CHEMISTRY FORM 6 SEM 3 02.pdf

CHEMISTRY FORM 6 ORGANIC CHEMISTRY CHAPTER 2 : HYDROCARBON

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

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

Name

Molecula Molecular structure r formula

Name

Molecula Molecular structure r formula

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 Place a prefix State the position Find the longest Identify the upon the similar where the ‘branch’ chain of carbon ‘branched’ carbon and name (alkyl group) that alkyl group (if any). is located at which If there is 2 similar carbon based on accordingly. (it attached to the the numbering does not has to be ‘main’ chain. Then, alkyl, prefix di is gave earlier. a straight chain). placed, if 3 similar name the alkyl accordingly alkyl, prefix tri is CH3– methyl placed. CH3CH2– ethyl CH3CH2CH2-propyl

3-methylpentane

3,5-dimethylheptane 3-ethyl-4-methylhexane

3-ethyl-3,5dimethyloctane

CH3(CH2)5CH3

CH(CH3)2C(CH3)3

CH3CH(CH2CH3)2

C(CH3)3CH2C(CH3)3

n-heptane

2,2,3-trimethylbutane

3-methylpentane

2,2,4,4tetramethylpentane

2,3-dimethylpentane

3-ethyl-3methylhexane

2,2,3-trimethylpentane

3,3-diethylhexane




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) Density trend Solubility

BOILING POINT INCREASE DOWN HOMOLOGOUS SERIES --

--

0.50

0.58

0.63

0.66

0.68

DENSITY INCREASE DOWN HOMOLOGOUS SERIES

water Not soluble in CCCCCCC.. organic solvent Soluble in CCCCCCCCCCCC

0.70

A) Boiling point of alkenes
The boiling point increase CCCC when going down to homologous series of alkane.
All alkane possessed the same intermolecular forces : weak Van Der Waals CCCCCCCCforces molecular mass stronger the CCCCCCCCC weak Van Der Waals’
Greater the CCCCCCCC, forces, increase CCCC the boiling point
Boiling point of isomers of the same molecular formula varies with the branched molecules H

H

H

H H

C

H

H H

C

C

C

C

H

H H

C

H H

H




H H

C

H H

H

H

C

C

C

C

C

H

H

H

H

H

H

H

H

higher boiling point compared to branched chain Straight chain has ..CC.. total surface area as straight chain molecule has higher CCCCCCCCCC 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

B) Solubility of alkane non-polar
All alkanes are often consider as CCCCCCCCC molecule as the dipole of moment created in molecule is very small. non-polar
Since alkane is CCCCCCC. Molecule, it dissolve easily in non-polar solvent such as benzene, and ether. hydrogen
Alkane does not form CCCCC bond in water, so it is insoluble in water. Thus, alkane is also described as CCCCC hydrophobic (water–hating). CCCCCC.
The longer the alkane chain, the more insoluble it is in water.

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 : CH3COOH + 2 NaOH
CH4 + H2O + Na2CO3
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

2 CH3CH2–Cl + 2 Na
CH3CH2CH2CH3 + 2 NaCl

2.5.2 Reaction of Alkane saturated hydrocarbon, so alkane is inert to
Since alkane is a CCCCC. most of the chemical reaction
Table below shows the description of reaction of ethane with other substances. 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






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.

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 y  C X H y +  x +  O 2 → x CO 2 + 4 

C2H6

y H 2O 2

∆H = − m kJ / mol

C2H6 + 7/2 O2
2 CO2 + 3 H2O

C5H12 C5H12 + 8 O2
5 CO2 + 6 H2O C8H18 C8H18 + 25/2 O2
8 CO2 + 9 H2O





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.

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

Since chlorine radical are highly reactive, when it collide with Cl – Cl
2 Cl• ∆H = +242 kJ/mol methane molecule forming HCl H3C–H
CH3• + H• and methyl radical ∆H = + 433 kJ/mol H C–H + •Cl
H C• + HCl

chlorine required lower Since ...................... energy to form radical, so the initiation will start off with

chlorine Gas CCCCC..

3

3

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

Step 3 : Termination

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

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 fossil fuel plants, so they are also known as CCCCCCCC 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), ethane while natural gas contain mainly methane C.CCC and some CCCCC The mixture in petroleum can be separated by using fractional distillation in oil refinery. Diagram below shows the CCCCCCCCCC. chamber and oil refinery used to separate the mixture of petroleum.

Fractional distillation

Products

Uses

Petrol gas

Use for house cooking gas

Gasoline

Use as fuel for automobile vehicle

Naphtha

Use to synthesis different petrochemical

Kerosene

Use as fuel for jet engine and oil stove

Diesel oil

Use as fuel of heavy vehicle such as bus or lorry

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  With the aid of zeolite as catalyst, (homolytic fission) take place and form carbon cracking can occur at lower various products of unbranched alkane temperature compare to thermal and alkene cracking.  Example, when breaking decane, C10H22  Products using catalytic cracking usually C10H22 → C3H6 + C7H16 contain branched alkane and alkene. C10H22 → C4H8 + C6H14 C10H22 →

2.7 Cycloalkane (alicyclic compound)
Cycloalkane has a general formula of CnH2n
Some examples of cycloalkane Cycloalkane

Molecular formula

Cyclopropane

C3H6

Cyclobutane

C4H8

Cyclopentane

C5H10

Cyclohexane

C6H12

Displayed formula

Skeletal formula

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 1,2,4-trimethylcyclohexane

3-ethyl-1-methylcyclopentane

3-ethyl-2-methyl-1propylcyclobutane

1,2,3-trimethylcyclooctane

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 Name

Molecular formula

Ethene

C2H4

Butene

C4H8

Molecular structure

Propene

But-2-ene Pentene

Molecular Molecular structure formula C3H6

But-1-ene

C5H10 pent-2-ene Hex-1-ene

Hexene

Name

C6H12

pent-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-methylpropene 2-methylbut-2-ene

2,3-dimethylpent-2-ene 2-ethyl-3-methylpent-1-ene

3,5-dimethylhept-3-ene 3,4-dimethylhex-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-diene

3-methylcyclopropene

2,5-dimethylhex-1,3-diene

3-ethyl-2-methylcyclohexene

oct-2,5-diene

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.

CH3




Isomers of pentene

CH2CH3

H

H

CH3

CH3

H

CH3

CH3

H

2.10

Physical Properties of Alkene

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)

A) Boiling Point of Alkene increase when going down to homologous series
The boiling point CCCC of alkane.
All alkane possessed the same intermolecular forces : weak Van Der Waals CCCCCCCforces weak Van Der Waals forces, molecular mass stronger the CCCCCCCCC
Greater the CCCCCC.., higher CCCC the boiling point

2.11 Preparation of Alkene
Alkene can be prepared in a few ways Name of reaction

Reagent used and condition

Ethanolic Dehydrohalogenation sodium from hydroxide (heat haloalkane & reflux)

Dehydration (removal of water) from alcohol

Excess conc. H2SO4 at 1800C or Alumina (Al2O3) at 350oC

Equation

2.12

Chemical reaction of alkene

Name of reaction

Reagent used and condition

Hydrogen gas CH3CH=CH2 + H2 (g) under propene -------------Nickel (Ni) at Hydrogenation 180oC @ Platinum (Pt) at room temperature cyclohexene Halogenation

Halogen gas, X2 (X2 = Cl2 ; Br2 ; I-2)

Addition of Hydrogen halide

Hydrogen halide (H–X) (X = Cl ; Br ; I)

Equation Ni

CH3CH2CH3 (g) propane

cyclohexane

Name of reaction

Reagent used and condition

Hydration

Steam (H2O) --------Phosphoric acid, (H3PO4 ) At 300oC ; 60 atm

Equation

Hydroxylation (cold, diluted KMnO4 (aq) / H+ (cold and diluted) acidified KMnO4) ethanal H

H

H

H

H

C

C

methanal H

+

Oxidation (under hot, concentrated acidified potassium manganate (VII)

H

C H

H+

KMnO4 (aq) / (hot & concentrated)

C

C + 2 [O]

KMnO4 / H

hot, concentrate

H

H

O

+ O

H

H

propene

[O]

[O]

H

C

H

OH

C

C

O

H

ethanoic acid

+

CO2 + H2O

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

δ+




δ–

Step 2 : Nucleophilic attack – the negative bromide ion react fast with the unstable carbocation.










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 electrondonating 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) 2-methylpropene



2-methylpropan-1-ol

(major) 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)

CH3CH2CH2CH2CH3 CH3CH2CHClCH2Cl CH3CHBrC(CH3)(OH)CH3 major CH3CHBrCBr(CH3)CH3 minor

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

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

CH2CH3

H

H

CH3

CH3

H

CH3

CH3

H

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 H

H H C H H

H H

C

C

C H

H

H

H

+

Cl

Cl

H

H C H

H H

C

C

C H

H

Cl

H

+ H

Cl