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Introduction to Organic chemistry  

Organic chemistry is the branch of chemistry that studies carbon compounds present in living things, once living things or synthetic/man-made.

Compounds that makes up living things whether alive or dead mainly contain carbon. Carbon is tetravalent.

It is able to form stable covalent bonds with itself and many non-metals like hydrogen, nitrogen ,oxygen and halogens to form a variety of compounds. This is because:

(i) carbon uses all the four valence electrons to form four strong covalent bond.

(ii)carbon can covalently bond to form a single, double or triple covalent bond with itself.

(iii)carbon atoms can covalently bond to form a very long chain or ring.

When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons

A.HYDROCARBONS (HCs)

Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only.

Depending on the type of bond that exist between the individual carbon atoms, hydrocarbon are classified as:

 (i) Alkanes

 (ii) Alkenes

 (iii) Alkynes


(i) Alkanes

(a)Nomenclature/Naming

These are hydrocarbons with a general formula CnH2n+2 where n is the number of Carbon atoms in a molecule.

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The carbon atoms are linked by single bond to each other and to hydrogen atoms.

They include:

 

 

n

General/

Molecular

formula

Structural formula

Name

1

CH4

H

 

H C H

 

H

Methane

2

C2H6

H H

 

H C C H

 

H H

Ethane


 

3

C3H8

H H H

 

H C C C H

 

H H H

Propane

4

C4H10

H H H H

 

H C C C C H

 

H H H H

Butane

5

C5H12

H H H H H

 

H C C C C C H CH3 (CH2) 6CH3

 

H H H H H

Pentane

6

C6H14

H H H H H H

 

H C C C C C C H CH3 (CH2) 6CH3

 

H H H H H H

Hexane

7

C7H16

H H H H H H H

 

H C C C C C C C H

 

H H H H H H H

Heptane

8

C8H18

H H H H H H H H

 

H C C C C C C C C H

 

H H H H H H H H

Octane

9

C9H20

H H 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 H H

Nonane

10

C10H22

H H 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 H H

decane

Note

1.The general formula/molecular formular of a compound shows the number of each atoms of elements making the compound e.g.

Decane has a general/molecular formula C10H22 this means there are 10 carbon atoms and 22 hydrogen atoms in a molecule of decane.

2.The structural formula shows the arrangement/bonding of atoms of each element making the compound e.g

Decane has the structural formula as in the table above this means the 1st carbon from left to right is bonded to three hydrogen atoms and one carbon atom.

The 2nd carbon atom is joined/bonded to two other carbon atoms and two Hydrogen atoms.

3.Since carbon is tetravalent ,each atom of carbon in the alkane MUST always be bonded using four covalent bond /four shared pairs of electrons.

4.Since Hydrogen is monovalent ,each atom of hydrogen in the alkane MUST always be bonded using one covalent bond/one shared pair of electrons.

5.One member of the alkane differ from the next/previous by a CH2 group.

e.g

Propane differ from ethane by one carbon and two Hydrogen atoms form ethane. Ethane differ from methane also by one carbon and two Hydrogen atoms

6.A group of compounds that differ by a CH2 group from the next /previous consecutively is called a homologous series.

7.A homologous series:

 (i) differ by a CH2 group from the next /previous consecutively

 (ii)have similar chemical properties

(iii)have similar chemical formula that can be represented by a general formula e.g alkanes have the general formula
CnH2n+2.

(iv)the physical properties (e.g.melting/boiling points)show steady gradual change)

8.The 1st four alkanes have the prefix meth_,eth_,prop_ and but_ to represent 1,2,3 and 4 carbons in the compound. All other use the numeral prefix pent_,Hex_,hept_ , etc to show also the number of carbon atoms.

9.If one hydrogen atom in an alkane is removed, an alkyl group is formed.e.g

 

Alkane name

molecular structure

CnH2n+2

Alkyl name

Molecula structure

CnH2n+1

methane

CH4

methyl

CH3

ethane

CH3CH3

ethyl

CH3 CH2

propane

CH3 CH2 CH3

propyl

CH3 CH2 CH2

butane

CH3 CH2 CH2 CH3

butyl

CH3 CH2 CH2 CH2

 

(b)Isomers of alkanes

Isomers are compounds with the same molecular general formula but different molecular structural formula.

Isomerism is the existence of a compounds having the same general/molecular formula but different structural formula.

The 1st three alkanes do not form isomers.Isomers are named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming.

The IUPAC system of nomenclature
uses the following basic rules/guidelines:

1.Identify the longest continuous carbon chain to get/determine the parent alkane.

2.Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible

3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens

4.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of branches attached to the parent alkane.

Practice on IUPAC nomenclature of alkanes

(a)Draw the structure of:

(i)2-methylpentane

Procedure

1. Identify the longest continuous carbon chain to get/determine the parent alkane.

 Butane is the parent name CH3 CH2 CH2 CH3

2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible

The methyl group is attached to Carbon “2”

3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e

Position of the branch at carbon “2”


Number of branches at carbon “1”

Type of the branch “methyl” hence

Molecular formula  

  CH3

 

CH3 CH CH2 CH3 //
CH3 CH (
CH3 ) CH2CH3

 

Structural formula

H H H H

 

H C C C C H

 

H H H

 

H C H

 

  H

(ii)2,2-dimethylpentane

Procedure

1. Identify the longest continuous carbon chain to get/determine the parent alkane.

 Butane is the parent name CH3 CH2 CH2 CH3

2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible

The methyl group is attached to Carbon “2”

3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e

Position of the branch at carbon “2”


Number of branches at carbon “2”

Type of the branch two”methyl” hence

Molecular formular

 
 

  CH3

 

CH3 C CH2 CH3 //
CH3 C (
CH3 )2 CH2CH3

 

  CH3


Structural formula

  H

 

H C H

 
 

H H H

 

H C C C C H

 

H H H

 

H C H

 

  H

(iii) 2,2,3-trimethylbutane

 

Procedure

1. Identify the longest continuous carbon chain to get/determine the parent alkane.

 Butane is the parent name CH3 CH2 CH2 CH3

2. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible

The methyl group is attached to Carbon “2 and 3”

3. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e

Position of the branch at carbon “2 and 3”


Number of branches at carbon “3”

Type of the branch three “methyl” hence

Molecular formular

 

 

 

 

  CH3

 

CH3 C CH CH3 //
CH3 C (
CH3 )3 CH2CH3

 

  CH3
CH3


Structural formula

  H

 

H C H

 
 

H H

 

H C C C H

 

H H


H

 


H
C C H

 

  H

H C   H

 

  H

(iv) 1,1,1,2,2,2-hexabromoethane

Molecular formula  

CBr3 CBr3


Structural formula

Br Br

 

Br C C Br

 

Br Br

(v) 1,1,1-tetrachloro-2,2-dimethylbutane

  CH3

 

CCl 3 C CH3 //
C Cl 3 C (
CH3 )2 CH3

 

  CH3


Structural formula

  Cl

 

Cl C Cl

 
 

H H

 

H C C C H

 

H H


H
C H

 

  H

 

(c)Occurrence and extraction

 

Crude oil ,natural gas and biogas are the main sources of alkanes:

(i)Natural gas is found on top of crude oil deposits and consists mainly of methane.

(ii)Biogas is formed from the decay of waste organic products like animal dung and cellulose. When the decay takes place in absence of oxygen , 60-75% by volume of the gaseous mixture of methane gas is produced.

(iii)Crude oil is a mixture of many flammable hydrocarbons/substances. Using fractional distillation, each hydrocarbon fraction can be separated from the other. The hydrocarbon with lower /smaller number of carbon atoms in the chain have lower boiling point and thus collected first.

As the carbon chain increase, the boiling point, viscosity (ease of flow) and colour intensity increase as flammability decrease. Hydrocarbons in crude oil are not pure. They thus have no sharp fixed boiling point.

Uses of different crude oil fractions

Carbon atoms in a molecule

Common name of fraction

Uses of fraction

1-4

Gas

L.P.G gas for domestic use

5-12

Petrol

Fuel for petrol engines

9-16

Kerosene/Paraffin

Jet fuel and domestic lighting/cooking

15-18

Light diesel

Heavy diesel engine fuel

18-25

Diesel oil

Light diesel engine fuel

20-70

Lubricating oil

Lubricating oil to reduce friction.

Over 70

Bitumen/Asphalt

Tarmacking roads

(d)School laboratory preparation of alkanes

In a school laboratory, alkanes may be prepared from the reaction of a sodium alkanoate with solid sodium hydroxide/soda lime.

Chemical equation:

Sodium alkanoate  +   soda lime  -> alkane + Sodium carbonate

CnH2n+1COONa(s) +   NaOH(s)  -> C n H2n+2 + Na2CO3(s)

The “H” in NaOH is transferred/moves to the CnH2n+1
in CnH2n+1COONa(s) to form C n H2n+2.

Examples

1. Methane is prepared from the heating of a mixture of sodium ethanoate and soda lime and collecting over water

Sodium ethanoate  +   soda lime  -> methane + Sodium carbonate

CH3COONa(s) +   NaOH(s)  -> C
H4 + Na2CO3(s)

The “H” in NaOH is transferred/moves to the CH3
in CH3COONa(s) to form CH4.

2. Ethane is prepared from the heating of a mixture of sodium propanoate and soda lime and collecting over water

Sodium propanoate  +   soda lime  -> ethane + Sodium carbonate

CH3 CH2COONa(s)  +   NaOH(s)  -> CH3 CH3 + Na2CO3(s)

The “H” in NaOH is transferred/moves to the CH3 CH2
in CH3 CH2COONa (s) to form CH3 CH3

3. Propane is prepared from the heating of a mixture of sodium butanoate and soda lime and collecting over water

Sodium butanoate  +   soda lime  -> propane + Sodium carbonate

CH3 CH2CH2COONa(s)  +   NaOH(s)  -> CH3 CH2CH3 + Na2CO3(s)

The “H” in NaOH is transferred/moves to the CH3 CH2 CH2
in CH3 CH2CH2COONa (s) to form CH3 CH2CH3

4. Butane is prepared from the heating of a mixture of sodium pentanoate and soda lime and collecting over water

Sodium pentanoate  +   soda lime  -> butane + Sodium carbonate

CH3 CH2 CH2CH2COONa(s)+NaOH(s) -> CH3 CH2CH2CH3 + Na2CO3(s)

The “H” in NaOH is transferred/moves to the CH3CH2 CH2 CH2
in CH3 CH2CH2 CH2COONa (s) to form CH3 CH2 CH2CH3

Laboratory set up for the preparation of alkanes

 

 

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(d)Properties of alkanes

I. Physical properties  

Alkanes are colourless gases, solids and liquids that are not poisonous.

They are slightly soluble in water.

The solubility decrease as the carbon chain and thus the molar mass increase

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 four straight chain alkanes (methane,ethane,propane and butane)are therefore gases ,the nect six(pentane ,hexane, heptane,octane,nonane, and decane) are liquids while the rest from unidecane(11 carbon atoms) are solids .

The density of straight chain alkanes increase with increasing carbon chain as the intermolecular forces increases.

This reduces the volume occupied by a given mass of the compound.

Summary of physical properties of alkanes

Alkane

General formula

Melting point(K)

Boiling point(K)

Density gcm-3

State at room(298K) temperature and pressure atmosphere (101300Pa)

Methane

CH4

90

112

0.424

gas

Ethane

CH3CH3

91

184

0.546

gas

Propane

CH3CH2CH3

105

231

0.501

gas

Butane

CH3(CH2)2CH3

138

275

0.579

gas

Pentane

CH3(CH2)3CH3

143

309

0.626

liquid

Hexane

CH3(CH2)4CH3

178

342

0.657

liquid

Heptane

CH3(CH2)5CH3

182

372

0.684

liquid

Octane

CH3(CH2)6CH3

216

399

0.703

liquid

Nonane

CH3(CH2)7CH3

219

424

0.708

liquid

Octane

CH3(CH2)8CH3

243

447

0.730

liquid

II.Chemical properties  

(i)Burning.

Alkanes burn with a blue/non-luminous non-sooty/non-smoky flame in excess air to form carbon(IV) oxide and water.

Alkane  +  Air  ->  carbon(IV) oxide  +   water (excess air/oxygen)

Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water.

Alkane  +  Air  ->  carbon(II) oxide  +   water (limited air)

Examples

1.(a) Methane when ignited burns with a blue
non
sooty flame in excess air to form carbon(IV) oxide and water.

Methane  +  Air  ->  carbon(IV) oxide  +   water (excess air/oxygen)

CH4(g)  +  2O2(g)  ->  CO2(g) +   2H2O(l/g)

(b) Methane when ignited burns with a blue
non
sooty flame in limited air to form carbon(II) oxide and water.

Methane  +  Air  ->  carbon(II) oxide  +   water (excess air/oxygen)

2CH4(g)  +  3O2(g)  ->  2CO(g) +   4H2O(l/g)

2.(a) Ethane when ignited burns with a blue
non
sooty flame in excess air to form carbon(IV) oxide and water.

Ethane  +  Air  ->  carbon(IV) oxide  +   water (excess air/oxygen)

2C2H6(g)  +  7O2(g)  ->  4CO2(g) +   6H2O(l/g)

(b) Ethane when ignited burns with a blue
non
sooty flame in limited air to form carbon(II) oxide and water.

Ethane  +  Air  ->  carbon(II) oxide  +   water (excess air/oxygen)

2C2H6(g)  +  5O2(g)  ->  4CO(g) +   6H2O(l/g)

3.(a) Propane when ignited burns with a blue
non
sooty flame in excess air to form carbon(IV) oxide and water.

Propane  +  Air  ->  carbon(IV) oxide  +   water (excess air/oxygen)

C3H8(g)  +  5O2(g)  ->  3CO2(g) +   4H2O(l/g)

(b) Ethane when ignited burns with a blue
non
sooty flame in limited air to form carbon(II) oxide and water.

Ethane  +  Air  ->  carbon(II) oxide  +   water (excess air/oxygen)

2C3H8(g)  +  7O2(g)  ->  6CO(g) +   8H2O(l/g)

ii)Substitution

Substitution reaction is one in which a hydrogen atom is replaced by a halogen in presence of ultraviolet light.

Alkanes react with halogens in presence of ultraviolet light to form halogenoalkanes.

During substitution:

(i)the halogen molecule is split into free atom/radicals.

(ii)one free halogen radical/atoms knock /remove one hydrogen from the alkane leaving an alkyl radical.

 (iii) the alkyl radical combine with the other free halogen atom/radical to form halogenoalkane.

 (iv)the chlorine atoms substitute repeatedly in the alkane. Each substitution removes a hydrogen atom from the alkane and form hydrogen halide.

(v)substitution stops when all the hydrogen in alkanes are replaced with halogens.

Substitution reaction is a highly explosive reaction in presence of sunlight / ultraviolet light that act as catalyst.

Examples of substitution reactions

Methane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of chlorine and methane explode to form colourless mixture of chloromethane and hydrogen chloride gas. The pale green colour of chlorine gas fades.

Chemical equation

1.(a)Methane  + chlorine  ->  Chloromethane  +  Hydrogen chloride

 

CH4(g)  +  Cl2(g)  ->  CH3Cl (g) +   HCl (g)

 

H H

 

H C H + Cl  Cl  -> H  C  Cl + H   Cl

 

H H

(b) Chloromethane + chlorine  ->  dichloromethane  +  Hydrogen chloride

 

CH3Cl (g)    +  Cl2(g)   ->   CH2Cl2 (g)   +   HCl (g)

H H

 

H C Cl + Cl  Cl  -> H  C  Cl + H   Cl

 


H Cl

(c) dichloromethane + chlorine  ->  trichloromethane  +  Hydrogen chloride

CH2Cl2 (g)   +  Cl2(g)   ->   CHCl3 (g)   +   HCl (g)

Cl H

 

H C Cl + Cl  Cl  -> Cl  C  Cl + H   Cl

 


H Cl

(c) trichloromethane + chlorine ->  tetrachloromethane + Hydrogen chloride

 

CHCl3 (g)   +  Cl2(g)   ->   CCl4 (g)   +   HCl (g)

H Cl

 


Cl C Cl + Cl  Cl  -> Cl  C  Cl + H   Cl

 


Cl Cl

Ethane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of bromine and ethane explode to form colourless mixture of bromoethane and hydrogen chloride gas. The red/brown colour of bromine gas fades.

Chemical equation

(a)Ethane  + chlorine  ->  Chloroethane  +  Hydrogen chloride

CH3CH3(g)  +  Br2(g)  ->  CH3CH2Br (g) +   HBr (g)

H H H   H

 

H C C H + Br
Br -> H C   C H + H
Br

 

H H H   Br

Bromoethane

H H H   Br

 

H C C H + Br
Br -> H C   C H + H
Br

 

H Br H   Br

 1,1-dibromoethane

H Br H   Br

 

H C C H + Br
Br -> H C   C Br + H
Br

 

H Br H   Br

 1,1,1-tribromoethane

H Br H   Br

 

H C C Br + Br
Br -> H C   C Br + H
Br

 

H Br Br   Br

 1,1,1,2-tetrabromoethane

H Br H   Br

 

H C C Br + Br
Br -> Br C   C Br + H
Br

 


Br
Br Br   Br

 1,1,1,2,2-pentabromoethane

H Br Br   Br

 

Br C C Br + Br
Br -> Br C   C Br + H
Br

 


Br
Br Br   Br

 1,1,1,2,2,2-hexabromoethane

Uses of alkanes

1.Most alkanes are used as fuel e.g. Methane is used as biogas in homes. Butane is used as the Laboratory gas.

2.On cracking ,alkanes are a major source of Hydrogen for the manufacture of ammonia/Haber process.

3.In manufacture of Carbon black which is a component in printers ink.

4.In manufacture of useful industrial chemicals like methanol, methanol, and chloromethane.

(ii) Alkenes

(a)Nomenclature/Naming

These are hydrocarbons with a general formula CnH2n 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 double bond to each other and single bonds to hydrogen atoms.

They include:

n

General/

Molecular

formula

Structural formula

Name

1

 

Does not exist

 

2

C2H6

 


 

H H

 

H C C H

 

CH2 CH2


 

Ethene

3

C3H8

H H H

 

H C C C H

 

H

 

CH2 CH CH3

Propene

4

C4H10

H H H H

 

H C C C C H

 

H H

 

CH2 CH CH2CH3

Butene

5

C5H12

H H H H H

 

H C C C C C H

 

H H H

CH2 CH (CH2)2CH3

Pentene

6

C6H14

H H H H H H

 

H C C C C C C H

 

H H H H

 

CH2 CH (CH2)3CH3

Hexene

7

C7H16

H H H H H H H

 

H C C C C C C C H

 

H H H H H H H

 

CH2 CH (CH2)4CH3

Heptene

8

C8H18

H H H H H H H H

 

H C C C C C C C C H

 

H H H H H H

 

CH2 CH (CH2)5CH3

Octene

9

C9H20

H H 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

 

CH2 CH (CH2)6CH3

Nonene

10

C10H22

H H 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

 

CH2 CH (CH2)7CH3

decene

Note

1.Since carbon is tetravalent ,each atom of carbon in the alkene MUST always be bonded using four covalent bond /four shared pairs of electrons including at the double bond.

2.Since Hydrogen is monovalent ,each atom of hydrogen in the alkene MUST always be bonded using one covalent bond/one shared pair of electrons.

3.One member of the alkene ,like alkanes,differ from the next/previous by a CH2 group.They also form a homologous series.

e.g

Propene differ from ethene by one carbon and two Hydrogen atoms from ethene. 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 represented by the general formula CnH2n

(iv)the physical properties also show steady gradual change

5.The = C= C = double bond in alkene is the functional group. A functional group is the reacting site of a molecule/compound.

6. The = C= C = double bond in alkene can easily be broken to accommodate more two more monovalent atoms. The = C= C = double bond in alkenes make it thus unsaturated.

7. An unsaturated hydrocarbon is one with a double =C=C= or triple – C C – carbon bonds in their molecular structure. Unsaturated hydrocarbon easily reacts to be saturated.

8.A saturated hydrocarbon is one without a double =C=C= or triple – C C – carbon bonds in their molecular structure.

Most of the reactions of alkenes take place at the = C = C =bond.

 

 

(b)Isomers of alkenes

Isomers are alkenes lie alkanes have the same molecular general formula but different molecular structural formula.

Ethene and propene do not form isomers. Isomers of alkenes 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 alkenes
uses the following basic rules/guidelines:

1.Identify the longest continuous/straight carbon chain which contains the =C = C= double bond get/determine the parent alkene.

2.Number the longest chain form the end of the chain which contains the =C = C= double bond so he =C = C= double bond lowest number possible.

3 Indicate the positions by splitting “alk-positions-ene” e.g. but-2-ene, pent-1,3-diene.

4.The position indicated must be for the carbon atom at the lower position in the =C = C= double bond.i.e

But-2-ene means the double =C = C= is between Carbon “2”and “3”

Pent-1,3-diene means there are two double bond one between carbon “1” and “2”and another between carbon “3” and “4”

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 alkene. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens

6.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of double C = C bonds and branches attached to the alkene.

7.Position isomers can be formed when the=C = C= double bond is shifted between carbon atoms e.g.

But-2-ene means the double =C = C= is between Carbon “2”and “3”

But-1-ene means the double =C = C= is between Carbon “1”and “2”

Both But-1-ene and But-2-ene are position isomers of Butene

8.Position isomers are molecules/compounds having the same general formular but different position of the functional group.i.e.

Butene has the molecular/general formular C4H8 position but can form both But-1-ene and But-2-ene as position isomers.

9. Like alkanes ,an alkyl group can be attached to the alkene. Chain/branch isomers are thus formed.

10.Chain/branch isomers are molecules/compounds having the same general formula but different structural formula e.g

Butene and 2-methyl propene both have the same general formualr but different branching chain.

Practice on IUPAC nomenclature of alkenes

Name the following isomers of alkene

 

H H H H

 

H C C C C H  But-1-ene

 

H H

H H H H

 

H C C C C H  But-2-ene

 

H H

  H H H H H H

 

H C C C C C C H  4-methylhex-1-ene

 

H H H

H   C H

 

H

H

 

  H   C  H

 

H H H H H

 

H C C C C C C H  4,4-dimethylhex-1-ene

 

H H H

H   C H

 

H

3. H

 

  H   C  H

 

H H H H

 

H C C C C C H   4,4-dimethylpent -1- ene

 

H H

H   C H

 

H

 

4. H

 

  H   C  H

 

H H H H

 

H C C C C C H   5,5-dimethylhex-2- ene

 

H C H H H

H   C H

H

H

5. H

 

  H   C  H

 

H H H

 

H C C C C H   2,2-dimethylbut -2- ene

 

H H

H   C H

 

H

 

8.H2C CHCH2 CH2 CH3  pent -1- ene

9.H2C C(CH3)CH2 CH2 CH3 2-methylpent -1- ene

10.H2C C(CH3)C(CH3)2 CH2 CH3 2,3,3-trimethylpent -1- ene

11.H2C C(CH3)C(CH3)2 C(CH3)2 CH3 2,3,3,4,4-pentamethylpent -1- ene

12.H3C C(CH3)C(CH3) C(CH3)2 CH3 2,3,4,4-tetramethylpent -2- ene

13. H2C C(CH3)C(CH3) C(CH3) CH3 2,3,4-trimethylpent -1,3- diene

14. H2C CBrCBr CBr CH3 2,3,4-tribromopent -1,3- diene

15. H2C CHCH CH2 But -1,3- diene

16. Br2C CBrCBr CBr2 1,1,2,3,4,4-hexabromobut -1,3- diene

17. I2C CICI CI2 1,1,2,3,4,4-hexaiodobut -1,3- diene

18. H2C C(CH3)C(CH3) CH2 2,3-dimethylbut -1,3- diene

 

 

 

(c)Occurrence and extraction

At indusrial level,alkenes are obtained from the cracking of alkanes.Cracking is the process of breaking long chain alkanes to smaller/shorter alkanes, an alkene and hydrogen gas at high temperatures.

Cracking is a major source of useful hydrogen gas for manufacture of ammonia/nitric(V)acid/HCl i.e.

 

Long chain alkane -> smaller/shorter alkane + Alkene + Hydrogen gas

Examples

1.When irradiated with high energy radiation,Propane undergo cracking to form methane gas, ethene and hydrogen gas.

Chemical equation

CH3CH2CH3 (g) -> CH4(g) + CH2=CH2(g) + H2(g)

2.Octane undergo cracking to form hydrogen gas, butene and butane gases

Chemical equation

CH3(CH2) 6 CH3 (g) -> CH3CH2CH2CH3(g) + CH3 CH2CH=CH2(g) + H2(g)

(d)School laboratory preparation of alkenes

In a school laboratory, alkenes may be prepared from dehydration of alkanols using:

(i) concentrated sulphuric(VI)acid(H2SO4).

(a) aluminium(III)oxide(Al2O3) i.e

Alkanol  –Conc. H2SO4 –>  Alkene   +   Water

 Alkanol   –Al2O3 –>  Alkene   +   Water e.g.

1.(a)At about 180oC,concentrated sulphuric(VI)acid dehydrates/removes water from ethanol to form ethene.

The gas produced contain traces of carbon(IV)oxide and sulphur(IV)oxide gas as impurities.

It is thus passed through concentrated sodium/potassium hydroxide solution to remove the impurities.

Chemical equation

CH3CH2OH (l) –conc H2SO4/180oC–> CH2=CH2(g) + H2O(l)

(b)On heating strongly aluminium(III)oxide(Al2O3),it dehydrates/removes water from ethanol to form ethene.

Ethanol vapour passes through the hot aluminium (III) oxide which catalyses the dehydration.

Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from ethanol to form ethene.

Chemical equation

CH3CH2OH (l) –(Al2O3/strong heat–> CH2=CH2(g) + H2O(l)

2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers).

Chemical equation

CH3CH2 CH2OH (l) — conc H2SO4/180oC –> CH3CH2=CH2(g) + H2O(l)

Propan-1-ol  Prop-1-ene

CH3CHOH CH3 (l) — conc H2SO4/180oC –> CH3CH2=CH2(g) + H2O(l)

Propan-2-ol  Prop-1-ene

(b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene

Chemical equation

CH3CH2 CH2OH (l) — Heat/Al2O3 –> CH3CH2=CH2(g) + H2O(l)

Propan-1-ol  Prop-1-ene

CH3CHOH CH3 (l) — Heat/Al2O3 –> CH3CH2=CH2(g) + H2O(l)

Propan-2-ol  Prop-1-ene

3(a) Butan-1-ol and Butan-2-ol(position isomers of butanol) are dehydrated by conc H2SO4 at about 180oC to But-1-ene and But-2-ene respectively

Chemical equation

CH3CH2 CH2 CH2OH (l) — conc H2SO4/180oC –>CH3 CH2CH2=CH2(g) + H2O(l)

Butan-1-ol  But-1-ene

CH3CHOH CH2CH3 (l)– conc H2SO4/180oC –>CH3CH=CH CH2(g) + H2O(l)

Butan-2-ol  But-2-ene

(b) Butan-1-ol and Butan-2-ol are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively.

Chemical equation

CH3CH2 CH2 CH2OH (l) — Heat/Al2O3 –> CH3 CH2CH2=CH2(g) + H2O(l)

Butan-1-ol  But-1-ene

CH3CHOH CH2CH3 (l) — Heat/Al2O3 –> CH3CH=CH CH2(g) + H2O(l)

Butan-2-ol  But-2-ene

Laboratory set up for the preparation of alkenes/ethene

Caution

(i)Ethanol is highly inflammable

(ii)Conc H2SO4 is highly corrosive on skin contact.

(iii)Common school thermometer has maximum calibration of 110oC and thus cannot be used. It breaks/cracks.

 

 

 

 

 

 

 

 

 

 

(i)Using conentrated sulphuric(VI)acid

 

Image From EcoleBooks.com

 

 

Some broken porcelain or sand should be put in the flask when heating to:

 (i)prevent bumping which may break the flask.

 (ii)ensure uniform and smooth boiling of the mixture

The temperatures should be maintained at above160oC.

At lower temperatures another compound –ether is predominantly formed instead of ethene gas.

(ii)Using aluminium(III)oxide

Image From EcoleBooks.com

 

(e)Properties of alkenes

I. Physical properties

Like alkanes, alkenes 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.

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 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)

Ethene

CH2CH2

-169

-104

gas

Propene

CH3 CHCH2

-145

-47

gas

Butene

CH3CH2 CHCH2

-141

-26

gas

Pent-1-ene

CH3(CH2 CHCH2

-138

30

liquid

Hex-1-ene

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.

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.

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.

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

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

 

      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.

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

 

 H H H H H H H H

3.Both But-1-ene and But-2-ene undergo halogenation with iodine to form 1,2-diiodobutane 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.

Chemical equation

H2C=CH2 + HBr
H3 C – CH2 Br

 

H H H H

 

 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

Image From EcoleBooks.comPropene  +   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

 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-1-ol.

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

 

 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-1-ol /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

 

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.

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

 

 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.

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

 

 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

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.

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 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

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.

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:

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

 

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

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.

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

(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.

 (ii)ceiling tiles

 (iii)clothe linings

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

– 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

 

 

 

 

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;

 

 

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.

Image From EcoleBooks.com 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

 

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 –

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.

(iii) Alkynes

(a)Nomenclature/Naming

These are hydrocarbons with a general formula CnH2n2 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

 

H C C C C C C H

 

H H H H

CH
C (CH2)3CH3

Hexyne

 

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.

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-1,3-diyne.

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”

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.

(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

Image From EcoleBooks.com

Chemical equation

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-1-yne)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.

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.

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.

Chemical equation


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

(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.

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

 

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.

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.

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

Image From EcoleBooks.comPropene  +   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

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


 




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EcoleBooks | Chemistry Form 3 Notes : Organic chemistry 1 - HYDROCARBONS (HCs)

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