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ALKANES,
ALKENES,
ALKYNES
NOMENCLATURE, STRUCTURES AND ISOMERISM
Alkanes = CnH2n+2
 Alkenes = CnH2n
Alkynes = CnH2n-2
 ALKANES, ALKENES, ALKYNES AND
CYCLOALKANES ARE HYDROCARBONS
(COMPOUNDS CONTAINING ONLY
CARBON AND HYDROGEN).
 EACH OF THESE FORM A HOMOLOGOUS
SERIES (A GROUP OF ORGANIC
COMPOUNDS HAVING A COMMON
GENERAL FORMULA/ OR IN WHICH EACH
MEMBER FIFFERS FROM THE NEXT BY A –
CH2)
 THE HYDROCARBONS MAY BE SATURATED
(CONTAINS ONLY SINGLE BONDS
BETWEEN CARBON-CARBON ATOMS/
CARBON ATOMS BONDED TO THE
MAXIMUM NUMBER OF HYDROGENS)
 OR UNSATURATED (CONTAINS AT LEAST
A DOUBLE BOND BETWEEN C-C ATOMS)
ALKANES: NOMENCLATURE
Also called paraffins.
A group of saturated hydrocarbons with the
general formula Cn H2n+2 .
They form a homologous series.
Straight chain alkanes have their carbon
atoms bonded together to give a single
chain
Alkanes may also be branched.
NAMING (GENERAL)
Hydrocarbon names are based on: 1) type, 2)
# of carbons, 3) side chain type and position
1) name will end in -ane, -ene, or -yne
2) the number of carbons is given by a “prefix”
1 meth- 2 eth- 3 prop- 4 but- 5 pent- 6
hex- 7 hept- 8 oct- 9 non- 10 dec-
Actually, all end in a, but a is dropped when next
to a vowel. E.g. a 6 C alkene is hexene
 Determine the longest continuous chain
(not always straight) in the molecule. The
base name of the hydrocarbon is the
name of the longest chain.
IUPAC system
IUPAC SYSTEM
 Name any chain branching off the longest
chain as an alkyl group (e.g., methyl,
ethyl etc)
 The complete name of a branch requires a
number that locates the branch on the
longest chain.
 Therefore number the chain in whichever
direction gives the smaller number for all
branches.
6. When two or more branches are identical, use
prefixes (di-, tri-, etc.) (e.g. 2,4-
dimethylhexane). Numbers are separated with
commas. Prefixes are ignored when
determining alphabetical order. (e.g. 2,3,5-
trimethyl-4-propylheptane)
7. When identical groups are on the same
carbon, repeat the number of this carbon in the
name. (e.g. 2,2-dimethylhexane)
Where there are two or more
different alkyl branches, the name
of each branch, with its position
number precedes the name. the
branch names are placed in
alphabetical order.
Alkenes and alkynes
Both groups are unsaturated hydrocarbons.
Each group is a homologous series.
 The main chain is defined as the chain
containing the greatest number of double/tripple
bonds
 We number the position of the double/tripple
bond so that it has the lowest numbers.
ALKENES
alkynes
Naming side chains
Example: name the following structure
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
Step 1 – Identify the correct functional
group
Naming side chains
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
Step 2 - find the longest chain
Naming side chains
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
Step 3 - add the prefix naming the longest
chain
Naming side chains
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
Step 4 - number the longest chain
with the lowest number closest to
the double bond
CH3 CH2 C
2
CH2
1
CH2
3
C
4
CH2
5
CH3
CH3
CH3
6
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
Naming side chains
Step 5 - add that number to the
name
CH3 CH2 C
2
CH2
1
CH2
3
C
4
CH2
5
CH3
CH3
CH3
6
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
Naming side chains
ethyl
methyl
methyl
Step 6 - Name the side chains
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
CH3 CH2 C
2
CH2
1
CH2
3
C
4
CH2
5
CH3
CH3
CH3
6
Naming side chains
ethyl
methyl
methyl
Step 7 - Place the side chains in
alphabetical order & name the
compound
name the following
CH3 CH2
CH CH3
CH2CH2
CH3
CH3 CH
CH
CH3
CH
CH3
CH2 CH2 CH3
CH2 CH3
CH3CH2CH CH CH CH2CH CH3
CH3
CH2CH3
CH3 CH3
CH3
CH2
CH2
CH2
CH2
CH2
CH2
CH3
CH3
CH
CH2
CH2
CH
CH2
CH2
CH3
CH3
CH3
CH2
CH
CH2
CHCH2
CH2
CH3
CH2
CH3
CH2 CH CH C CH3CH3
CH3
CH3
1
2
3
4
CH3 CH2
CH CH3
CH2CH2
CH3
CH3 CH
CH
CH3
CH
CH3
CH2 CH2 CH3
CH2 CH3
CH3CH2CH CH CH CH2CH CH3
CH3
CH2CH3
CH3 CH3
9 1
0
1
1
CH3 C CH CH CH3
CH2 CH2
CH3
CH3
CH CH
CH2
CH
CH3
CH3
CH3
CH2 C C
CH2
CH3
CH3
ISOMERS
 A GOOD TIME TO INTRODUCE ISOMERS
(COMPOUNDS WITH THE SAME
MOLECULAR FORMULA BUT DIFFERENT
STRUCTURAL FORMULAE)
 TRY THE FOLLOWING:
Reactions of alkanes & alkenes
We study three particular reaction
cases:
Substitution
Addition
Elimination
Combustion
Reactions of alkanes
Substitution (of H,
commonly by Cl or Br)
Combustion (conversion to
CO2 & H2O)
 Combustion
 When alkanes are heated in a plentiful
supply of air, combustion occurs
 Alkanes are energetically unstable with
respect to water and carbon dioxide
 They only burn when they are in the
gaseous state

 Explain what happens when a candle
burns!
 2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10
H2O(g)
 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18
H2O(g)

SUBSTITUTION
 Reactions with chlorine
 Alkanes only react with chlorine when a
mixture of the two is exposed to sunlight
or ultraviolet light
 The light provides the energy required to
break the very strong bonds
 This is an example of a substitution
reaction
 In the presence of light, or at high
temperatures, alkanes react with
halogens to form alkyl halides. Reaction
with chlorine gives an alkyl chloride.
CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)
Cracking
 Cracking happens when alkanes are
heated in the absence of air
 The products of the cracking of long-chain
hydrocarbons are shorter chain molecules
 Ethane is cracked industrially to produce
ethene
PHYSICAL PROPERTIES
 Alkanes are non polar so they are
insoluble in water but soluble in each
other.
 Low molecular alkanes are gases.
 Boiling points increase with increasing
chain length (molecular weight) for the
first few members
 Boiling points decrease with increasing
number of branches.(Explain this in terms
of Van der Waals’ forces and surface area.
 Melting and boiling points increase with
increased molecular weight (Methane bp.
 -164°C, decane bp. 174°C)
 While boiling point decrease with chain
branching (decrease in surface area),
melting
points increase
 · Alkanes are less dense than water and
swim on top of water
alkenes: preparation and reactions
Alkenes: Preparation and
reactions
Two ways of making alkenes:
 1. Heat a concentrated solution of
potasium /sodium hydroxide in alcohol
(alcoholic KOH) with a haloalkane
(halogenoalkane)
This is dehydrohalogenation (removal of
hydrogen and halogen)
2. Heat concentrated sulphuric acid with
the alcohol- dehydration. THE ACID IS A
DEHYDRATING AGENT
i) Dehydration of alcohols
conc. H2SO4
R-CH2-CH2-
OH
R-CH=CH2 +
H2O
ii) Dehydrohalogenation of haloalkanes
NaOH/ethan
olR-CH2-CH2-X
reflux
R-CH=CH2 + HX
NaOH can be replaced by KOH
LEARNERS MUST KNOW MAJOR PRODUCTS IN ALL CASES AND
REACTION CONDITIONS
CH3CH2-CH-CH3
OH
H+
H+
CH3CH=CH-CH3 + H2O
CH3CH2-CH=CH2 + H2O
2-butanol
2-butene
major product
1-butene
Dehydration of alcohols
Dehydrohalogenation of haloalkanes
CH3CH-CH-CH2
BrH H
KOH CH3CH=CH-CH3 CH3CH2CH=CH2
alcohol
reflux
2-bromobutane
2-butene
(major product)
1-butene
REACTIONS OF ALKENES (VIP)
Catalytic hydrogenation:
- hydrogenation: addition of hydrogen
to a double bond and triple bond to yield
saturated product.
- alkenes will combine with hydrogen in
the present to catalyst to form alkanes.
C C H H C C
H H
Pt or Pd
25-90o
C
- Plantinum (Pt) and palladium (Pd) – Catalysts
- Pt and Pd: temperature 25-90oC
- Nickel can also used as a catalyst, but a higher
temperature of 140oC – 200oC is needed.
Addition of halogens:
i) In inert solvent:
- alkenes react with halogens at room temperature
and in dark.
- the halogens is usually dissolved in an inert solvent
such as dichloromethane (CH2Cl2) and
tetrachloromethane (CCl4).
- Iodine will not react with alkenes because it is less
reactive than chlorine and bromine.
- Fluorine is very reactive. The reaction will produce
explosion.C C X X C C
X X
inert solvent
X X = halogen such as Br2 or Cl2
Inert solvent = CCl4 or CH2Cl2
EXAMPLES:
C C
HH
H H Br Br
Br2
Br
Br
CCl4
CH3CH=CH2 Cl2
CCl4
CH3CH
Cl
CH2
Cl
C C
Br
H H
Br
H H
inert solvent (CCl4)
ethene
1,2-dibromoethane
* the red-brown colour of the bromine solution will fade and the
solution becomes colourless.
cyclohexene 1,2-dibromocyclohexane
propene 1,2-dichloropropane
Hydrohalogenation
MARKOVNIKOV’S RULE ( A statement
of the rule is not needed)
 There are 2 possible products when hydrogen
halides react with an unsymmetrical alkene.
 It is because hydrogen halide molecule can add to
the C=C bond in two different ways.
C C
H
HCH3
H
H-I
C C
H
HCH3
H
H-I
C C
H
HCH3
H
H I
C C
H
HCH3
H
I H
1-iodopropane
2-iodopropane
(major product)
Markovnikov’s rules (Not for
examination)
- the addition of HX to an
unsymmetrical alkene, the
hydrogen atom attaches itself to the
carbon atom (of the double bond)
with the larger number of hydrogen
atoms.
Addition reaction with concentrated
sulfuric acid: hydration of alkenes
- the alkene is absorbed slowly when it
passed through concentrated sulfuric
acid in the cold (0-15oC
Addition reaction with acidified water
(H3O+): hydration of alkenes
• Hydration: The addition of H atoms and
–OH groups from water molecules to a
multiple bond.
• Reverse of the dehydration reaction.
• Direct hydration of ethene:
 - passing a mixture of ethene and
steam over phosphoric (v) acid (H3PO4)
absorbed on silica pellets at 300oC and a
pressure of 60 atmospheres.
 - H3PO4 is a catalyst.
CH2=CH2 H2O
H3PO4
CH3CH2OH(g) (g)
300 o
C, 60 atm
(g)
ethene ethanol
C C H2O C C
H OH
alkene alcohol
Oxidation (Combustion of alkenes)
 The alkenes are highly flammable and
burn readily in air, forming carbon
dioxide and water.
 For example, ethene burns as follows
:
C2H4 + 3O2 → 2CO2 + 2H2O
HALOGENOALKANES
Halogenoalkanes are compounds in which one
or more hydrogen atoms in an alkane have
been replaced by halogen atoms (fluorine,
chlorine, bromine or iodine).
 Functional group = halogen
◦ Ex. Fluorine = fluoro
 Number by which carbon attached to, put
in alphabetical order
 Ex.
Bromoethane
Halogenoalkanes fall into different classes
depending on how the halogen atom is
positioned on the chain of carbon atoms.
There are some chemical differences between
the various types.
• Primary
• Secondary
• Tertiary
◦ Primary (1°) – carbon carrying halogen is
attached to only one carbon alkyl group
◦ Secondary (2°)– carbon carrying halogen is
attached to two other alkyl groups
◦ Tertiary (3°) – carbon carrying halogen is
attached to three alkyl groups
Reactions of the halogenoalkanes
 Substitution:In a substitution reaction,
one atom or group of atoms, takes the
place of another in a molecule.
Elimination: Halogenoalkanes also undergo
elimination reactions in the presence of
sodium or potassium hydroxide which is
dissolved in ethanol.
Example of substitution
When an aqueous solution of NaOH or KOH is
added to haloalkane an alcohol is produced.
propan-2-ol
Example of elimination
what conditions are needed?
ALCOHOLS (CnH2n+1OH)
Preparation and properties
nomenclature
 Select the longest chain which contains
the OH group and number so that the OH
group has the smallest number. See the
examples below
Classification
 In a primary (1°) alcohol, the carbon
which carries the -OH group is only
attached to one alkyl group.
 In a secondary (2°) alcohol, the carbon
with the -OH group attached is joined
directly to two alkyl groups, which may
be the same or different.
 In a tertiary (3°) alcohol, the carbon atom
holding the -OH group is attached directly
to three alkyl groups, which may be any
combination of same or different.
 See the examples below
Alcohols are classified as primary,
secondary or Tertiary
CH3 CH2 CH2 CH CH3
OH
CH3 CH2 CH2 C OH
CH3
CH3
CH3 CH2 CH2 CH2 OH
Reactions of alcohols
 Alcohols contain an –OH group covalently
bonded to a carbon atom.
 We need know:
 the esterification reaction
 Substitution and
 elimination
Preparation and reactions
 1. By hydration of alkanes
 The acid is absorbed in conc sulphuric
acid and then the acid is diluted.
 2. Hydrolysis of halogenoalkanes
 The halogen of the halogenoalkane is
replaced by an OH group Refer to
Halogenoalkanes
Classic example for learners to
write
CH3CHCH3
OH
H2SO4
CH2 CHCH3
H2
Pt
CH3CH2CH3
alcohol alkene alkane
esterification
 Acid + Alcohol yields Ester + Water
 Sulfuric acid is a catalyst.
 Each step is reversible.
CH3 C OH
O
+ CH2CH2CHCH3
CH3
OH
H
+
CH3C
O
OCH2CH2CHCH3
CH3
+ HOH
=>
 Acid + Alcohol yields Ester + Water
 Sulfuric acid is a catalyst.
 Each step is reversible.
Chapter 11 72
CH3 C OH
O
+ CH2CH2CHCH3
CH3
OH
H
+
CH3C
O
OCH2CH2CHCH3
CH3
+ HOH
=>
Aldehydes and
Ketones (Know the functional
groups)
 Nomenclature of Aldehydes:
 Select the longest carbon chain containing the
 carbonyl carbon.
 • The -e ending of the parent alkane name is
replaced
 by the suffix -al.
 • The carbonyl carbon is always numbered “1.” (It is
 not necessary to include the number in the name.)
 • Name the substituents attached to the chain in the
 usual way
Nomenclature of Ketones
No reactions. Just naming
SOME FUNCTIONAL GROUPS TO
KNOW
Alkanes, alkenes &alkynes

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Alkanes, alkenes &alkynes

  • 2. Alkanes = CnH2n+2  Alkenes = CnH2n Alkynes = CnH2n-2
  • 3.  ALKANES, ALKENES, ALKYNES AND CYCLOALKANES ARE HYDROCARBONS (COMPOUNDS CONTAINING ONLY CARBON AND HYDROGEN).  EACH OF THESE FORM A HOMOLOGOUS SERIES (A GROUP OF ORGANIC COMPOUNDS HAVING A COMMON GENERAL FORMULA/ OR IN WHICH EACH MEMBER FIFFERS FROM THE NEXT BY A – CH2)
  • 4.  THE HYDROCARBONS MAY BE SATURATED (CONTAINS ONLY SINGLE BONDS BETWEEN CARBON-CARBON ATOMS/ CARBON ATOMS BONDED TO THE MAXIMUM NUMBER OF HYDROGENS)  OR UNSATURATED (CONTAINS AT LEAST A DOUBLE BOND BETWEEN C-C ATOMS)
  • 5.
  • 6. ALKANES: NOMENCLATURE Also called paraffins. A group of saturated hydrocarbons with the general formula Cn H2n+2 . They form a homologous series. Straight chain alkanes have their carbon atoms bonded together to give a single chain Alkanes may also be branched.
  • 7. NAMING (GENERAL) Hydrocarbon names are based on: 1) type, 2) # of carbons, 3) side chain type and position 1) name will end in -ane, -ene, or -yne 2) the number of carbons is given by a “prefix” 1 meth- 2 eth- 3 prop- 4 but- 5 pent- 6 hex- 7 hept- 8 oct- 9 non- 10 dec- Actually, all end in a, but a is dropped when next to a vowel. E.g. a 6 C alkene is hexene
  • 8.  Determine the longest continuous chain (not always straight) in the molecule. The base name of the hydrocarbon is the name of the longest chain.
  • 10. IUPAC SYSTEM  Name any chain branching off the longest chain as an alkyl group (e.g., methyl, ethyl etc)  The complete name of a branch requires a number that locates the branch on the longest chain.  Therefore number the chain in whichever direction gives the smaller number for all branches.
  • 11.
  • 12. 6. When two or more branches are identical, use prefixes (di-, tri-, etc.) (e.g. 2,4- dimethylhexane). Numbers are separated with commas. Prefixes are ignored when determining alphabetical order. (e.g. 2,3,5- trimethyl-4-propylheptane) 7. When identical groups are on the same carbon, repeat the number of this carbon in the name. (e.g. 2,2-dimethylhexane)
  • 13. Where there are two or more different alkyl branches, the name of each branch, with its position number precedes the name. the branch names are placed in alphabetical order.
  • 14. Alkenes and alkynes Both groups are unsaturated hydrocarbons. Each group is a homologous series.  The main chain is defined as the chain containing the greatest number of double/tripple bonds  We number the position of the double/tripple bond so that it has the lowest numbers.
  • 17. Naming side chains Example: name the following structure CH3 CH2 C CH2 CH2 C CH2 CH3 CH3 CH3 Step 1 – Identify the correct functional group
  • 18. Naming side chains CH3 CH2 C CH2 CH2 C CH2 CH3 CH3 CH3 Step 2 - find the longest chain
  • 19. Naming side chains CH3 CH2 C CH2 CH2 C CH2 CH3 CH3 CH3 Step 3 - add the prefix naming the longest chain
  • 20. Naming side chains CH3 CH2 C CH2 CH2 C CH2 CH3 CH3 CH3 Step 4 - number the longest chain with the lowest number closest to the double bond
  • 21. CH3 CH2 C 2 CH2 1 CH2 3 C 4 CH2 5 CH3 CH3 CH3 6 CH3 CH2 C CH2 CH2 C CH2 CH3 CH3 CH3 Naming side chains Step 5 - add that number to the name
  • 22. CH3 CH2 C 2 CH2 1 CH2 3 C 4 CH2 5 CH3 CH3 CH3 6 CH3 CH2 C CH2 CH2 C CH2 CH3 CH3 CH3 Naming side chains ethyl methyl methyl Step 6 - Name the side chains
  • 23. CH3 CH2 C CH2 CH2 C CH2 CH3 CH3 CH3 CH3 CH2 C 2 CH2 1 CH2 3 C 4 CH2 5 CH3 CH3 CH3 6 Naming side chains ethyl methyl methyl Step 7 - Place the side chains in alphabetical order & name the compound
  • 24. name the following CH3 CH2 CH CH3 CH2CH2 CH3 CH3 CH CH CH3 CH CH3 CH2 CH2 CH3 CH2 CH3 CH3CH2CH CH CH CH2CH CH3 CH3 CH2CH3 CH3 CH3
  • 26. CH3 CH2 CH CH3 CH2CH2 CH3 CH3 CH CH CH3 CH CH3 CH2 CH2 CH3 CH2 CH3 CH3CH2CH CH CH CH2CH CH3 CH3 CH2CH3 CH3 CH3 9 1 0 1 1
  • 27. CH3 C CH CH CH3 CH2 CH2 CH3 CH3 CH CH CH2 CH CH3 CH3 CH3 CH2 C C CH2 CH3 CH3
  • 28. ISOMERS  A GOOD TIME TO INTRODUCE ISOMERS (COMPOUNDS WITH THE SAME MOLECULAR FORMULA BUT DIFFERENT STRUCTURAL FORMULAE)  TRY THE FOLLOWING:
  • 29. Reactions of alkanes & alkenes We study three particular reaction cases: Substitution Addition Elimination Combustion
  • 30. Reactions of alkanes Substitution (of H, commonly by Cl or Br) Combustion (conversion to CO2 & H2O)
  • 31.  Combustion  When alkanes are heated in a plentiful supply of air, combustion occurs  Alkanes are energetically unstable with respect to water and carbon dioxide  They only burn when they are in the gaseous state   Explain what happens when a candle burns!
  • 32.  2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(g)  2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) 
  • 33. SUBSTITUTION  Reactions with chlorine  Alkanes only react with chlorine when a mixture of the two is exposed to sunlight or ultraviolet light  The light provides the energy required to break the very strong bonds  This is an example of a substitution reaction
  • 34.  In the presence of light, or at high temperatures, alkanes react with halogens to form alkyl halides. Reaction with chlorine gives an alkyl chloride. CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)
  • 35. Cracking  Cracking happens when alkanes are heated in the absence of air  The products of the cracking of long-chain hydrocarbons are shorter chain molecules  Ethane is cracked industrially to produce ethene
  • 36. PHYSICAL PROPERTIES  Alkanes are non polar so they are insoluble in water but soluble in each other.  Low molecular alkanes are gases.  Boiling points increase with increasing chain length (molecular weight) for the first few members  Boiling points decrease with increasing number of branches.(Explain this in terms of Van der Waals’ forces and surface area.
  • 37.  Melting and boiling points increase with increased molecular weight (Methane bp.  -164°C, decane bp. 174°C)  While boiling point decrease with chain branching (decrease in surface area), melting points increase  · Alkanes are less dense than water and swim on top of water
  • 39. Alkenes: Preparation and reactions Two ways of making alkenes:  1. Heat a concentrated solution of potasium /sodium hydroxide in alcohol (alcoholic KOH) with a haloalkane (halogenoalkane) This is dehydrohalogenation (removal of hydrogen and halogen) 2. Heat concentrated sulphuric acid with the alcohol- dehydration. THE ACID IS A DEHYDRATING AGENT
  • 40. i) Dehydration of alcohols conc. H2SO4 R-CH2-CH2- OH R-CH=CH2 + H2O ii) Dehydrohalogenation of haloalkanes NaOH/ethan olR-CH2-CH2-X reflux R-CH=CH2 + HX NaOH can be replaced by KOH
  • 41. LEARNERS MUST KNOW MAJOR PRODUCTS IN ALL CASES AND REACTION CONDITIONS CH3CH2-CH-CH3 OH H+ H+ CH3CH=CH-CH3 + H2O CH3CH2-CH=CH2 + H2O 2-butanol 2-butene major product 1-butene Dehydration of alcohols Dehydrohalogenation of haloalkanes CH3CH-CH-CH2 BrH H KOH CH3CH=CH-CH3 CH3CH2CH=CH2 alcohol reflux 2-bromobutane 2-butene (major product) 1-butene
  • 43. Catalytic hydrogenation: - hydrogenation: addition of hydrogen to a double bond and triple bond to yield saturated product. - alkenes will combine with hydrogen in the present to catalyst to form alkanes. C C H H C C H H Pt or Pd 25-90o C - Plantinum (Pt) and palladium (Pd) – Catalysts - Pt and Pd: temperature 25-90oC - Nickel can also used as a catalyst, but a higher temperature of 140oC – 200oC is needed.
  • 44. Addition of halogens: i) In inert solvent: - alkenes react with halogens at room temperature and in dark. - the halogens is usually dissolved in an inert solvent such as dichloromethane (CH2Cl2) and tetrachloromethane (CCl4). - Iodine will not react with alkenes because it is less reactive than chlorine and bromine. - Fluorine is very reactive. The reaction will produce explosion.C C X X C C X X inert solvent X X = halogen such as Br2 or Cl2 Inert solvent = CCl4 or CH2Cl2
  • 45. EXAMPLES: C C HH H H Br Br Br2 Br Br CCl4 CH3CH=CH2 Cl2 CCl4 CH3CH Cl CH2 Cl C C Br H H Br H H inert solvent (CCl4) ethene 1,2-dibromoethane * the red-brown colour of the bromine solution will fade and the solution becomes colourless. cyclohexene 1,2-dibromocyclohexane propene 1,2-dichloropropane
  • 47. MARKOVNIKOV’S RULE ( A statement of the rule is not needed)  There are 2 possible products when hydrogen halides react with an unsymmetrical alkene.  It is because hydrogen halide molecule can add to the C=C bond in two different ways. C C H HCH3 H H-I C C H HCH3 H H-I C C H HCH3 H H I C C H HCH3 H I H 1-iodopropane 2-iodopropane (major product)
  • 48. Markovnikov’s rules (Not for examination) - the addition of HX to an unsymmetrical alkene, the hydrogen atom attaches itself to the carbon atom (of the double bond) with the larger number of hydrogen atoms.
  • 49. Addition reaction with concentrated sulfuric acid: hydration of alkenes - the alkene is absorbed slowly when it passed through concentrated sulfuric acid in the cold (0-15oC
  • 50. Addition reaction with acidified water (H3O+): hydration of alkenes • Hydration: The addition of H atoms and –OH groups from water molecules to a multiple bond. • Reverse of the dehydration reaction. • Direct hydration of ethene:  - passing a mixture of ethene and steam over phosphoric (v) acid (H3PO4) absorbed on silica pellets at 300oC and a pressure of 60 atmospheres.  - H3PO4 is a catalyst.
  • 51.
  • 52. CH2=CH2 H2O H3PO4 CH3CH2OH(g) (g) 300 o C, 60 atm (g) ethene ethanol C C H2O C C H OH alkene alcohol
  • 53. Oxidation (Combustion of alkenes)  The alkenes are highly flammable and burn readily in air, forming carbon dioxide and water.  For example, ethene burns as follows : C2H4 + 3O2 → 2CO2 + 2H2O
  • 54. HALOGENOALKANES Halogenoalkanes are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine).
  • 55.  Functional group = halogen ◦ Ex. Fluorine = fluoro  Number by which carbon attached to, put in alphabetical order  Ex. Bromoethane
  • 56. Halogenoalkanes fall into different classes depending on how the halogen atom is positioned on the chain of carbon atoms. There are some chemical differences between the various types. • Primary • Secondary • Tertiary
  • 57. ◦ Primary (1°) – carbon carrying halogen is attached to only one carbon alkyl group ◦ Secondary (2°)– carbon carrying halogen is attached to two other alkyl groups ◦ Tertiary (3°) – carbon carrying halogen is attached to three alkyl groups
  • 58. Reactions of the halogenoalkanes  Substitution:In a substitution reaction, one atom or group of atoms, takes the place of another in a molecule. Elimination: Halogenoalkanes also undergo elimination reactions in the presence of sodium or potassium hydroxide which is dissolved in ethanol.
  • 59. Example of substitution When an aqueous solution of NaOH or KOH is added to haloalkane an alcohol is produced. propan-2-ol
  • 60. Example of elimination what conditions are needed?
  • 62. nomenclature  Select the longest chain which contains the OH group and number so that the OH group has the smallest number. See the examples below
  • 63. Classification  In a primary (1°) alcohol, the carbon which carries the -OH group is only attached to one alkyl group.  In a secondary (2°) alcohol, the carbon with the -OH group attached is joined directly to two alkyl groups, which may be the same or different.
  • 64.  In a tertiary (3°) alcohol, the carbon atom holding the -OH group is attached directly to three alkyl groups, which may be any combination of same or different.  See the examples below
  • 65.
  • 66. Alcohols are classified as primary, secondary or Tertiary CH3 CH2 CH2 CH CH3 OH CH3 CH2 CH2 C OH CH3 CH3 CH3 CH2 CH2 CH2 OH
  • 67. Reactions of alcohols  Alcohols contain an –OH group covalently bonded to a carbon atom.  We need know:  the esterification reaction  Substitution and  elimination
  • 68. Preparation and reactions  1. By hydration of alkanes  The acid is absorbed in conc sulphuric acid and then the acid is diluted.
  • 69.  2. Hydrolysis of halogenoalkanes  The halogen of the halogenoalkane is replaced by an OH group Refer to Halogenoalkanes
  • 70. Classic example for learners to write CH3CHCH3 OH H2SO4 CH2 CHCH3 H2 Pt CH3CH2CH3 alcohol alkene alkane
  • 71. esterification  Acid + Alcohol yields Ester + Water  Sulfuric acid is a catalyst.  Each step is reversible. CH3 C OH O + CH2CH2CHCH3 CH3 OH H + CH3C O OCH2CH2CHCH3 CH3 + HOH =>
  • 72.  Acid + Alcohol yields Ester + Water  Sulfuric acid is a catalyst.  Each step is reversible. Chapter 11 72 CH3 C OH O + CH2CH2CHCH3 CH3 OH H + CH3C O OCH2CH2CHCH3 CH3 + HOH =>
  • 73. Aldehydes and Ketones (Know the functional groups)  Nomenclature of Aldehydes:  Select the longest carbon chain containing the  carbonyl carbon.  • The -e ending of the parent alkane name is replaced  by the suffix -al.  • The carbonyl carbon is always numbered “1.” (It is  not necessary to include the number in the name.)  • Name the substituents attached to the chain in the  usual way
  • 74.