The document discusses different types of hydrocarbons including alkanes, alkenes, alkynes and cycloalkanes. It provides their general formulas and describes them as saturated or unsaturated. The document also discusses hydrocarbon naming conventions including identifying the functional group, number of carbons, side chains and their positions. Homologous series and structural isomers are introduced. Methods for producing alkenes from alkanes like dehydration of alcohols and dehydrohalogenation of haloalkanes are outlined.

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)

6. 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. Hydrocarbon names are based on:
1) Type
2)No. 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

9.  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. 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)

12. 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.

13. 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.

16. Example: name the following structure
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
Step 1 – Identify the correct functional
group

17. CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
Step 2 - find the longest chain

18. CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
Step 3 - add the prefix naming the longest
chain

19. 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

20. 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
Step 5 - add that number to the
name

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
ethyl
methyl
methyl
Step 6 - Name the side chains

22. 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
6ethyl
methyl
methyl
Step 7 - Place the side chains in
alphabetical order & name the
compound

23. 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

24. 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

25. 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 10
11

26. CH3 C CH CH CH3
CH2 CH2
CH3
CH3
CH CH
CH2
CH
CH3
CH3
CH3
CH2 C C
CH2
CH3
CH3

27.  A GOOD TIME TO INTRODUCE ISOMERS
(COMPOUNDS WITH THE SAME
MOLECULAR FORMULA BUT DIFFERENT
STRUCTURAL FORMULAE)
 TRY THE FOLLOWING:

28. We study three particular reaction
cases:
Substitution
Addition
Elimination
Combustion

29. Substitution (of H,
commonly by Cl or Br)
Combustion (conversion to
CO2 & H2O)

30.  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!

31.  2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10
H2O(g)
 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18
H2O(g)


32.  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

33.  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)

34.  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

35.  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.

36.  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

38. 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

39. i) Dehydration of alcohols
conc. H2SO4
R-CH2-CH2-OH R-CH=CH2 + H2O
ii) Dehydrohalogenation of haloalkanes
NaOH/ethanol
R-CH2-CH2-X
reflux
R-CH=CH2 + HX
NaOH can be replaced by KOH

40. 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

42. 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
25-90
o
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.

43. 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

44. 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

46.  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)

47. 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.

48. - the alkene is absorbed slowly when it
passed through concentrated sulfuric
acid in the cold (0-15oC

49. • 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. CH2=CH2 H2O
H3PO4
CH3CH2OH(g) (g)
300 o
C, 60 atm
(g)
ethene ethanol
C C H2O C C
H OH
alkene alcohol

52.  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.  Functional group = halogen
› Ex. Fluorine = fluoro
 Number by which carbon attached to,
put in alphabetical order
 Ex.
Bromoethane

55. 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

56. › 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

57.  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.

58. When an aqueous solution of NaOH or KOH is added
to haloalkane an alcohol is produced.
propan-2-ol

61.  Select the longest chain which contains
the OH group and number so that the
OH group has the smallest number. See
the examples below

62.  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.

63.  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. CH3 CH2 CH2 CH CH3
OH
CH3 CH2 CH2 C OH
CH3
CH3
CH3 CH2 CH2 CH2 OH

66.  Alcohols contain an –OH group covalently
bonded to a carbon atom.
 We need know:
 the esterification reaction
 Substitution and
 elimination

67.  1. By hydration of alkanes
 The acid is absorbed in conc sulphuric
acid and then the acid is diluted.

68.  2. Hydrolysis of halogenoalkanes
 The halogen of the halogenoalkane is
replaced by an OH group Refer to
Halogenoalkanes

69. CH3CHCH3
OH
H2SO4
CH2 CHCH3
H2
Pt
CH3CH2CH3
alcohol alkene alkane

70.  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
=>

71.  Acid + Alcohol yields Ester + Water
 Sulfuric acid is a catalyst.
 Each step is reversible.
Chapter 11 71
CH3 C OH
O
+ CH2CH2CHCH3
CH3
OH
H
+
CH3C
O
OCH2CH2CHCH3
CH3
+ HOH
=>

72.  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