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Part 7.4
The chemistry of
AROMATIC COMPOUNDS
Some
free radical addition reactions of
benzene and methylbenzene - hydrogenation and halogenation
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aromatic
chemistry revision notes on the free
radical reactions of benzene & methylbenzene
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Part 7.4
Free radical addition reactions of benzene and methylbenzene with hydrogen
and
chlorination substitution of the methyl group of
methylbenzene
Sub-index for this page
7.4.1 Hydrogenation of benzene and
methylbenzene
7.4.2
Free radical addition of chlorine to
benzene
7.4.3
Free radical substitution
by chlorine of the methyl
group in methylbenzene
7.4.1 Hydrogenation of benzene and
methylbenzene
If benzene or methylbenzene is mixed with hydrogen
and passed over a heated nickel catalyst addition of three
moles of hydrogen per arene molecule takes place.
The higher temperature and Ni catalyst are needed because of the stability of
the benzene ring in arenes.
The hydrogen molecules are adsorbed on the Ni
surface and split into atoms.
These atoms are effectively very reactive
hydrogen radicals (H•)
which can break open the pi bond system of the arene and form new C-H
bonds until the totally saturated cycloalkane is formed.
(1)
+ 3H2 ===>
or
+ 3H2 ===>
hydrogenation of
the benzene to cyclohexane.
ΔHhydrogenation(benzene)
= -208 kJ mol-1
(2)
+ 3H2 ===>
or
+ 3H2 ===>
hydrogenation of
methylbenzene to methylcyclohexane
ΔHhydrogenation(methylbenzene)
= -205 kJ mol-1
Note the similarity of the enthalpies of
hydrogenation.
The mechanism is very complicated,
but the very simplified diagram above gives you some about it.
Hydrogen free radicals (hydrogen
atoms with their electron) are formed on the nickel surface, and
then attack the pi electron cloud of the 'benzene' ring forming new
C-H bonds.
A series of free radical chain
reactions ensue until all the pi electrons have been used in forming
C-H bonds and the fully saturated cycloalkane is formed.
TOP OF PAGE and
sub-index
7.4.2 Free radical addition of chlorine to
benzene
If chlorine is bubbled into boiling liquid benzene as
ultraviolet light is shone on the mixture you get the addition of three
chlorine molecules per benzene molecule.
Unlike alkenes, addition will not take place without the
input of high energy uv light to trigger a free radical reaction.
You do get addition under these particular reaction conditions.
The reaction yield yields
1,2,3,4,5,6-hexachlorocyclohexane.
C6H6
+ 3Cl2 ===> C6H6Cl6
Initiation
Cl2 == uv ==> 2Cl•
Propagation steps
??
C6H6 + Cl•
===> C6H6Cl•
C6H6Cl• + Cl2
===> C6H6Cl2 + Cl•
etc. but not sure of mechanism? (can't find it on
the internet)
until the final step is maybe?
C6H6Cl5•
+ Cl2 ===> C6H6Cl6
+ Cl•
or
C6H6Cl5•
+ Cl• ===> C6H6Cl6
TOP OF PAGE and
sub-index
7.4.3 Free radical substitution
by chlorine of the methyl
group in methylbenzene
When methylbenzene is reacted with chlorine and the mixture
irradiated with uv light, you do NOT get addition like you do with benzene!
Instead, you get a free radical substitution reaction
involving the methyl group - a side-chain substitution.
The initial product is (chloromethyl)benzene (benzyl
chloride).
(i) methylbenzene + chlorine ===>
(chloromethyl)benzene + hydrogen chloride
C6H5CH3 +
Cl2 == uv ==> C6H5CH2Cl
+ HCl
+ Cl2 ===>
+ HCl
(chloromethyl)benzene is not an aryl halide, but a
primary aliphatic halogenoalkane.
With excess chlorine, the free radical chain reaction
can continue to form two other substitution products.
(ii) (chloromethyl)benzene + chlorine
===> (dichloromethyl)benzene + hydrogen chloride
C6H5CH2Cl
+ Cl2 ===> C6H5CHCl2
+ HCl
(iii) (dichloromethyl)benzene + chlorine
===> (trichloromethyl)benzene + hydrogen chloride
C6H5CHCl2
+ Cl2 ===> C6H5CCl3
+ HCl
Again, these are all aliphatic halogenoalkanes, NOT
aryl halides.
A similar reaction will happen with bromine to give three
possible bromo-substituted products and via similar free radical chain
mechanism outlined below..
The free radical mechanism of the side-chain chlorination of methylbenzene
(•
= unpaired electron on a free radical)
Initiation step
Cl2 == uv ==> 2Cl•
Chain propagation steps
C6H5CH3 +
Cl• ===> C6H5CH2•
+ HCl
C6H5CH2• +
Cl2 ===> C6H5CH2Cl
+ Cl•
(1st product, monochloro...)
Termination steps
2Cl• ===> Cl2
C6H4CH2• +
Cl• ===> C6H5CH2Cl
2C6H4CH2• ===>
C6H5CH2C6H5
Further propagation steps to form the 2nd and 3rd
substitution products.
C6H5CH2Cl +
Cl• ===> C6H5CHCl• +
HCl
C6H5CHCl• + Cl2
===> C6H5CHCl2 +
Cl• (2nd product, dichloro...)
C6H5CHCl2
+ Cl• ===> C6H5CCl2•
+ HCl
C6H5CCl2•
+ Cl2 ===> C6H5CCl3
+ Cl• (3rd product,
trichloro...)
As far as I know its the same set of reactions for
bromine.
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notes on free radical mechanism of chlorination of the methyl group in
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