CHEMISTRY FORM 6 SEM 3 02.pdf

CHEMISTRY FORM 6

ORGANIC CHEMISTRY

Hydrocarbon which contain only carbon-carbon single bond, C–C

Hydrocarbon which contain at least one carbon-carbon double 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 C_nH_2n+2
Table below shows the naming of straight chain of alkane

carbon-carbon single bond, C–C one carbon-carbon double bond, C=C

Name Molecula

r formula Molecular structure Name

Molecula

r formula Molecular structure

Methane CH₄ Ethane C₂H₆

Propane C₃H₈ Butane C₄H₁₀

Pentane C₅H₁₂

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

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– methyl CH₃CH₂– ethyl CH₃CH₂CH₂-propyl placed.

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

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

3-ethyl-3,5-dimethyloctane

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

2,2,4,4-2,3-dimethylpentane 3-ethyl-3-methylhexane

2,2,3-trimethylpentane _{3,3-diethylhexane}

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

Step 3

Complete the structure by placing one hydrogen

2.4 Physical properties of alkane Alkane CH₄ C₂H₆ C₃H₈ C₄H₁₀ C₅H₁₂ C₆H₁₄ C₇H₁₆ C₈H₁₈ Boiling point o_C – 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

BOILING POINT INCREASE DOWN HOMOLOGOUS SERIES

Density trend

Solubility Not soluble in CCCCCCC.. Soluble in CCCCCCCCCCCC

DENSITY INCREASE DOWN HOMOLOGOUS SERIES

water

A) Boiling point of alkenes

The boiling point CCCC when going down to homologous series

of alkane.

All alkane possessed the same intermolecular forces : weak

CCCCCCCCforces

Greater the CCCCCCCC, stronger the CCCCCCCCC

forces, CCCC the boiling point

Boiling point of isomers of the same molecular formula varies with

the branched molecules

increase

Van Der Waals

molecular mass weak Van Der Waals’

increase

the branched molecules

Straight chain has ..CC.. boiling point compared to branched chain

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

C C C C C H H H H H H H H H H H C H H H C C C C H H H H H H H C H H H H C H H H higher

B)

Solubility of alkane

All alkanes are often consider as CCCCCCCCC

molecule as the dipole of moment created in molecule is

very small.

Since alkane is CCCCCCC. Molecule, it dissolve

easily in non-polar solvent such as benzene, and ether.

non-polar

non-polar

easily in non-polar solvent such as benzene, and ether.

Alkane does not form CCCCC bond in water, so it is

CCCCC in water. Thus, alkane is also described as

CCCCCC. (water–hating).

The longer the alkane chain, the more insoluble it is in

water.

hydrogen

insoluble

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

-CH

₃

_{COOH + 2 NaOH
CH}

₄

+ H

₂

O + Na

₂

CO

₃

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

2.5.2 Reaction of Alkane

Since alkane is a CCCCC. hydrocarbon, so alkane is inert to

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

saturated

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.

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

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 mol / kJ m H O H 2 y CO x O 4 y x H C_{X y}  ₂ → ₂ + ₂ ∆ = −      + +

C

₂

H

₆

+ 7/2 O

₂

_{2 CO}

₂

+ 3 H

₂

O

C

₅

H

₁₂

+ 8 O

₂

_{5 CO}

₂

+ 6 H

₂

O

C₅H₁₂ C₈H₁₈

Note that the reaction is exothermic for all hydrocarbons. Equation

above is also known for ∆H_c. 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.

C

₅

H

₁₂

+ 8 O

₂

_{5 CO}

₂

+ 6 H

₂

O

C

₈

H

₁₈

+ 25/2 O

₂

_{8 CO}

₂

+ 9 H

₂

O

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) + Cl_{2 (g)
CH3}Cl (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 H_3C–H
CH3• + H•

∆H = + 433 kJ/mol Since ... required lower energy to form radical, so the

initiation will start off with CCCCC.. Gas

Since chlorine radical are highly reactive, when it collide with methane molecule forming HCl and methyl radical

H_{3C–H + •Cl
H3}C• + HCl Methyl radical will propagate with other chlorine molecule and

forming back chlorine radical H_{3C• + Cl–Cl
H3}CCl + •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

H_{3C• + •Cl
H3}C–Cl

∆H = -349 kJ/mol H₃C• + •CH_3
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

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 CCCCCCCC

Coal is complex mixture consisting mainly hydrocarbons, which is

mainly made up from dead plaints in swamp.

fossil fuel

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

while natural gas contain mainly C.CCC and some CCCCC

The mixture in petroleum can be separated by using

CCCCCCCCCC. in oil refinery. Diagram below shows the chamber and oil refinery used to separate the mixture of petroleum.

methane

ethane

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

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

 With the aid of zeolite as catalyst, carbon cracking can occur at lower temperature compare to thermal various products of unbranched alkane

and alkene

 Example, when breaking decane, C10H22

C₁₀H₂₂ → C₃H₆ + C₇H₁₆ C₁₀H₂₂ → C₄H₈ + C₆H₁₄

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 C_nH_2n
Some examples of cycloalkane

Cycloalkane Molecular

formula Displayed formula Skeletal formula

Cyclopropane

C

₃

H

₆ Cyclobutane Cyclopentane Cyclohexane

C

₄

H

₈

C

₅

H

₁₀

C

₆

H

₁₂

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 200o_C

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

/ bromine gas under sunlight, substitution reaction take place

Mechanism :

Initiation

Propagation

2.8 Alkene – Nomenclature of alkenes and cycloalkenes

The homologous series of alkenes has general formula of C_nH_2n.

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

with its name end with –ene

Name Molecular

formula Molecular structure Name

Molecular

formula Molecular structure

Ethene C₂H₄ Propene C₃H₆ Butene C₄H₈ Pentene C₅H₁₀ Hexene C₆H₁₂ But-2-ene But-1-ene pent-2-ene pent-1-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.

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 C_nH_2n–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 2,5-dimethylhex-1,3-diene oct-2,5-diene

3-methylcyclopropene 3-ethyl-2-methylcyclohexene

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

CH₂CH₃ CH₃ CH₃ H H CH₃ H H CH₃

2.10 Physical Properties of Alkene

A) Boiling Point of Alkene

Alkene C₂H₄ C₃H₆ C₄H₈ C₅H₁₀ C₆H₁₂ C₇H₁₄ C₈H₁₆ C₉H₁₈ Boiling point o_C – 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

The boiling point CCCC when going down to homologous series

of alkane.

All alkane possessed the same intermolecular forces : weak

CCCCCCCforces

Greater the CCCCCC.., stronger the CCCCCCCCC forces,

CCCC the boiling point

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 condition Equation

Dehydro-halogenation from haloalkane Ethanolic sodium hydroxide (heat & reflux) Dehydration (removal of water) from alcohol Excess conc. H₂SO₄ at 1800_C or Alumina (Al₂O₃) at 350o_C

2.12 Chemical reaction of alkene

Name of reaction

Reagent used and

condition Equation Hydrogenation Hydrogen gas under ---Nickel (Ni) at 180o_C @ Platinum (Pt) at room temperature CH₃CH=CH₂ + H₂ (g) CH₃CH₂CH₃ (g) propene propane cyclohexene cyclohexane Halogen gas, X Ni

Halogenation Halogen gas, X2

(X₂ = Cl₂ ; Br₂ ; I-₂) Addition of Hydrogen halide Hydrogen halide ( H – X ) (X = Cl ; Br ; I)

Name of reaction

Reagent used and

condition Equation Hydration Steam (H₂O) ---Phosphoric acid, (H₃PO₄ ) At 300o_{C ; 60 atm} Hydroxylation (cold, diluted acidified KMnO₄ (aq) / H+

(cold and diluted) acidified

KMnO₄)

(cold and diluted)

Oxidation (under hot, concentrated acidified potassium manganate (VII) KMnO₄ (aq) / H+ (hot & concentrated) + C C C H H H H H H 2 [O] KMnO4/ H + hot, concentrate propene C C O H H H H C H H O + [O] [O] C C O H H OH H + CO₊ 2 H₂O ethanal methanal ethanoic acid

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 (180o_C)

(180 C)

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

(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 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

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 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=CH_{2 + H–Br
CH3}CH₂CH₂Br + 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 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

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 (1}0₎

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
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.

(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
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

(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 KMnO ) 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

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

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 CH₃₃CH(CH₃₃)CHClCH₃₃ major CH₃CH(CH₃)CH₂CH₂Cl minor

4-ethyl-2,2,4-trimethylhexane

2,2,4,5-tetramethylhexane

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

7-ethyl-1,3-dimethylcyloheptene

C(CH₃)₂=C(CH₂CH₃)CH(CH₃)CH(CH₃)₂

Isomers of pentene

CH₂CH₃ CH₃ CH₃ H H CH₃ H H CH₃

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

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+ _{→ CH}

3CH2CH(OH)CH(OH)CH2CH3

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

CH

₃

C(CH

₃

)=CHCH

₂

CH

₂

CH

₃

+ KMnO

₄

/H

+

_→

CH C(CH )(OH)CH(OH)CH CH CH

CH

₃

C(CH

₃

)(OH)CH(OH)CH

₂

CH

₂

CH

₃

12. Propane react with fluorine under the presence of sunlight

CH₃CH₂CH₃ + F₂ → CH₃CH₂CH₂F + HF

13. Propene is polymerized at 2000_{C 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 C H C H C H H H C H H H H Cl Cl + H C H C H C H H H C H H H Cl H Cl +