During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles) H H H H H H H H C = C + C = C + C = C + C = C + … H H H H H H H H Ethene + Ethene + Ethene + Ethene + … (ii)the double bond joining the ethane molecule break to free readicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H H H H H H H H Ethene radical + Ethene radical + Ethene radical + Ethene radical + … (iii)the free radicals collide with each other and join to form a larger molecule
39 H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H H H H H H H H Lone pair of electrons can be used to join more monomers to form longer polyethene. Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. 40 It is an elastic, tough, transparent and durable plastic.


Polyethene molecule can be represented as: H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- H H H H H H H H Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H H Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. 40 It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.


The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship: Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer Examples Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 ) Number of monomers/repeating units in polyomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760 Substituting 4760 = 170 ethene molecules 28 The commercial name of polyethene is polythene. 40 It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene.


40 It is an elastic, tough, transparent and durable plastic. Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as:
41 H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC).


Polythene is used: (i)in making plastic bag (ii)bowls and plastic bags (iii)packaging materials 2.Formation of Polychlorethene Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as:
41 H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic.


During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H Cl H Cl H Cl H Cl chloroethene + chloroethene + chloroethene + chloroethene + … (ii)the double bond joining the chloroethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H Cl H Cl H Cl H Cl (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons •C – C - C – C - C – C - C - C• + … H Cl H Cl H Cl H Cl Lone pair of electrons can be used to join more monomers to form longer polychloroethene. Polychloroethene molecule can be represented as:
41 H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.


Polychloroethene molecule can be represented as:
41 H H H H H H H H extension of molecule/polymer - C – C - C – C - C – C - C – C- + … H Cl H Cl H Cl H Cl Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H Cl Examples Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H
42 C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.


It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H
42 C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples
43 Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number) 104 The commercial name of polyphenylethene is polystyrene.


PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 3.Formation of Polyphenylethene Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H
42 C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples
43 Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number) 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic.


During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H
42 C = C + C = C + C = C + C = C + … H C6H5 H C6H5 H C6H5 H C6H5 phenylethene + phenylethene + phenylethene + phenylethene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H C6H5 H C6H5 H C6H5 H C6H5 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H C6H5 H C6H5 H C6H5 H C6H5 Lone pair of electrons can be used to join more monomers to form longer polyphenylethene. Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples
43 Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number) 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators.


Polyphenylethene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H C6H5 H C6H5 H C6H5 H C6H5 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H C6H5 Examples
43 Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760 Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number) 104 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 4.Formation of Polypropene Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.


It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings 4.Formation of Polypropene Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H CH3 H CH3 H CH3 H CH3 propene + propene + propene + propene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H CH3 H CH3 H CH3 H CH3
44 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H CH3 H CH3 H CH3 H CH3 Lone pair of electrons can be used to join more monomers to form longer propene. propene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic.


During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) H H H H H H H H C = C + C = C + C = C + C = C + … H CH3 H CH3 H CH3 H CH3 propene + propene + propene + propene + … (ii)the double bond joining the phenylethene molecule break to free radicals H H H H H H H H •C – C• + •C - C• + •C - C• + •C - C• + … H CH3 H CH3 H CH3 H CH3
44 (iii)the free radicals collide with each other and join to form a larger molecule H H H H H H H H lone pair of electrons • C – C - C – C - C – C - C - C • + … H CH3 H CH3 H CH3 H CH3 Lone pair of electrons can be used to join more monomers to form longer propene. propene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators.


propene molecule can be represented as: H H H H H H H H - C – C - C – C - C – C - C - C - H CH3 H CH3 H CH3 H CH3 Since the molecule is a repetition of one monomer, then the polymer is: H H ( C – C )n H CH3 Examples Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760 Substituting 4760 = 108.1818 =>108 propene molecules(whole number) 44 The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings
45 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.


It is a very light durable plastic. Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings
45 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.


Polystyrene is used: (i)in making packaging material for carrying delicate items like computers, radion,calculators. (ii)ceiling tiles (iii)clothe linings
45 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer
46 - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E).


(ii)ceiling tiles (iii)clothe linings
45 5.Formation of Polytetrafluorothene Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure. During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer
46 - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic.


During polymerization: (i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles) F F F F F F F F C = C + C = C + C = C + C = C + … F F F F F F F F tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + … (ii)the double bond joining the tetrafluoroethene molecule break to free radicals F F F F F F F F •C – C• + •C - C• + •C - C• + •C - C• + … F F F F F F F F (iii)the free radicals collide with each other and join to form a larger molecule F F F F F F F F lone pair of electrons •C – C - C – C - C – C - C - C• + … F F F F F F F F Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene. polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer
46 - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 6.Formation of rubber from Latex Natural rubber is obtained from rubber trees.


polytetrafluoroethene molecule can be represented as: F F F F F F F F extension of molecule/polymer
46 - C – C - C – C - C – C - C – C- + … F F F F F F F F Since the molecule is a repetition of one monomer, then the polymer is: F F ( C – C )n F F Examples Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 ) Number of monomers/repeating units in monomer = Molar mass polymer Molar mass monomer => Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760 Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number) 62.5 The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used: (i)in making plastic rope (ii)water pipes (iii)crates and boxes 6.Formation of rubber from Latex Natural rubber is obtained from rubber trees. During harvesting an incision is made on the rubber tree to produce a milky white substance called latex. Latex is a mixture of rubber and lots of water. The latex is then added an acid to coagulate the rubber. Natural rubber is a polymer of 2-methylbut-1,3-diene ; H CH3 H H CH2=C (CH3) CH = CH2 H - C = C – C = C - H During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus;
47 H CH3 H H H CH3 H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H CH3 H H -(- C - C = C - C -)n- H H Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.


The latex is then added an acid to coagulate the rubber. Natural rubber is a polymer of 2-methylbut-1,3-diene ; H CH3 H H CH2=C (CH3) CH = CH2 H - C = C – C = C - H During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus;
47 H CH3 H H H CH3 H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H CH3 H H -(- C - C = C - C -)n- H H Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher. During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer. H CH3 H H H CH3 H H - C - C - C - C - C - C - C - C - H S H H S H H CH3 S H H CH3 S H - C - C - C - C - C - C - C - C - H H H H H H Vulcanized rubber is used to make tyres, shoes and valves. 7.Formation of synthetic rubber Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot. Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ; H Cl H H Sulphur atoms make cross link between polymers
48 CH2=C (Cl CH = CH2 H - C = C – C = C - H During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus; H Cl H H H Cl H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber. (c)Test for the presence of – C = C – double bond.


Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ; H Cl H H Sulphur atoms make cross link between polymers
48 CH2=C (Cl CH = CH2 H - C = C – C = C - H During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus; H Cl H H H Cl H H - C - C = C - C - C - C = C - C - H H H H Generally the structure of rubber is thus; H Cl H H -(- C - C = C - C -)n- H H Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber. (c)Test for the presence of – C = C – double bond. (i)Burning/combustion All unsaturated hydrocarbons with a – C = C – or – C = C – bond burn with a yellow sooty flame. Experiment Scoop a sample of the substance provided in a clean metallic spatula. Introduce it on a Bunsen burner. Observation Inference Solid melt then burns with a yellow sooty flame – C = C –, – C = C – bond (ii)Oxidation by acidified KMnO4/K2Cr2O7 Bromine water ,Chlorine water and Oxidizing agents acidified KMnO4/K2Cr2O7 change to unique colour in presence of – C = C –
49 or – C = C – bond. Experiment Scoop a sample of the substance provided into a clean test tube. Add 10cm3 of distilled water. Shake. Take a portion of the solution mixture. Add three drops of acidified KMnO4/K2Cr2O7 . Observation Inference Acidified KMnO4 decolorized Orange colour of acidified K2Cr2O7 turns green Bromine water is decolorized Chlorine water is decolorized – C = C – – C = C – bond (d)Some uses of Alkenes 1. In the manufacture of plastic 2. Hydrolysis of ethene is used in industrial manufacture of ethanol. 3. In ripening of fruits. 4. In the manufacture of detergents.


In ripening of fruits. 4. In the manufacture of detergents. (iii) Alkynes (a)Nomenclature/Naming
50 These are hydrocarbons with a general formula CnH2n-2 and C C double bond as the functional group . n is the number of Carbon atoms in the molecule. The carbon atoms are linked by at least one triple bond to each other and single bonds to hydrogen atoms. They include: n General/ Molecular formula Structural formula Name 1 Does not exist - 2 C2H2 H C C H CH CH Ethyne 3 C3H4 H H C C C H H CH C CH3 Propyne 4 C4H6 H H H C C C C H H H CH C CH2CH3 Butyne 5 C5H8 H H H H C C C C C H H H H CH C (CH2)2CH3 Pentyne 6 C6H10 H H H H Hexyne
51 H C C C C C C H H H H H CH C (CH2)3CH3 7 C7H12 H H H H H H C C C C C C C H H H H H H H H CH C (CH2)4CH3 Heptyne 8 C8H14 H H H H H H H C C C C C C C C H H H H H H H CH C (CH2)5CH3 Octyne 9 C9H16 H H H H H H H H C C C C C C C C C H H H H H H H H CH C (CH2)6CH3 Nonyne 10 C10H18 H H H H H H H H H C C C C C C C C C C H H H H H H H H H CH C (CH2)7CH3 Decyne Note 1. Since carbon is tetravalent ,each atom of carbon in the alkyne MUST always be bonded using four covalent bond /four shared pairs of electrons including at the triple bond. 2. Since Hydrogen is monovalent ,each atom of hydrogen in the alkyne MUST always be bonded using one covalent bond/one shared pair of electrons. 52 3.


2. Since Hydrogen is monovalent ,each atom of hydrogen in the alkyne MUST always be bonded using one covalent bond/one shared pair of electrons. 52 3. One member of the alkyne ,like alkenes and alkanes, differ from the next/previous by a CH2 group(molar mass of 14 atomic mass units).They thus form a homologous series. e.g Propyne differ from ethyne by (14 a.m.u) one carbon and two Hydrogen atoms from ethyne. 4.A homologous series of alkenes like that of alkanes: (i) differ by a CH2 group from the next /previous consecutively (ii) have similar chemical properties (iii)have similar chemical formula with general formula CnH2n-2 (iv)the physical properties also show steady gradual change 5.The - C = C - triple bond in alkyne is the functional group. The functional group is the reacting site of the alkynes. 6. The - C = C - triple bond in alkyne can easily be broken to accommodate more /four more monovalent atoms. The - C = C - triple bond in alkynes make it thus unsaturated like alkenes. 7. Most of the reactions of alkynes like alkenes take place at the - C = C- triple bond. (b)Isomers of alkynes Isomers of alkynes have the same molecular general formula but different molecular structural formula. Isomers of alkynes are also named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming. The IUPAC system of nomenclature of naming alkynes uses the following basic rules/guidelines: 1.Identify the longest continuous/straight carbon chain which contains the - C = C- triple bond to get/determine the parent alkene. 2. Number the longest chain form the end of the chain which contains the -C = C- triple bond so as - C = C- triple bond get lowest number possible. 3 Indicate the positions by splitting “alk-positions-yne” e.g. but-2-yne, pent-1,3diyne. 4.The position indicated must be for the carbon atom at the lower position in the -C = C- triple bond.


3 Indicate the positions by splitting “alk-positions-yne” e.g. but-2-yne, pent-1,3diyne. 4.The position indicated must be for the carbon atom at the lower position in the -C = C- triple bond. i.e But-2-yne means the triple -C = C- is between Carbon “2”and “3” Pent-1,3-diyne means there are two triple bonds; one between carbon “1” and “2”and another between carbon “3” and “4”
53 5. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of alkyl carbon chains attached to the alkyne. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens 6.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of triple - C = C- bonds and branches attached to the alkyne. 7.Position isomers can be formed when the - C = C- triple bond is shifted between carbon atoms e.g. But-2-yne means the double - C = C- is between Carbon “2”and “3” But-1-yne means the double - C = C- is between Carbon “1”and “2” Both But-1-yne and But-2-yne are position isomers of Butyne. 9. Like alkanes and alkynes , an alkyl group can be attached to the alkyne. Chain/branch isomers are thus formed. Butyne and 2-methyl propyne both have the same general formular but different branching chain. (More on powerpoint) (c)Preparation of Alkynes. Ethyne is prepared from the reaction of water on calcium carbide. The reaction is highly exothermic and thus a layer of sand should be put above the calcium carbide to absorb excess heat to prevent the reaction flask from breaking. Copper(II)sulphate(VI) is used to catalyze the reaction Chemical equation
54 CaC2(s) + 2 H2O(l) -> Ca(OH) 2 (aq) + C2H2 (g) (d)Properties of alkynes I.


Ethyne is prepared from the reaction of water on calcium carbide. The reaction is highly exothermic and thus a layer of sand should be put above the calcium carbide to absorb excess heat to prevent the reaction flask from breaking. Copper(II)sulphate(VI) is used to catalyze the reaction Chemical equation
54 CaC2(s) + 2 H2O(l) -> Ca(OH) 2 (aq) + C2H2 (g) (d)Properties of alkynes I. Physical properties Like alkanes and alkenes,alkynes are colourles gases, solids and liquids that are not poisonous. They are slightly soluble in water. The solubility in water decrease as the carbon chain and as the molar mass increase but very soluble in organic solvents like tetrachloromethane and methylbenzene. Ethyne has a pleasant taste when pure. The melting and boiling point increase as the carbon chain increase. This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st three straight chain alkynes (ethyne,propyne and but-1yne)are gases at room temperature and pressure. The density of straight chain alkynes increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkyne. Summary of physical properties of the 1st five alkenes Alkyne General formula Melting point(oC) Boiling point(oC) State at room(298K) temperature and pressure atmosphere (101300Pa) Ethyne CH CH -82 -84 gas Propyne CH3 C CH -103 -23 gas Butyne CH3CH2 CCH -122 8 gas Pent-1-yne CH3(CH2) 2 CCH -119 39 liquid Hex-1-yne CH3(CH2) 3C CH -132 71 liquid II. Chemical properties (a)Burning/combustion Alkynes burn with a yellow/ luminous very sooty/ smoky flame in excess air to form carbon(IV) oxide and water. Alkyne + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkenes burn with a yellow/ luminous verysooty/ smoky flame in limited air to form carbon(II) oxide/carbon and water.


Summary of physical properties of the 1st five alkenes Alkyne General formula Melting point(oC) Boiling point(oC) State at room(298K) temperature and pressure atmosphere (101300Pa) Ethyne CH CH -82 -84 gas Propyne CH3 C CH -103 -23 gas Butyne CH3CH2 CCH -122 8 gas Pent-1-yne CH3(CH2) 2 CCH -119 39 liquid Hex-1-yne CH3(CH2) 3C CH -132 71 liquid II. Chemical properties (a)Burning/combustion Alkynes burn with a yellow/ luminous very sooty/ smoky flame in excess air to form carbon(IV) oxide and water. Alkyne + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkenes burn with a yellow/ luminous verysooty/ smoky flame in limited air to form carbon(II) oxide/carbon and water. 55 Alkyne + Air -> carbon(II) oxide /carbon + water (limited air) Burning of alkynes with a yellow/ luminous sooty/ smoky flame is a confirmatory test for the presence of the - C = C – triple bond because they have very high C:H ratio. Examples of burning alkynes 1.(a) Ethyne when ignited burns with a yellow very sooty flame in excess air to form carbon(IV) oxide and water. Ethyne + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(l/g) (b) Ethyne when ignited burns with a yellow sooty flame in limited air to form a mixture of unburnt carbon and carbon(II) oxide and water. Ethyne + Air -> carbon(II) oxide + water (limited air ) C2H2(g) + O2(g) -> 2CO2(g) + C + 2H2O(l/g) 2.(a) Propyne when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.


Examples of burning alkynes 1.(a) Ethyne when ignited burns with a yellow very sooty flame in excess air to form carbon(IV) oxide and water. Ethyne + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(l/g) (b) Ethyne when ignited burns with a yellow sooty flame in limited air to form a mixture of unburnt carbon and carbon(II) oxide and water. Ethyne + Air -> carbon(II) oxide + water (limited air ) C2H2(g) + O2(g) -> 2CO2(g) + C + 2H2O(l/g) 2.(a) Propyne when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water. Propyne + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H4(g) + 4O2(g) -> 3CO2(g) + 2H2O(l/g) (a) Propyne when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water. Propene + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C3H4(g) + 5O2(g) -> 6CO(g) + 4H2O(l/g) (b)Addition reactions An addition reaction is one which an unsaturated compound reacts to form a saturated compound. Addition reactions of alkynes are also named from the reagent used to cause the addition/convert the triple - C = C- to single C- C bond. (i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at 150oC temperatures react with alkynes to form alkenes then alkanes. Examples 1.During hydrogenation, two hydrogen atom in the hydrogen molecule attach itself to one carbon and the other two hydrogen to the second carbon breaking the triple bond to double the single.


Addition reactions of alkynes are also named from the reagent used to cause the addition/convert the triple - C = C- to single C- C bond. (i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at 150oC temperatures react with alkynes to form alkenes then alkanes. Examples 1.During hydrogenation, two hydrogen atom in the hydrogen molecule attach itself to one carbon and the other two hydrogen to the second carbon breaking the triple bond to double the single. Chemical equation
56 HC = CH + H2 -Ni/Pa -> H2C = CH2 + H2 -Ni/Pa -> H2C - CH2 H H H H H H C = C + H – H - Ni/Pa -> H - C = C – H + H – H - Ni/Pa -> H - C - C – H H H H H H H 2.Propyne undergo hydrogenation to form Propane Chemical equation H3C CH = CH2 + 2H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + 2H – H - Ni/Pa-> H - C – C - C- H H H H H H 3(a) But-1-yne undergo hydrogenation to form Butane Chemical equation But-1-yne + Hydrogen –Ni/Pa-> Butane H3C CH2 C = CH + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H C C - C = C + 2H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H (b) But-2-yne undergo hydrogenation to form Butane Chemical equation But-2-yne + Hydrogen –Ni/Pa-> Butane H3C C = C CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H C C = C - C H + 2H – H- Ni/Pa-> H - C- C – C - C- H H H H H H H
57 (ii) Halogenation.


(i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at 150oC temperatures react with alkynes to form alkenes then alkanes. Examples 1.During hydrogenation, two hydrogen atom in the hydrogen molecule attach itself to one carbon and the other two hydrogen to the second carbon breaking the triple bond to double the single. Chemical equation
56 HC = CH + H2 -Ni/Pa -> H2C = CH2 + H2 -Ni/Pa -> H2C - CH2 H H H H H H C = C + H – H - Ni/Pa -> H - C = C – H + H – H - Ni/Pa -> H - C - C – H H H H H H H 2.Propyne undergo hydrogenation to form Propane Chemical equation H3C CH = CH2 + 2H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + 2H – H - Ni/Pa-> H - C – C - C- H H H H H H 3(a) But-1-yne undergo hydrogenation to form Butane Chemical equation But-1-yne + Hydrogen –Ni/Pa-> Butane H3C CH2 C = CH + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H C C - C = C + 2H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H (b) But-2-yne undergo hydrogenation to form Butane Chemical equation But-2-yne + Hydrogen –Ni/Pa-> Butane H3C C = C CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H C C = C - C H + 2H – H- Ni/Pa-> H - C- C – C - C- H H H H H H H
57 (ii) Halogenation. Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkyne to form an alkene then alkane. The reaction of alkynes with halogens with alkynes is faster than with alkenes.


Chemical equation
56 HC = CH + H2 -Ni/Pa -> H2C = CH2 + H2 -Ni/Pa -> H2C - CH2 H H H H H H C = C + H – H - Ni/Pa -> H - C = C – H + H – H - Ni/Pa -> H - C - C – H H H H H H H 2.Propyne undergo hydrogenation to form Propane Chemical equation H3C CH = CH2 + 2H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + 2H – H - Ni/Pa-> H - C – C - C- H H H H H H 3(a) But-1-yne undergo hydrogenation to form Butane Chemical equation But-1-yne + Hydrogen –Ni/Pa-> Butane H3C CH2 C = CH + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H C C - C = C + 2H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H (b) But-2-yne undergo hydrogenation to form Butane Chemical equation But-2-yne + Hydrogen –Ni/Pa-> Butane H3C C = C CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H C C = C - C H + 2H – H- Ni/Pa-> H - C- C – C - C- H H H H H H H
57 (ii) Halogenation. Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkyne to form an alkene then alkane. The reaction of alkynes with halogens with alkynes is faster than with alkenes. The triple bond in the alkyne break and form a double then single bond. The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases. Two bromine atoms bond at the 1st carbon in the triple bond while the other two goes to the 2nd carbon.


The triple bond in the alkyne break and form a double then single bond. The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases. Two bromine atoms bond at the 1st carbon in the triple bond while the other two goes to the 2nd carbon. Examples 1Ethyne reacts with brown bromine vapour to form 1,1,2,2-tetrabromoethane. Chemical equation HC = CH + 2Br2 H Br2 C - CH Br2 H H H H C = C + 2Br – Br Br - C – C - Br Br Br Ethyne + Bromine 1,1,2,1-tetrabromoethane 2.Propyne reacts with chlorine to form 1,1,2,2-tetrachloropropane.


Two bromine atoms bond at the 1st carbon in the triple bond while the other two goes to the 2nd carbon. Examples 1Ethyne reacts with brown bromine vapour to form 1,1,2,2-tetrabromoethane. Chemical equation HC = CH + 2Br2 H Br2 C - CH Br2 H H H H C = C + 2Br – Br Br - C – C - Br Br Br Ethyne + Bromine 1,1,2,1-tetrabromoethane 2.Propyne reacts with chlorine to form 1,1,2,2-tetrachloropropane. Chemical equation H3C C = CH + 2Cl2 H3C CHCl2 - CHCl2 Propyne + Chlorine 1,1,2,2-tetrachloropropane H H Cl H H C C = C + 2Cl – Cl H - C – C - C- Cl H H H Cl Cl Propyne + Iodine 1,1,2,2-tetraiodopropane H3C C = CH + 2I2 H3C CHI2 - CHI2 H H H H H I H H C C - C = C + 2I – I H - C- C – C - C- I H H H H I I
58 3(a)But-1-yne undergo halogenation to form 1,1,2,2-tetraiodobutane with iodine Chemical equation But-1-yne + iodine 1,1,2,2-tetrabromobutane H3C CH2 C = CH + 2I2 H3C CH2C I2 - CHI2 H H H H I I H C C - C = C -H + 2I – I H - C- C – C - C- H H H H H H I I (b) But-2-yne undergo halogenation to form 2,2,3,3-tetrafluorobutane with fluorine But-2-yne + Fluorine 2,2,3,3-tetrafluorobutane H3C C = C -CH2 + 2F2 H3C CF2CF2 - CH3 H H H H H H H H H C C = C - C -H + F – F H - C- C – C - C- H H H H H H H 4.


Examples 1Ethyne reacts with brown bromine vapour to form 1,1,2,2-tetrabromoethane. Chemical equation HC = CH + 2Br2 H Br2 C - CH Br2 H H H H C = C + 2Br – Br Br - C – C - Br Br Br Ethyne + Bromine 1,1,2,1-tetrabromoethane 2.Propyne reacts with chlorine to form 1,1,2,2-tetrachloropropane. Chemical equation H3C C = CH + 2Cl2 H3C CHCl2 - CHCl2 Propyne + Chlorine 1,1,2,2-tetrachloropropane H H Cl H H C C = C + 2Cl – Cl H - C – C - C- Cl H H H Cl Cl Propyne + Iodine 1,1,2,2-tetraiodopropane H3C C = CH + 2I2 H3C CHI2 - CHI2 H H H H H I H H C C - C = C + 2I – I H - C- C – C - C- I H H H H I I
58 3(a)But-1-yne undergo halogenation to form 1,1,2,2-tetraiodobutane with iodine Chemical equation But-1-yne + iodine 1,1,2,2-tetrabromobutane H3C CH2 C = CH + 2I2 H3C CH2C I2 - CHI2 H H H H I I H C C - C = C -H + 2I – I H - C- C – C - C- H H H H H H I I (b) But-2-yne undergo halogenation to form 2,2,3,3-tetrafluorobutane with fluorine But-2-yne + Fluorine 2,2,3,3-tetrafluorobutane H3C C = C -CH2 + 2F2 H3C CF2CF2 - CH3 H H H H H H H H H C C = C - C -H + F – F H - C- C – C - C- H H H H H H H 4. But-1,3-diyne should undergo halogenation to form 1,1,2,3,3,4,4 octaiodobutane.


Chemical equation HC = CH + 2Br2 H Br2 C - CH Br2 H H H H C = C + 2Br – Br Br - C – C - Br Br Br Ethyne + Bromine 1,1,2,1-tetrabromoethane 2.Propyne reacts with chlorine to form 1,1,2,2-tetrachloropropane. Chemical equation H3C C = CH + 2Cl2 H3C CHCl2 - CHCl2 Propyne + Chlorine 1,1,2,2-tetrachloropropane H H Cl H H C C = C + 2Cl – Cl H - C – C - C- Cl H H H Cl Cl Propyne + Iodine 1,1,2,2-tetraiodopropane H3C C = CH + 2I2 H3C CHI2 - CHI2 H H H H H I H H C C - C = C + 2I – I H - C- C – C - C- I H H H H I I
58 3(a)But-1-yne undergo halogenation to form 1,1,2,2-tetraiodobutane with iodine Chemical equation But-1-yne + iodine 1,1,2,2-tetrabromobutane H3C CH2 C = CH + 2I2 H3C CH2C I2 - CHI2 H H H H I I H C C - C = C -H + 2I – I H - C- C – C - C- H H H H H H I I (b) But-2-yne undergo halogenation to form 2,2,3,3-tetrafluorobutane with fluorine But-2-yne + Fluorine 2,2,3,3-tetrafluorobutane H3C C = C -CH2 + 2F2 H3C CF2CF2 - CH3 H H H H H H H H H C C = C - C -H + F – F H - C- C – C - C- H H H H H H H 4. But-1,3-diyne should undergo halogenation to form 1,1,2,3,3,4,4 octaiodobutane. The reaction uses four moles of iodine molecules/eight iodine atoms to break the two(2) triple double bonds at carbon “1” and “2”.


Chemical equation H3C C = CH + 2Cl2 H3C CHCl2 - CHCl2 Propyne + Chlorine 1,1,2,2-tetrachloropropane H H Cl H H C C = C + 2Cl – Cl H - C – C - C- Cl H H H Cl Cl Propyne + Iodine 1,1,2,2-tetraiodopropane H3C C = CH + 2I2 H3C CHI2 - CHI2 H H H H H I H H C C - C = C + 2I – I H - C- C – C - C- I H H H H I I
58 3(a)But-1-yne undergo halogenation to form 1,1,2,2-tetraiodobutane with iodine Chemical equation But-1-yne + iodine 1,1,2,2-tetrabromobutane H3C CH2 C = CH + 2I2 H3C CH2C I2 - CHI2 H H H H I I H C C - C = C -H + 2I – I H - C- C – C - C- H H H H H H I I (b) But-2-yne undergo halogenation to form 2,2,3,3-tetrafluorobutane with fluorine But-2-yne + Fluorine 2,2,3,3-tetrafluorobutane H3C C = C -CH2 + 2F2 H3C CF2CF2 - CH3 H H H H H H H H H C C = C - C -H + F – F H - C- C – C - C- H H H H H H H 4. But-1,3-diyne should undergo halogenation to form 1,1,2,3,3,4,4 octaiodobutane. The reaction uses four moles of iodine molecules/eight iodine atoms to break the two(2) triple double bonds at carbon “1” and “2”. But-1,3-diene + iodine 1,2,3,4-tetraiodobutane H C = C C = C H + 4I2 H C I2 C I2 C I2 C H I2 I I I I H C C - C = C -H + 4(I – I) H - C- C – C - C- H I I I I (iii) Reaction with hydrogen halides.


But-1,3-diyne should undergo halogenation to form 1,1,2,3,3,4,4 octaiodobutane. The reaction uses four moles of iodine molecules/eight iodine atoms to break the two(2) triple double bonds at carbon “1” and “2”. But-1,3-diene + iodine 1,2,3,4-tetraiodobutane H C = C C = C H + 4I2 H C I2 C I2 C I2 C H I2 I I I I H C C - C = C -H + 4(I – I) H - C- C – C - C- H I I I I (iii) Reaction with hydrogen halides. Hydrogen halides reacts with alkyne to form a halogenoalkene then halogenoalkane. The triple bond in the alkyne break and form a double then single bond. 59 The main compound is one which the hydrogen atom bond at the carbon with more hydrogen . Examples 1. Ethyne reacts with hydrogen bromide to form bromoethane. Chemical equation H C = C H + 2HBr H3 C - CH Br2 H H H H C = C + 2H – Br H - C – C - Br H Br Ethyne + Bromine 1,1-dibromoethane 2. Propyne reacts with hydrogen iodide to form 2,2-diiodopropane (as the main product ) Chemical equation H3C C = CH + 2HI H3C CHI2 - CH3 Propene + Chlorine 2,2-dichloropropane H H I H H C C = C + 2H – I H - C – C - C- H H H H I H 3.


Ethyne reacts with hydrogen bromide to form bromoethane. Chemical equation H C = C H + 2HBr H3 C - CH Br2 H H H H C = C + 2H – Br H - C – C - Br H Br Ethyne + Bromine 1,1-dibromoethane 2. Propyne reacts with hydrogen iodide to form 2,2-diiodopropane (as the main product ) Chemical equation H3C C = CH + 2HI H3C CHI2 - CH3 Propene + Chlorine 2,2-dichloropropane H H I H H C C = C + 2H – I H - C – C - C- H H H H I H 3. Both But-1-yne and But-2-yne reacts with hydrogen bromide to form 2,2- dibromobutane Chemical equation But-1-ene + hydrogen bromide 2,2-dibromobutane H3C CH2 C = CH + 2HBr H3C CH2CHBr -CH3 H H H H Br H H C C - C = C + 2H – Br H - C- C – C - C- H H H H H H Br H Carbon atom with more Hydrogen atoms gets extra hydrogen
60 But-2-yne + Hydrogen bromide 2,2-dibromobutane H3C C = C -CH3 + 2HBr H3C CBr2CH2 - CH3 H H H Br H H H C C = C - C -H + 2Br – H H - C- C – C - C- H H H H Br H H 4. But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses four moles of hydrogen iodide molecules/four iodine atoms and two hydrogen atoms to break the two double bonds.


Both But-1-yne and But-2-yne reacts with hydrogen bromide to form 2,2- dibromobutane Chemical equation But-1-ene + hydrogen bromide 2,2-dibromobutane H3C CH2 C = CH + 2HBr H3C CH2CHBr -CH3 H H H H Br H H C C - C = C + 2H – Br H - C- C – C - C- H H H H H H Br H Carbon atom with more Hydrogen atoms gets extra hydrogen
60 But-2-yne + Hydrogen bromide 2,2-dibromobutane H3C C = C -CH3 + 2HBr H3C CBr2CH2 - CH3 H H H Br H H H C C = C - C -H + 2Br – H H - C- C – C - C- H H H H Br H H 4. But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses four moles of hydrogen iodide molecules/four iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diyne + iodine 2,2,3,3-tetraiodobutane H C = C C = C H + 4HI H3C C I2 C I2 CH3 H H H I I H H C C - C = C -H + 4(H – I) H - C- C – C - C- H H I I H B.ALKANOLS(Alcohols)
61 (A) INTRODUCTION.


But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses four moles of hydrogen iodide molecules/four iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diyne + iodine 2,2,3,3-tetraiodobutane H C = C C = C H + 4HI H3C C I2 C I2 CH3 H H H I I H H C C - C = C -H + 4(H – I) H - C- C – C - C- H H I I H B.ALKANOLS(Alcohols)
61 (A) INTRODUCTION. Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH C2H5 OH H H H C – C –O - H │ H H Ethanol 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ Hexanol
62 H H H H H H 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group.


The reaction uses four moles of hydrogen iodide molecules/four iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diyne + iodine 2,2,3,3-tetraiodobutane H C = C C = C H + 4HI H3C C I2 C I2 CH3 H H H I I H H C C - C = C -H + 4(H – I) H - C- C – C - C- H H I I H B.ALKANOLS(Alcohols)
61 (A) INTRODUCTION. Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH C2H5 OH H H H C – C –O - H │ H H Ethanol 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ Hexanol
62 H H H H H H 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group. (iv) each member differ by –CH2 group from the next/previous.


But-1,3-diyne + iodine 2,2,3,3-tetraiodobutane H C = C C = C H + 4HI H3C C I2 C I2 CH3 H H H I I H H C C - C = C -H + 4(H – I) H - C- C – C - C- H H I I H B.ALKANOLS(Alcohols)
61 (A) INTRODUCTION. Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH C2H5 OH H H H C – C –O - H │ H H Ethanol 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ Hexanol
62 H H H H H H 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group. (iv) each member differ by –CH2 group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g.


Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include n General / molecular formular Structural formula IUPAC name 1 CH3OH H – C –O - H │ H Methanol 2 CH3 CH2OH C2H5 OH H H H C – C –O - H │ H H Ethanol 3 CH3 (CH2)2OH C3H7 OH H H H H C – C - C –O - H │ H H H Propanol 4 CH3 (CH2)3OH C4H9 OH H H H H H C – C - C - C –O - H │ H H H H Butanol 5 CH3(CH2)4OH C5H11 OH H H H H H H C – C - C- C- C –O - H │ H H H H H Pentanol 6 CH3(CH2)5OH C6H13 OH H H H H H H H C – C - C- C- C– C - O - H │ Hexanol
62 H H H H H H 7 CH3(CH2)6OH C7H15 OH H H H H H H H H C – C - C- C- C– C –C- O - H │ H H H H H H H Heptanol 8 CH3(CH2)7OH C8H17 OH H H H H H H H H H C – C - C- C- C– C –C- C -O - H │ H H H H H H H H Octanol 9 CH3(CH2)8OH C9H19 OH H H H H H H H H H H C – C - C- C- C– C –C- C –C- O - H │ H H H H H H H H H Nonanol 10 CH3(CH2)9OH C10H21 OH H H H H H H H H H H H C – C - C- C- C– C –C- C –C- C-O - H │ H H H H H H H H H H Decanol Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where: (i)general name is derived from the alkane name then ending with “-ol” (ii)the members have –OH as the fuctional group (iii)they have the same general formula represented by R-OH where R is an alkyl group. (iv) each member differ by –CH2 group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g. boiling and melting points.


(iv) each member differ by –CH2 group from the next/previous. (v)they show a similar and gradual change in their physical properties e.g. boiling and melting points. (vi)they show similar and gradual change in their chemical properties. B. ISOMERS OF ALKANOLS. 63 Alkanols exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines: (i)Like alkanes , identify the longest carbon chain to be the parent name. (ii)Identify the position of the -OH functional group to give it the smallest /lowest position. (iii) Identify the type and position of the side branches. Practice examples of isomers of alkanols (i)Isomers of propanol C3H7OH CH3CH2CH2OH - Propan-1-ol OH CH3CHCH3 - Propan-2-ol Propan-2-ol and Propan-1-ol are position isomers because only the position of the –OH functional group changes. (ii)Isomers of Butanol C4H9OH CH3 CH2 CH3 CH2 OH Butan-1-ol CH3 CH2 CH CH3 OH Butan-2-ol CH3 CH3 CH3 CH3 OH 2-methylpropan-2-ol Butan-2-ol and Butan-1-ol are position isomers because only the position of the -OH functional group changes. 2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes.


Practice examples of isomers of alkanols (i)Isomers of propanol C3H7OH CH3CH2CH2OH - Propan-1-ol OH CH3CHCH3 - Propan-2-ol Propan-2-ol and Propan-1-ol are position isomers because only the position of the –OH functional group changes. (ii)Isomers of Butanol C4H9OH CH3 CH2 CH3 CH2 OH Butan-1-ol CH3 CH2 CH CH3 OH Butan-2-ol CH3 CH3 CH3 CH3 OH 2-methylpropan-2-ol Butan-2-ol and Butan-1-ol are position isomers because only the position of the -OH functional group changes. 2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes. (iii)Isomers of Pentanol C5H11OH CH3 CH2 CH2CH2CH2 OH Pentan-1-ol (Position isomer)
64 CH3 CH2 CH CH3 OH Pentan-2-ol (Position isomer) CH3 CH2 CH CH2 CH3 OH Pentan-3-ol (Position isomer) CH3 CH3 CH2 CH2 C CH3 OH 2-methylbutan-2-ol (Position /structural isomer) CH3 CH3 CH2 CH2 C CHOH CH3 2,2-dimethylbutan-1-ol (Position /structural isomer) CH3 CH3 CH2 CH C CH3 CH3 OH 2,3-dimethylbutan-1-ol (Position /structural isomer) (iv)1,2-dichloropropan-2-ol CClH2 CCl CH3 OH (v)1,2-dichloropropan-1-ol CClH2 CHCl CH2 OH
65 (vi) Ethan1,2-diol H H HOCH2CH2OH H-O - C - C – O-H H H (vii) Propan1,2,3-triol H OH H HOCH2CHOHCH2OH H-O - C- C – C – O-H H H H C. LABORATORY PREPARATION OF ALKANOLS.


2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes. (iii)Isomers of Pentanol C5H11OH CH3 CH2 CH2CH2CH2 OH Pentan-1-ol (Position isomer)
64 CH3 CH2 CH CH3 OH Pentan-2-ol (Position isomer) CH3 CH2 CH CH2 CH3 OH Pentan-3-ol (Position isomer) CH3 CH3 CH2 CH2 C CH3 OH 2-methylbutan-2-ol (Position /structural isomer) CH3 CH3 CH2 CH2 C CHOH CH3 2,2-dimethylbutan-1-ol (Position /structural isomer) CH3 CH3 CH2 CH C CH3 CH3 OH 2,3-dimethylbutan-1-ol (Position /structural isomer) (iv)1,2-dichloropropan-2-ol CClH2 CCl CH3 OH (v)1,2-dichloropropan-1-ol CClH2 CHCl CH2 OH
65 (vi) Ethan1,2-diol H H HOCH2CH2OH H-O - C - C – O-H H H (vii) Propan1,2,3-triol H OH H HOCH2CHOHCH2OH H-O - C- C – C – O-H H H H C. LABORATORY PREPARATION OF ALKANOLS. For decades the world over, people have been fermenting grapes juice, sugar, carbohydrates and starch to produce ethanol as a social drug for relaxation. In large amount, drinking of ethanol by mammals /human beings causes mental and physical lack of coordination. Prolonged intake of ethanol causes permanent mental and physical lack of coordination because it damages vital organs like the liver. Fermentation is the reaction where sugar is converted to alcohol/alkanol using biological catalyst/enzymes in yeast. It involves three processes: (i)Conversion of starch to maltose using the enzyme diastase.


Prolonged intake of ethanol causes permanent mental and physical lack of coordination because it damages vital organs like the liver. Fermentation is the reaction where sugar is converted to alcohol/alkanol using biological catalyst/enzymes in yeast. It involves three processes: (i)Conversion of starch to maltose using the enzyme diastase. (C6H10O5)n (s) + H2O(l) --diastase enzyme --> C12H22O11(aq) (Starch) (Maltose) (ii)Hydrolysis of Maltose to glucose using the enzyme maltase. C12H22O11(aq)+ H2O(l) -- maltase enzyme -->2 C6H12O6(aq) (Maltose) (glucose) (iii)Conversion of glucose to ethanol and carbon(IV)oxide gas using the enzyme zymase. C6H12O6(aq) -- zymase enzyme --> 2 C2H5OH(aq) + 2CO2(g) (glucose) (Ethanol) At concentration greater than 15% by volume, the ethanol produced kills the yeast enzyme stopping the reaction. 66 To increases the concentration, fractional distillation is done to produce spirits (e.g. Brandy=40% ethanol). Methanol is much more poisonous /toxic than ethanol. Taken large quantity in small quantity it causes instant blindness and liver, killing the consumer victim within hours. School laboratory preparation of ethanol from fermentation of glucose Measure 100cm3 of pure water into a conical flask. Add about five spatula end full of glucose. Stir the mixture to dissolve. Add about one spatula end full of yeast. Set up the apparatus as below. Preserve the mixture for about three days. D.PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOLS Use the prepared sample above for the following experiments that shows the characteristic properties of alkanols (a) Role of yeast
67 Yeast is a single cell fungus which contains the enzyme maltase and zymase that catalyse the fermentation process. (b) Observations in lime water. A white precipitate is formed that dissolve to a colourless solution later. Lime water/Calcium hydroxide reacts with carbon(IV)0xide produced during the fermentation to form insoluble calcium carbonate and water.


(b) Observations in lime water. A white precipitate is formed that dissolve to a colourless solution later. Lime water/Calcium hydroxide reacts with carbon(IV)0xide produced during the fermentation to form insoluble calcium carbonate and water. More carbon (IV)0xide produced during fermentation react with the insoluble calcium carbonate and water to form soluble calcium hydrogen carbonate. Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) H2O(l) + CO2 (g) + CaCO3(s) -> Ca(HCO3) 2 (aq) (c)Effects on litmus paper Experiment Take the prepared sample and test with both blue and red litmus papers. Repeat the same with pure ethanol and methylated spirit. Sample Observation table Substance/alkanol Effect on litmus paper Prepared sample Blue litmus paper remain blue Red litmus paper remain red Absolute ethanol Blue litmus paper remain blue Red litmus paper remain red Methylated spirit Blue litmus paper remain blue Red litmus paper remain red Explanation Alkanols are neutral compounds/solution that have characteristic sweet smell and taste. They have no effect on both blue and red litmus papers. (d)Solubility in water. Experiment Place about 5cm3 of prepared sample into a clean test tube Add equal amount of distilled water. Repeat the same with pure ethanol and methylated spirit. Observation No layers formed between the two liquids. Explanation Ethanol is miscible in water.Both ethanol and water are polar compounds . 68 The solubility of alkanols decrease with increase in the alkyl chain/molecular mass. The alkyl group is insoluble in water while –OH functional group is soluble in water. As the molecular chain becomes longer ,the effect of the alkyl group increases as the effect of the functional group decreases. e)Melting/boiling point. Experiment Place pure ethanol in a long boiling tube .Determine its boiling point. Observation Pure ethanol has a boiling point of 78oC at sea level/one atmosphere pressure. Explanation The melting and boiling point of alkanols increase with increase in molecular chain/mass . This is because the intermolecular/van-der-waals forces of attraction between the molecules increase. More heat energy is thus required to weaken the longer chain during melting and break during boiling.


Explanation The melting and boiling point of alkanols increase with increase in molecular chain/mass . This is because the intermolecular/van-der-waals forces of attraction between the molecules increase. More heat energy is thus required to weaken the longer chain during melting and break during boiling. f)Density Density of alkanols increase with increase in the intermolecular/van-der-waals forces of attraction between the molecule, making it very close to each other. This reduces the volume occupied by the molecule and thus increase the their mass per unit volume (density). Summary table showing the trend in physical properties of alkanols Alkanol Melting point (oC) Boiling point (oC) Density gcm-3 Solubility in water Methanol -98 65 0.791 soluble Ethanol -117 78 0.789 soluble Propanol -103 97 0.803 soluble Butanol -89 117 0.810 Slightly soluble Pentanol -78 138 0.814 Slightly soluble Hexanol -52 157 0.815 Slightly soluble Heptanol -34 176 0.822 Slightly soluble Octanol -15 195 0.824 Slightly soluble Nonanol -7 212 0.827 Slightly soluble Decanol 6 228 0.827 Slightly soluble
69 g)Burning Experiment Place the prepared sample in a watch glass. Ignite. Repeat with pure ethanol and methylated spirit. Observation/Explanation Fermentation produce ethanol with a lot of water(about a ratio of 1:3)which prevent the alcohol from igniting. Pure ethanol and methylated spirit easily catch fire / highly flammable. They burn with an almost colourless non-sooty/non-smoky blue flame to form carbon(IV) oxide (in excess air/oxygen)or carbon(II) oxide (limited air) and water. Ethanol is thus a saturated compound like alkanes.


Pure ethanol and methylated spirit easily catch fire / highly flammable. They burn with an almost colourless non-sooty/non-smoky blue flame to form carbon(IV) oxide (in excess air/oxygen)or carbon(II) oxide (limited air) and water. Ethanol is thus a saturated compound like alkanes. Chemica equation C2 H5OH(l) + 3O2 (g) -> 3H2O(l) + 2CO2 (g) ( excess air) C2 H5OH(l) + 2O2 (g) -> 3H2O(l) + 2CO (g) ( limited air) 2CH3OH(l) + 3O2 (g) -> 4H2O(l) + 2CO2 (g) ( excess air) 2 CH3OH(l) + 2O2 (g) -> 4H2O(l) + 2CO (g) ( limited air) 2C3 H7OH(l) + 9O2 (g) -> 8H2O(l) + 6CO2 (g) ( excess air) C3 H7OH(l) + 3O2 (g) -> 4H2O(l) + 3CO (g) ( limited air) 2C4 H9OH(l) + 13O2 (g) -> 20H2O(l) + 8CO2 (g) ( excess air) C4 H9OH(l) + 3O2 (g) -> 4H2O(l) + 3CO (g) ( limited air) Due to its flammability, ethanol is used; (i) as a fuel in spirit lamps (ii) as gasohol when blended with gasoline (h)Formation of alkoxides Experiment Cut a very small piece of sodium. Put it in a beaker containing about 20cm3 of the prepared sample in a beaker. Test the products with litmus papers. Repeat with absolute ethanol and methylated spirit.


Put it in a beaker containing about 20cm3 of the prepared sample in a beaker. Test the products with litmus papers. Repeat with absolute ethanol and methylated spirit. Sample observations Substance/alkanol Effect of adding sodium Fermentation prepared sample (i)effervescence/fizzing/bubbles
70 (ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Pure/absolute ethanol/methylated spirit (i)slow effervescence/fizzing/bubbles (ii)colourless gas slowly produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Explanations Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas. If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e.


Repeat with absolute ethanol and methylated spirit. Sample observations Substance/alkanol Effect of adding sodium Fermentation prepared sample (i)effervescence/fizzing/bubbles
70 (ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Pure/absolute ethanol/methylated spirit (i)slow effervescence/fizzing/bubbles (ii)colourless gas slowly produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Explanations Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas. If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e. Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
71 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube.


Sample observations Substance/alkanol Effect of adding sodium Fermentation prepared sample (i)effervescence/fizzing/bubbles
70 (ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Pure/absolute ethanol/methylated spirit (i)slow effervescence/fizzing/bubbles (ii)colourless gas slowly produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue Explanations Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas. If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e. Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
71 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube. Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid.


If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e. Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
71 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube. Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid. Warm/Heat gently.


Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas Sodium + Water -> Sodium hydroxides + Hydrogen gas Potassium + Water -> Potassium hydroxides + Hydrogen gas Examples 1.Sodium metal reacts with ethanol to form sodium ethoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 2.Potassium metal reacts with ethanol to form Potassium ethoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 3.Sodium metal reacts with propanol to form sodium propoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
71 4.Potassium metal reacts with propanol to form Potassium propoxide Potassium metal reacts with water to form Potassium Hydroxide 2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s) 2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s) 5.Sodium metal reacts with butanol to form sodium butoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) 6.Sodium metal reacts with pentanol to form sodium pentoxide Sodium metal reacts with water to form sodium Hydroxide 2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s) 2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s) (i)Formation of Esters/Esterification Experiment Place 2cm3 of ethanol in a boiling tube. Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid. Warm/Heat gently. Pour the mixture into a beaker containing about 50cm3 of cold water.


Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid. Warm/Heat gently. Pour the mixture into a beaker containing about 50cm3 of cold water. Smell the products. Repeat with methanol Sample observations Substance/alkanol Effect on adding equal amount of ethanol/concentrated sulphuric(VI)acid Absolute ethanol Sweet fruity smell Methanol Sweet fruity smell Explanation Alkanols react with alkanoic acids to form a group of homologous series of sweet smelling compounds called esters and water. This reaction is catalyzed by concentrated sulphuric(VI)acid in the laboratory. Alkanol + Alkanoic acid –Conc. H2SO4-> Ester + water Naturally esterification is catalyzed by sunlight. Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids that
72 create a variety of known natural(mostly in fruits) and synthetic(mostly in juices) esters . Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g. Ethanol + Ethanoic acid -> Ethylethanoate + Water Ethanol + Propanoic acid -> Ethylpropanoate + Water Ethanol + Methanoic acid -> Ethylmethanoate + Water Ethanol + butanoic acid -> Ethylbutanoate + Water Propanol + Ethanoic acid -> Propylethanoate + Water Methanol + Ethanoic acid -> Methyethanoate + Water Methanol + Decanoic acid -> Methyldecanoate + Water Decanol + Methanoic acid -> Decylmethanoate + Water During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol. R1 -COOH + R2 –OH -> R1 -COO –R2 + H2O e.g. 1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water. Ethanol + Ethanoic acid --Conc.


1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water. Ethanol + Ethanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C2H5OH (l) + CH3COOH(l) --Conc. H2SO4 --> CH3COO C2H5(aq) +H2O(l) CH3CH2OH (l)+ CH3COOH(l) --Conc. H2SO4 --> CH3COOCH2CH3(aq) +H2O(l) 2. Ethanol reacts with propanoic acid to form the ester ethylpropanoate and water. Ethanol + Propanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C2H5OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 -->CH3CH2COO C2H5(aq) +H2O(l) CH3CH2OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COOCH2CH3(aq) +H2O(l) 3. Methanol reacts with ethanoic acid to form the ester methyl ethanoate and water. Methanol + Ethanoic acid --Conc. H2SO4 -->Methylethanoate + Water CH3OH (l) + CH3COOH(l) --Conc. H2SO4 --> CH3COO CH3(aq) +H2O(l) 4. Methanol reacts with propanoic acid to form the ester methyl propanoate and water. Methanol + propanoic acid --Conc. H2SO4 -->Methylpropanoate + Water CH3OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COO CH3(aq) +H2O(l) 5. Propanol reacts with propanoic acid to form the ester propylpropanoate and water. 73 Propanol + Propanoic acid --Conc.


H2SO4 --> CH3 CH2COO CH3(aq) +H2O(l) 5. Propanol reacts with propanoic acid to form the ester propylpropanoate and water. 73 Propanol + Propanoic acid --Conc. H2SO4 -->Ethylethanoate + Water C3H7OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 -->CH3CH2COO C3H7(aq) +H2O(l) CH3CH2 CH2OH (l)+ CH3 CH2COOH(l) --Conc. H2SO4 --> CH3 CH2COOCH2 CH2CH3(aq) +H2O(l) (j)Oxidation Experiment Place 5cm3 of absolute ethanol in a test tube.Add three drops of acidified potassium manganate(VII).Shake thoroughly for one minute/warm.Test the solution mixture using pH paper. Repeat by adding acidified potassium dichromate(VII). Sample observation table Substance/alkanol Adding acidified KMnO4/K2Cr2O7 pH of resulting solution/mixture Nature of resulting solution/mixture Pure ethanol (i)Purple colour of KMnO4decolorized (ii) Orange colour of K2Cr2O7turns green. pH= 4/5/6 pH = 4/5/6 Weakly acidic Weakly acidic Explanation Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds.


Repeat by adding acidified potassium dichromate(VII). Sample observation table Substance/alkanol Adding acidified KMnO4/K2Cr2O7 pH of resulting solution/mixture Nature of resulting solution/mixture Pure ethanol (i)Purple colour of KMnO4decolorized (ii) Orange colour of K2Cr2O7turns green. pH= 4/5/6 pH = 4/5/6 Weakly acidic Weakly acidic Explanation Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds. They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour: (i) Purple KMnO4 is reduced to colourless Mn2+ (ii)Orange K2Cr2O7is reduced to green Cr3+ The pH of alkanoic acids show they have few H+ because they are weak acids i.e Alkanol + [O] -> Alkanal + [O] -> alkanoic acid NB The [O] comes from the oxidizing agents acidified KMnO4 or K2Cr2O7 Examples 1.When ethanol is warmed with three drops of acidified KMnO4 there is decolorization of KMnO4 Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH 2.When methanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. 74 methanol + [O] -> methanal + [O] -> methanoic acid CH3OH + [O] -> CH3O + [O] -> HCOOH 3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.


pH= 4/5/6 pH = 4/5/6 Weakly acidic Weakly acidic Explanation Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds. They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour: (i) Purple KMnO4 is reduced to colourless Mn2+ (ii)Orange K2Cr2O7is reduced to green Cr3+ The pH of alkanoic acids show they have few H+ because they are weak acids i.e Alkanol + [O] -> Alkanal + [O] -> alkanoic acid NB The [O] comes from the oxidizing agents acidified KMnO4 or K2Cr2O7 Examples 1.When ethanol is warmed with three drops of acidified KMnO4 there is decolorization of KMnO4 Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH 2.When methanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. 74 methanol + [O] -> methanal + [O] -> methanoic acid CH3OH + [O] -> CH3O + [O] -> HCOOH 3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Propanol + [O] -> Propanal + [O] -> Propanoic acid CH3CH2 CH2OH + [O] -> CH3CH2 CH2O + [O] -> CH3 CH2COOH 4.When butanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.


They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour: (i) Purple KMnO4 is reduced to colourless Mn2+ (ii)Orange K2Cr2O7is reduced to green Cr3+ The pH of alkanoic acids show they have few H+ because they are weak acids i.e Alkanol + [O] -> Alkanal + [O] -> alkanoic acid NB The [O] comes from the oxidizing agents acidified KMnO4 or K2Cr2O7 Examples 1.When ethanol is warmed with three drops of acidified KMnO4 there is decolorization of KMnO4 Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH 2.When methanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. 74 methanol + [O] -> methanal + [O] -> methanoic acid CH3OH + [O] -> CH3O + [O] -> HCOOH 3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Propanol + [O] -> Propanal + [O] -> Propanoic acid CH3CH2 CH2OH + [O] -> CH3CH2 CH2O + [O] -> CH3 CH2COOH 4.When butanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Butanol + [O] -> Butanal + [O] -> Butanoic acid CH3CH2 CH2 CH2OH + [O] ->CH3CH2 CH2CH2O +[O] -> CH3 CH2COOH Air slowly oxidizes ethanol to dilute ethanoic acid commonly called vinegar. If beer is not tightly corked, a lot of carbon(IV)oxide escapes and there is slow oxidation of the beer making it “flat”.


Propanol + [O] -> Propanal + [O] -> Propanoic acid CH3CH2 CH2OH + [O] -> CH3CH2 CH2O + [O] -> CH3 CH2COOH 4.When butanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green. Butanol + [O] -> Butanal + [O] -> Butanoic acid CH3CH2 CH2 CH2OH + [O] ->CH3CH2 CH2CH2O +[O] -> CH3 CH2COOH Air slowly oxidizes ethanol to dilute ethanoic acid commonly called vinegar. If beer is not tightly corked, a lot of carbon(IV)oxide escapes and there is slow oxidation of the beer making it “flat”. (k)Hydrolysis /Hydration and Dehydration I. Hydrolysis/Hydration is the reaction of a compound/substance with water. Alkenes react with water vapour/steam at high temperatures and high pressures in presence of phosphoric acid catalyst to form alkanols.i.e. Alkenes + Water - H3PO4 catalyst-> Alkanol Examples (i)Ethene is mixed with steam over a phosphoric acid catalyst at 300oC temperature and 60 atmosphere pressure to form ethanol Ethene + water ---60 atm/300oC/ H3PO4 --> Ethanol H2C =CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2OH(l) This is the main method of producing large quantities of ethanol instead of fermentation (ii) Propene + water ---60 atm/300oC/ H3PO4 --> Propanol CH3C =CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2 CH2OH(l) (iii) Butene + water ---60 atm/300oC/ H3PO4 --> Butanol CH3 CH2 C=CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2 CH2 CH2OH(l) II.


Hydrolysis/Hydration is the reaction of a compound/substance with water. Alkenes react with water vapour/steam at high temperatures and high pressures in presence of phosphoric acid catalyst to form alkanols.i.e. Alkenes + Water - H3PO4 catalyst-> Alkanol Examples (i)Ethene is mixed with steam over a phosphoric acid catalyst at 300oC temperature and 60 atmosphere pressure to form ethanol Ethene + water ---60 atm/300oC/ H3PO4 --> Ethanol H2C =CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2OH(l) This is the main method of producing large quantities of ethanol instead of fermentation (ii) Propene + water ---60 atm/300oC/ H3PO4 --> Propanol CH3C =CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2 CH2OH(l) (iii) Butene + water ---60 atm/300oC/ H3PO4 --> Butanol CH3 CH2 C=CH2 (g) + H2O(l) --60 atm/300oC/ H3PO4 --> CH3 CH2 CH2 CH2OH(l) II. Dehydration is the process which concentrated sulphuric(VI)acid (dehydrating agent) removes water from a compound/substances. Concentrated sulphuric(VI)acid dehydrates alkanols to the corresponding alkenes at about 180oC. i.e Alkanol --Conc. H2 SO4/180oC--> Alkene + Water Examples
75 1. At 180oC and in presence of Concentrated sulphuric(VI)acid, ethanol undergoes dehydration to form ethene. Ethanol ---180oC/ H2SO4 --> Ethene + Water CH3 CH2OH(l) --180oC/ H2SO4 --> H2C =CH2 (g) + H2O(l) 2. Propanol undergoes dehydration to form propene.


At 180oC and in presence of Concentrated sulphuric(VI)acid, ethanol undergoes dehydration to form ethene. Ethanol ---180oC/ H2SO4 --> Ethene + Water CH3 CH2OH(l) --180oC/ H2SO4 --> H2C =CH2 (g) + H2O(l) 2. Propanol undergoes dehydration to form propene. Propanol ---180oC/ H2SO4 --> Propene + Water CH3 CH2 CH2OH(l) --180oC/ H2SO4 --> CH3CH =CH2 (g) + H2O(l) 3. Butanol undergoes dehydration to form Butene. Butanol ---180oC/ H2SO4 --> Butene + Water CH3 CH2 CH2CH2OH(l) --180oC/ H2SO4 --> CH3 CH2C =CH2 (g) + H2O(l) 3. Pentanol undergoes dehydration to form Pentene. Pentanol ---180oC/ H2SO4 --> Pentene + Water CH3 CH2 CH2 CH2 CH2OH(l)--180oC/ H2SO4-->CH3 CH2 CH2C =CH2 (g)+H2O(l) (l)Similarities of alkanols with Hydrocarbons I. Similarity with alkanes Both alkanols and alkanes burn with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water. This shows they are saturated with high C:H ratio. e.g. Both ethanol and ethane ignite and burns in air with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water.


This shows they are saturated with high C:H ratio. e.g. Both ethanol and ethane ignite and burns in air with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water. CH2 CH2OH(l) + 3O2(g) -Excess air-> 2CO2 (g) + 3H2 O(l) CH2 CH2OH(l) + 2O2(g) -Limited air-> 2CO (g) + 3H2 O(l) CH3 CH3(g) + 3O2(g) -Excess air-> 2CO2 (g) + 3H2 O(l) 2CH3 CH3(g) + 5O2(g) -Limited air-> 4CO (g) + 6H2 O(l) II. Similarity with alkenes/alkynes Both alkanols(R-OH) and alkenes/alkynes(with = C = C = double and – C = C- triple ) bond: (i)decolorize acidified KMnO4 (ii)turns Orange acidified K2Cr2O7 to green. Alkanols(R-OH) are oxidized to alkanals(R-O) ant then alkanoic acids(R-OOH). Alkenes are oxidized to alkanols with duo/double functional groups. Examples 1.When ethanol is warmed with three drops of acidified K2Cr2O7 the orange of acidified K2Cr2O7 turns to green. Ethanol is oxidized to ethanol and then to ethanoic acid. 76 Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH 2.When ethene is bubbled in a test tube containing acidified K2Cr2O7 ,the orange of acidified K2Cr2O7 turns to green. Ethene is oxidized to ethan-1,2-diol. Ethene + [O] -> Ethan-1,2-diol. H2C=CH2 + [O] -> HOCH2 -CH2OH III. Differences with alkenes/alkynes Alkanols do not decolorize bromine and chlorine water.


Ethene + [O] -> Ethan-1,2-diol. H2C=CH2 + [O] -> HOCH2 -CH2OH III. Differences with alkenes/alkynes Alkanols do not decolorize bromine and chlorine water. Alkenes decolorizes bromine and chlorine water to form halogenoalkanols Example When ethene is bubbled in a test tube containing bromine water,the bromine water is decolorized. Ethene is oxidized to bromoethanol. Ethene + Bromine water -> Bromoethanol. H2C=CH2 + HOBr -> BrCH2 -CH2OH IV. Differences in melting and boiling point with Hydrocarbons Alkanos have higher melting point than the corresponding hydrocarbon (alkane/alkene/alkyne) This is because most alkanols exist as dimer.A dimer is a molecule made up of two other molecules joined usually by van-der-waals forces/hydrogen bond or dative bonding. Two alkanol molecules form a dimer joined by hydrogen bonding. Example In Ethanol the oxygen atom attracts/pulls the shared electrons in the covalent bond more to itself than Hydrogen. This creates a partial negative charge (δ-) on oxygen and partial positive charge(δ+) on hydrogen. Two ethanol molecules attract each other at the partial charges through Hydrogen bonding forming a dimmer. H H H H C C O H H H H H O C C H Hydrogen bonds Covalent bonds
77 H H Dimerization of alkanols means more energy is needed to break/weaken the Hydrogen bonds before breaking/weakening the intermolecular forces joining the molecules of all organic compounds during boiling/melting. E.USES OF SOME ALKANOLS (a)Methanol is used as industrial alcohol and making methylated spirit (b)Ethanol is used: 1. as alcohol in alcoholic drinks e.g Beer, wines and spirits. 2.as antiseptic to wash woulds 3.in manufacture of vanishes, ink ,glue and paint because it is volatile and thus easily evaporate 4.as a fuel when blended with petrol to make gasohol. B.ALKANOIC ACIDS (Carboxylic acids) (A) INTRODUCTION.


as alcohol in alcoholic drinks e.g Beer, wines and spirits. 2.as antiseptic to wash woulds 3.in manufacture of vanishes, ink ,glue and paint because it is volatile and thus easily evaporate 4.as a fuel when blended with petrol to make gasohol. B.ALKANOIC ACIDS (Carboxylic acids) (A) INTRODUCTION. Alkanoic acids belong to a homologous series of organic compounds with a general formula CnH2n +1 COOH and thus -COOH as the functional group .The 1st ten alkanoic acids include:
78 Alkanoic acids like alkanols /alkanes/alkenes/alkynes form a homologous series where: (i)the general name of an alkanoic acids is derived from the alkane name then ending with “–oic” acid as the table above shows. (ii) the members have R-COOH/R C-O-H as the functional group.