1. Free Radical Chain Reaction: Alkane + Halogen
1) Initiation: photochemical homolysis
RDS
2) Propagation
Bond stability depends on size of atom X H C
Methyl free radical
fast
slow X X
2 X
UV light
Halogen free radicals
Homolytic fission
3) Termination
Reaction comes to a stop if 2 free radicals pair up X
fast
Additional halogens can add on during propagation… H C X
Further reaction will result in the substitution of all hydrogens with halogens C H
fast
Ethane (bi-product) C H X
fast

2. + + – + – + – Electrophilic Addition: Alkene + Halogen C C X X X
Dipole induced
Bond lengthens and weakens
Halogen approaches alkene:
δ – δ+ X
Heterolytic Fission
(electrophile)
slow C + X
Carbocation
Anion (nucleophile) will attack carbocation through backside attack
Anion depends on solution
Inert solution
Dissolved in NaCl
Dissolved in water Cl – C + X
fast X – C +
fast
Anions
Cl–
X--
Anion
X– (halide anion) OH – C + X
fast
Anions
OH– X–
Example - Bromine and Ethene (X = Br2)
Inert Solution: ,2 – dibromoethane Bromine Dissolved in Water: 2-bromo ethan-1-ol
Bromine Dissolved in NaCl: bromo, 2-chloro ethane

3. Electrophilic Addition: Alkene + Hydrogen Halide
Hydrogen Halide approaches alkene C + H
Carbocation X –
Halide ion (Nucleophile)
Backside attack
fast
Halo alkane C
slow
δ – δ+ X H
(electrophile)
Heterolytic Fission
Stability of Carbocations
Existing dipole strengthened
Bond lengthens and weakens
Tertiary (3°)
Secondary (2°)
Primary (1°) R C + R C + H H C + R
>
>
Note: for asymmetric alkenes, Markovnikov’s rule is used
Hydrogen will bond to the carbon that is richest in hydrogens already -This allows for the most stable carbocation to form
Hydrogen will be abstracted from the Carbon atom that has the least Hydrogens already
*Positive inductive effect : alkyl groups donate electron density to centre carbon -Carbon with the most electron density is the most stable
Example – Hydrogen iodide and 2-methyl pent-2-ene(X = I2) C C I C + H I –
slow
fast C
δ – δ+ I H
Heterolytic Fission
(electrophile) C
Iodine ion (nucleophile)
Tertiary carbocation
2-iodo 2-methyl pentane

4. Electrophilic Addition: Alkene + Dilute Sulphuric Acid (with silver catalyst)
Sulphuric Acid approaches alkene C + H
Carbocation
OSO3H –
Hydrogen sulphate ion
Backside attack
fast
Ethyl hydrogen sulphate O
SO3H C
slow
δ – δ+
HSO3O H
(electrophile)
Heterolytic Fission
Existing dipole strengthened
Bond lengthens and weakens C O
SO3H H
Water (warm)
fast
Hydrolysis
Ethanol
H2SO4
Net Reaction: Hydration of an Alkene C O H
H+ / H2SO4
Alkene
Alcohol
This reaction is industrially important in the preparation of alcohols

5. Ethane 1,2-diol (antifreeze)
Oxidation of Alkenes: Alkene + Oxidizing Agent
KMnO4
MnO4--
Cold, dilute
Potassium Permanganate
Oxidizing Agent
Permanganate ion
Purple
Oxid # Mn = +7
Oxidizing Agent: Mn O __ Mn O C O Mn C OH C
Alkene
Ethane 1,2-diol (antifreeze)
Manganese (IV) Oxide
Oxid # Mn = +4
Permanganate Ion
Oxidizing Agent
Purple
Transition state
Alkene + Hydrogen : Reduction of Alkenes/ Hydrogenation C H
This method is used in the food industry to solidify oils into edible fats (ex: margarine)
High Temperature C H
(g)
Alkene
Catalyst: Transition Metal
Pt, Ni, Pd
Heterogeneous Catalysis -Absorption, reaction, desorption
Alkane

6. Addition Polymerization: Alkene + Free Radical
Alkyl free radical
Free radical is reproduced to continue the polymerization R
Free Radical C
Alkene
High pressure
High temperature
Catalyst C R
Alkyl free radical C
Free radical R C
Alkene
Termination occurs when 2 free radicals combine
Polyethene C
Product: n
Note: The degree of polymerization can be altered by changing temperatures, pressures, catalysts.
This will result in many polyethenes with a variety of different characteristics

7. Effect of Steric Hindrance on Secondary Haloalkane:
SN2: Haloalkane Nucleophilic Substitution
Rate Determining Step = k [R-X]1 [Nucleophile]1 H C X R Nu C H X R C H Nu R Nu _
slow
fast H
Nucleophile
Only strong bases can displace the halide ion (OH--, CN--) H
Primary Haloalkane
Transition State
Trigonal bipyramidal
Bonds must flatten themselves for a planar arrangement
High Potential Energy, unstable compound
Note: Inversion of configuration
Possibility of optical isomers
Secondary Haloalkanes will undergo SN2 when steric hindrance does not occur (when alkyl groups are not too large):
Effect of Steric Hindrance on Secondary Haloalkane: R’ C H X R Nu _
Big and bulky alkyl groups
Nucleophile does not have space to attack carbon
long chains/ branching
Nucleophile’s path is blocked

8. SN1: Haloalkane Nucleophilic Substitution
Note: Polar solvents will hydrate the carbocation and stabilize it
This also provides energy for Step I R’ C
R’’ + R O H
δ –
Rate Determining Step = k [R-X]1 R’ C
R’’ X R R’ C
R’’ + R X _
Step I :
slow
Heterolytic Fission
Tertiary Carbocation
Intermediate
Most stable due to positive inductive effect of alkyl groups
Transient existence
Halide Ion
Tertiary Haloalkane
Secondary Alkanes will undergo SN1 if alkyl groups are too large hence, steric hindrance is present
Nucleophilic Substitution Nu _ R’ C
R’’ Nu R R’ C
R’’ + R
fast
Nucleophile
Does not require a strong base
Step II:
Note: possibility of optical isomers
Racemic Mixture

9. E2: Haloalkane Elimination Reaction
Elimination occurs instead of substitution when pathway of nucleophile is blocked, hence the Lewis base will remove an electrophile from the haloalkane instead OH _ C H X R O H
slow H C
δ +
δ –
Heterolytic Fission
Alkene
Water
Primary Haloalkane
Secondary Haloalkanes will undergo E2when steric hindrance does not occur (when alkyl groups are not too large)
Note: Elimination requires more energy than substitution
The elimination reaction is favoured if: a) the base is strong b) the temperature is high

10. Step I : Step II : E1: Haloalkane Elimination Reaction _ X _ OH O C
Heterolytic Fission
Tertiary Haloalkane R’ C
R’’ + R X _
slow
Step I :
Tertiary Carbocation
Stable due to positive inductive effect of alkyl groups
Halogen Ion (product)
Nucleophilic Elimination OH _
Hydroxide Ion
Nucleophile
fast C
R’’ + R H O
Alkene
Water
Step II :
Note: Markovnikov’s rule still applies -When adding Hydrogen, adds to carbon that is richest in Hydrogens already
-When eliminating Hydrogen, removes from carbon that has least Hydrogens
Example – 2-bromo 2-methyl butane OH _ Br _
C2H5 C
CH3 C
CH3 H
slow O H Br
fast C
CH3 + H
Heterolytic Fission
Water
2-methyl but-2-ene

11. Excess Acid+ High Temperature:
Dehydration of Alcohols: Elimination Reaction H C O C H O H
OPO3H2 +
OPO3H2 __
δ +
δ –
δ +
δ –
Dihydrogen Phosphate ion
Phosphoric Acid (dehydrating agent)
Protonated Alcohol
Alcohol
OPO3H2 __ H +
H3 PO4
Acid regenerated
Excess Acid+ High Temperature: C H +
Alkene C H +
Heterolytic Fission O H
Carbocation
Water
Excess Alcohol + Low Temperature: C H + O
Ether
Alcohol
Heterolytic Fission +

12. K2 Cr2O7 Cr2O72-- [O] Oxidation of Alcohols: Alcohol + Oxidizing Agent
Potassium Dichromate
Oxidizing Agent
Dichromate ion
Orange
Oxid # Cr= +6
Source of Oxygen
Primary Alcohol
Secondary Alcohol
Tertiary Alcohol C R H O C R R’ H O
[O] C R R’
R’’ O H
[O] C R H O
Aldehyde C R R’ H O
Ketone
No Oxidation Occurs
Central carbon does not have any Hydrogens to be oxidized O H O H
Excess Oxidizing Agent [O]
Reflux/Bone C R H OH O
Carboxylic Acid

13. Esterification (condensation reaction): Alcohol + Carboxylic Acid
Water
Catalyst: concentrated sulphuric acid
Homogeneous catalysis
Dehydrating Agent (loss of water) -shifts equilibrium right, produces more product
Note: Water is a product, not a solvent
So it is included in the equilibrium calculation
𝐾 𝑐 = 𝐻 2 𝑂 [𝐸𝑠𝑡𝑒𝑟] 𝐴𝑙𝑐𝑜ℎ𝑜𝑙 [𝐴𝑐𝑖𝑑]
This is an example of condensation polymerization:  A reaction in which two molecules join together to form a larger molecule with the loss of a smaller molecule
Another example of condensation polymerization is the formation of peptide linkages in proteins
Condensation Polymerization N C O R’ H
Carboxylic Acid C R’ H O
Amine N R H
Peptide Linkage

14. Alcohol + Concentrated Halide Acid
Heterolytic Fission C R R’
R’’ O H H C R R’
R’’ O + H X X –
Halide ion (nucleophile)
δ +
δ –
Strong Acid
HCl
HBr HI
Protonated Alcohol (intermediate)
Tertiary Alcohol
Reacts most readily
Alcohols are soluble in the concentrated halide acid
Therefore, the reactant solution is colourless C R R’
R’’ +
Carbocation
Tertiary carbocations are the most stable O H
Water C R R’
R’’ X
Halogenoalkane
Useful method to prepare alkyl halides C R R’
R’’ + X –
Note: This is an important test for alcohols, tertiary alcohols will react rapidly whereas primary and secondary alcohols will react more slowly or not at all
Therefore, reaction can be used to distinguish between 1°, 2°, and 3° alcohols
Halogenoalkanes are not soluble in the concentrated halide acid
Therefore, the product solution is cloudy