Free Radical Chain Reaction: Alkane + Halogen
1) Initiation: photochemical homolysis
Bond stability
depends on size of atom
2) Propagation
RDS
H
slow
X
X
X
2 X
UV light
3) Termination
fast
H
H
C
H
H
C
H
H
fast
C
X
C
H
H
X
H
fast
X
C
H
X
Additional halogens can add on during propagation…
H
H
H
C
C
H
C
H
H
Ethane (bi-product)
H
C
H
X
X
H
H
H
H
H
H
fast
H
H
H
H
C
X
H
X
H
H
• Methyl free radical
Reaction comes to a stop if 2 free radicals pair up
X
fast
H
H
• Halogen free radicals
Homolytic fission
- electron pair
splits to form two
free radicals
X
C
H
H
C
H
X
H
X
X
H
X
C
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
d+
d–
X (electrophile)
Anion (nucleophile) will
attack carbocation through
backside attack
C+
C
X
Anion depends on solution
Carbocation
Heterolytic Fission
X
• Dipole induced
• Bond lengthens and weakens
C
X
C+
X–
fast
Anion
• X– (halide anion)
C
Inert solution
Dissolved
in NaCl
Dissolved
in water
X
C
C
X
X
C
X
C+
C+
–
OH
Anions
• OH–
• X–
OH
fast
C
Cl
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
d+
d–
H (electrophile)
• Existing dipole strengthened
• Bond lengthens and weakens
C
H
C
Halo alkane
Halide ion
(Nucleophile)
Carbocation
Heterolytic Fission
X
fast
X–
C+
C
X
Stability of Carbocations
Tertiary (3°)
R
Note: for asymmetric alkenes, Markovnikov’s rule is used
• Hydrogen will bond to the carbon richest in hydrogens
-This allows for the most stable carbocation to form
• Hydrogen will be abstracted from the Carbon atom bonded to the
least hydrogen atoms
• Tend to create alkyl groups, not destroy them
R
>
C+
Primary (1°)
Secondary (2°)
R
R
R
R
>
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)
d+
d–
I
Heterolytic Fission
slow
I
C
C
C
C
C+
H
Tertiary carbocation
C
I
–
Iodine ion
(nucleophile)
fast
C
C
C
C
C
2-iodo 2-methyl pentane
C
Electrophilic Addition: Alkene + Conc. Sulphuric Acid (with silver catalyst) + Water
C
slow
H (electrophile)
d+
C
C
d–
• Existing dipole strengthened
• Bond lengthens and weakens
–
C+
OSO3H
Carbocation
Hydrogen sulphate ion
O
H
O
H
fast
C
Water (warm)
Alkene
H+ / H 2SO4
H2SO4
Ethanol
H
O
C
C
Hydrolysis
Net Reaction: Hydration of an Alkene
C
C
H
O
C
C
Ethyl hydrogen sulphate
SO3H
C
O
fast
H
Heterolytic Fission
HSO3O
SO3H
Backside attack
Sulphuric Acid approaches alkene
C
C
Alcohol
This reaction is industrially
important in the preparation of
alcohols
Cold, dilute
Oxidation of Alkenes: Alkene + Oxidizing Agent
KMnO4
Oxidizing Agent:
MnO4--
• Potassium
Permanganate
• Oxidizing Agent
O
Mn
C
O
R
O
• Permanganate Ion
• Oxidizing Agent
• Purple
__
O
• Permanganate ion
• Purple
• Oxid # Mn = +7
O
C
C
C
O
O
C
C
Mn
OH OH
Mn
O
Alkene
O
O
Ethane 1,2-diol
(antifreeze)
Transition state
Manganese (IV)
Oxide
• Oxid # Mn = +4
Alkene + Hydrogen : Reduction of Alkenes/ Hydrogenation
C
C
H
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
Free Radical
R
C
C
Alkyl free radical
C
C
R
Alkene
C
R
• High pressure
• High temperature
• Catalyst
C
Alkyl free radical
• Free radical is reproduced to continue
the polymerization
R
C
C
Alkene
C
C
C
Free radical
Termination occurs when 2 free radicals combine
Product:
C
C
n
Polyalkene
Note: The degree of polymerization can be altered by changing temperatures, pressures, catalysts.
This will result in many polyalkenes with a variety of different characteristics
C
SN2: Haloalkane Nucleophilic Substitution
Rate Determining Step = k [R-X]1 [Nucleophile]1
Backside attack
Nucleophile
• Only strong bases can
displace the halide ion
(CN-, OH-, NH3)
Concerted reaction:
everything happens all
at once, there is no
intermediate
R
H
C
H
_
R
_
Nu
Heterolytic Fission
slow
Nu
X
Nonpolar
solvent
Low temp.
Primary Haloalkane
X
C
H
R
fast
H
• Note: Inversion of configuration
• Possibility of optical isomers
X
Effect of Steric Hindrance on Secondary Haloalkane:
Big bulky alkyl groups
Nucleophile does not have
space to attack carbon
long chains/ branching
R
_
R’
C
H
X
_
Halide Ion
Secondary Haloalkanes will undergo SN2 when steric hindrance
does not occur (when alkyl groups are not too large):
Nu
Nu
H
H
Transition State
• Trigonal bipyramidal
• Bonds must flatten themselves
for a planar arrangement
• High Potential Energy, unstable
compound
Nucleophile’s path
is blocked
C
SN1: Haloalkane Nucleophilic Substitution
Note: Polar solvents will hydrate the carbocation and stabilize it
R
Non-concerted reaction: contains multiple steps and has an
intermediate
Rate Determining Step = k
R
Step I :
R’
C
R’’
C+
R’
Heterolytic Fission
Step II:
H
The ion-dipole attraction provides energy for Step I
R
slow
R’
Warm temperature
Dilute polar solvent
Alkali
C+
X
_
R’’
•
•
Tertiary Haloalkane
Secondary Alkanes will undergo SN1 if
alkyl groups are too large hence, steric
hindrance is present
O
d–
R’’
[R-X]1
X
H
•
•
Halide Ion
Tertiary Carbocation
Intermediate
Most stable due to positive
inductive effect of alkyl groups
Transient existence
Trigonal planar , 120o , equal
opportunity for nucleophile to
come in from any side
Nucleophilic Substitution
Nu
_
Nucleophile
• Does not require a
strong base
R
R’
C+
R’’
R
fast
R’
C
Nu
R’’
Forms racemic mixture:
Products optically inactive
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
Heterolytic Fission
_
H
H
C
H
H
C
d+
H
Primary Haloalkane
X
d–
slow
Hot,
Concentrated
alkali
H
C
C
Alkene
O
H
Water
X
_
Halide Ion
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
Step I :
slow
R’
C
X
Hot, polar,
concentrated, alkali
R’’
R’
X
C+
R’’
Tertiary Carbocation
• Stable due to positive inductive
effect of alkyl groups
• Intermediate
Heterolytic Fission
Tertiary Haloalkane
_
Halide Ion (product)
Nucleophilic Elimination
Step II :
OH
_
H
Hydroxide Ion
• Nucleophile
C
H
H
R
C+
fast
C
R’’
H
R
C
O
H
R’’
Alkene
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
CH3
C2H5
C
Br
slow
OH
_
H
CH3
Heterolytic Fission
H
_
Br
H
fast
H
C
C
H
H
CH3
C
CH3
H
+
CH3
O
H
Water
H
C
C
C
H
H
CH3
2-methyl but-2-ene
Dehydration of Alcohols: Elimination Reaction
H
Alcohol
Excess Acid+ High Temperature:
C
C+
H
C
H
C
Excess Alcohol + Low Temperature:
C+
H
H
H
C
O
H
H
H
+
H
H
H
C
C
O+
H
H
H
H
H
C
C+
H
H
__
H
OPO3H2
unstable
Dihydrogen Phosphate ion
Protonated Alcohol
Alkene
Heterolytic Fission
H
Phosphoric Acid
(dehydrating agent,
homogeneous catalyst)
__
O
H
OPO3H2
H
H
Water
Carbocation
H3 PO4
• Acid regenerated
Heterolytic Fission
C
C
H
H
d–
H
H
H
H
d+
H
H
H
d–
d+
H
H
O
OPO3H2
H
C
C
H
H
H
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
K2 Cr2O7
Primary Alcohol
H
C
O
H
H
H+/Cr2O72-
[O]
C
R
O
H
Aldehyde
O
H
H
Excess Oxidizing Agent [O]
Reflux/Boil
R
C
O
OH
Carboxylic Acid
[O]
2 Cr3+
• Chromium (III) ions
• Green
• Oxid # Cr= +3
• Dichromate (VI) ion
• Orange
• Oxid # Cr= +6
• Acidified
Potassium
Dichromate
• Oxidizing Agent
R
reduced
Cr2O72--
Source of Oxygen
Secondary Alcohol
Tertiary Alcohol
R’
R
C
R’
O
H
R
H
H+/Cr2O72-
O
R’’
[O]
R’
R
C
C
Ketone
O
O
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
O
H
R’
O
R
R’
C
O
R
H
Catalyst:
concentrated sulphuric acid
• Homogeneous catalysis, participates in reaction and is
regenerated afterwards
• Dehydrating agent (loss of water)
-shifts equilibrium right, produces more product
This is an example of condensation polymerization:
à 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
O
H
C
O
Carboxylic Acid
H
N
R
R’
O
H
C
N
H
R’
Amine
H
Water
Ester
Alcohol
Carboxylic Acid
O
H
Peptide Linkage
R
Alcohol + Concentrated H-X Acid
Heterolytic Fission
R’
C
O
H
H
R
d+
R’’
X
R
d–
R
C
+
R’’
X–
R
C
O+
Protonated Alcohol
(intermediate)
Alcohols are soluble in the concentrated halide acid
Therefore, the reactant solution is colourless
R’
H
R’’
Strong Acid
• HCl
• HBr
• HI
Tertiary Alcohol
• Reacts most
readily
R’
R’
R
C
H
X–
Halide ion
(nucleophile)
O
+
R’’
Carbocation
• Tertiary carbocations
are the most stable
H
H
Water
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, the reaction can be used to
distinguish between 1°, 2°, and 3° alcohols