Free Radical Chain Reaction: Alkane + Halogen
1) Initiation: photochemical homolysis
Bond stability
depends on size of atom
2) Propagation
RDS
H
X
slow
X
X
2 X
UV light
C
H
fast
H
H
• Methyl free radical
3) Termination
C
Reaction comes to a stop if 2 free radicals pair up
fast
X
X
H
X
H
fast
X
C
H
H
X
H
C
H
H
C
H
fast
H
Additional halogens can add on during propagation…
H
C
H
H
C
C
H
C
H
H
Ethane (bi-product)
H
H
fast
H
H
H
X
H
H
H
H
C
H
X
X
H
H
H
H
H
H
X
C
X
H
• Halogen free radicals
Homolytic fission
H
C
H
X
X
X
H
C
X
H
H
X
X
X
X
C
X
H
Further reaction will result in the substitution of all
hydrogens with halogens
X
X
Electrophilic Addition: Alkene + Halogen
Halogen approaches alkene:
C
C
slow
Carbocation
Anion depends on solution
C
X (electrophile)
X
δ+
δ–
C+
Anion (nucleophile) will
attack carbocation through
backside attack
Heterolytic Fission
X
Inert solution
• Dipole induced
• Bond lengthens and weakens
Dissolved
in NaCl
Dissolved
in water
Cl
X
C
X
C+
X–
fast
Anion
• X– (halide anion)
C
C
C
C+
X
X
OH
C
X
C+
–
OH
Anions
• OH–
• X–
fast
C
Cl
–
fast
Anions
• Cl–
• X--
C
C
X
C
X
Example - Bromine and Ethene (X = Br2)
Inert Solution:
Bromine Dissolved in Water:
Bromine Dissolved in NaCl:
1,2 – dibromoethane
2-bromo ethan-1-ol
1-bromo, 2-chloro ethane
Electrophilic Addition: Alkene + Hydrogen Halide
Backside attack
Hydrogen Halide approaches alkene
C
C
slow
C+
C
H (electrophile)
X
C
C
Halo alkane
Halide ion
(Nucleophile)
Carbocation
Heterolytic Fission
X
fast
–
H
δ+
δ–
X
Stability of Carbocations
• Existing dipole strengthened
• Bond lengthens and weakens
Tertiary (3°)
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
R
R
R
>
C+
R
R
• Hydrogen will be abstracted from the Carbon atom that has the least
Hydrogens already
Primary (1°)
Secondary (2°)
>
C+
H
C+
H
H
*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
C
C
C
C
H (electrophile)
δ+
δ–
I
I
C
Heterolytic Fission
slow
C
C
C
C+
H
Tertiary carbocation
C
I
–
fast
C
C
C
C
C
Iodine ion
(nucleophile)
2-iodo 2-methyl pentane
C
Electrophilic Addition: Alkene + Dilute Sulphuric Acid (with silver catalyst)
C
C
slow
C
H (electrophile)
–
C+
OSO3H
O
fast
C
C
H
δ+
Carbocation
Heterolytic Fission
δ–
SO3H
Backside attack
Sulphuric Acid approaches alkene
Hydrogen sulphate ion
Ethyl hydrogen sulphate
HSO3O
• Existing dipole strengthened
• Bond lengthens and weakens
H
SO3H
H
O
C
C
O
H
C
fast
O
Ethanol
Net Reaction: Hydration of an Alkene
H
O
Alkene
C
H2SO4
Hydrolysis
Water (warm)
C
C
H+ / H2SO4
C
C
Alcohol
This reaction is industrially
important in the preparation of
alcohols
Oxidation of Alkenes: Alkene + Oxidizing Agent
Cold, dilute
KMnO4
MnO4--
• Potassium
Permanganate
• Oxidizing Agent
• Permanganate ion
• Purple
• Oxid # Mn = +7
Oxidizing Agent:
__
O
O
Mn
O
C
C
C
C
O
O
O
C
C
Mn
OH OH
Mn
O
Alkene
O
O
• Permanganate Ion
• Oxidizing Agent
• Purple
O
Ethane 1,2-diol
(antifreeze)
Transition state
Manganese (IV)
Oxide
• Oxid # Mn = +4
Alkene + Hydrogen : Reduction of Alkenes/ Hydrogenation
H
C
C
H
H(g)
High Temperature
C
C
H
Alkene
Catalyst: Transition Metal
• Pt, Ni, Pd
• Heterogeneous Catalysis
-Absorption, reaction, desorption
Alkane
This method is used
in the food industry
to solidify oils into
edible fats (ex:
margarine)
Addition Polymerization: Alkene + Free Radical
C
R
Free Radical
R
C
C
Alkyl free radical
C
Alkene
C
R
• High pressure
• High temperature
• Catalyst
C
C
Alkyl free radical
• Free radical is reproduced to continue
the polymerization
R
C
Alkene
C
C
C
Free radical
Termination occurs when 2 free radicals combine
Product:
C
C
n
Polyethene
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
C
SN2: Haloalkane Nucleophilic Substitution
_
Nu
Nucleophile
• Only strong bases can
displace the halide ion
(OH--, CN--)
Rate Determining Step = k [R-X]1 [Nucleophile]1
R
R
R
slow
H
C
Nu
X
H
H
X
C
fast
Effect of Steric Hindrance on Secondary Haloalkane:
Nu
Big and bulky alkyl groups
Nucleophile does not have
space to attack carbon
long chains/ branching
_
R
R’
C
H
X
Nu
• 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):
Nucleophile’s path
is blocked
C
H
H
Transition State
• Trigonal bipyramidal
• Bonds must flatten themselves
for a planar arrangement
• High Potential Energy, unstable
compound
Primary Haloalkane
H
SN1: Haloalkane Nucleophilic Substitution
Note: Polar solvents will hydrate the carbocation and stabilize it
R
C+
R’
O
δ–
R’’
Rate Determining Step = k
[R-X]1
R’
C
H
This also provides energy for Step I
R
Step I :
H
R
slow
X
R’
R’’
_
C+
X
R’’
Heterolytic Fission
Tertiary Carbocation
• Intermediate
• Most stable due to positive
inductive effect of alkyl groups
• Transient existence
Tertiary Haloalkane
Secondary Alkanes will undergo SN1 if
alkyl groups are too large hence, steric
hindrance is present
Halide Ion
Nucleophilic Substitution
Nu
Step II:
_
Nucleophile
• Does not require a
strong base
R
R’
C+
R’’
R
fast
R’
C
Nu
R’’
Note: possibility of optical isomers
• Racemic Mixture
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
_
R
H
C
C
δ+
slow
X
δ–
H
Heterolytic Fission
C
C
Alkene
O
H
H
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
E1: Haloalkane Elimination Reaction
R
R
slow
R’
Step I :
C
R’
X
_
X
C+
R’’
R’’
Tertiary Carbocation
• Stable due to positive inductive
effect of alkyl groups
Heterolytic Fission
Tertiary Haloalkane
Halogen Ion
(product)
Nucleophilic Elimination
Step II :
OH
_
H
H
Hydroxide Ion
• Nucleophile
C
H
R
C+
fast
C
R’’
H
O
C
Alkene
H
Water
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
CH3
C2H5
C
Br
_
Br
fast
Heterolytic Fission
H
H
C
C
H
H
CH3
C
CH3
H
slow
H
CH3
_
+
CH3
O
H
H
Water
H
C
C
C
H
H
CH3
2-methyl but-2-ene
Dehydration of Alcohols: Elimination Reaction
H
H
H
C
C
H
H
H
H
O
δ–
OPO3H2
δ+
H
δ–
δ+
Excess Acid+ High Temperature:
H
H
H
C
C+
H
C
H
C
H
H
C
C
O
H
H
+
H
+
H
OPO3H2
Dihydrogen Phosphate ion
__
H
H
C
C+
H
H
H
Water
H3 PO4
Heterolytic Fission
• Acid regenerated
Excess Alcohol + Low Temperature:
Heterolytic Fission
H
H
C
C+
C
C
H
H
H
H
H
H
H
H
O
H
H
H
OPO3H2
H
O
Carbocation
Alkene
__
H
Protonated Alcohol
Phosphoric Acid
(dehydrating agent)
Alcohol
H
H
H
H
O
C
C
H
H
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
C
C
H
+
H
Alcohol
H
O
Ether
H
H
+
+
Oxidation of Alcohols: Alcohol + Oxidizing Agent
[O]
Cr2O72--
K2 Cr2O7
• Potassium
Dichromate
• Oxidizing Agent
• Dichromate ion
• Orange
• Oxid # Cr= +6
Source of Oxygen
Primary Alcohol
Secondary Alcohol
R’
H
R
C
O
R
H
[O]
H
H
C
R
O
H
O
H
H
Excess Oxidizing Agent [O]
R
H
C
C
O
OH
Carboxylic Acid
R’
O
H
R’
R
C
H
R
[O]
H
Ketone
Aldehyde
Reflux/Bone
Tertiary Alcohol
O
O
C
O
R’’
H
H
No Oxidation Occurs
Central carbon does not
have any Hydrogens to
be oxidized
H
Esterification (condensation reaction): Alcohol + Carboxylic Acid
O
O
C
H
O
R’
R
O
R’
C
O
R
O
H
Ester
Alcohol
Carboxylic Acid
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
𝐻2 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
H
N
C
O
Carboxylic Acid
H
R
R’
O
H
C
N
H
R’
Amine
H
Water
𝐾𝑐 =
O
H
Peptide Linkage
R’
Alcohol + Concentrated Halide Acid
Heterolytic Fission
R’
R
C
O
H
H
X
δ+
δ–
R
R’’
R’
H
C
O
R’’
Strong Acid
• HCl
• HBr
• HI
Tertiary Alcohol
• Reacts most
readily
Protonated Alcohol
(intermediate)
Alcohols are soluble in the concentrated halide acid
Therefore, the reactant solution is colourless
R’
R
C
H
+
X–
Halide ion
(nucleophile)
O
+
H
H
R’’
Carbocation
• Tertiary carbocations
are the most stable
R’
R
C
+
R’’
X–
Water
R’
R
C
X
R’’
Halogenoalkane
• Useful method to prepare
alkyl halides
Halogenoalkanes are not soluble in the concentrated halide acid
Therefore, the product solution is cloudy
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