CHEMISTRY FORM 6

ORGANIC CHEMISTRY

CHAPTER 2 : HYDROCARBON

2.1 Nomenclature of ALKANE

� Alkane is a saturated hydrocarbon as it contain only single bond in its

molecule

� General formula for homologous series of alkane is CnH2n+2

� Table below shows the naming of straight chain of alkane

Hydrocarbon which contain only

carbon-carbon single bond, C–C

Hydrocarbon which contain at least one

carbon-carbon double bond, C=C or triple bond, C≡C

NameMolecular

formulaMolecular structure Name

Molecular

formulaMolecular structure

Methane CH4 Ethane C2H6

Propane C3H8 Butane C4H10

Pentane C5H12

Hexane C6H14

2.2 Naming alkane according IUPAC

Step 1 Step 2 Step 3 Step 4

Find the longest

chain of carbon and

name accordingly. (it

does not has to be a

straight chain).

Identify the

‘branched’ carbon

(alkyl group) that

attached to the

‘main’ chain. Then,

name the alkyl

accordingly

CH3– methyl

CH3CH2– ethyl

CH3CH2CH2-propyl

Place a prefix upon

the similar alkyl

group (if any). If

there is 2 similar

alkyl, prefix di is

placed, if 3 similar

alkyl, prefix tri is

placed.

State the position

where the ‘branch’ is

located at which

carbon based on the

numbering gave

earlier.

CH3(CH2)5CH3 CH(CH3)2C(CH3)3 CH3CH(CH2CH3)2 C(CH3)3CH2C(CH3)3

2,3-dimethylpentane 3-ethyl-3-

methylhexane

2,2,3-trimethylpentane 3,3-diethylhexane

3-methylpentane 3,5-dimethylheptane 3-ethyl-4-methylhexane3-ethyl-3,5-

dimethyloctane

n-heptane 2,2,3-trimethylbutane 3-methylpentane 2,2,4,4-

tetramethylpentane

� Step 3 Complete the structure by placing one hydrogen (H)

atom at each of single bonds.

Isomer of hexane, C6H14

2.4 Physical properties of alkane

Alkane CH4 C2H6 C3H8 C4H10 C5H12 C6H14 C7H16 C8H18

Boiling

point oC– 162 – 8.6 – 42.2 – 0.5 36.3 68.7 98.4 126

Boiling point

trend

Density

(g/cm3)-- -- 0.50 0.58 0.63 0.66 0.68 0.70

Density trend

SolubilityNot soluble in AAAAAAA..

Soluble in AAAAAAAAAAAA

BOILING POINT INCREASE DOWN HOMOLOGOUS SERIES

DENSITY INCREASE DOWN HOMOLOGOUS SERIES

water

organic solvent

A) Boiling point of alkenes

� The boiling point AAAA when going down to homologous series of

alkane.

� All alkane possessed the same intermolecular forces : weak

AAAAAAAAforces

� Greater the AAAAAAAA, stronger the AAAAAAAAA forces,

AAAA the boiling point

� Boiling point of isomers of the same molecular formula varies with the

branched molecules

� Straight chain has ..AA.. boiling point compared to branched chain as

straight chain molecule has higher AAAAAAAAAA compared to a

branched chain. The positioning of alkyl and number of alkyl also effect

the boiling point of alkane. 2-methylpentane as a higher boiling point than

3-methylpentane as it has a greater exposure of intermolecular forced

increase

Van Der Waals

molecular mass weak Van Der Waals’increase

higher

total surface area

B) Solubility of alkane

� All alkanes are often consider as AAAAAAAAA

molecule as the dipole of moment created in molecule is very

small.

� Since alkane is AAAAAAA. Molecule, it dissolve easily in

non-polar solvent such as benzene, and ether.

� Alkane does not form AAAAA bond in water, so it is

AAAAA in water. Thus, alkane is also described as

AAAAAA. (water–hating).

� The longer the alkane chain, the more insoluble it is in water.

non-polar

non-polar

hydrogen

insoluble

hydrophobic

2.5 Chemical Properties of Alkane

2.5.1 Preparation of Alkane

� Alkane can be prepared using the following methods :

� Decarboxylation of sodium salt of a carboxylic acid

R–COOH + NaOH → R–H + Na2CO3

Example :

� Kolbe’s method : electrolysing concentrated sodium ethanoate

� Cathode : 2 H2O + 2 e- � H2 + 2 OH-

� Anode : 2 CH3COO- � C2H6 + 2 CO2 + 2 e-

� Wurtz reaction : reaction of sodium on alkyl halide in ether.

2 R–X + 2 Na � R – R + 2 NaX

Example

CH3COOH + 2 NaOH � CH4 + H2O + Na2CO3

2 CH3CH2–Cl + 2 Na � CH3CH2CH2CH3 + 2 NaCl

2.5.2 Reaction of Alkane

� Since alkane is a AAAAA. hydrocarbon, so alkane is inert to most of

the chemical reaction

� Table below shows the description of reaction of ethane with other

substances.

� From the series of reaction above it can be conclude that

� Ethane does no react with polar or ionic substances

� Ethane react with non-polar substances such as Cl2 , Br2 and O2 and

energies are required for reaction to occur.

Reagents Effect on ethane

Sodium hydroxide aqueous No effect on hot or cold condition

Concentrated hydrochloric acid No effect on hot or cold condition

Acidified potassium manganate (VII) No effect on hot or cold condition

Air (oxygen) No effect under room condition. Burns when heated

Bromine water No effect on dark. Decolourised slowly under sunlight

Chlorine gas No effect on dark. Reaction occur under sunlight

saturated

1. Combustion of alkanes

� All hydrocarbon react with oxygen to form carbon dioxide and water.

� The equation for a complete combustion for all hydrocarbons can be

represented by the equation

C2H6

C5H12

C8H18

� Note that the reaction is exothermic for all hydrocarbons. Equation above

is also known for ∆Hc. Higher the number of carbon, the more exothermic

the reaction.

� Under limited supply of air (oxygen), sometimes, carbon monoxide (CO) is

produced instead of CO2.

mol/kJmHOH2

yCOxO

4

yxHC 222yX −=∆+→

++

C2H6 + 7/2 O2 � 2 CO2 + 3 H2O

C5H12 + 8 O2 � 5 CO2 + 6 H2O

C8H18 + 25/2 O2 � 8 CO2 + 9 H2O

2. Halogenation of alkanes

� When alkane is run together with chlorine gas under the presence of

ultraviolet ray (which comes naturally from sunlight)

Example : CH4 (g) + Cl2 (g) � CH3Cl (g) + HCl (g)

C2H6 (g) + Cl2 (g)

� The mechanism for the reaction of chlorination of alkane can be explained

using the following steps

Step 1 : Initiation Step 2 : Propagation Step 3 : Termination

Cl – Cl � 2 Cl• ∆H = +242 kJ/mol

H3C–H � CH3• + H•

∆H = + 433 kJ/mol

Since ...................... required lower

energy to form radical, so the initiation

will start off with AAAAA.. Gas

Since chlorine radical are highly

reactive, when it collide with methane

molecule forming HCl

and methyl radical

H3C–H + •Cl � H3C• + HCl

Methyl radical will propagate with other

chlorine molecule and

forming back chlorine radical

H3C• + Cl–Cl � H3CCl + •Cl

Under such propagation reaction

thousands of methane and chlorine

molecules will react continuously

When 2 free radicals collide with each

other and combined, the reaction stops.

This reaction is highly exothermic,

where

H3C• + •Cl � H3C–Cl

∆H = -349 kJ/mol

H3C• + •CH3 � H3C–CH3

(H = -368 kJ/mol

Usually, termination will occur when

[radical] > [molecule], which is after

thousands of propagation.

The presence of small amount of

ethane may also present due to the

collision between 2 methyl radicals

chlorine

chlorine

2.1.1 Sources of hydrocarbon� The main sources of hydrocarbons are :

a) crude oil b) coal

c) natural gas

� Since all these main sources are made up from dead animals and plants,

so they are also known as AAAAAAAA

� Coal is complex mixture consisting mainly hydrocarbons, which is mainly

made up from dead plaints in swamp.

� Petroleum is a mixture of hydrocarbons (alkanes, alkenes, alkyne), while

natural gas contain mainly A.AAA and some AAAAA

� The mixture in petroleum can be separated by using

AAAAAAAAAA. in oil refinery. Diagram below shows the chamber

and oil refinery used to separate the mixture of petroleum.

fossil fuel

methane ethane

fractional distillation

Fractional

distillationProducts Uses

Petrol gas Use for house cooking gas

Gasoline Use as fuel for automobile vehicle

Naphtha Use to synthesis different petrochemical

Diesel oil Use as fuel of heavy vehicle such as bus

or lorry

Kerosene Use as fuel for jet engine and oil stove

Lubricant Oil Use for lubrication, making wax and polish

Fuel Oil Fuel for ship and power station

Bitumen (asphalt) Use as tar for paving road surface and

coating underground water pipe

� The separation does not end with fractional distillation. They are then

treated with various ways to improve the quality and quantity of useful

hydrocarbon. One of the major treatments gives after fractional distillation

is cracking process.

Cracking of hydrocarbon

Thermal cracking (Pyrolysis) Catalytic cracking

� Using high temperature, bond breaking

(homolytic fission) take place and form

various products of unbranched alkane

and alkene

� Example, when breaking decane, C10H22

C10H22 → C3H6 + C7H16

C10H22 → C4H8 + C6H14

� With the aid of zeolite as catalyst, carbon

cracking can occur at lower temperature

compare to thermal cracking.

� Products using catalytic cracking usually

contain branched alkane and alkene.

C10H22 →

2.7 Cycloalkane (alicyclic compound)

� Cycloalkane has a general formula of CnH2n

� Some examples of cycloalkane

Cycloalkane Molecular formula Displayed formula Skeletal formula

Cyclopropane

Cyclobutane

Cyclopentane

Cyclohexane

C3H6

C4H8

C5H10

C6H12

2.7.1 Naming cycloalkane

� The way of naming cyclolalkane is more or less the same with naming

alkane. If theirs is one alkyl attached to the cycle, it will be automatically

become ‘1’ by itself. E.g. methylcyclobutane

(not “1-methylcyclobutane)

� If there’s more than one “group” attaching the cycle, only then numbering

will be given to the particular number of C that it is attached.

methylcyclopropane

3-ethyl-1-methylcyclopentane

1,2,4-trimethylcyclohexane

1,2,3-trimethylcyclooctane

3-ethyl-2-methyl-1-

propylcyclobutane

2.7.2 Preparation and Reaction of Cycloalkane

� Cycloalkane can be prepared by catalytic hydrogenation of benzene at

200oC

� Reaction of cycloalkane is similar to alkane. When react with chlorine /

bromine gas under sunlight, substitution reaction take place

Mechanism :

� Initiation

� Propagation

� Termination

2.8 Alkene – Nomenclature of alkenes and cycloalkenes

� The homologous series of alkenes has general formula of CnH2n.

� The significance of alkene is all of them have C=C in their molecules with

its name end with –ene

NameMolecular

formulaMolecular structure Name

Molecular

formulaMolecular structure

Ethene C2H4 Propene C3H6

Butene C4H8

Pentene C5H10

Hexene C6H12

But-2-ene But-1-ene

pent-2-ene pent-1-ene

Hex-1-ene Hex-2-ene Hex-3-ene

� In naming alkene, the following steps are given

Step 1 : Find the longest C – C chain which contain double bond in it

(parent chain) and name them

Step 2 : Find and name the alkyls attached to the parent chain.

Step 3 : If there are more than 2 of the same type alkyls, prefix are put

accordingly.

Step 4 : Put the number of the alkyl that attached to the particular

carbon atom.

Example : Name the following alkenes accordingly

2-methylbut-2-ene 2-ethyl-3-methylpent-1-ene 3,4-dimethylhex-3-ene

2-methylpropene 2,3-dimethylpent-2-ene 3,5-dimethylhept-3-ene

2.8.1 Naming alkene with more than one single bond & cycloalkene

� A “diene” (alkene with 2 C=C bond) and cycloalkene has general formula

of CnH2n–2.

� In diene, the position of both C=C in parent chain has to be stated in alkan-

x,y-diene, whereas in cycloalkene, C=C is always place as C1=C2. So the

numbering is fixed for naming.

� Example, name the following diene / cycloalkene below

2-methylbut-1,3-diene2,5-dimethylhex-1,3-diene oct-2,5-diene

3-methylcyclopropene3-ethyl-2-methylcyclohexene

3,4,5-trimethylcyclopentene

2.9 Isomerism in alkene.

� Alkenes which contain at least 4 Carbon atoms may exhibit 2 isomerism,

structural and stereoisomerism.

� For example, butane (C4H8) contain 5 isomers.

� Isomers of pentene

2.10 Physical Properties of Alkene

A) Boiling Point of Alkene

� The boiling point AAAA when going down to homologous series of

alkane.

� All alkane possessed the same intermolecular forces : weak

AAAAAAAforces

� Greater the AAAAAA.., stronger the AAAAAAAAA forces,

AAAA the boiling point

Alkene C2H4 C3H6 C4H8 C5H10 C6H12 C7H14 C8H16 C9H18

Boiling

point oC– 164 – 12.0 – 5.8 – 0.5 38.0 72.07 96.5 117

Boiling

point trend

Solubility in

water

Boiling point increase

Insoluble in water (solubulity decrease)

increase

Van Der Waals

molecular mass weak Van Der Waals

higher

2.11 Preparation of Alkene

� Alkene can be prepared in a few ways

Name of

reaction

Reagent used

and conditionEquation

Dehydro-

halogenation

from

haloalkane

Ethanolic

sodium

hydroxide (heat

& reflux)

Dehydration

(removal of

water)

from

alcohol

Excess conc.

H2SO4

at 1800C

or

Alumina (Al2O3)

at 350oC

2.12 Chemical reaction of alkene

Name of

reaction

Reagent used and

conditionEquation

Hydrogenation

Hydrogen gas under

--------------

Nickel (Ni) at 180oC

@

Platinum (Pt) at

room temperature

CH3CH=CH2 + H2 (g) CH3CH2CH3 (g)

propene propane

cyclohexene cyclohexane

HalogenationHalogen gas, X2

(X2 = Cl2 ; Br2 ; I2)

Addition of

Hydrogen

halide

Hydrogen halide

( H – X )

(X = Cl ; Br ; I)

Ni

Name of

reaction

Reagent used and

conditionEquation

Hydration

Steam (H2O)

---------

Phosphoric acid,

(H3PO4 )

At 300oC ; 60 atm

Hydroxylation

(cold, diluted

acidified

KMnO4)

KMnO4 (aq) / H+

(cold and diluted)

Oxidation

(under hot,

concentrated

acidified

potassium

manganate

(VII)

KMnO4 (aq) / H+

(hot &

concentrated)

2.12 Chemical reaction

(1) Hydrogenation of alkene

� Carry out under mixture of alkene and hydrogen over a finely divided

transition metal as a catalyst.

� 2 catalysts can be used in hydrogenation

i) Platinum : ~ can react even under room condition. Longer alkene required

some heat

ii) Nickel : ~ required high temperature to allow hydrogenation to occur

(180oC)

� Hydrogenation is an exothermic reaction and its ∆H is about –120 kJ / mol

� CH3CH=CH2 (g) + H2 (g) � CH3CH2CH3 ∆H = –124 kJ / mol

� Catalytic hydrogenation is important in food industries especially in

hardening unsaturated fats and oil to make margarine. Unsaturated

hydrocarbon makes them too soft for commercial use.

� CH3(CH2)7CH=CH(CH2)7COOH + H2 (g) � CH3(CH2)16COOH

� In industries, a special “Raney Catalyst” is used to replace platinum as it is

EXPENSIVE!!!

(2) Halogenation of alkene

� Chlorine and bromine react readily with alkene and form dichloroalkane

and dibromoalkane respectively. Cl2 and Br2 gas are add across double

bond.

� CH3CH=CH2 (g) + Cl2 (g) � CH3CH(Cl)CH2Cl

� The mechanism of halogenation can be explained by a few steps describe

below :

� Step 1 : Formation of carbocation – propene has region of high electron

density because of the π electron. When Cl2 approaches, molecule is

strongly polarised by region and consequently formed an induce dipole.

The positive charge end of Cl2 molecule act as electrophile and bond to

C=C via electroplilic addition and caused Clδ+–Clδ− repelled. As a result,

carbocation & chloride ion are formed.

� Step 2 : Nucleophilic attack to form addition product – carbocation

formed is very unstable. It quickly combines with Cl− ion to produce by

heterolytic fission of Cl2 molecule to give 1,2-dichloropropane.

� However, if bromine water is used instead of bromine gas, the results of

products are not as same as in bromine gas. When bromine water is

reacted with propene

(3) Addition of hydrogen halide

� Unlike addition of halogen, addition of hydrogen halide produced 2

products. For example, when propene react with hydrogen bromide (H–Br)

CH3CH=CH2 + H–Br � CH3CH2CH2Br + CH3CH(Br)CH3

Propene 1-bromopropane 2-bromopropane

(minor) (major)

� The major / minor product of the reaction can be predicted using

Markovnikoff’s Rule where it stated when an unsymmetrically substituted

alkene reacts with a hydrogen halide, the hydrogen adds to the carbon

that has the greater number of hydrogen substituents, and the halogen

adds to the carbon having fewer hydrogen substituents.

� Step 1 : Electrophilic attack – when the polar hydrogen bromide

approaches propene, the positively charged hydrogen ion is polarising

C=C, and caused Br− to form

� Step 2 : Nucleophilic attack – the negative bromide ion react fast with the

unstable carbocation.

δ+ δ–

� Relative stability of carbocation can be explained using Markovnikoff’s

Rule. According to the rule, a tertiary (30) carbocation is more stable than

a secondary (20) carbocation than a primary (10) carbocation. this is due to

the inductive effect of the electron-donating alkyl group.

� In the example above, there are 2 methyl group donating electron to

positive charged carbon electron at 20 carbocation whereas there are 1

ethyl group in 10 carbocation donating electron to the positively charged

electron.

� As a result, 20 carbocation are more stable as the 2 alkyl group tend to

decrease the charge density of C, making the cation more stable.

stability of carbocation increase.

(4) Hydration (addition of water) in alkene

� Using phosphoric acid as acidic medium, hydration of alkene can be

represent by equation :

CH3C(CH3)=CH2 + H–OH CH3CH(CH3)CH2OH + CH3C(CH3)(OH)CH3

(minor) (major)

2-methylpropene 2-methylpropan-1-ol 2-methylpropan-2-ol

� Similar to hydrogen halide, hydration of alkene follows Markovnikoff’s Rule.

� The mechanism of hydration of alkene is slightly different from addition of

hydrogen halide

Step 1 : Protonation of the carbon–carbon double bond in the

direction that leads to the more stable carbocation

Step 2 : Water acts as a nucleophile to capture carbocation

Step 3 : Deprotonation of tert-butyloxonium ion. Water acts as a Brønsted –Lowry base:

� Other than using diluted acid medium, sometimes, hydration of alcohol is prepared by adding concentrated sulphuric acid to alkene.

� When H2SO4 (conc) is added to alkene under room condition, it give an alkyl hydrogensulphate

� Hydrolysis of alkyl hydrogensulphate will convert into alcohol

(5) Oxidation of alkene using acidified potassium manganate (VII)

� Alkene are readily oxidised by acidified KMnO4 (decolourised the purple

colour of KMnO4) and give different products under different condition

� If cold diluted acidified KMnO4 is used, a diol is given as a product.

� If hot concentrated acidified KMnO4 is used, a ketone or an aldehyde

is formed which will further oxidised to become a carboxylic acid or into

carbon dioxide and water depend on alkene.

a) Hydroxylation of alkene (react under cold dilute acidified KMnO4)

� The product of this reaction is a diol (di-alcohol) – which contain 2

hydroxyl group.

� This reaction is often used to distinguish between saturated hydrocarbon

and unsaturated hydrocarbon (alkane and alkene)

b) Oxidation of alkene using hot, concentrated acidified potassium

manganate (VII)

� When alkene react with hot concentrated acidified potassium manganate

(VII), it will oxidise immediately to form aldehyde or ketone, depend on the

type of alkene

� Using this method, the position of C=C in alkene can be deduced. If the

alkene is a 10 alkene, it will turn lime water chalky when the particular

alkene is reacted with hot concentrated acidified potassium manganate

(VII)

Alkene Products

methanal

Methanoic acid

a. CH3CH2CH=CHCH3 + H2 (g)

b. CH3CH2CH=CH2 + Cl2 (g)

c. CH3CH=C(CH3)CH3 + Br2 (l)

d. CH3CH(CH3)CH=CH2 + HCl (g)

CH3CH2CH2CH2CH3

CH3CH2CHClCH2Cl

CH3CHBrC(CH3)(OH)CH3 major

CH3CHBrCBr(CH3)CH3 minor

CH3CH(CH3)CHClCH3 major

CH3CH(CH3)CH2CH2Cl minor

4-ethyl-2,2,4-trimethylhexane

2,2,4,5-tetramethylhexane

5-ethyl-3,4-dimethyloctane

2,3,4,6,6-pentamethyl-3-heptene

7-ethyl-1,3-dimethylcyloheptene

C(CH3)2=C(CH2CH3)CH(CH3)CH(CH3)2

CH2=CHC(CH3)(CH2CH3)C(CH3)=CH2

� Isomers of pentene

Practice : Write the chemical equation for the following reaction

1. Butane react with chlorine under the presence of sunlight

CH3CH2CH2CH3 + Cl2 → CH3CH2CH2CH2Cl + HCl

2. Pentane burned with excess air

C5H12 + 8 O2 → 5 CO2 + 6 H2O

3. Octane burned with excess air

C8H18 + 25 / 2 O2 → 8 CO2 + 9 H2O

4. Propene reacts with hydrogen gas using platinum as catalyst

CH3CH=CH2 + H2 → CH3CH2CH3

5. 1-hexene burned with excess air

C6H12 + 9 O2 → 6 CO2 + 6 H2O

6. 2-heptene reacts with bromine water

CH3CH2CH2CH2CH2CH=CH2 + Br2 + H2O →

CH3CH2CH2CH2CH2CH(OH)CH2Br + CH3CH2CH2CH2CH2CHBrCH2Br

7. Propene reacts with hydrogen chloride

CH3CH=CH2 + HCl → CH3CH2CH2Cl (min) + CH3CHClCH3 (maj)

8. 1-Butene react with excess oxygen

C4H8 + 6 O2 → 4 CO2 + 4 H2O

9. 2-Pentene reacts with steam catalysed by sulphuric acid

CH3CH=CHCH2CH3 + H2O →CH3CH(OH)CH2CH2CH3 CH3CH2CH(OH)CH2CH3

10. 3-Hexene reacts with cold dilute acidified KMnO4

CH3CH2CH=CHCH2CH3 + KMnO4/H+ → CH3CH2CH(OH)CH(OH)CH2CH3

11. 2-methylhex-2-ene reacts with cold dilute acidified KMnO4

CH3C(CH3)=CHCH2CH2CH3 + KMnO4/H+ →

CH3C(CH3)(OH)CH(OH)CH2CH2CH3

12. Propane react with fluorine under the presence of sunlight

CH3CH2CH3 + F2 → CH3CH2CH2F + HF

13. Propene is polymerized at 2000C and 1200 atm

14. 2-methylbut-2-ene react with bromine water under the presence of

sunlight.

4. Proposed the mechanism for the following reaction below