JK-TK/Newdata huggingface dataset:
This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st four straight chain alkenes (ethene,propane,but-1-ene and pent-1-ene)are gases at room temperature and pressure. The density of straight chain alkenes,like alkanes, increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkene. Summary of physical properties of the 1st five alkenes Alkene General formula Melting point(oC) Boiling point(K) State at room(298K) temperature and pressure atmosphere (101300Pa)
27 Ethene CH2CH2 -169 -104 gas Propene CH3 CHCH2 -145 -47 gas Butene CH3CH2 CHCH2 -141 -26 gas Pent-1ene CH3(CH2 CHCH2 -138 30 liquid Hex-1ene CH3(CH2) CHCH2 -98 64 liquid II. Chemical properties (a)Burning/combustion Alkenes burn with a yellow/ luminous sooty/ smoky flame in excess air to form carbon(IV) oxide and water. Alkene + Air -> carbon(IV) oxide + water (excess air/oxygen) Alkenes burn with a yellow/ luminous sooty/ smoky flame in limited air to form carbon(II) oxide and water. Alkene + Air -> carbon(II) oxide + water (limited air) Burning of alkenes with a yellow/ luminous sooty/ smoky flame is a confirmatory test for the presence of the =C=C= double bond because they have higher C:H ratio. A homologous series with C = C double or C C triple bond is said to be unsaturated. A homologous series with C C single bond is said to be saturated.Most of the reactions of the unsaturated compound involve trying to be saturated to form a C C single bond . Examples of burning alkenes 1.(a) Ethene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.
A homologous series with C = C double or C C triple bond is said to be unsaturated. A homologous series with C C single bond is said to be saturated.Most of the reactions of the unsaturated compound involve trying to be saturated to form a C C single bond . Examples of burning alkenes 1.(a) Ethene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water. Ethene + Air -> carbon(IV) oxide + water (excess air/oxygen) C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l/g) (b) Ethene when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water. 28 Ethene + Air -> carbon(II) oxide + water (limited air ) C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l/g) 2.(a) Propene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water. Propene + Air -> carbon(IV) oxide + water (excess air/oxygen) 2C3H6(g) + 9O2(g) -> 6CO2(g) + 6H2O(l/g) (a) Propene 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) C3H6(g) + 3O2(g) -> 3CO(g) + 3H2O(l/g) (b)Addition reactions An addition reaction is one which an unsaturated compound reacts to form a saturated compound.Addition reactions of alkenes are named from the reagent used to cause the addtion/convert the double =C=C= to single C-C bond. (i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at high temperatures react with alkenes to form alkanes. Examples 1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed.
Propene + Air -> carbon(IV) oxide + water (excess air/oxygen) C3H6(g) + 3O2(g) -> 3CO(g) + 3H2O(l/g) (b)Addition reactions An addition reaction is one which an unsaturated compound reacts to form a saturated compound.Addition reactions of alkenes are named from the reagent used to cause the addtion/convert the double =C=C= to single C-C bond. (i)Hydrogenation Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at high temperatures react with alkenes to form alkanes. Examples 1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed. Hydrogenation is thus used to harden oils to solid fat especially margarine. During hydrogenation, one hydrogen atom in the hydrogen molecule attach itself to one carbon and the other hydrogen to the second carbon breaking the double bond to single bond.
Examples 1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed. Hydrogenation is thus used to harden oils to solid fat especially margarine. During hydrogenation, one hydrogen atom in the hydrogen molecule attach itself to one carbon and the other hydrogen to the second carbon breaking the double bond to single bond. Chemical equation H2C=CH2 + H2 -Ni/Pa-> H3C - CH3 H H H H C = C + H – H - Ni/Pa -> H - C – C - H H H H H 2.Propene undergo hydrogenation to form Propane
29 Chemical equation H3C CH=CH2 + H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + H – H - Ni/Pa-> H - C – C - C- H H H H H H 3.Both But-1-ene and But-2-ene undergo hydrogenation to form Butane Chemical equation But-1-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 CH=CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H H But-2-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 =CH CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C = C - C -H + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H 4. But-1,3-diene should undergo hydrogenation to form Butane. The reaction uses two moles of hydrogen molecules/four hydrogen atoms to break the two double bonds.
Chemical equation H2C=CH2 + H2 -Ni/Pa-> H3C - CH3 H H H H C = C + H – H - Ni/Pa -> H - C – C - H H H H H 2.Propene undergo hydrogenation to form Propane
29 Chemical equation H3C CH=CH2 + H2 -Ni/Pa-> H3C CH - CH3 H H H H H H H C C = C + H – H - Ni/Pa-> H - C – C - C- H H H H H H 3.Both But-1-ene and But-2-ene undergo hydrogenation to form Butane Chemical equation But-1-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 CH=CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H H But-2-ene + Hydrogen –Ni/Pa-> Butane H3C CH2 =CH CH2 + H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C = C - C -H + H – H - Ni/Pa-> H - C- C – C - C- H H H H H H H 4. But-1,3-diene should undergo hydrogenation to form Butane. The reaction uses two moles of hydrogen molecules/four hydrogen atoms to break the two double bonds. But-1,3-diene + Hydrogen –Ni/Pa-> Butane H2C CH CH=CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C -H + 2(H – H) - Ni/Pa-> H - C- C – C - C- H
30 H H H H (ii) Halogenation. Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkene to form an alkane. The double bond in the alkene break and form a single bond.
But-1,3-diene + Hydrogen –Ni/Pa-> Butane H2C CH CH=CH2 + 2H2 -Ni/Pa-> H3C CH2CH - CH3 H H H H H H H H H C C - C = C -H + 2(H – H) - Ni/Pa-> H - C- C – C - C- H
30 H H H H (ii) Halogenation. Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkene to form an alkane. The double bond in the alkene break and form a single bond. The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases/reduces. One bromine atom bond at the 1st carbon in the double bond while the other goes to the 2nd carbon. Examples 1Ethene reacts with bromine to form 1,2-dibromoethane. Chemical equation H2C=CH2 + Br2 H2 Br C - CH2 Br H H H H C = C + Br – Br Br - C – C - Br H H H H Ethene + Bromine 1,2-dibromoethane 2.Propene reacts with chlorine to form 1,2-dichloropropane.
One bromine atom bond at the 1st carbon in the double bond while the other goes to the 2nd carbon. Examples 1Ethene reacts with bromine to form 1,2-dibromoethane. Chemical equation H2C=CH2 + Br2 H2 Br C - CH2 Br H H H H C = C + Br – Br Br - C – C - Br H H H H Ethene + Bromine 1,2-dibromoethane 2.Propene reacts with chlorine to form 1,2-dichloropropane. Chemical equation H3C CH=CH2 + Cl2 H3C CHCl - CH2Cl Propene + Chlorine 1,2-dichloropropane H H H H H H H C C = C + Cl – Cl H - C – C - C- Cl H H H Cl H H H H H H H H H H C C - C = C + I – I H - C- C – C - C- I
31 H H H H H H H H 3.Both But-1-ene and But-2-ene undergo halogenation with iodine to form 1,2diiodobutane and 2,3-diiodobutane Chemical equation But-1-ene + iodine 1,2 diiodobutane H3C CH2 CH=CH2 + I2 H3C CH2CH I - CH2I But-2-ene + Iodine 2,3-diiodobutane H3C CH= CH-CH2 + F2 H3C CHICHI - CH3 H H H H H H H H H C C = C - C -H + I – I H - C- C – C - C- H H H H I I H 4. But-1,3-diene should undergo halogenation to form Butane. The reaction uses two moles of iodine molecules/four iodine atoms to break the two double bonds.
Chemical equation H3C CH=CH2 + Cl2 H3C CHCl - CH2Cl Propene + Chlorine 1,2-dichloropropane H H H H H H H C C = C + Cl – Cl H - C – C - C- Cl H H H Cl H H H H H H H H H H C C - C = C + I – I H - C- C – C - C- I
31 H H H H H H H H 3.Both But-1-ene and But-2-ene undergo halogenation with iodine to form 1,2diiodobutane and 2,3-diiodobutane Chemical equation But-1-ene + iodine 1,2 diiodobutane H3C CH2 CH=CH2 + I2 H3C CH2CH I - CH2I But-2-ene + Iodine 2,3-diiodobutane H3C CH= CH-CH2 + F2 H3C CHICHI - CH3 H H H H H H H H H C C = C - C -H + I – I H - C- C – C - C- H H H H I I H 4. But-1,3-diene should undergo halogenation to form Butane. The reaction uses two moles of iodine molecules/four iodine atoms to break the two double bonds. But-1,3-diene + iodine 1,2,3,4-tetraiodobutane H2C= CH CH=CH2 + 2I2 H2CI CHICHI - CHI H H H H H H H H H C C - C = C -H + 2(I – I) H - C- C – C - C- H I I I I (iii) Reaction with hydrogen halides. Hydrogen halides reacts with alkene to form a halogenoalkane. The double bond in the alkene break and form a single bond. The main compound is one which the hydrogen atom bond at the carbon with more hydrogen . Examples 1. Ethene reacts with hydrogen bromide to form bromoethane.
The main compound is one which the hydrogen atom bond at the carbon with more hydrogen . Examples 1. Ethene reacts with hydrogen bromide to form bromoethane. Chemical equation H2C=CH2 + HBr H3 C - CH2 Br H H H H
32 C = C + H – Br H - C – C - Br H H H H Ethene + Bromine bromoethane 2. Propene reacts with hydrogen iodide to form 2-iodopropane. Chemical equation H3C CH=CH2 + HI H3C CHI - CH3 Propene + Chlorine 2-chloropropane H H H H H H H C C = C + H – Cl H - C – C - C- H H H H Cl H 3. Both But-1-ene and But-2-ene reacts with hydrogen bromide to form 2- bromobutane Chemical equation But-1-ene + hydrogen bromide 2-bromobutane H3C CH2 CH=CH2 + HBr H3C CH2CHBr -CH3 H H H H H H H H H C C - C = C + H – Br H - C- C – C - C- H H H H H H Br H But-2-ene + Hydrogen bromide 2-bromobutane H3C CH= CH-CH2 + HBr H3C CHBrCH2 - CH3 H H H H H H H H H C C = C - C -H + Br – 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 two moles of hydrogen iodide molecules/two iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diene + iodine 2,3-diiodobutane Carbon atom with more Hydrogen atoms gets extra hydrogen
33 H2C= CH CH=CH2 + 2HI2 H3CCHICHI - CH3 H H H H H H H H H C C - C = C -H + 2(H – I) H - C- C – C - C- H H I I H (iv) Reaction with bromine/chlorine water.
But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses two moles of hydrogen iodide molecules/two iodine atoms and two hydrogen atoms to break the two double bonds. But-1,3-diene + iodine 2,3-diiodobutane Carbon atom with more Hydrogen atoms gets extra hydrogen
33 H2C= CH CH=CH2 + 2HI2 H3CCHICHI - CH3 H H H H H H H H H C C - C = C -H + 2(H – I) H - C- C – C - C- H H I I H (iv) Reaction with bromine/chlorine water. Chlorine and bromine water is formed when the halogen is dissolved in distilled water.Chlorine water has the formular HOCl(hypochlorous/chloric(I)acid) .Bromine water has the formular HOBr(hydrobromic(I)acid). During the addition reaction .the halogen move to one carbon and the OH to the other carbon in the alkene at the =C=C= double bond to form a halogenoalkanol. Bromine water + Alkene -> bromoalkanol Chlorine water + Alkene -> bromoalkanol Examples 1Ethene reacts with bromine water to form bromoethanol. Chemical equation H2C=CH2 + HOBr H2 Br C - CH2 OH H H H H C = C + Br – OH Br - C – C - OH H H H H Ethene + Bromine water bromoethanol 2.Propene reacts with chlorine water to form chloropropan-2-ol / 2-chloropropan-1ol.
During the addition reaction .the halogen move to one carbon and the OH to the other carbon in the alkene at the =C=C= double bond to form a halogenoalkanol. Bromine water + Alkene -> bromoalkanol Chlorine water + Alkene -> bromoalkanol Examples 1Ethene reacts with bromine water to form bromoethanol. Chemical equation H2C=CH2 + HOBr H2 Br C - CH2 OH H H H H C = C + Br – OH Br - C – C - OH H H H H Ethene + Bromine water bromoethanol 2.Propene reacts with chlorine water to form chloropropan-2-ol / 2-chloropropan-1ol. Chemical equation I.H3C CH=CH2 + HOCl H3C CHCl - CH2OH Propene + Chlorine water 2-chloropropane H H H H H H H C C = C + HO – Cl H - C – C - C- OH
34 H H H Cl H II.H3C CH=CH2 + HOCl H3C CHOH - CH2Cl Propene + Chlorine chloropropan-2-ol H H H H H H H C C = C + HO – Cl H - C – C - C- Cl H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form 2-bromobutan-1ol /3-bromobutan-2-ol respectively Chemical equation I.But-1-ene + bromine water 2-bromobutan-1-ol H3C CH2 CH=CH2 + HOBr H3C CH2CH Br - CH2OH H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H II.But-2-ene + bromine water 3-bromobutan-2-ol H3C CH= CHCH3 + HOBr H3C CH2OHCH Br CH3 H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H 4.
Bromine water + Alkene -> bromoalkanol Chlorine water + Alkene -> bromoalkanol Examples 1Ethene reacts with bromine water to form bromoethanol. Chemical equation H2C=CH2 + HOBr H2 Br C - CH2 OH H H H H C = C + Br – OH Br - C – C - OH H H H H Ethene + Bromine water bromoethanol 2.Propene reacts with chlorine water to form chloropropan-2-ol / 2-chloropropan-1ol. Chemical equation I.H3C CH=CH2 + HOCl H3C CHCl - CH2OH Propene + Chlorine water 2-chloropropane H H H H H H H C C = C + HO – Cl H - C – C - C- OH
34 H H H Cl H II.H3C CH=CH2 + HOCl H3C CHOH - CH2Cl Propene + Chlorine chloropropan-2-ol H H H H H H H C C = C + HO – Cl H - C – C - C- Cl H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form 2-bromobutan-1ol /3-bromobutan-2-ol respectively Chemical equation I.But-1-ene + bromine water 2-bromobutan-1-ol H3C CH2 CH=CH2 + HOBr H3C CH2CH Br - CH2OH H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H II.But-2-ene + bromine water 3-bromobutan-2-ol H3C CH= CHCH3 + HOBr H3C CH2OHCH Br CH3 H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H 4. But-1,3-diene reacts with bromine water to form Butan-1,3-diol. The reaction uses two moles of bromine water molecules to break the two double bonds.
Chemical equation I.H3C CH=CH2 + HOCl H3C CHCl - CH2OH Propene + Chlorine water 2-chloropropane H H H H H H H C C = C + HO – Cl H - C – C - C- OH
34 H H H Cl H II.H3C CH=CH2 + HOCl H3C CHOH - CH2Cl Propene + Chlorine chloropropan-2-ol H H H H H H H C C = C + HO – Cl H - C – C - C- Cl H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form 2-bromobutan-1ol /3-bromobutan-2-ol respectively Chemical equation I.But-1-ene + bromine water 2-bromobutan-1-ol H3C CH2 CH=CH2 + HOBr H3C CH2CH Br - CH2OH H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H II.But-2-ene + bromine water 3-bromobutan-2-ol H3C CH= CHCH3 + HOBr H3C CH2OHCH Br CH3 H H H H H H H H H C C - C = C + HO– Br H - C- C – C - C- OH H H H H H Br H 4. But-1,3-diene reacts with bromine water to form Butan-1,3-diol. The reaction uses two moles of bromine water molecules to break the two double bonds. But-1,3-diene + bromine water 2,4-dibromobutan-1,3-diol H2C= CH CH=CH2 + 2HOBr H2COH CHBrCHOH CHBr H H H H H H H H
35 H C C - C = C -H + 2(HO – Br) H - C- C – C - C- H HO Br HO Br (v) Oxidation. Alkenes are oxidized to alkanols with duo/double functional groups by oxidizing agents.
The reaction uses two moles of bromine water molecules to break the two double bonds. But-1,3-diene + bromine water 2,4-dibromobutan-1,3-diol H2C= CH CH=CH2 + 2HOBr H2COH CHBrCHOH CHBr H H H H H H H H
35 H C C - C = C -H + 2(HO – Br) H - C- C – C - C- H HO Br HO Br (v) Oxidation. Alkenes are oxidized to alkanols with duo/double functional groups by oxidizing agents. When an alkene is bubbled into orange acidified potassium/sodium dichromate (VI) solution,the colour of the oxidizing agent changes to green. When an alkene is bubbled into purple acidified potassium/sodium manganate(VII) solution, the oxidizing agent is decolorized. Examples 1Ethene is oxidized to ethan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution. The purple acidified potassium/sodium manganate(VII) solution is decolorized. The orange acidified potassium/sodium dichromate(VI) solution turns to green. Chemical equation H2C=CH2 [O] in H+/K2Cr2O7 HO CH2 - CH2 OH H H H H C = C+ [O] in H+/KMnO4 H - C – C - H H H OH OH Ethene + [O] in H+/KMnO4 ethan-1,2-diol 2. Propene is oxidized to propan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution. The purple acidified potassium/sodium manganate(VII) solution is decolorized. The orange acidified potassium/sodium dichromate(VI) solution turns to green.
Propene is oxidized to propan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution. The purple acidified potassium/sodium manganate(VII) solution is decolorized. The orange acidified potassium/sodium dichromate(VI) solution turns to green. Chemical equation H3C CH=CH2 [O] in H+/KMnO4 H3C CHOH - CH2OH Propene [O] in H+/KMnO4 propan-1,2-diol H H H H H H H C C = C [O] in H+/KMnO4 H - C – C - C- OH
36 H H H OH H 3.Both But-1-ene and But-2-ene react with bromine water to form butan-1,2-diol and butan-2,3-diol Chemical equation I.But-1-ene + [O] in H+/KMnO4 butan-1,2-diol H3C CH2 CH=CH2 + [O] H3C CH2CHOH - CH2OH H H H H H H H H H C C - C = C + [O] H - C- C – C - C- OH H H H H H OH H (v) Hydrolysis. Hydrolysis is the reaction of a compound with water/addition of H-OH to a compound. Alkenes undergo hydrolysis to form alkanols . This takes place in two steps: (i)Alkenes react with concentrated sulphuric(VI)acid at room temperature and pressure to form alkylhydrogen sulphate(VI). Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI) (ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed. alkylhydrogen sulphate(VI) + water -warm-> Alkanol.
This takes place in two steps: (i)Alkenes react with concentrated sulphuric(VI)acid at room temperature and pressure to form alkylhydrogen sulphate(VI). Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI) (ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed. alkylhydrogen sulphate(VI) + water -warm-> Alkanol. Examples (i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII) Chemical equation H2C=CH2 + H2SO4 CH3 - CH2OSO3H H H H O-SO3H C = C + H2SO4 H - C – C - H
37 H H H H Ethene + H2SO4 ethylhydrogen sulphate(VI) (ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H OSO3H H OH H - C - C - H + H2O H - C – C - H + H2SO4 H H H H ethylhydrogen sulphate(VI) + H2O Ethanol 2.
Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI) (ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed. alkylhydrogen sulphate(VI) + water -warm-> Alkanol. Examples (i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII) Chemical equation H2C=CH2 + H2SO4 CH3 - CH2OSO3H H H H O-SO3H C = C + H2SO4 H - C – C - H
37 H H H H Ethene + H2SO4 ethylhydrogen sulphate(VI) (ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H OSO3H H OH H - C - C - H + H2O H - C – C - H + H2SO4 H H H H ethylhydrogen sulphate(VI) + H2O Ethanol 2. Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII) Chemical equation CH3H2C=CH2 + H2SO4 CH3CH2 - CH2OSO3H H H H H H O-SO3H C = C - C - H + H2SO4 H - C - C – C - H H H H H H H Propene + H2SO4 propylhydrogen sulphate(VI) (ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H H OSO3H H H OH H - C - C - C - H + H2O H - C - C – C - H + H2SO4 H H H H H H propylhydrogen sulphate(VI) + H2O propanol (vi) Polymerization/self addition
38 Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization.
Examples (i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII) Chemical equation H2C=CH2 + H2SO4 CH3 - CH2OSO3H H H H O-SO3H C = C + H2SO4 H - C – C - H
37 H H H H Ethene + H2SO4 ethylhydrogen sulphate(VI) (ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H OSO3H H OH H - C - C - H + H2O H - C – C - H + H2SO4 H H H H ethylhydrogen sulphate(VI) + H2O Ethanol 2. Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII) Chemical equation CH3H2C=CH2 + H2SO4 CH3CH2 - CH2OSO3H H H H H H O-SO3H C = C - C - H + H2SO4 H - C - C – C - H H H H H H H Propene + H2SO4 propylhydrogen sulphate(VI) (ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H H OSO3H H H OH H - C - C - C - H + H2O H - C - C – C - H + H2SO4 H H H H H H propylhydrogen sulphate(VI) + H2O propanol (vi) Polymerization/self addition
38 Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization. Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene During addition polymerization (i)the double bond in alkenes break (ii)free radicals are formed (iii)the free radicals collide with each other and join to form a larger molecule.
Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII) Chemical equation CH3H2C=CH2 + H2SO4 CH3CH2 - CH2OSO3H H H H H H O-SO3H C = C - C - H + H2SO4 H - C - C – C - H H H H H H H Propene + H2SO4 propylhydrogen sulphate(VI) (ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol Chemical equation CH3 - CH2OSO3H + H2O CH3 - CH2OH + H2SO4 H H OSO3H H H OH H - C - C - C - H + H2O H - C - C – C - H + H2SO4 H H H H H H propylhydrogen sulphate(VI) + H2O propanol (vi) Polymerization/self addition
38 Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization. Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene During addition polymerization (i)the double bond in alkenes break (ii)free radicals are formed (iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule. Examples of addition polymerization 1.Formation of Polyethene Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.