Observation: immediate decolourisation of Br2 occurs. A white ppt of 2,4,6tribromophenol.
Differences between phenols and alcohol
a) With Br (l)- phenols react but alcohols do not
b) Phenol does not react with phosphorus halides but alcohols do
c) Phenol does nor undergo dehydration but alcohols do
d) No oxidation with Cr2O72-/H+
e) No reaction with carboxylic acids but alcohols do
f) Phenol dissolves/reacts with NaOH but alcohols do not
g) Phenol reacts with HNO3
Reaction similarities
a) With Na
b) With acid chlorides (esterification)
Comparing acidity of water, phenol and ethanol.
H2O ⇌ H+ + OHCH3CH2OH ⇌ CH3CH2
O- + H+
OH
O⇌
+ H+
Ethanol and other aliphatic alcohols are weaker acids than water and phenol,
presence of alkyl group i.e. the ethyl group (other alkyl groups in alcohols)
which is electron releasing i.e. has a positive inductive effect. It increases the
negative charge on the ethoxide ion (CH3CH2O-) rendering it unstable and a
reactive conjugate base i.e. it becomes proton accepting shifting equilibrium
much to the left. In water the basic OH- has no electron releasing group hence
negative charge is smaller and less basic than the ethoxide ion.
Phenol is a stronger acid than both water and ethanol because the phenoxide
ion is more stable than both the OH- and CH3CH2O- conjugate bases. This is so
because the negative charge on the phenoxide ion is partially delocalised due
to interaction with the π electrons of the benzene ring thus stabilising the
phenoxide ion. That’s why phenol dissolves in NaOH whereas alcohols do not.
Phenol dissolves in NaOH due to an acid-base reaction i.e.
NaOH + C6H5OH ⟶ C6H5O-Na+ + H2O
Hence order of decreasing acidity is:
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OH
>
H2O
>
CH3CH2OH
Comparison of acidity between phenol and
substituted phenol.
The presence of electron releasing groups decreases the acidity of phenol
because they increase the negative charge density on the phenoxide ion and
destabilises it making it a reactive base/more proton accepting:
O-
OH
⇌
OH
CH3
+ H2O
O-
CH3
+ H+
⇌
Less acidic
reactive conjugate base
Electron withdrawing groups/deactivating groups e.g. NO2, X- groups are
electron withdrawing. They reduce the negative charge on the phenoxide ion
thereby stabilising it i.e. it becomes a less reactive conjugate base or less
proton accepting.
O-
OH
NO2 + H+
NO2 ⇌
More acidic
O-
OH
+ H+
⇌
stable base
less acidic
less stable base
e.g. 2,4,6-trinitrophenol is a stronger acid than phenol and nitrophenol i.e.
O2N
O-
OH
NO2 ⇌ O2N
NO2 + H+
NO2
NO2
Very stable/unreactive base
Order of decreasing reactivity
O-
OH
O2N
NO2 >
NO2
NO2 >
NO2
ORGANIC AND
OH
OH
>
NO2
INORGANIC CHEMISTRY
>
OH
CH3
90
J. MASAITI
Carbonyl
compounds
Ketones and aldehydes are collectively called carbonyl compounds. They have
the carbonyl group Cƍ+=Oƍ- as their functional group. Functional group of
aldehydes is called the aldehyde group (R-CHO) and functional group of
ketones is called the ketone group R
C=O
R
H
General structure of the aldehyde group is
Aldehydes: Methanal, HCHO
R-C=O
-Ethanal, CH3CHO
-Propanal, CH3CH2CHO
-Benzaldehyde,
CHO
-Phenylethanal,
CH2CHO
Positional isomers
Ketones: Propanone, CH3COCH3(CH3)2CO
-Pentan-2-one, CH3COCH2CH2CH3
-Pentan-3-one, CH3CH2COCH2CH3
Functional group is the carbonyl group i.e. Cƍ+=Oƍ-
Nucleophiles attack the partial positive C atom of the carbonyl group adding
across to the C to O double bond hence reaction is called nucleophilic addition
e.g.
CNCƍ+=Oƍ-
CN
C-O
Aldehydes are more reactive than ketones because:
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a) Steric hindrance: large alkyl group offers steric hindrance to small partial
positive C atom of the carbonyl group to attacking nucleophiles.
b) The electron releasing alkyl groups reduces the partial positive charge on
the C atom of the carbonyl group making it a less reactive nucleophilic
site.
Formation of aldehydes: 3 ways
1. By controlled oxidation of primary alcohols using Cr2O72-/H+ under cold
conditions to avoid further oxidation to carboxylic acids.
CH3CH2OH Cr2O72-/H+, cool CH3CHO + H2O
CH2OH
CHO
Cr2O72-/H+, cool
+ H2O
2. By catalytic dehydrogenation of alcohols using Cu catalyst (ketones also
produced if alcohol is secondary) i.e. passing the alcohol vapour of Cu
catalyst at 300℃.
CH3CH2OH Cu catalyst, 300℃ CH3CHO + H2O
CH2OH
CHO
Cu catalyst, 300℃
+ H2O
3. By hydrolysis of primary dihalogenoalkanes. When dihalogenoalkanes
(containing 2 halogens bonded to the same C atoms) are heated under
reflux with NaOH they are first hydrolysed to diols (in which 2 OHgroups are attached to the same C atom). The diols with the 2 hydroxyl
groups attached to the same C atom are unstable and as a result a water
molecule is lost forming aldehydes (or ketones) i.e. internal
condensation occurs:
Cl
OH
H3C-C-Cl + 2OH- heat, reflux H3C-C-OH + 2ClH
H
-H2O
CH3CHO
OH
CH2CHCl2
+ 2OH-
CH2CHO
heat, reflux
CH2-C-OH
H
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Formation of ketones: 3 ways
1. By the oxidation of secondary alcohols using Cr2O72-/H+ or MnO4-/H+.
CH3CH(OH)CH3
(CH3)2CO, propanone
Cr2O72-/H+ warming, MnO4-/H+
OH
O
cyclohexanone
Cr2O72-/H+ warming, MnO4-/H+
CH(OH)CH3
COCH3
phenylethanone
Cr2O72-/H+ warming, MnO4-/H+
2. By catalytic dehydration of secondary alcohols using a Cu catalyst i.e. by
passing the alcohol vapour over the Cu catalyst at 300℃.
Dehydrogenation takes place, forming ketones.
CH3CH(OH)CH3 Cu catalyst, 300℃ (CH3)2CO + H2
CH2CHOHCH3
COCH3
+ H2
Cu catalyst, 300℃
3. By hydrolysis of secondary dihalogenoalkanes. When secondary
dihalogenoalkanes are heated under reflux with NaOH they are first
hydrolysed to diols. The diols are unstable and undergo condensation
forming ketones i.e.
Cl
Cl-C-CH3 + 2OH-
Cl
CH3-C-CH3 + 2OHCl
OH
HO-C-CH3
heat, reflux
heat, reflux
OH
CH3-C-OH
O
C-CH3 + H2O
O
CH3-C-CH3 + H2O
CH3
Reduction of aldehydes
Aldehydes are reduced to primary alcohols by the reducing agents LiAlH4 in dry
ether or NaBH4 in methanol/H2O alternatively H2 gas Pt/Ni catalyst on heating
at 200℃ can also reduce aldehydes to primary alcohols.
CH3CHO
LiAlH4 in dry ether or NaBH4 in methanol
ORGANIC AND
CH3CH2OH
INORGANIC CHEMISTRY
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CHO
CH2OH
LiAlH4 in dry ether or NaBH4 in methanol
CH2CHO
CH2CH2OH
LiAlH4 in dry ether or NaBH4 in methanol
Reduction of ketones
Ketones are also reduced to secondary alcohols using the same reducing
agents LiAlH4 in dry ether or NaBH4 in methanol or even H2 in the presence of
Pt/Ni catalyst on heating under pressure.
(CH3)2CO LiAlH4 in dry ether or NaBH4 in methanol CH3CH(OH)CH3 propan-2-ol
O
OH
LiAlH4 in dry ether or NaBH4 in methanol
Cyclohexanol
COCH3
CH(OH)CH3
1-hydroxy-1-phenylethane
LiAlH4 in dry ether or NaBH4 in methanol
Reactions of ketones and aldehydes with HCN
Carbonyl compounds i.e. ketones and aldehydes react with HCN (since HCN is
poisonous it is generated in the reaction mixture by the action of dilute H2SO4
+ KCN) under cold conditions (10-20℃) in the presence of a little base (e.g.
NaOH or KCN) to increase the c(CN-) since HCN partially dissociates giving a low
c(CN-). The reaction produces a class of compounds. The type of reaction is
called nucleophilic addition since HCN is added across the C=O double bond of
the carbonyl group with CN- ion acting as a nucleophile attacking the Cƍ+ atom.
The little base has 2 functions i.e.
1. To increase the CN- ion concentration i.e.
HCN ⇌ H+ + CNWhen it reacts with H+ the above equilibrium shifts to the right thus increasing
c(CN-).
2. As a catalyst to increase the rate of reaction.
CH3CHO HCN, little base, 10-20℃ CH3CH(OH)CN, a cyanohydrin.
Note: KCN/NaCN/CN- used to start the reaction because c(CN-) from HCN is too
low.
Acid hydrolysis of nitrile groups gives carboxylic acids e.g.
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CH3CH(OH)CN
CH3CH(OH)COOH
H+ (aq), boiling
O
CH3-C-CH3 HCN (g), little base
OH
CH3-C-CN
2-hydroxyl propanoic acid
2-hydroxyl-2-methyl
propane nitrile
CH3
CHO
H
hydroxyl-phenyl
C-CN ethane nitrile
OH (optically active)
HCN (g), little base 10-20℃
O
HO
CN
hydroxyl cyclo-hexane
nitrile (optically active)
HCN (g), little base 10-20℃
OH
COCH3
H3C-C-CN
1-hydroxyl-1-phenyl propane nitrile
HCN (g), little base 10-20℃
(optically active)
The reaction becomes slow on adding an acid.
Acid hydrolysis of cyanohydrins
Acid hydrolysis of cyanohydrins by boiling with aq acid converts the nitrile
group to a carboxylic acid.
OH
CH3-C-CN
CH3
OH
H+ (aq), boil
OH
3HC-C-CN
CH3-C-COOH + NH4+
CH3
OH
CH3-C-CN 2-hydroxyl-2-phenyl propanoic acid
H+ (q), boil
Optically active
Alkaline hydrolysis on boiling the cyanohydrin in NaOH converts the
cyanohydrin into an organic salt i.e.
CH3CH(OH)CN OH- (aq), boil CH3CH(OH)COO- + NH3 (g)
H
CN
H
OH- (aq), boil
COO+ NH3
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Mechanism of nucleophilic addition of HCN
HCN ⇌ H+ + CN- / HCN + H2O ⇌ H3O+ + CNMechanism
H
CH3-Cƍ+=OƍCN-
OCH3-C-CN
H
H+
OH
CH3-C-CN
H
Note: HCN reacts in place of H2O to regenerate CN- catalyst. The addition of
HCN to carbonyl compounds is an important method of increasing the number
of C atoms in a chain (step-up process). Benzaldehyde does not use the above
mechanism.
Distinguishing reaction between ketones
and aldehydes
For aldehydes
1.
Oxidation: aldehydes are oxidised to carboxylic acids by Cr2O72-/H+ or
MnO4-/H+ because they contain a free H atom bonded to the C atom of
the carbonyl group. Observation: With Cr2O72-/H+, a green solution of
Cr3+. With MnO4-/H+, decolourisation, pale pink solution
Ketones can’t be oxidised because of the absence of a H atom on the carbonyl
group. The product, a carboxylic acid is tested using Na2CO3 or NaHCO3.
Effervescence occurs.
Note: oxygen in the presence of a Pt/Cu catalyst on heating also gives
carboxylic acids.
2. Reaction with Tollens reagent (silver mirror test).
The aldehyde is a reducing group. Tollens reagent, a mixture of AgNO3 + NH3
react to give the reacting species [Ag(NH3)2]+. On heating with aldehydes it
gives a characteristic shiny layer of silver metal or a grey/black ppt hence it is
called the silver mirror test. Ketones give a negative test with Tollens reagent.
Type of reaction is redox.
CH3CHO + 2[Ag(NH3)4]+ + 3OH- → CH3COO- + 2Ag (s) + 4NH3 + 2H2O
3. Reaction with Fehling’s solution
When aldehydes are boiled in Fehling’s solution (an alkaline solution of CuSO4),
a red-brown (brick-red) ppt of Cu2O is observed. The reducing aldehyde group
reduces Cu2+ to Cu+ (in the form of Cu2O brick-red ppt).
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CH3CHO + 2Cu2+ + 5OH- → RCOO- + Cu2O + 3H2O
Ketones give a negative test with Fehling’s solution since ketones have no free
H atom attached to the C atom of the carbonyl group i.e. they are nonreducing.
4. Triiodomethane/iodoform reaction.
Aldehydes and ketones which contain the group CH3CO-/ CH3-C=O react with
alkaline iodine (NaOH + I2 react to give the oxidant IO- in the reaction mixture)
to give a characteristic yellow ppt of Triiodomethane commonly called
iodoform. The reaction occurs on warming. The reaction is a test for the
presence of the group CH3CO- e.g.
H
CH3-C=O + 3I2 + 4NaOH warming HCOO- + CHI3 + 3NaI + 3H2O
Yellow ppt
O
CH3-C-CH3 + 3I2 + 4NaOH warming CH3COO- + CHI3 + 3NaI + 3H2O
Aldehydes and ketones which give a positive iodoform test are:
H
- For aldehydes ethanal only, CH3-C=O
- For ketones dimethyl ketones e.g. propanone, butanone,
phenylethanone, etc.
Note: some alcohols containing CH3CH(OH) also give a positive iodoform test.
CARBOXYLIC ACIDS
AND DERIVATIVES
Carboxylic acids are organic acids which have a functional group –COOH e.g.
HCOOH- Methanoic acid
CH3COOH- Ethanoic acid
CH3CH2COOH- Propanoic acid
CH3CH(OH)COOH- 2-hydroxyl propanoic acid (lactic acid)
CH3CH=CHCH2COOH- Pent-3-enoic acid
HO2C-CO2H – Ethan-dioc acid (oxalic acid)
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COOH
COOH
COOH
Benzene-1,4-dioc acid
Benzoic acid
4-nitrobenzoic acid
COOH
NO2
Physical properties
Aliphatic acids C5-C10 are liquids. B.p increases with increasing Mr Value. VDW
forces and H-bonds bond the molecules together. In solution in organic
solvents dimerization occurs through H-bonds e.g.
O‖‖‖‖‖‖‖‖‖H-O
CH3-C
C-CH3
In non-polar solvents e.g. benzene
O-H‖‖‖‖‖‖‖‖‖O
pV = nRT =
Mr =
𝑚
𝑀𝑟
𝑚 × 𝑅𝑇
RT
𝑝𝑉
Mr value doubles showing dimerization.
Acidity
Carboxylic acids are weak acids, they undergo partial/incomplete dissociation
in water. This is because of their very reactive conjugate bases i.e.
RCOOH ⇌ RCOO- + H+
They have electron releasing alkyl groups that increase the negative charge
density on the carboxylate ion i.e. the polar Cƍ+=Oƍ- group attracts electrons
away from the –O-H bond making it easier for H atom to ionise than is the case
in the –O-H bond in alcohols.
The flow of electrons from the hydroxyl group towards the carbonyl carbon
atom reduces the partial positive charge on the carbonyl carbon atom. For this
reason, it is not attacked by nucleophiles e.g. CN- that attack carbonyl
compounds. When the carboxylate ion, RCO2- is formed, the negative charge
on the ion is shared (delocalised) between the 2 oxygen atoms i.e. the
delocalisation of the charge
O
makes the carboxylate ion stable/less
R-C
ready to accept a proton than is an
O
alkoxide e.g. ethoxide ion CH3CH2O-.
The carboxylate ion is therefore a weaker base than the alkoxide ion thus
carboxylic acids are stronger acids than alcohol i.e.
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O
+ H+
CH3COOH (aq) ⇌ CH3C
O
CH3CH2OH (aq) ⇌ CH3CH2-O- + H+
Uses
Ethanoic acid is converted into ethanoic anhydride which is used in the
manufacture of the fabric, acetate rayon and the drug Aspirin. Chloroethanoic
acid is used in the manufacture of weed killer 2,4D. Benzoic acid and benzoates
are used as food preservatives.
Formation of carboxylic acids
1. By the oxidation of primary alcohols using Cr2O72-/H+ on warming or
MnO4-/H+.
Note: aldehydes are formed as intermediates.
Observations: green solution (Cr3+) for Cr2O72-/H+
-Decolourisation of MnO4-/pale-pink colour
e.g. CH3CH2OH Cr2O72-/H+ on warming or MnO4-/H+ CH3COOH + 2H+ + 2e2. By the oxidation of aldehydes using Cr2O72-/H+ on warming or MnO4-/H+.
Observations: green solution (Cr3+) for Cr2O72-/H+
Decolourisation of MnO4-/pale-pink colour
-
e.g. CH3CHO + H2O → CH3COOH + 2H+ + 2eCHO
COOH
Cr2O72-/H+ warming or MnO4-/H+
3. By acid hydrolysis of nitriles.
Nitriles are produced by nucleophilic substitution of haloalkanes using
NaCN/KCN in little ethanol. Acid hydrolysis of nitriles gives carboxylic acids.
CH3CH2Cl KCN/NaCN in little ethanol CH3CH2CN dil H2SO4, heat CH3CH2COOH
CH2Cl
CH2CN
Reflux with KCN/NaCN
dil H2SO4
In little alcohol
heat
CH2COOH
+ NH4+
Note: alkaline hydrolysis does not give a carboxylic acid but an organic salt and
ammonia.
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CH3CH2CN OH-, heat, reflux CH3CH2COO- + NH3 (g)
Reactions of carboxylic acids
Formation of salts
Carboxylic acids react with bases to form salts (organic salts). Type of reaction
is called neutralisation.
CH3COOH + NaOH → CH3COO-Na+ + H2O
2 CH3COOH + Na2CO3 → CH3COO-Na+ + CO2 + H2O
2 CH3COOH + CuO → Cu(CH3COO-)2 + H2O
Formation of esters: esterification
Carboxylic acids react with alcohols to form esters on heating and in the
presence of little conc H2SO4 catalyst. Type of reaction is called condensation.
Observation: sweet/fruity smell of ester.
The conc H2SO4 has 2 functions:
1. Acts as a catalyst i.e. supplies H+ ions to catalyse the reaction.
2. Absorbs water produced in the reaction thereby increasing the yield of
ester i.e. the forward reaction is promoted in the equilibrium system e.g.
CH3COOH + CH3CH2OH ⇌ CH3CO2CH2CH3 + H2O
COOH
CH3OH +
CO2CH3
little conc, H2SO4 cat, boil
+ H2O
Formation of acid/acyl chlorides
With PCl5
Carboxylic acids react with PCl5 to give acid chlorides. Type of reaction is
nucleophilic substitution.
Observation: white fumes of HCl gas.
CH3COOH (l) + PCl5 (l) distil
CH3COCl (l) + POCl3 (l) + HCl (g)
COOH
COCl
+ PCl5
distil
+ POCl3 + HCl (g)
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Carboxylic acids also react with SOCl2 to produce acid chlorides. It is a more
convenient method than that which uses PCl5 because the by-products are
gases, no distillation is required. Type of reaction is nucleophilic substitution.
Observation: white fumes of HCl and a pungent irritating smell of SO2.
COOH
COCl
+ SOCl2
+ SO2 (g) + HCl (g)
CH3COOH + SOCl2 → CH3COCl (l) + SO2 (g) + HCl (g)
Acidity of carboxylic acids
Acid
HCOOH
pKa
Dissociated form
3.75
HCOO- + H+
O
CH3COOH
4.76
+ H+
O
CH3 C
O
CH3CH2COOH
4.87
CH3CH2
C
+ H+
O
HCOOH is a stronger acid (though a weak acid) due to the absence of alkyl
electron releasing groups which make the conjugate base reactive. Benzoic
acid is a stronger acid than ethanoic acid because the negative charge on the
benzoate ion is partially delocalised due to interaction with π electrons of the
benzene ring thereby stabilising it.
Comparing the acidity of carboxylic acids
and chlorine-substituted ethanoic acid
Acid name
Formula
Ethanoic acid
CH3COOH
Chloroethanoic acid
CH2ClCOOH
Dichloroethanoic acid CHCl2COOH
Trichloroethanoic acid CCl3COOH
*show dissociation of each acid*
pKa
4.76
2.86
1.29
0.65
Relative acid
strength
Acid strength
increases
Electron releasing alkyl groups (activating groups) have a positive inductive
effect i.e. they increase the negative charge on the carboxylate ion rendering it
unstable i.e. reactive, thus ethanoic acid is the weakest acid when compared to
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chloro-substituted acids. Electron withdrawing groups e.g. halogen groups (are
deactivating) hence they have a negative inductive effect. They reduce the
negative charge on the carboxylate ion stabilising it i.e. it becomes stable and
less reactive (less proton accepting) e.g. chloroethanoic acid is a stronger acid
than ethanoic acid because the electron withdrawing Cl group reduces the
negative charge on the chloroethanoic conjugate base rendering it stable and
less reactive i.e. less proton accepting. As the number of electron withdrawing
halogen groups increases, the stability of their respective conjugate bases
increases and reactivity decreases leading to an increase in acidity, thus the
order of decreasing acidity is:
CCl3COOH > CHCl2COOH > CH2ClCOOH > CH3COOH
Acid strength is also determined by the nature of the halogen atom. Small and
more electronegative halogens e.g. F2 have the greatest electron withdrawing
tendency hence flourocarboxylate ions (bases) are more stable and less
reactive bases causing flouro-substituted organic acids to be stronger than
chloro, bromo and iodo-substituted organic acids. Thus the order of acidity
decreases as follows:
CH2FCOOH > CH2ClCOOH > CH2BrCOOH > CH2ICOOH
Acid
Ethanoic acid
Bromo-ethanoic
acid
Chloro-ethanoic
acid
Flouro-ethanoic
acid
Formula
CH3COOH
CH2BrCOOH
pKa
4.76
2.90
CH2ClCOOH
2.86
CH2FCOOH
2.66
Relative strength
Acid strength
increases
Comparing organic acids, halo-substituted organic
acids and acyl chlorides.
Acyl groups are the most acidic because they are hydrolysed giving a very
strong acidic solution due to the presence of HCl- a strong acid. Thus, ethanoyl
chloride is a very much stronger acid than ethanoic, chloro-ethanoic and
dichloroethanoic and Trichloroethanoic acid. Order of decreasing acidity is:
CH3COCl > CCl3COOH > CHCl2COOH > CH2ClCOOH > CH3COOH
Acid/acyl chlorides
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Oƍ-
Functional group is
ƍ+
C
Reaction of acid chlorides
depend upon the attack by
Cl
nucleophiles on the partial positive
carbon atom- a very reactive nucleophilic site since this C atom is bonded to 2
very electronegative elements. General structure is O
where R can be
aliphatic, aromatic or H atoms.
R-C-Cl
HCOCl- methanoyl chloride.
CH3COCl- ethanoyl chloride
CH3CH2COCl- propanoyl chloride
CH2COCl
Phenyl-ethanoyl chloride
Reactions with water/hydrolysis
Acyl chlorides react rapidly/vigorously in water to form a solution of two acids
i.e. ethanoyl chloride on hydrolysis gives ethanoic acid and hydrochloric acid (a
strong acid).
CH3COCl + H2O → CH3COOH + HCl
This reaction has three important features:
1. The reaction is rapid and vigorous (heat evolved)
2. A solution of strong HCl acid is produced pH ≈ 1
3. The products give an immediate white ppt of AgCl with AgNO3.
Acyl chlorides are rapidly hydrolysed or they fume in moist/humid air
because they possess a C atom bonded to 2 highly electronegative
elements chlorine and oxygen which pull electrons away from the C
atom causing a very large positive charge on the C atom hence it
becomes very reactive towards nucleophiles e.g. H2O, CN-, etc.
OƍOO
H3C-Cƍ+ Cl
H3C-C-Cl
CH3-C O-H + H+ + ClOƍO
H
H
H
H
Reaction with alcohols and phenols: esterification
Acyl chlorides react with alcohols and phenols to produce esters. Reaction is
rapid and does not require heating. They react with alcohols in the presence of
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pyridine, a base, to absorb acidic fumes of HCl gas. Phenol reacts with alkaline
solution of NaOH (because it dissolves in NaOH through an acid-base reaction).
CH3COCl + CH3OH pyridine CH3CO2CH3 + HCl
OH
OCOCH3
+ CH3COCl
OH
+ HCl
Phenyl ethanoate
COCl
+
-OCO-
+ HCl
Phenyl benzoate
Note: whenever an ester has an oxygen atom of the ester group directly
bonded to the benzene ring it means that one of the reactants in the formation
of that ester is phenol.
Note: phenol can only form esters when it reacts with acid chlorides.
Reaction with primary amides: amide formation
Acid chlorides react with amines e.g. primary amines to form amides at room
temperature.
Observation: white fumes of HCl gas. Type of reaction is called nucleophilic
substitution.
CH3NH2 + CH3COCl
r.t
CH3CH2NH2 + CH3COCl
CH3CONHCH3 + HCl
rt
CH3CONHCH2CH3 + HCl (g)
COCl
CONHCH2CH3
+ CH3CH2NH2
r.t
NH2
+ HCl
ethyl benzamide
NHCOCH3
+ CH3COCl
r.t
+ HCl
phenyl ethanamide
Comparing ease of hydrolysis between primary, secondary and tertiary
haloalkanes, aryl halides and acid chlorides (notes given already).
Esters
O
General structure R-C-O-R’
Where R: alkyl group and the R bonded to C=O can also be H.
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O
Functional group is called the ester group –C-O-/-COOHCO2CH3- methylmethanoate
CH3CO2CH3- methylethanoate
CH3CO2CH2CH3- ethylethanoate
CH3CO2CH3OCO-
-phenylethanoate
-methylbenzoate
CH3CH2OCO-CO2-
-ethylbenzoate
-phenylbenzoate
Esters are neutral compounds. They show functional group isomerism with
carboxylic acids.
Formation of esters
1. By the reaction between organic acids (or salt of organic acid) and
aliphatic alcohols in the presence of a little conc H2SO4 catalyst on
heating under reflux.
CH3COOH + CH3CH2OH ⇌ CH3CO2CH2CH3 + H2O
COOH
CO2CH3
+ CH3OH ⇌
+ H2 O
Type of reaction is called condensation i.e. water removed.
Observation: sweet/fruity smell of ester.
Note: H2SO4 has two functions:
a) Acts as a catalyst i.e. supplies H+ ions to catalyse the redaction.
b) Promotes the forward reaction by removing water.
Phenol does not react with carboxylic acids because its more acidic (than
alcohols).
2. Between alcohols and acid chlorides.
Alcohols and phenols react with acid chlorides to form esters. Type of reaction
is esterification. Observation: fruity/sweet smell of ester and white fumes of
HCl.
CH3OH + CH3COCl → CH3CO2CH3 + HCl
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COCl
+ CH3CH2OH
--CO2CH2CH3 + HCl
pyridine
ethylethanoate
OH
OCOCH3
+ CH3COCl
OH
+ HCl
pyridine
phenylethanoate
COCl
+
--CO2--
pyridine
+ HCl
phenylbenzoate
Note: ethylethanoate is formed in 2 ways i.e.
CH3CH2OH + CH3COOH little conc H2SO4 cat, heat CH3CO2CH2CH3 + H2O
CH3CH2OH + CH3COCl
CH3CO2CH2CH3 + H2O
pyridine
Rule: whenever an ester is composed of an Oxygen atom directly bonded to
the benzene ring it means one of the reactants is phenol and phenol can only
form esters when it reacts with acid chlorides e.g.
OCOCH3
OH
Formed from
& CH3COCl
Qn:
-CH2OCOCH2CH3 formed in 2 ways. Show that phenylethanoate is
formed in one way.
Hydrolysis of esters
Hydrolysis of esters occurs when the ester is boiled with acid (acid hydrolysis)
or with alkali (alkaline hydrolysis) to give an equilibrium mixture of carboxylic
acid (or organic salt) together with an alcohol or phenol.
Acid hydrolysis of Esters
Acid hydrolysis of esters occurs when esters are boiled/heated with an acid
giving a carboxylic acid, alcohol or phenol. The acid acts as a catalyst i.e.
equilibrium is quickly attained when an acid e.g. HCl is used.
CH3CO2CH2CH3 + H2O ⇌ CH3COOH + CH3CH2OH
-CO2-
+H2O ⇌ OH
COOH
+
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OCOCH3
OH
⇌
+ H2O
+ CH3COOH
HCO2CH3 + H2O H+ cat, boil, reflux HCOOH + CH3OH
+ H2O ⇌ OH
-OCO-
COOH
+
NB: for all reactions, include H+ cat, boil/heat, reflux.
Alkaline hydrolysis
Occurs when esters are boiled/heated with alkali giving alcohols together with
organic salt or phenoxide ion. The reaction proceeds faster with alkali i.e. alkali
catalyse the reaction and also moves the position of equilibrium in favour of
the products.
CH3CO2CH2CH3 + H2O ⇌ CH3COO- + CH3CH2OH
OCOCH3
+ H2O ⇌
O+ CH3CH2OH
COO+H2O ⇌
-CO2-
O-
+
CH3OCOCH2CH2CO2CH3 ⇌2CH3OH + -OOCCH2CH2COOCO2CH2CH3
COO-
+ H2O ⇌
O+
NB: include OH- cat, heat/boil reflux for every reaction.
Supponification
Like all esters, fats undergo (alkaline) hydrolysis when boiled with alkaline
NaOH giving propan-1,2,3-triol (glycerol) together with sodium stereate
(commonly called soap) i.e.
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CH2OOC(CH2)16CH3
CH2OOC(CH2)16CH3 + 3NaOH boil, reflux 3CH3(CH2)16COO-Na+
sodium stereate
H2OOC(CH2)16CH3
Propan-1,2,3-triol
This process is called soap making or saponification.
Aspirin
Aspirin is an important analgesic (pain killer) and it has two functional groups
OCOCH3
i.e. ester group –OCO and the carboxylic group -COOH
COOH.
2-hydroxy benzoic acid commonly called salicyclic acid
is the active analgesic (aspirin is
OH
formed
from salicyclic acid) but is sufficiently a strong acid
COOH which
irritates the stomach. Its ester aspirin is less acidic and
less irritating
hence it’s preferably used. Aspirin passes the acidic stomach contents
unchanged. It is hydrolysed in the alkali conditions of the intestines to form
sodium-2-hydroxy benzoic acid i.e. sodium salicyclic which dissolves in the
blood stream.
Alkaline hydrolysis
O-
OCOCH3
COOH + OH-
COO- + CH3COO-
heat, boil
Acid hydrolysis
OCOCH3
OH
COOH + H+ heat, boil
COOH + CH3COOH
Uses of esters
a) As solvents for paints and varnishes
b) As artificial flavouring for food (food flavours)
c) In perfumes due to sweet smell
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NITROGEN
COMPOUNDS
Amines
General structure R-NH2 where R is alkyl or aliphatic or aryl (phenyl group).
NH2- amine group is the functional group.
Classification: classification according to the number of C atoms bonded to the
N atom of the amine group into:
(1)
(2)
(3)
(4)
Primary amines- nitrogen atom of amine group is bonded to only one
C atom e.g. CH3CH2NH2.
Secondary amines- nitrogen atom of amine group is bonded to two C
atom e.g. (CH3)2NH (dimethyl amine), CH3CH2NHCH3,
N
&
NHCH3
Tertiary amines- nitrogen atom of amine group is bonded to 3 C
atoms e.g. (CH3)3N (trimethyl amine),
-N(CH3)2
Tertiary amines are alkyl substituted ammines e.g. (CH3)4N+Cl(tetramethyl ammonium chloride).
Physical properties
1. For the primary amines e.g. CH3NH2, CH3CH2NH2, etc the m.p and b.p
increases due to increase in VDW forces as molecular size/electron cloud
increases.
Note: extra strong H-bonds bond the amines in liquid and solid state.
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2. Amines are soluble in water because they associate with the water
molecules through H-bonding e.g.
H
H O
CH3CH2-N
H
H
H ‖‖‖‖O
O
H
H
H
Comparing the m.p & b.p of primary,
secondary, tertiary
and quaternary amines.
CH3CH2NH2- ethylamine- primary amine
(CH3)2NH- dimethylamine- secondary amine
(CH3)3N- Trimethylamine- tertiary amine
(CH3)4N+Cl-- tetramethyl ammonium chloride- quaternary amine
Quaternary amine has the highest m.p and b.p because of strong ionic bonds.
Primary amines have greater m.p and b.p than secondary amines because they
have a greater number of H-bonds per molecule hence greater m.p and b.p
e.g. ethylamine has 3 H-bonds per molecule whilst secondary amine
dimethylamine has only 2 H-bonds. The tertiary amines have weak
intermolecular VDW forces. Order of decreasing m.p and b.p is:
(CH3)4N+Cl- > CH3CH2NH2 > (CH3)2NH > (CH3)3N
Note: alcohols have greater m.p and b.p than amines e.g. CH3CH2OH because
they have stronger H-bonds than those in amines since O is smaller and more
electronegative than N hence the permanent dipoles are larger hence stronger
H-bonds are formed in alcohols.
Smell/odour
Amines have a characteristic smell. Lower members resemble NH3. Higher
members have a ‘fishy’ odour. Dimethylamine and Trimethylamine are found
in rotting fish (amines are produced when protein material decomposes).
Reactivity of amines page 579 Understanding Chemistry.
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Formation of ethylamine (amino-ethane)
It is formed in 2 ways:
1. By nitrile reduction i.e. by the reduction of ethanenitrile using LiAlH4 in
dry ether.
CH3CN LiAlH4 in dry ether CH3CH2NH2
Other reducing agents like Na + ethanal or H2 with a catalyst can be used.
Note: the nitrile is produced by reacting the haloalkane with KCN/NaCN in
ethanal on heating (step-up process- nucleophilic substitution). Amines can
also be produced on reacting haloalkanes with conc NH3 in a sealed
tube/bomb but the method is not suitable since it produces a mixture of
primary, secondary, tertiary and quaternary amines which need fractional
distillation to separate them.
Reactivity of amines
The lone pair of electrons on the N atom of the amine group are used to form
bonds with:
a) An H+ ion i.e. to accept a proton hence amines are bases e.g.
CH3CH2NH2 + H+ → CH3CH2NH3+
b) The electron deficient C atom (bonded to the small electronegative Cl
atom in haloalkanes) can be attacked by amines hence amines act as
nucleophiles i.e. nucleophilic substitution reactions.
e.g. CH3CH2Cl + CH3NH2 → CH3NHCH2CH3 + HCl
c) As ligands in complex ion formation with transition metal ions i.e.
[Cr(H2O)6]3+ + 3RN-CH2CH2NH2 → [Cr(RN-CH2CH2NH2)3]3+ + 3H2O
Phenylamine, C6H5NH2- is an aromatic amine. It is less soluble/almost insoluble
in water because of:
1. The large hydrophobic benzene ring which is non-polar and cannot
interact with the water molecules. It also offers steric hindrance to the
formation of H-bonds between NH2 group and water molecules.
2. The lone pair of electrons on the amine group are less available to form
H-bonds as it gets partially delocalised due to interaction with π
electrons of the benzene ring.
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Formation of Phenylamine
Phenylamine is produced by the reduction of nitrobenzene using Sn and
concentrated HCl (this generates the Sn(II)Cl2) which reduces nitrobenzene.
-NO2
-NH2 + 2H2O
Sn + conc HCl
OR
-NO2 + 12H+ + 3Sn → 2
2
-NH2 + 4H2O + 3Sn4+
Note: nitrobenzene is produced by the reaction between benzene and conc
H2SO4 + HNO3 at 60℃ i.e.
Conc H2SO4 + conc HNO3, 60℃
-NO2
In practice, since the Phenylamine is a base, as soon as it is formed it reacts
with the HCl acid to produce phenyl ammonium chloride (salt) i.e.
NH3+Cl-
NH2
(l) + HCl (aq)
(aq)
Sodium hydroxide is then added to the solution of the salt to produce
Phenylamine formed as a separate layer by steam distillation.
Basicity of amines
Amines are weak bases (like NH3). The N atom of the amine group uses its lone
pair of electrons to form a dative bond with a proton. Due to this amines are
proton acceptors e.g.
CH3CH2NH2 + H+ → CH3CH2NH3+ ethylammonium
NH3+
NH2
+ H+ →
phenylammonium ion
Amines are weak alkalis in water e.g.
CH3CH2NH2 + H2O ⇌ CH3CH2NH3+ + OHAmines react with acids to form salts e.g.
CH3CH2NH2 + HCl → CH3CH2NH3+Cl- (ethyl ammonium chloride)
Comparing relative basicities of ammonia,
ethylamine and phenylamine.
NH3 + H2O ⇌ NH4+ + OHCH3CH2 NH2 + H2O ⇌ CH3CH2 NH3+ + OH-
ORGANIC AND
pKb= 4.74
pKb= 3.28
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NH3+
NH2
+ H2O ⇌
pKb= 9.30
+ OH-
Ethylamine and other aliphatic amines are stronger bases than ammonia and
phenylamine because of the presence of the electron releasing ethyl group
(and other alkyl groups e.g. –CH3 group) all have a positive inductive effect
which increases the electron density of the basic amine group thus making
ethylamine more basic i.e. more proton accepting. Also, ethylamine (and other
aliphatic amines) are stronger bases than ammonia and C6H5-NH2 because the
conjugate acid i.e. ethylammonium ion is stabilised by the electron releasing
ethyl group which reduces the positive charge thus stabilising the base i.e. it
becomes less acidic or rather the positive charge on the ethylammonium ion is
distributed between N and the C atom bonded to the N atom thereby
stabilising the conjugate base. Phenylamine is a weaker base than NH3 and
ethylamine/aliphatic amines because the lone pair of electrons is delocalised
due to interaction with π electrons of the benzene ring, as a result of this
partial delocalisation, the lone pair is less available to form dative bonds with
protons. The lone pair of electrons on the N atom in NH3 is not
delocalised/mobile hence it is always available for coordination with a proton.
The order of decreasing basicity is:
CH3CH2NH2 (aliphatic amines) > NH3 >
-NH2
Reaction of phenylamine
The amine group is a very strong activating group. It is ortho-para directing and
makes the benzene ring reactive hence electrophiles attack the benzene ring
on the ortho and para positions giving positional isomers.
Reaction of phenylamine with aq Br2
The reaction with aq Br2 is used as a test for phenylamine (similar to phenol).
Phenylamine reacts with aq Br2 or Cl2 to produce a white ppt of 2,4,6tribromophenylamine and white fumes of HBr (two observations). The type of
reaction is called electrophilic substitution i.e.
NH2
NH2
+ 3Br2 (aq) → Br
Br + 3HBr (g)
Br
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In this reaction, Br2 is decolourised. For mono-bromination to occur the amine
group is made less activating by ethanoylation i.e. by converting it to the less
activating amide group NHCOCH3. The phenylethanamide then reacts with Br2
(aq) to give two positional isomers 2-bromophenylethanamide and 4bromophenylethanamide.
NH2
NHCOCH3
+ CH3COCl2
phenylethanamide (amide group
(or (CH3CO)2O
is less activating than the –NH2)
NHCOCH3
+ Br2 (aq) →
NHCOCH3
Br &
NHCOCH3
positional isomers
Br
Alkaline hydrolysis of the amides restores the amine group.
NHCOCH3
+ H2O
NH2
Br + CH3COO-
OH- (aq), boil
Reaction with nitrous acid: formation of
diazonium salts
Phenylamine (aromatic primary amine) reacts with nitrous acid which is
produced in the reaction mixture by the reaction between NaNO2 and dil HCl
acid at a temperature below 5/10℃ to produce benzene diazonium chloride
(salt). The reaction is carried out below 5/10℃ to avoid the decomposition of
both nitrous acid and the benzene diazonium chloride (benzene diazonium is
thermally unstable).
NH2
N≡N+Cl(l) + NaNO2 (aq) + 2HCl (aq)
temp below 5/10℃
+ NaCl (aq) + 2H2O
Benzene diazonium chloride
Formation of phenol
When the benzene diazonium chloride is warmed with dil H2SO4, phenol is
produced.
N2+ClOH
+ H2O
+ N2 (g) + HCl (aq)
warm T>10℃ with dill H2SO4
Observation: bubbles of colourless gas, N2.
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To avoid the reaction between the phenol formed and the diazonium
compound (i.e. benzene diazonium chloride), the diazonium compound is
added slowly to a large excess of boiling dilute H2SO4, the volatile phenol is
distilled.
Formation of azo-dyes: coupling reactions
The benzene diazonium ion is an electrophile. The benzene diazonium cation
from the benzene diazonium chloride (salt) will attack reactive nucleophilic
sites such as the ortho and para positions in phenol and aromatic amines
(phenylamine) in which two aromatic rings are joined together by an azo group
-N≡N-, this is called azo-coupling. Benzene diazonium chloride (electrophilic
reagent which provides the electrophile
-N2+) reacts with phenol and
phenylamine to form deeply coloured compounds called azo-dyes which are
used in dyes. Reaction takes place below 5/10℃ to avoid decomposition of the
electrophile benzene diazonium ion i.e.
N2+Cl-
OH
+
-N=N-
temp < 5/10℃
-OH
4-hydroxyphenyl-azo-benzene
&
-N=N-
2-positional isomers
OH
2-hydroxyphenyl-azo-benzene
These azo-dyes are deeply coloured due to extensive delocalisation of π
electrons of the benzene ring. Due to this deep colour they are used as dyes
for clothes, indicators (e.g. methyl orange) and ink. The type of reaction is
called electrophilic substitution.
OH + Cl-N2+-
N2+Cl-
-NO2 →
OH
-N=N-
NH2
+
OH
ORGANIC AND
-NO2
OH
temp <5/10℃
-N=N-
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CH3
-N2+Cl- +
-N
CH3
temp 5/10℃ NaO3
S-
-N=N-
CH3
-N
CH3
Amides
General structure RCONHR’ where R is –H, alkyl/aryl groups. Functional group
is -CONH-, amide group. Amides are solids, they dissolve in water to give
neutral solutions.
HCONH2- methanamide
CH3CONH2- ethanamide
CH3CH2CONH2- propanamide
CH3CONHCH3- methylethanamide
-CONH2- benzanamide
N – Cyclic amide
O
O
H2N-C-NH2- carbamide commonly called Urea is also an amide- a diamide of
carbonic acid.
Formation of amides
Are formed by reacting acid/acyl chlorides with ammonia solution and amines.
CH3COCl + NH3 → CH3CONH2 + HCl
CH3COCl + CH3NH2 → CH3CONHCH3 + HCl
NH2
+ CH3COCl
NHCOCH3
+ HCl
-COCl + NH3
-CONH2 + HCl
Type of reaction is called nucleophilic substitution because ammonia and the
amine act as nucleophiles.
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Note: substituted amides are formed when acid chlorides react with amines as
given above.
Amide hydrolysis
Amides are hydrolysed by boiling with dilute acid (acid hydrolysis) or by boiling
with alkalis (alkaline hydrolysis).
Acid hydrolysis of amides
Amides are hydrolysed by boiling with dilute acid (H2SO4 or HCl) to give
carboxylic acids together with an ammonium salt (alkyl ammonium salts). NH3
produced reacts with the acid to form ammonium salts.
CH3CONH2 + H2O H+ (aq), boil CH3COOH + NH4+
CH3CONHCH3 + H2O H+ (aq), boil CH3COOH + CH3NH3+
CONHCH3
COOH
+ H2O H+ (aq), boil
+ CH3NH3+
Type of reaction: acid hydrolysis.
Observation: pungent irritating smell of ammonia and amine. ‘Fishy’ smell of
amine.
Alkali hydrolysis of amides
Amides are also hydrolysed on boiling with alkali (i.e. NaOH) to give an organic
salt together with ammonia gas.
CH3CONH2 + H2O OH- (aq), boil CH3COO- + NH3
CH3CONHCH3 + H2O OH- (aq), boil CH3COO- + CH3NH2
NHCOCH3
+ H2O
NH2
OH- (aq), boil
+ CH3COO-
Type of reaction: alkaline hydrolysis
Observation: pungent irritating smell of ammonia and amines. ‘Fishy’ smell of
amine.
Paracetamol has 3 functional groups:
HO-
-NHCOCH3
1. The phenolic group
2. The amide group
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3. The benzene ring
It reacts with (i) CH3 (ii) Na (iii) NaOH (iv) HCl (aq) (v) Br2 (aq)
Amino acids
H
General structure H2N-C-COOH where R is H, alkyl or aryl group.
R
The –NH2 group is on the C atom adjacent to the –COOH group hence amino
acids of this type are called α amino-acids (both the –NH2 and –COOH groups
are bonded to the same C atom) e.g.
H
H
H
H2N-C-COOH H2N-C-COOH H2N-C-COOH
H
CH3
CH2COOH
Glycine
Alanine
Aspartic acid
Amino-acids have two functional groups:
a) The basic amine group- a proton acceptor
b) The acidic carboxyl group- a proton donor
Since they contain both the basic amine group and acidic carboxyl group amino
acids can therefore react with both bases and acids or have both acidic and
basic properties and are therefore amphoteric e.g. Glycine:
H2NCH2COOH + H+ → H3N+CH2COOH
H2NCH2COOH + OH- → H3N+CH2COO- + H2O
Reactions with acids: basic properties
RCH(NH2)COOH + H+ → RCH(NH3+)COOH
H2NCH2COOH + H+ → H2N+CH2COOH
Reactions with bases: acid properties
RCH(NH2)COOH + OH- → RCH(NH2)COO- + H2O
e.g. CH3CH(NH2)COOH + OH- → CH3CH(NH2)COO- + H2O
The above reactions occur when the pH of the media increases i.e. at higher
c(OH-). Since amino acids can accept both H+ and OH- ions, they exert a
buffering effect i.e. act as buffers.
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Formation of Zwitterions
Zwitterions are dipolar ions formed at neutral pH due to internal acid-base
reaction i.e. the acidic –COOH group donates a proton to the basic –NH2 group.
General structure is:
H
H
H3N+-C-COO- for example H3N+-C-COO- for Glycine
R
H
Amino-acids exist as crystalline solids of high m.p and b.p because of the
presence of zwitterions bound by strong ionic bonds. Amino acids dissolve in
water because they are ionic, ion-dipole interactions cause dissolving.
They are good conductors of electricity (in aq solution) because of the
presence of zwitterions which are charge carriers (they are poor conductors in
solid state because ions are bound by strong ionic bonds).
Amino acids also form H-bonds with water molecules leading to dissolving i.e.
they dissolve due to formation of H-bonds with water molecules.
H3N+-CH2COOH pH < 7 H3N+-CH2-COO- pH > 7 H2N-CH2-COOAcidic media
Basic media
Behaviour of amino-acids in acidic and basic media
In acidic solutions (pH < 7) e.g. in dil HCl, the amino acid molecules are positive
ions because the basic NH2 group accepts a proton e.g.
H2NCH2COOH + H+ ⟶ H3N+CH2COOH
In basic/alkaline solution (pH > 7) in NaOH, the amino acid donates a proton to
the base.
H2NCH2COOH + OH- ⟶ H2NCH2COO- + H2O
Formation of peptides: dipeptides & tripeptides
Amino acids (the monomers in polyamides) undergo condensation reactions
under the action of enzymes to form dipeptides, tripeptides and polypeptides
by condensation reaction. The amino-acids are joined together/linked through
–CONH group called the peptide linkage/bond.
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H
H O H H O
H HOHH O
N-C-C +
N-C-C enzyme N-C-C-N-C-C
H H OH H CH3 OH
H H
CH3 OH
H2O
Amide linkage
Type of reaction is called condensation because a water molecule is removed
as the amino acids join together through the peptide/amide linkage/bond.
Formation of proteins
The dipeptides formed/amino acids can join by the action of enzyme through
the peptide linkage –CONH- group to form a long chain of amino acid residues
forming protein.
Type of reaction is called condensation polymerisation.
When the number of amino acid residues reaches 100 or more the polymer is
called a protein. Proteins are a diverse group of polymers:
a) Fibrous proteins have linear molecules and are insoluble in water. They
are resistant to acids and alkalis e.g. keratin (in hair, nails, feathers and
horns), collagen (in muscles and tendons, elastic in arteries and tendons
and fibrous in silk). Other types of proteins are:
b) Globular proteins: e.g. albumen in eggs and casein in milk, enzymes are
also globular proteins).
H HO
H H O
H H O H H O
N-C-C OH +
N-C-C
nH2O -N-C-C-N-C-CH H
H CH3 OH
H
CH3
Glycine
Alanine
protein segment
H2O
Hydrolysis of peptides and proteins
Proteins are hydrolysed by heating with alkali (alkaline hydrolysis) or with acid
(acid hydrolysis) to produce amino acid monomers.
During this process the amide linkage/group is hydrolysed e.g. for the protein:
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H+ (aq) / OH- (aq), boil
O
NH-CH2CONHCHCONHCH(CH3)CONHCH-CCH3
CH2
CH2
COOH
CH2COOH
H2NCH2COOH + H2NCHCOOH + H2NCHCOOH + H2NCHCOOH
CH3
CH2
CH2
COOH
CH2COOH
Polymerisation
Polymers are long chain molecules formed by reactions between smaller
molecules known as monomers. Polymers are very large molecules made up of
many repeat units. There are two main classes of polymers: natural polymers
(starch, fat, protein, DNA, etc.) and synthetic polymers (polyesters,
polyethenes, nylons, etc.)…synthetic polymers are classified as:
-addition polymers
-condensation polymers
Two main processes used to convert monomers to polymers are:
a) Addition polymerisation known as chain growth polymerisation
b) Condensation polymerisation known as step growth polymerisation
Addition polymerisation:
characteristics
Is a process during which a large number of unsaturated hydrocarbons
(alkanes or their derivatives) combine together to give only one product
namely the addition polymer. The reactive group (functional group) in the
alkene or its derivative (monomers) is the C to C double bond.
Addition polymerisation occurs through a free radical mechanism i.e. it is a
free radical polymerisation process. Addition polymerisation is generally fast
and it generates lots of energy. Examples of common addition polymers are:
poly (ethene), poly (vinyl chloride) PVC, poly (styrene) and poly aerylontrile,
PTFE (polytetra-flouroethene).
nCH2 = CH2 1000-2000atm, 100-300℃ ( CH2-CH2 ) n
LDPE
(Low density polymer)
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High density polyethene is made at low pressure (5-25atm) at 25-30℃. ZigglaNatta catalyst.
Copy table on page 560 Briggs.
LDPE is used for making plastic bags, collapsible squizzy bottles. HDPE is used
to make rigid bottles and bottle crates.
Polyvinylchloride, PVC
nCH2=CHCl
( CH2-CHCl ) n
PVC
Uses: PVC is more rigid than polyethene. It is used in glittering and electrical
insulation. With added plasticisers PVC is used for wellingtons, etc.
Polytetra-flouroethene, PTFE: Teflon
nCF2=CF2
( CF2-CF2 ) n
Used for coating surfaces to reduce friction e.g. non-stick frying pans. Page 583
Ramsden for more examples.
Disposal of plastics
These plastics are non-biodegradable (and resistant to most chemicals and
bacteria) hence they cause water and soil pollution.
Disposal by burning produces harmful products:
1. CO2 causes global warming
2. CO is poisonous
3. Oxides of chlorine cause ozone depletion
Proper disposals is using them as fuels on burning.
Condensation polymerisation: characteristics
In condensation polymerisation, reactive groups on both ends of each
monomer react with each other. The chain starts smaller with reactive groups
on each end and the chain grows steadily as the small chains join. Long chains
formed by condensation process i.e. by step growth polymerisation. Two
different functional groups are involved in condensation polymerisation. The
two functional groups may be on two different monomer species (e.g. nylon
1,6) or on the same monomer e.g. amino acids (protein formation). During
condensation polymerisation a small molecules e.g. water or HCl is eliminated.
Hydrogen bonds form between the different polymer chains. Examples of
condensation polymers are polyesters e.g. Terylene, nylon e.g. nylon 6.6 and
nylon 6, polyethene, etc.
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Nylon 6.6, a synthetic fibre, is formed from two different monomers hexane
1.6 dioc and hexane 1.6 diamine, condensation polymerisation occurs in which
the monomers are joined by a peptic/peptide/amide linkage/bond i.e.
O
O
H
H
C-(CH2)4-C
+
N-(CH2)6-N
HO
OH H
H
Hexane 1.6 dioc acid
O
hexane 1.6 diamine
O H
H
C-(CH2)4-C−N-(CH2)6-N
+ 2n H2O
n
nylon 6.6
Nylon 6 is formed by a simple type of monomer, ClOC-(CH2)5-NH2 i.e.
O
H
O
C-(CH2)5-N
Cl
+
H
H
C-(CH2)5-N
Cl
H
O
H O
H
C-(CH2)5-N-C-(CH2)5-N
+ HCl
n
The hydrogen bonds present between the different polymers provide strength
and elasticity of nylon.
Uses: in making fabrics
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Formation of polyesters
Polyesters are formed when a diol reacts with a dicarboxylic acid. Terylene, a
polyester is formed by condensation polymerisation reaction between the
monomers ethan-1,2-diol and benzene 1,4 dioc acid i.e.
O
H-O C
O
C
O
C
+
O-H
HO-CH2CH2-OH
O
C-O-CH2CH2-O
+ 2n H2O
Terylene and nylon are used for making fibres.
The end
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ORGANIC AND
INORGANIC CHEMISTRY