a power point on alkanes and alkenes.

2. Homologous Groups
Alkanes
Alkenes
Alcohols
Halogenoalkanes
Carbonyl Compounds
Carboxylic acids and Derivatives
Benzene
Phenols

3. Bond Cleavage and Reaction
Mechanisms

4. Bond Cleavage
All single covalent bonds contain 2 electrons
Bond cleavage (or bond scission or fission) is the
splitting of chemical bonds
Bonds can be cleaved or broken via:
Heterolytic cleavage (heterolysis)
Homolytic cleavage (homolysis)

5. Heterolytic Cleavage (Heterolysis)
Both electrons of the covalent bond go to only one
atom during bond breaking
The atom which receives both electron now has a
negative charge (anion)
The atom which did not receive any electrons is
positively charged (cation)

6. Heterolytic Bond Cleavage
 Double-headed arrow shows the
movement of an electron pair
(two electrons)
 Heterolytic bond cleavage always
produces charged species (or
ions)
 Carbocation (or carbonium ion)
- An ion which contains a
positively charged carbon atom
The carbon in the carbocation
has only three bonds

7. Homolytic Cleavage (Homolysis)
 Symmetrical cleavage – each atom of the bond receives
one electron each
 Energy required to homolytically cleave a bond is called
the Bond Dissociation Energy (BDE)
 Homolytic cleavage produces uncharged , neutral, highly
reactive species called radicals

8. Homolytic Bond Cleavage
 Single-headed arrow (or fish
hooks or curly arrows) show the
movement of a single electron
 Homolytic bond cleavage always
produces reactive species known
as radicals
Radicals (or free radicals)
 are molecules with an
unpaired electron
 are usually highly reactive
because of unpaired e-
 May be electrically neutral

9. Classwork
Identify the family to which each of the following belong to:

10. Classwork
a. Alkanes
b. Alkenes
c. Alcohols
d. Halogenoalkanes
e. Aldehydes
f. Ketones
g. Carboxylic acids
h. Esters
i. Acid chlorides
j. Amides
k. Amines
l. Nitriles
Write the functional group and general formula for EACH of
the following family of compounds:

11. Homework
Define the following terms:
1. Electrophile
2. Nucleophile
3. Inductive Effect
a) Negative Inductive Effect
b) Positive Inductive Effect
4. Conjugative effect
5. Reaction Mechanisms

12. Reaction Mechanisms
In organic reactions, electrons flow from “electron rich”
areas to “electron poor” areas
Reaction mechanism
the step by step sequence of elementary reactions by which
overall chemical change occurs
Review the following terms:
Rate determining step
Elementary reactions (unimolecular, bimolecular)

13. Alkanes

14. Alkanes
Contains carbon and hydrogen only (hydrocarbons)
Contains no double bonds; no other atoms can be
added to carbon (saturated)
Chemically unreactive due to strong C–C and C–H
bonds (σ bonds)
Do not react with acids, bases, or strong oxidizing and
reducing agents
Used as fuels - hydrocarbons with small molecules
make good because they are volatile, flow easily and
are easily ignited

15. What is meant by the following?
1. Aliphatic alkane
2. Acyclic alkane
3. Cycloalkane
4.

16. Preparation of Alkanes
 Petroleum is a source of
hydrocarbons by fractional
distillation and cracking
 Crude petroleum is a
complex mixture of gaseous,
liquid and solid
hydrocarbons
 The crude petroleum mixture
may contain alkanes,
cycloalkanes, aromatics
(benzene) and some alkenes
as well as some compounds
of oxygen, nitrogen and
sulphur

17. http://www.bbc.co.uk/schools/gcsebitesize/science/aqa_pre_2011/rocks/fuelsrev3.shtml
Fractional Distillation of Crude Oil

18. Fraction Size
Boiling-Point
Range (0C)
Uses
Gas C1 to C5 -160 to 30
Gaseous fuel,
production of H2
Gasoline C5 to C7 25 to 75 Motor fuel
Naphtha C6 to C10 75 to 190
cracked to make
more petrol and
alkenes
Kerosene, fuel-oil C12 to C18 180 to 400
Diesel fuel,
furnace fuel,
cracking
Lubricants C16 and up 350 and up Lubricants
Paraffins C20 and up Low-melting solids Candles, matches
Asphalt C36 and up Gummy residues Surfacing roads

19. Review: Nomenclature of Alkanes
(IUPAC Rules)
1. Locate the longest continuous chain of carbon
atoms. This chain determines the parent name for the
alkane
C
H3 CH2 CH2 CH2 CH
CH2
CH3
CH3

20. 2. Number the longest chain beginning
with the end of the chain nearer the
substituent
3. Use the numbers obtained to
designate the location of the
substituent(s)
4. If the same substituent occurs more
than once use the prefixes;- di-, tri-,
tetra-, etc
C
H3 CH2 CH2 CH2 CH
CH2
CH3
CH3
1
2
3
4
5
6
7
3-Methylheptane
C
H3 CH2 CH2 CH CH
CH2
CH3
CH3
CH3
3,4-dimethylheptane

21. 5. If different substituents are
present then they are written
alphabetically (ignoring
prefixes;- di-, tri- etc)
6. If two substituents are present
on the same carbon use that
number twice
C
H3 CH2 CH2 CH CH
C
H
CH3
CH3
CH2
H3C
CH3
4-ethyl-2,3-dimethylheptane
C
H3 CH2 CH2 CH CH2
C
CH3
CH2
H3C
CH3
C
H3
4-ethyl-2,2-dimethylheptane

22. 7. When two chains can be used as
the base chain (i.e. they have the
same length), the one with the
most substituent is used.
8. When branching occurs at an
equal distance from either end of
the chain, choose the name that
gives the lower number at the first
point of difference
C
H3 CH CH CH2 CH2
C
CH3
CH3
C
H3
CH3
C
H2
CH3
5-ethyl-2,2,6-trimethylheptane
Not
3-Ethyl-2,6,6-trimethylheptane

23. 2.2 Describe selected chemical
reactions of alkanes
2.3 Explain the steps involved
in the mechanism of free
radical substitution

24. Reactions of Alkanes
1.Combustion
Alkanes undergo combustion in air; good fuels
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
0H = –2855
kJ/mol
Complete combustion occurs in excess O2 and
produces CO2 and H2O
Incomplete produces a combination of CO and
Carbon in addition to water and carbon dioxide
Carbon dioxide contributes to global warming;
Carbon monoxide is toxic

25. Reactions of Alkanes (Cont'd)
2. Free Radical Substitution Reactions
In the presence of light alkanes undergo
substitution reactions with halogens
R-H + Cl2 → R-Cl + H-Cl
In substitution reactions, one atom of a
molecule is removed and replaced or
substituted by another atom or group of atoms
Mechanism of substitution reaction for alkanes
involves free radicals

26. Free Radical Substitution of Alkanes
For reaction to occur
between alkane and
chlorine, C-H and Cl-Cl
bonds must break
The C-H bond is stronger
than Cl-Cl bond,
therefore, the reaction
proceeds by first breaking
the Cl-Cl bond

27. Free Radical Substitution: Mechanism
Step 1: Initiation
 UV light provides enough energy to break the halogen bond
 Cl-Cl bond is broken via homolytic fission to produce radicals
 Reactivity of halogens increases in the order of Br < Cl < F

28. Free Radical Substitution: Mechanism
The chlorine radical abstracts a hydrogen atom
from the alkane forming hydrogen chloride and a
free radical of the alkane
The alkyl free radical goes on to react with a
molecule of chlorine and regenerate another
chlorine free radical, and so on...

29. Free Radical Substitution of Alkanes
Step 3: Termination
 Two radicals combine to form neutral molecule
 These steps stop the reaction (no new radicals formed to
propagate the reaction)

30. Free Radical Substitution of Alkanes
Once initiated, the reaction can keep itself going
and is referred to as a chain reaction
The reaction gives a mixture of halogenated
products (halogenoalkanes/alkylhalides)
because more than one hydrogen can be
substituted

31. Reactions of Alkanes (Cont'd)
3. Cracking of Alkanes
Cracking is the name given to breaking up large
hydrocarbon molecules into smaller, more useful
molecules
Cracking can be achieved by:
– using high pressures and temperatures without a
catalyst (thermal cracking) or
– lower temperatures and pressures in the presence
of a catalyst (catalytic cracking)
http://www.chemguide.co.uk/organicprops/alkanes/cracking.html

32. Cracking of Alkanes (Cont'd)
Hydrocarbons are broken up in a random manner
to produce mixtures of smaller hydrocarbons (eg.
alkanes and alkenes)

33. Cracking of Alkanes (Cont'd)
Thermal cracking
 High temperatures
(typically in the range 450
to 750 0C) and pressures
(up to about 70
atmospheres) are used
 Gives mixtures of
products containing high
proportions of
hydrocarbons with double
bonds (alkenes)
Catalytic cracking
 Modern cracking uses zeolites
(alumino silicates) as the
catalyst
 Zeolites are chosen to give high
percentages of hydrocarbons
with between 5 and 10 carbon
atoms - particularly useful for
gasoline
 produces high proportions of
branched alkanes and aromatic
hydrocarbons like benzene
http://www.chemguide.co.uk/organicprops/alkanes/cracking.html

34. Homework
What are the intermolecular forces found in alkanes?
Explain how these forces arise.
For the following pairs of compounds, state which will have
a higher boiling point and explain why.
Ethane and pentane
Pentane and 2-methylbutane
Pentane and pentanol

35. Homework: Alkanes
a) Do you expect methane to react with
bromine in the dark?
b) Write a mechanism to show how methane
would react with bromine.

36. Review on own: Empirical and Molecular
Formulae
1. Define the following: a) Empirical Formula b) Molecular
Formulae c) Structural formula
2. 14.4 g of an alkane, when analyzed, is found to contain 12 g carbon.
What is its molecular formula?
3. A compound containing 85.71% C and 14.29% H has a relative
molecular mass of 56. Find its molecular formula.
4. 0.00382 mol of a hydrocarbon X was burnt in oxygen and 1.68 g of
carbon dioxide was formed and 0.756 g of water was formed.
a) Calculate the empirical formula of hydrocarbon X
b) Suggest a molecular formula for hydrocarbon X

37. Alkenes

38. Alkenes
Alkenes are hydrocarbons containing a C=C double bond
Alkenes are called unsaturated molecules because atoms
can be added if double bond breaks; carbon atoms are not
bonded to the maximum number of atoms they can be
Alkenes are extremely reactive and important compounds in
chemical industry and are converted into useful compounds
e.g. plastics and alcohols

39. 1. Find the longest carbon chain that contains the C=C double
bond
2. Number the main chain to give the lowest possible number to
the carbon-carbon double bond
3. Do not need to number cycloalkenes because it is understood
that the double bond is in the one position
4. Add substituents and their position to the alkene as prefixes. Of
course remember to give the lowest numbers possible. And
remember to name them in alphabetical order when writing
them.
5. Identifying any stereoisomers and use cis and trans to name
them where appropriate
Review: Alkene Nomenclature

40. Alkenes
 Alkene contains at least one
double bond which forms through
one σ bond and one π bond
 Pi electrons are loosely held and
readily react with electrophiles or
free radical
 Alkenes are more reactive
compared to alkanes
 Usually undergo addition
reactions in which the pi bond is
broken and two new groups are
attached to give a saturated
compound

41. Alkenes undergo Electrophilic Addition reactions
Key:
E+
= electrophile
N
-
= nucleophile

42. Double bonds and Electrophiles
The double bond of an alkene is an area of high electron density, and
therefore an area of high negative charge
The negative charge of the double bond may be attacked by electron-
deficient species (electrophiles), which will accept a lone pair of
electrons
These species have either a full positive charge or slight positive
charge on one or more of their atoms. They are called electrophiles,
meaning ‘electron loving’
Alkenes undergo addition reactions when attacked by electrophiles.
This is called electrophilic addition

43. 2.4 Describe selected chemical
reactions of alkenes
2.5 Explain the steps involved in
the mechanism of selected
chemical reactions of alkene
functional group

44. Reactions of Alkenes
1. Reaction with bromine liquid (Br2 (l))
Alkenes are unsaturated molecules
Bromine adds across the C=C double bond
Test used to distinguish between alkenes and alkanes; Alkenes
decolourise bromine liquid (in CCl4); red-brown colour
disappears
Product: dihalogenoalkane
Conditions: room temperature
http://www.chemguide.co.uk/mechanisms/eladd/symbr2.html

45. 1. Reactions of alkenes with bromine
As Br2 approaches the alkene it
becomes polarised by the pi
electron cloud; a dipole is
induced in the Br2 molecule
What is the name of the
product that is formed?
The double bond attacks the
electrophilic end of the
molecule to form a
bromonium ion
A carbon in the bromonium ion
intermediate is now attacked
from the back by the bromide
ion
δ+
δ-

46. Examples:
C C
H
H
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. Reactions of Alkenes
2. Reaction with aqueous bromine (Br2 (aq))
Aqueous bromine contains Br2 and H2O
The major product of the test is not 1,2-dibromoethane
(CH2BrCH2Br) but 2-bromoethan-1-ol (CH2BrCH2OH)
CH2=CH2 + Br2 + H2O  CH2BrCH2OH + HBr
Bromine water used to test for unsaturation eg. alkenes

48. Reaction with aqueous bromine (Br2 (aq))
 Water is present in such a large amount (greater than Br-),
a water molecule (which has a lone pair) adds to one of
the carbon atoms, followed by the loss of a H+ ion.
 The major product of the test is not 1,2-dibromoethane
(CH2BrCH2Br) but 2-bromoethan-1-ol (CH2BrCH2OH).
Ethene is a symmetrical alkene – it
gives the same carbocation regardless
which carbon of the double bond,
bromine attaches to
What if the alkene is unsymmetrical?
Which carbon does bromine add to?

49. Reactions of Alkenes
3. Reaction with hydrogen halides (eg. HBr )
Alkenes react with hydrogen halides (in gaseous state or in
aqueous solution) to form addition products
The hydrogen and halogen atoms add across the double bond to
form haloalkanes (alkyl halides)
Reactivity of hydrogen halides : HF < HCl < HBr < HI

50. Reaction of Ethene with HBr

51. Addition to unsymmetrical alkenes eg. prop-1-ene
When an electrophile (e.g. HBr)
attacks an alkene with three or
more carbon atoms (e.g. propene),
a mix of products is formed. This is
because these alkenes are
unsymmetrical
The major product can be predicted using Markovnikov’s rule.
This states that when a substance H–X reacts with an alkene, the major product is the one
in which the hydrogen atom is bonded to the carbon atom which bears the greatest
number of hydrogen atoms.
minor product:
1-bromopropane
major product:
2-bromopropane
+ HBr
Unequal amounts of each
product are formed due to the
relative stabilities of the
carbocation intermediates.

52. A chain of carbon atoms can be represented by R when drawing
organic structures. This is an alkyl group (general formula
CnH2n+1).
 Primary (1°) carbocations have
one alkyl group attached to the
positively-charged carbon.
 Secondary (2°) carbocations have
two alkyl groups attached to the
positively-charged carbon.
 Tertiary (3°) carbocations have
three alkyl groups attached to the
positively-charged carbon.
Classification of carbocations

53. The stability of carbocations increases as the number of alkyl groups on the
positively-charged carbon atom increases.
The stability increases because alkyl groups contain a greater electron
density than hydrogen atoms. This density is attracted towards, and
reduces, the positive charge on the carbon atom, which has a
stabilizing effect.
Stability of Carbocations
increasing stability
tertiary
primary secondary

54. Addition of hydrogen halides to unsymmetrical
alkenes and Markovnikov’s rule
CH3CH=CH2 HCl
CH3CHCH2
H Cl
CH3CHCH2
Cl H
1-chloropropane
2-chloropropane
(major product)
according to Markovnikov's
rules
1
2
3
Propene

55. MECHANISM:
Step 1: Formation of carbocation
C
C
H H
H
CH3
H Cl C
C
H H
H
C
H
H
H
H
C
C
H H
H
C
H
H
H H
or
less stable carbocation
(1o
carbocation)
more stable carbocation
(2o
carbocation)
Cl-
- 2o
carbocation is more stable than 1o
carbocation.
- 2
o
carbocation tends to persist longer, making it more likely to combine with
Cl
-
ion to form 2-chloromethane (basis of Markovnikov's rule).
C
C
H H
H
C
H
H
H H
Cl-
Step 2: Rapid reaction with a negative ion
C
C
H H
H
C
H
H
H H
Cl
2-chloromethane (major product)

56. When HBr is added to an alkene in the absence of peroxides it obey
Markovnikov’s rule
When HBr (not HCl or HI) reacts with unsymmetrical alkene in the
presence of peroxides or oxygen, HBr adds in the opposite direction
to that predicted by Markovnikov’s rule
The product between propene and HBr under these conditions is 1-
bromopropane and not 2-bromopropane.
How would you get the anti-markovnikov product as
the major product?
CH3CH=CH2 HBr CH3CH2CH2Br
peroxide
1-bromopropane
(major product)
anti-Markovnikov's orientation

57. Reactions of Alkenes
4. Reaction with concentrated H2SO4
Hydration of alkenes to produce alcohols
What are the necessary conditions for the rxn to proceed?
Alkenes react with concentrated sulphuric acid in the cold
to produce alkyl hydrogensulphates
Follow by addition of hot water to form alcohol
For unsymmetrical alkenes, follow Markovnikov’s rule
http://www.chemguide.co.uk/organicprops/alkenes/h2so4.html

58. 4. Reaction of alkene with conc. H2SO4

59. 5. Hydrogenation of Alkenes
Hydrogen can be added to the carbon–carbon double bond using a nickel
catalyst in a process called hydrogenation.
C2H4 + H2  C2H6
Vegetable oils are unsaturated and
may be hydrogenated to make
margarine, which has a higher
melting point.
As well as a nickel catalyst, this
requires a temperature of 200 °C
and a pressure of 1000 kPa.

60. Hydrogenation of alkenes
1. What are the conditions under which alkenes are
hydrogenated?
2. What is meant by “hardening” of fats?
3. Give one benefit of hydrogenating fats?
4. What are “trans fats”? How do they arise?
5. In some parts of the world “trans fats” have been
banned in restaurants. Why is this?

61. 6. Oxidation of Alkenes
a) Reaction between alkenes and cold, acidified KMnO4
 Solution changes from purple to colourless
 Used as a test for alkenes

62. 6. Oxidation of Alkenes
b) Reaction between alkenes and hot, acidified KMnO4
 The double bond of the alkene is broken
 Products may be carboxylic acids, ketones or CO2 depends on the
alkene
C C
R
R
R'
H
C C H
R'
R
OH
R
OH
KMnO4/H+
C O
R
R C
O
H
R'
C O
R
R C
O
OH
R'
ketone acid ketone aldehyde
Example:
KMnO4/H+
C
O
O
C
HO
4-methyl-4-octene 2-pentanone butanoic acid
Under vigorous
conditions:
 Diol is oxidized
further (C=C is
broken)
 Aldehydes are quickly
oxidized to carboxylic
acids

63. For example, milder oxidation of an alkene produces
methanal and propanone. What is the structure of the
alkene?
Determination of the position of the double bond
C O
methanal
H
H C
O CH3
CH3
propanone
C
H
H C CH3
CH3
C
C CH3
CH3
H
H
remove the oxygen atoms from the carbonyl compounds and
joining the carbon atoms with a double bond.
2-methylpropene

64. Example:
An alkene with the molecular formula C6H12 is oxidised with hot
KMnO4 solution. The carboxylic acids, butanoic acid
(CH3CH2CH2COOH) and ethanoic acid (CH3COOH), are produced.
Identify the structural formula of the alkene.
C C
H
R
H
R'
CH3CH2CH2COOH and CH3COOH
C O
OH
R C
O
OH
R'
CH3CH2CH2CH=CHCH3
KMnO4/H+
i) cleavage of the double bond gives a mixture of carboxylic acids
ii) location of the double bond is done by taking away the oxygen atoms from the
carboxylic acids and then joining the carbon atoms by the double bond.
RCOOH and R'COOH RCH=CHR'
butanoic acid ethanoic acid 2-hexene

65. Electrophiles: true or false?

66. Checkpoint: Structure of Carbocations

67. Homework
1. Show the mechanism for the reaction between
cyclohexene and bromine in CCl4.
What is the name of the product formed?
2. Write a mechanism to show how prop-1-ene
would react with
a) bromine water
b) concentrated sulphuric acid
Name the major and minor products in each case.