1. Chapter HYDROCARBONS

2. contain only carbon and
12.1 : ALKANES
are compounds which
contain only carbon and
hydrogen atoms.
Hydrocarbon

3. CYCLO-ALKANES (saturated) Alkanes which C atoms are
HIDROCARBONS
ALIPHATIC
AROMATIC
(contain one or more benzene ring)
ALKANES
(saturated)
contain only
single bond
ALKENES
(unsaturated)
contain
C=C
ALKYNES
(unsaturated)
contain
C≡C
CYCLO-ALKENES
(unsaturated)
CYCLO-ALKANES (saturated)
Alkanes which C atoms are
join in rings

4. Saturated hydrocarbons
- compound with C-C
- Example : alkanes and cycloalkanes
Contain the maximum number of hydrogen atoms that the carbon compound can possess.

5. Unsaturated hydrocarbons
-compounds with multiple bonds
-Example : alkenes, cycloalkenes, alkynes and aromatic hydrocarbons
They posses fewer than the maximum number of hydrogen atoms.

6. ALKANES
Alkanes are known as saturated hydrocarbon which contain only single covalent bonds.
General formula for straight chain of alkanes is CnH2n+2
where n ≥ 1
General formula for cycloalkanes is CnH2n
where n ≥ 3

7. Each carbon atom in alkanes is
- sp3 hybridised
- tetrahedral with four sigma bond (formed by the four sp3 hybrid orbitals.)
- all bond angles are close to 109.5o
Alkanes IUPAC names have the –ane suffix.

8. The First Ten Unbranched Alkanes
Molecular formula
Structural formula
No .of C atoms
Name
CH4 1
Methane
C2H6
CH3—CH3 2
Ethane
C3H8
CH3—CH2—CH3 3
Propane
C4H10
CH3—(CH2)2—CH3 4
Butane
C5H12
CH3—(CH2)3—CH3 5
Pentane
C6H14
CH3—(CH2)4—CH3 6
Hexane
C7H16
CH3—(CH2)5—CH3 7
Heptane
C8H18
CH3—(CH2)6—CH3 8
Octane
C9H20
CH3—(CH2)7—CH3 9
Nonane
C10H22
CH3—(CH2)8—CH3 10
Decane

10. Starting from C4H10 onwards, the alkanes show the phenomenon of chain isomerism.
They can exist as linear or branched alkanes.

11. Example
C4H10
CH3(CH2)2CH CH3CH(CH3)2 2-
isobutane

12. C5H12

13. IUPAC NOMENCLATURE
IUPAC – International Union of Pure and Applied Chemistry
Branched - chain alkanes are named according to the following rules:

14. STEP 1
Choose the longest continuous chain of carbon atoms; this chain determines the parent name for alkanes.
Examples:
CH3CH2CH2CH2CHCH3 |
CH3
Parent name: hexane

15. CH3CH2CH2CH2CHCH3
CH2
CH3
Parent name : heptane

16. STEP 2
Number the longest chain beginning with the end of the chain nearer the substituent.
Example: 6 5 4 3 2 1
CH3CH2CH2CH2CHCH3
CH3
substituent

17. 7 6 5 4 3
CH3CH2CH2CH2CH |
CH2
CH3
CH3
substituent 2 1

18. STEP 3 Use rule number 2 to locate the position of the substituent.
The position and the name of the substituent must be written in front of the parent chain.

19. CH3CH2CH2CH2CHCH3 CH3 2-methylhexane Examples: Substituent6 5 4 3 2 1
CH3CH2CH2CH2CHCH3
Substituent
-methyl at C-2
CH3
2-methylhexane

20. CH3CH2CH2CH2CH CH3 | CH2 CH3 3-methylheptane Substituent
-methyl at C-3 7 6 5 4 3
CH3CH2CH2CH2CH CH3 |
CH2
CH3 2 1
3-methylheptane

21. Some Common Substituent Groups
Alkane
name
substituent
methyl
CH3
methane
ethyl
CH2CH3
ethane
propyl
CH2CH2CH3
propane
CHCH3
isopropyl
CH3
butyl
butane
CH2CH2CH2CH3

22. isobutyl
sec-butyl
tert-butyl
neopentyl
CH2CHCH3
CH3
CHCH2CH3
CH3
CH3
CCH3
CH3
CH3
CH2CCH3
CH3

23. cyclopropyl
cyclobutyl
phenyl
benzyl
C6H5 or

24. name
substituent
Bromo
-Br
Chloro
-Cl
Flouro -F
Iodo -I
Hydroxyl
-OH
Amino
-NH2
Cyano
-CN
Nitro
-NO2

25. STEP 4
If two or more substituents are present, give each substituent a number corresponding to its location on the longest chain.
the substituent should be listed alphabetically.
In alphabetizing, the prefixes di, tri, tetra, sec-, tert- are ignored except iso and neo.

26. Example:
CH3CHCH2CHCH2CH3
| |
CH CH2 |
CH3
4-ethyl-2-methylhexane

27. STEP 5
If two substituents are present on the same carbon atom, use that number twice
Example:
CH3 |
CH3CH2CCH2CH2CH3 |
2 CH2
1 CH3
3-ethyl-3-methylhexane

28. STEP 6
If two or more identical substituents are present, use prefixes di-(2 identical substituents),tri-(3 identical substituents), tetra-(4 identical substituents).
Commas are used to separate numbers from each other.
Example:
CH3CH―CH CH3
| |
CH3 CH3
2,3-dimethylbutane

29. STEP 7
If there are two chains of equal length as the parent chain, choose the chain with the greater number of substituents.

30. 4-sec-butyl-2,3-dimethylheptane
CH3CH2-CH CH CH CHCH3
| | | |
CH3 CH2 CH3 CH3 |
CH2
CH3 5 6 7
2,3,5-trimethyl-4-propylheptane
(four substituents)
4-sec-butyl-2,3-dimethylheptane
(three substituents)

31. (NOT 2,4,5-trimethylhexane)
STEP 8
If branching occurs at an equal distance from either end of the longest chain, choose the name that gives the lower number at the first point of difference.
CH3CHCH2CH CHCH3
| | |
CH CH3 CH3
2,3,5-trimethylhexane
(NOT 2,4,5-trimethylhexane)

32. CYCLOALKANES
Cycloalkanes – alkanes which carbon atoms are joined in rings.
Cycloalkanes are known as saturated hydrocarbon, because it has the maximum number of bonded hydrogen ( only has single bonds).
General formula:
CnH2n where n = 3, 4, 5, ……

33. NOMENCLATURE OF CYCLOALKANES
Cycloalkanes with only one ring are named with the prefix cyclo- to the names of the alkanes (contain the same number of carbon atoms)
STEP 1

34. Examples: Monocyclic compounds
C3H6 cyclopropane
C4H8 cyclobutane
C5H cyclopentane

35. If only one substituent is present, it is not necessary to designate its position.
Examples:
Chlorocyclopropane
STEP 2 Cl
Methylcyclohexane
CH3

36. If two substituents are present, number carbon in the ring beginning with the substituent according to the alphabetical order
and
number in the direction that gives the next substituent the lowest number possible.
STEP 3

37. 1-ethyl-2-methylcyclohexane NOT 1-ethyl-6-methylcyclohexane
Examples:
CH3 3 2
CH2CH3 4 1 6 5
1-ethyl-2-methylcyclohexane
NOT
1-ethyl-6-methylcyclohexane

38. 2 Cl 3 1 4 5
1,3-dichlorocyclopentane
(NOT 1,5-dichlorocyclopentane)

39. When three or more substituents are present, begin at the carbon with substituent that leads to the lowest set of locants.
STEP 4

40. 1-chloro-3-ethyl-4-methylcyclohexane
Locants
chloro
ethyl
methyl
Example:
CH2CH3 1 4 3 2 3 2 3 2 1 4 Cl
CH3 4 1 1 4 5 6 6 5
1-chloro-3-ethyl-4-methylcyclohexane
4- chloro-2-ethyl-1-methylcyclohexane

41. 1-ethyl-1,3-dimethylcyclopentane
H3C
CH2CH3
CH3
1-ethyl-1,3-dimethylcyclopentane
(NOT 3-ethyl-1,3-dimethylcyclopentane)

42. When a single ring system is attached to a single chain with a greater number of carbon atomsor
when more than one ring system is attached to a single chain, then it is appropriate to name the compounds as cycloalkylalkane.
STEP 5
Number of C at linear chain
Number of C at ring

43. 1,3-dicyclohexylpropane2
1,3-dicyclohexylpropane

44. Examples:
CH2CH2CH2CH2CH3
1-cyclobutylpentane

45. Physical properties of alkanes & cycloalkanes
Physical state
Boiling Point
Solubility

46. Physical state
At room temperature (25oC) and atmospheric pressure (1 atm), for unbranched alkanes,
C1 – C4 : gases
C5 – C17 : liquids
C18 - more : solids

47. Mr ↑ boiling point ↑ Boiling points
The boiling points of the straight alkanes show a regular increase with increasing molecular weight.
Branching of the alkanes chain, lower the boiling point.
Mr ↑ boiling point ↑

49. Mr ↑ boiling point ↑ C-H is non polar bond
Intermolecular forces exist – London dispersion
The London dispersion forces increase :
As molecular weight increases,
Molecular size increases
Molecular surface area increase
Therefore, more energy is required to separate molecules from one another
Result - a higher boiling point.
Mr ↑ boiling point ↑

51. Chain branching :
makes a molecule more compact
Surface area reduces
The strength of the London dispersion forces reduce
Lower boiling points.

52. Have different boiling point due to branching
Isomeric Alkanes
Have different boiling point due to branching
Pentane
CH3CH2CH2CH2CH3
37oC
2-methylbutane
CH3CHCH2CH
CH3
28.5oC
2,2-dimethylpropane
CH3CCH3
9oC

53. For example, butane and 2-methylpropane both have a molecular formula C4H10, but the atoms are arranged differently
CH3CH2CH2CH3
butane
CH3CH2CH2CH3
CH3CHCH3
CH3
CH3CHCH3
CH3
2-methylpropane
In butane the C atoms are arranged in a single chain, but 2-methylpropane is a shorter chain with a branch

54. Name
Molecular structure
Boiling point
Butane has a higher boiling point because the dispersion forces are greater.
The molecules are longer (and so set up bigger temporary dipoles) and can lie closer together than the shorter, fatter 2-methylpropane molecules.

55. Cycloalkanes
The boiling points of cycloalkanes are 10oC to 15oC higher than the corresponding straight chain alkanes.
Cycloalkane
Boiling point
alkane
Cyclobutane
13oC
Butane
-0.5oC
Cyclopentane
49oC
Pentane
36..3oC

56. Solubility Alkanes – less dense than water
Alkanes and cycloalkanes are almost totally insoluble in water
They are non-polar molecule
Unable to form hydrogen bond with H2O

57. Liquid alkanes and cycloalkanes are soluble in one another, and they generally dissolve in non-polar solvents.
Good solvents for alkanes are benzene, carbon tetrachloride, chloroform, and other hydrocarbons.

58. Nature sources of alkanes;
Natural gas
oil
Natural gas contains primarily methane and ethane, with some propane and butane
Oil is a mixture of liquid alkanes and other hydrocarbons

59. CHEMICAL REACTION OF ALKANES
Non-reactivity of alkanes
Alkanes are generally inert towards many chemical reagents (bases, acids, dehydrating agents and aqueous oxidizing agents)
C and H has nearly the same electronegativity, the C-H bonds of alkanes are only slightly polarised (non polar)

60. Reaction of alkanes
Combustion
Halogenation

61. Combustion of alkanes CxHy+ O2 → x CO2 + H2O + Heat
Alkanes burnt in air ( oxygen ) to give carbon dioxide gas, water and heat.
CxHy O2 → x CO H2O
+ Heat

62. C4H8+ 3O2 → 4CO2 + 4H2O + Heat Examples:
Combustion of alkane in excess oxygen
C4H8+ 3O2 → 4CO H2O
+ Heat

63. C4H8+ 4O2 → 4CO + 4H2O + Heat Examples:
Combustion of alkane in limited oxygen
C4H8+ 4O2 → 4CO + 4H2O
+ Heat

64. Alkanes are unreactive towards polar or ionic reagents but can react with non-polar reagents such as oxygen and bromine.

65. The low reactivity of alkanes toward many reagents explain why alkanes were originally called paraffins.
(Latin : parum affinis= low affinity).

66. Halogenation: a free radical substitution reaction
Alkanes react with halogen (chlorine & bromine) to produce haloalkanes in the presence of light or temperature >> 100 oC.
R–H + X R–X + HX
With methane, the reaction produces a mixture of halomethane and a hydrogen halide.

67. Examples: i. CH4 + Cl2 CH3Cl + CH2Cl2 + CHCl3 + CCl4 + HCl
ii. CH3CH3 + Cl2
CH3CH2Cl + HCl

68. CH CH3
| |
iii. CH3 CCH3 + Cl2
CH3 CCH2Cl + HCl
| |
CH CH3
(ii) & (iii): 1 product only because all the
hydrogen atoms are identical

69. iv. CH3CH2CH3 + Br2
CH3CH2CH2Br
( minor)
+ CH3CHCH3 | Br
( major)
+ HBr

70. Reaction mechanism Cl – Cl 2Cl• CH4 + Cl2 CH3Cl + HCl
Chain initiation step
Cl – Cl Cl•

71. Chain propagation steps
3HC• + Cl Cl_  CH3Cl + Cl• H
H-C H + •Cl  •CH3 + HCl

72. Chain termination step.
•CH •CH3  CH3CH3
•Cl + •CH3  CH3Cl
•Cl + •Cl  Cl2

73. Example The bromination of 2-methylbutane yields a mixture of isomers.
Increasing % yield

74. Exercise
Chlorination reaction of certain alkanes can be used for laboratory preparations, for example in the preparation of chlorocyclopentane from cyclopentane. Give the mechanism for the reaction.

75. ALKENES
12.2

76. General formula CnH2n , n  2.
Functional group  double bond C=C
Unsaturated hydrocarbon
Each carbon atom ( C=C ) is sp2 hybridized.
Restricted rotation of carbon-carbon double bond causes cis-trans isomerism

77. IUPAC Nomenclature Step 1
Determine the parent name by selecting the longest chain that contains the double bond and change the ending ‘-ane’ in alkane to ‘-ene’.

78. Step 2
When the chain contains more than three carbon atoms, a number is need to indicate the location of the double bond.
The chain is numbered starting from the end closest to the double bond..

79. 3-octene (not 5-octene)

80. (not 3-methyl-2-butene)
Step 3
Indicate the position of the substituent by the number of the carbon atoms to which they are attached.
2-methyl-2-butene
(not 3-methyl-2-butene)

81. (not 2,5-dimethyl-4-hexene)

82. Step 4
The ending of the alkenes with more than one double bond should be change from - ene to
diene – if there are two double bonds
triene – if there are three double bonds

83. 1,3-butadiene
1,3,5-heptatriene
MORE

84. Step 5
In cycloalkenes :
Number the carbon atoms with a double bond as 1 and 2, in the direction that gives the substituent encountered first with a small number.

85. (not 2-methylcyclopentene)

86. 3,5-dimethylcyclohexene (not 4,6-dimethylcyclohexene)1 6 2 5 3 4
3,5-dimethylcyclohexene
(not 4,6-dimethylcyclohexene)
MORE

87. Step 6
Two frequently encountered alkenyl groups are vinyl group and allyl group.
RCH=CHR RCH=CH-
alkene alkenyl
CH2=CH CH2=CHCH2-
vinyl group allyl group -H

88. Step 7 When two identical groups are attached:
a) on the same side of the double bond, the compound is cis
b) on the opposite sides of double bond, it is trans.
cis / trans

89. cis-1,2-dichloroethene
trans-1,2-dichloroethene

90. Give IUPAC names for the following alkenes1)
CH3 2)
CH3

91. 3) 4)
CH2CH3
CH3
CH3CH=CHCH2C(CH3)2CH3 5)

92. Preparation of alkenes
a) Dehydration of alcohols
b) Dehydrohalogenation of alkyl halide

93. Dehydration of alcohols
Alcohols react with strong acids in the presence of heat to form alkenes and water.

94. Concentrated sulphuric acid (H2SO4) or phosphoric acid (H3PO4) : as acidic catalysts and dehydrating agents.
The major product is the most stable alkene, which has greater number of alkyl group attach to C=C is follow Saytzeff’s rule.

95. Saytzeff’s Rule
An elimination usually gives the most stable alkene product, commonly the most highly substituted alkene.

96. + + Examples CH OH H SO (conc.) D C O (1) (2) C H CH OH SO (conc.) D O3 2 OH H SO 4
(conc.) D C O
(1) C H 3 CH OH 2 SO 4
(conc.) D + O
(2)

97. Mechanism for the dehydration of alcohol2 SO 4
(conc.) D CH 3
CH=CHCH
(major product) 3 2 3 OH + CH 2
=CHCH 3
(minor product) + H 2 O

98. Step 1: Protonation of alcohol.3 2 OH : .. H + + : O H H C H 3 2 OH : + O ..

99. Step 2: Formation of carbocation3 2 H C + H H C C C H C H 3 2 3 + : OH H + O H : ..

100. Step 3: Formation of alkenesH : .. C 3 H + C O H + b a
Root a : major product
Root b : minor product

101. C H 3 H O : + +
stable alkenes + CH 2
=CHCH 3
(minor product)

102. Rearrangement During Dehydration of Alcohol
Example 1: C H 3 CH OH 2 SO 4
(conc.) D

103. (major product) C H 3 +
(minor product) C H 3 2

104. Step 1 and Step 2 are similar to the previous example.

105. Step 3:
Now the rearrangement occurs.
The less stable, secondary carbocation rearrange to form more stable tertiary carbocation.

106. 2o carbocation 3o carbocation (less stable) (more stable)H 3 +
rearrangement
2o carbocation o carbocation
(less stable) (more stable)

107. The rearrangement occurs through the migration of an alkyl group (methyl) from the carbon atom adjacent to the one with the positive charge.
Because a group migrates from the one carbon to the next, this kind of rearrangement is often called a 1,2 shift.

108. + C H 3 2 (minor product) less stable alkene (a) C H 3 (major product): O H 3 .. H C H H 3 C H 3
(major product)
more stable alkene
(b)
(b)

109. The final step can occur in two ways:
Path (a): leads to less stable, disubstituted alkene and produces the minor product of the reaction.
Path (b): leads to highly stable tetrasubstituted alkene and produces the major product according to the Saytzeff’s rule.

110. Try this! C H H C H 3 3 H SO (conc) 2 4 H C C C C H H C C C H C H 3 3D 3 3 H OH
(major product) C H 3 + H C C C H C H 2 2 3
(minor product)

111. + (2) Dehydrohalogenation of Alkyl Halides
The elimination of a hydrogen and a halogen from an alkyl halide to form an alkene.
Saytzeff’s rule is used to determine the major product C H X +
KOH
alcohol
reflux HX

112. + Examples: C H Cl KOH alcohol reflux (major product) (minor product)
(1) C H 2 3 Cl +
KOH
alcohol
reflux
(major product)
(minor product)

113. (2)

114. (3) Br alcohol C H C H + KOH reflux C H C H 3 3 (minor product) C H +2 C H 3
(major product)

115. (4) CH3 H CH3 C C CH3 + KOH alcohol CH3 Cl reflux CH3 H CH3
CH C C CH CH3 C C CH3
CH CH3
2,3-dimethyl-1-butene ,3-dimethyl-2-butene
(minor product) ( major product)
2 alkyl groups alkyl groups

116. Chemical Reaction of Alkenes
Comparison of The Reactivity Between
Alkanes and Alkenes
Alkenes are more reactive compared to alkanes.
Alkanes have carbon-carbon single bonds (σ bonds) while alkenes have carbon-carbon double bonds (π bonds).
The double bond is a site of high electron density (nucleophilic).
Therefore most alkenes reactions are electrophilic additions.

117. Electrophilic Additions Mechanism+ H X
slow + C C X Y H H - + Y

118. H C H X Y + -
fast + H C C H Y X H

119. + + Example: H C C H slow C C H Cl H C C H C H C H H C C C C H Cl H i.3 3
slow + C C H Cl H C C H 3 3 C H C H 3 3 + + - H C C C C H Cl 3 3 H

120. + - + C H C H fast H C C C C H Cl H C H C H H C C C C H H Cl 3 3 3 3 3

121. Addition Reaction of Alkenes
(1) Hydrogenation
The reaction of an alkene with hydrogen in the presence of catalyst such as platinum, nickel and palladium to form alkane.

122. Examples:
(1) + H 2
Pt /Ni / Pd C 3 + H 2
Pt /Ni / Pd
(2)

123. + (2) Halogenation of Alkenes C X CH2Cl2 (i) In inert solvent (CH2Cl2)
Alkenes react rapidly with chlorine or bromine in CH2Cl2 at room temperature to form vicinal dihalides. C + X 2
CH2Cl2

124. Example: C H 3 + Cl 2
CH2Cl2 C H 3 Cl

125. When bromine is used for this reaction, it can serve as a test for the presence of carbon-carbon double bonds.
If bromine is added to alkene, the reddish brown color of the bromine disappears almost instantly as long as the alkene is present in excess.

126. + Unsaturation test CH2Cl2 Br C room temperature C Br
Observation: The reddish brown bromine decolourised

127. + C H X HX (3) Hydrohalogenation : Markovnikov’s rule
Hydrogen halides (HI, HBr, HCl and HF) add to the double bond of alkenes to form haloalkanes.
The addition of HX to an unsymmetrical alkenes, follows Markovnikov’s rule. + C H X HX

128. Markovnikov’s Rule:
In the addition of HX to an alkenes, the hydrogen atom adds to the carbon atom of the double bond that already has the greater number of hydrogen atoms.

129. The addition of HBr to propene, could conceivably lead to either 1-bromopropane or 2-bromopropane.
The main product, however, is 2-bromopropane C H 2 3 +
HBr Br

130. When 2-methylpropane reacts with HBr, the main product is tert-butyl bromide, not isobutyl bromide.
The addition of HX to an unsymmetrical alkenes, yield the main product according to the Markovnikov’s rule. C H 3 2
HBr + Br

131. Mechanism CH 3 C + H Br
slow
+ Br -
30 Carbocation

132. fast
+ Br -

133. (4) Addition of HBr to Alkenes in The
Presence of Peroxides
(Anti Markovnikov’s rule)
When alkenes are treated with HBr in the presence of peroxides, ROOR (eg: H2O2) the addition occurs in an anti-Markovnikov manner
The hydrogen atom of HBr attached to the doubly bonded carbon with fewer hydrogen atoms.

134. Example: C H 2 3
HBr
ROOR + C H 2 3 Br

135. Complete the following reactions:
(1)
(2)

136. + C X H O OH = Cl or Br halohydrin
(ii) In aqueous (Halohydrin Formation)
If the halogenation of an alkene is carried out in aqueous solution, the major product of the overall reaction is a haloalcohol called a halohydrin. C + X 2 H O OH
= Cl or Br
halohydrin

137. Example: C H + Br 2 O OH
2-bromobutanol
Unsaturated test
Observation : the reddish brown bromine decolourised
If the alkenes is unsymmetrical, the halogen ends up on the carbon atom with the greater number of hydrogen atoms.

138. Example: C H 3 2 + Br O
1-bromo-2-methyl-2-propane C H 3 OH Br

139. Complete the following reaction:

140. (5) Hydration of Alkene
The acid-catalyzed addition of water to the double bond of an alkene (hydration of an alkene) is a method for the preparation of low molecular weight alcohols.
The acid most commonly used to catalyse the hydration of alkenes are dilute solution of sulphuric acid and phosphoric acid.

141. The addition of water to the double bond follows Markovnikov’s rule.C + H 2 O 3 OH

142. Example: C H 3 2 O + C H 3 OH

143. Complete the following reactions:
(1)
(2)

144. Mechanism
STEP 1 CH 3 C
+ H O H + ..

145. STEP 2 CH 3 C H +
+ H 2O .. .. ..

146. STEP 3 .. ..
H2O ..

147. (6) Addition of Sulphuric Acid to Alkenes
Alkenes dissolve in concentrated sulphuric acid to form alkyl hydrogen sulphates.
Alkyl hydrogen sulphates can be easily hydrolysed to alcohols by heating them with water.
The overall result of the addition of sulphuric acid to alkenes followed by hydrolysis is the Markovnikov addition of -H and -OH.

148. alkyl hydrogen sulphate H C C H conc. H SO C C3
OSO
alkyl hydrogen sulphate H C C H 3 3
conc. H 2 SO 4 C C H C C H 3 3 C H 3 OH H 2 O
heat

149. Example: C H 2 3
conc.H SO 4
OSO H 2 O
heat C 3 OH

150. 7 ) Oxidation of Alkenes
Alkenes undergo a number of reactions in which the C=C is oxidized
KMnO4
ozonolysis
basic,cold,dilute
acidic,hot,concentrated

151. With cold and dilute potassium permanganate, KMnO4
Potassium permanganate in alkaline solution can be used to oxidise alkenes to 1,2-diols (glycols).

152. + + C KMnO (purple) OH ,cold C OH MnO (brown precipitate) - 4 2
Observation: Purple colour of KMnO4 decolourised and brown precipitate formed.

153. This reaction is called Baeyer’s test.
It is a test for the presence of C=C where the purple colour of the KMnO4 decolourised, and brown precipitate of MnO2 is formed.

154. Example: C H 2 +
KMnO 4
cold OH - ,H O C H OH +
MnO 2

155. ii) With hot potassium permanganate solutions to alkenes
When oxidation of the alkene is carried out in acidic solution of KMnO4, cleavage of the double bond occurs and carbonyl-containing products are obtained.

156. If the double bond is tetrasubstituted, the two carbonyl-containing products are ketones3 +
KMnO 4
(ii) H O
(i) OH -
,heat C O H 3 +

157. If a hydrogen is present at double bond, one of the carbonyl-containing products is a carboxylic acid;
If two hydrogens are present on one carbon, CO2 is formed.

158. + + + C H KMnO (i) OH ,heat (ii) H O O CO H O C HO C H 2) 2 3 4 - 2 2

159. The oxidative cleavage of alkenes can be used to establish the location of the double bond in an unknown alkene.

160. Example:
An unknown alkene with the formula C7H14 undergoes oxidation with hot basic potasium permanganate solution to form propanoic acid and butanoic acid. What is the structure of this alkene?

161. + + C H KMnO (i) OH-,heat (ii) H O O O H C C C H C C H OH H C OH C H 714 +
KMnO 4
(i) OH-,heat
(ii) H 3 O O O + H C C C H C 3 2 C H OH H C OH C H 2 3 2

162. Answer: C O OH H 2 3
propanoic acid
butanoic acid C H 2 3
-heptene

163. Example
An unknown alkene undergoes oxidation in hot basic KMnO4 followed by acidific to give the following product:
CH3CCH2CH2CH2CH2C OH
Deduce the structural formula for the unknown alkene. O O

164. Ozonolysis of Alkenes
A more widely used method for locating the double bond of an alkene is the use of ozone (O3).
Ozone reacts vigorously with alkenes to form unstable compounds called molozonides, which rearrange spontaneously to form compounds called ozonides.

165. C + O 3
ozonide

166. Ozonides:
very unstable compounds
can easily explode violently
they are not usually isolated but are reduced directly by treatment with water and in the presence of zinc and acid (normally acetic acid) to give carbonyl compounds (either aldehydes or ketones).

167. C O + Zn H 2
O,H R
ozonide
C = O +
O = C R H

168. Example: C H 3
(i) O
(ii) Zn,H 2
O/H + C O H 3 +

169. Exercise
Write the structure of alkene that would produce the following products when treated with ozone followed by water, zinc and acid
CH3COCH3 and CH3CH(CH3)CHO

170. Example
Deduce the structural formula of an alkene that gives the following compound when it reacts with ozone in the presence of Zn / H+.
O=CH-CH2-CH2-CH(CH3)CH=O

171. 2. Acid-catalyzed dehydration of neopentyl alcohol, (CH3)3CCH2OH, yields 2-methyl-2-butene as the major product. Outline a mechanism showing all steps in its formation.

172. Exercise
Compounds A, B, C and D are isomers with the molecular formula C4H8. A and B give a positive Baeyer test, while C and D do not. A exists as cis- and trans- isomers, while B does not have geometrical isomers. C has only secondary hydrogen, while D has primary, secondary and tertiary hydrogen. Give the IUPAC names for A, B, C, and D.