5.3 (a) (i) Reactions of benzene
Benzene structure
Benzene can be represented as C6H6 or 
.
Task 5.3(a)(i).1
Draw fully displayed structural formulae for
cyclohexane C6H12, cyclohexene C6H10,
and benzene. Label sigma and pi bonds between carbon atoms.
Story 5.3(a)(i)
A Dream about Benzene
Benzene is more thermodynamically stable than expected if it contained 3 C-C single bonds and 3 C-C double bonds. All C-C bonds are the same. Each pair of carbon atoms form a normal covalent bond. This leaves 1 electron left over for each C atom. These 6 electrons are delocalised across all C atoms.
Task
5.3(a)(i).2
Benzene with nitrating mixture
Benzene can be nitrated if it is boiled under reflux at 60oC
with concentrated nitric and sulphuric acids.
C6H6 (l) + HNO3 (l)--------------->
C6H5NO2
(l) + H2O (l)
benzene
nitrobenzene
Benzene with bromine, chloroalkanes and acid chlorides
Bromination: Conditions:
(dry) Anhydrous AlCl3
C6H6(l) + Br2(l)
---------------> C6H5Br(l) + HBr(g)
bromine
bromobenzene
Alkylation: Conditions: (dry) Anhydrous AlCl3
C6H6(l) + C2H5Cl(l)
------------------> C6H5C2H5(l) +
HCl(g)
chloroethane
ethylbenzene
Acylation: Conditions: (dry) Anhydrous AlCl3
C6H6(l) + CH3COCl (l)
----------------> C6H5COCH3(l) + HCl(g)
acid chloride
phenylethanone
Task 5.3(a)(i).3 Classify each reaction as an
addition or substitution reaction.
Task 5.3(a)(i).4 Draw fully displayed structural formulae for each of the
products.
Task 5.3(a)(i).5 Suggest safety precautions for each reactions.
Task 5.3(a)(i).6 Draw suitable apparatus for these
reactions.
5.3 (a) (ii) Reactions of aromatic compounds with
side chains
The side chain is oxidised if boiled under reflux with
alkaline potassium manganate(VII) KMnO4.
C6H5CH3 + 3O ---> C6H5COOH
+ H2O
methylbenzene benzoic acid
Task 5.3(a)(ii)
experimental method
5.3 (a) (iii) Reactions of phenol
Phenol with sodium hydroxide
Phenol dissolves in aqueous sodium hydroxide because phenol behaves
as an acid and gives up its proton to the hydroxide ion which is a base. A
soluble ionic product is formed.
C6H5OH(aq) + NaOH(aq) ----> C6H5O-Na+(aq)
+ H2O(l)
phenol
sodium phenoxide
Phenol with bromine
This is a substitution reaction in which an antiseptic smelling
compound is formed. Note 2,4,6-trichlorophenol is called TCP. OH is
electron releasing so activates the ring and speeds up the electrophilic substitution.
C6H5OH(aq) + 3Br2(aq)
----> C6H2Br3OH(aq) + 3HBr(aq)
2,4,6-tribromophenol
Phenol with ethanoyl chloride
This reaction forms an ester. Conditions: dry
C6H5OH + CH3COCl ---> CH3COOC6H5
+ HCl
phenol ethanoyl chloride phenylethanoate
Task 5.3(a)(iii).1 Draw
fully displayed structural formulae for the products above.
Task 5.3(a)(iii).2 Describe how KOH reacts with phenol.
Task 5.3(a)(iii).3 Describe how TCP can be prepared.
5.3 (a) (iv) Aromatic nitro compounds
Nitrobenzene is heated under reflux with tin in conc.
HCl as a reducing agent.
C6H5NO2 + 6H+ +
6e- -------> C6 H5NH2 + 2H2O
nitrobenzene
aminobenzene (phenylamine)
Task 5.3(a)(iv).1 Draw
fully displayed structural formulae for the organic reactant and product.
Task 5.3(a)(iv).2 Write a chemical equation for Tin and HCl and so
classify the organic reaction.
Task 5.3(a)(iv).3 Describe the purification of aminobenzene after
preparation as above.
5.3 (a) (v) Reactions of phenylamine
HNO2 is made in situ from HCl and NaNO2
at 5oC
C6H5NH2 + HNO2
+ HCl --------> 2H2O + C6H5N+=
N + Cl-
diazonium ion
Diazo-coupling reaction
in NaOH at 5oC
C6H5N+ =
N + C6H5OH ------------> C6
H5N
= NC6H4-OH + H+
a yellow azo dye
(4-hydroxyphenyl)azobenzene
C6H5N+ =
N + C6H5NH2 ----> C6
H5N
= NC6H4-NH2 + H+
an orange azo dye
(4-aminophenyl)azobenzene
Task 5.3(a)(v).1 Draw fully displayed
structural formulae for the azo dyes above.
Task 5.3(a)(v).2 Write an equation to show diazonium chloride reacting
with C6H5N(CH3)2
5.3 (b) (i) Homolytic, free radical substitution
(alkanes with chlorine)
Presentation of all mechanisms - (needs
Powerpoint)
Substitution reaction is defined as the replacement of
one atom (or group of atoms) in a molecule by another atom ( or group ).
E.g. Reaction between methane and chlorine to form hydrogen chloride &
chloromethane. The reaction is exothermic but energy in the form
of U.V. light must be supplied to initiate the reaction. Chlorine absorbs
the U.V light, the energy which is equal to approx. 400 kJ/mol. This is
greater than the bond strength of the Chlorine molecule which then splits
into Cl atoms. This is homolytic fission as the same number of electrons
from the bond are passed to each atom. Free radical substitution
has 3 steps. The half arrow or is the movement of one
electron.
step 1: Initiation Cl---Cl ----> 2Cl.
Each atom retains one electron from the covalent bond
between the atoms. An atom or group of atoms which posses an unpaired electron
is called a free radical.
step 2: Propagation
Each chlorine atom then reacts with a molecule of methane
by abstracting a hydrogen atom to form hydrogen chloride and a methyl radical.
H3C---H + Cl. -----> CH3.
+ HCl
The methyl radical reacts with a molecule of Chlorine
to form chloromethane and a Cl atom:
H3C. +
Cl---Cl ----> CH3Cl
+ Cl.
step 3: Termination
Radicals combine
H3C. +
Cl. ---> CH3Cl
In reactions 1 and 2 covalent bonds are broken so that
one electron of the pair in each bond becomes associated with each of the
atoms or groups. These are examples of homolysis or homolytic fission.
see an animation of homolytic, free radical substitution
Task5.3(a)(i) Draw out the mechanism for the reactions of bromine
with ethane, and chlorine with dichloromethane.
5.3 (b) (ii) Homolytic, free radical addition
(polymerisation of ethene)
At high pressure and temperature low density polythene is made by free
radical addition polymerisation. This reaction can proceed in the gas
phase, liquid phase or in solution.
Initiation
An agent such as benzoyl peroxide splits to form radicals.
(C6H5COO)2 ----> 2C6H5COO.
-----> 2C6H5.
+ 2CO2
Propagation
C6H5.
+ H2C=CH2 ---> C6H5CH2CH2.
C6H5CH2CH2.
+ H2C=CH2 ---> C6H5CH2CH2CH2CH2.
Termination
Termination results in chains several hundred monomer units long with
a lot of branches.
C6H5(CH2CH2)nCH2CH2.
+ C6H5(CH2CH2)mCH2CH2.
---> C6H5CH2CH2(CH2CH2)n+mCH2CH2H5C6
Task5.3(b)(ii).1 Redraw the reactions above
showing the movement of electrons in bonds.
Task5.3(b)(ii).2 Draw a mechanism for the polymerisation of propene.
5.3 (b) (iii) Heterolytic, electrophilic addition
(alkenes)
Heterolytic fission- In this type of fission the two
shared electrons in the bond are split unequally between the two atoms.
One of the atoms keeps both electrons. As a result ions are formed.
An electrophile is a species which attacks a carbon atom
by accepting an electron pair. It is thus a Lewis acid
Ethene reacts with bromine to form 1,2-dibromoethane:
This is an addition reaction
The cation C2H5 +, and other cations in which a carbon atom bears the positive charge, are given the general name carbonium ions. In the first step the two electrons which form the new bond are both provided by the alkene: thus, the reagent ( Br or Cl) is described as electrophilic ( i.e `electron-seeking').
CH2
CH2+
||
+ Brd+-Brd- ---------> |
+ Br -
CH2
CH2Br
CH2+ 
CH2Br
|
+ Br ----------> |
CH2Br CH2Br
means
the movement of a pair of electrons.
see and animation of
heterolytic, electrophilic addition
Ethene reacts with hydrogen chloride to give chloroethane:
This is also an addition reaction. The first step is
the formation of two ions, the ethyl cation and the chloride anion:
The two ions then rapidly combine to form the product.
In this reaction, the partially charged atom in hydrogen chloride is the
electrophilic reagent. Hd+ -Cld-
Since organic reactions involving heterolytic fission
of bonds produce ions, these reactions tend to take place in polar solvents.
Task 5.3(b)(iii) Describe the mechanism for the
reaction of HCl with ethene and Chlorine with propene.
5.3 (b) (iv) Heterolytic, electrophilic
substitution (benzene)
Electrophilic substitution is also possible on benzene
rings. In this type of substitution two of the delocalised [pi] electrons
on the benzene ring are donated to the electrophile. An unstable [pi] complex
containing both an electrophile and a leaving group is formed as an intermediate.
The nitration is carried out under reflux at 55-60oC using a
nitrating mixture. This contains equal amounts of concentrated nitric acid
and sulphuric acid. The two acids react to generate the nitryl cation NO2+
or nitronium ion.
HNO3 + H2SO4 -------> NO2+ + H2O + HSO4-
step 1
-----> 
+ H+ step 2
see animation of heterolytic, electrophilic
substitution
The reaction with bromine also follows this mechanism.
Aluminium chloride is used as a catalyst which helps form the electrophile.
Br-Br + AlCl3 ---> BrD+---Br---AlCl3D-
Alkylation follows the same mechanism. Halogenoalkanes are weak
electrophiles because their polar bond. e.g.CH3D+-BrD-
. The catalyst AlCl3 makes the halogenoalkane a better
electrophile.
CH3-Br + AlCl3 ---> D+CH3---Br---AlCl3D-
Acylation also follows the same mechanism. The electrophile is again
improved by AlCl3.
CH3COCl + AlCl3 ---> CH3C+=O
+ Cl-AlCl3-
Task
5.3(b)(iv).1 Write balanced chemical equations
for the mechanisms shown above.
Task 5.3(b)(iv).2
Describe a mechanism for the formation of ethylbenzene.
Task 5.3(b)(iv).3 Iron is sometimes
added to bromine to catalyse the bromination of
benzene. Explain
how this works.
5.3 (b) (v) Heterolytic, nucleophilic
substitution (SN1 and SN2)
A nucleophile is a species which attacks a carbon atom
with a partial positive charge by donating an electron pair. An example
of a nucleophilic substitution is provided by the hydrolysis of 2-methyl-2-bromopropane,
to form 2-methylpropan-2-ol. The hydroxide ion is a nucleophile. This is
known as nucleophilic substitution. The bromide ion is known as the leaving
group.
(CH3)3CBr + NaOH ------> (CH3)3COH
+ NaBr
Rate studies of this type show that;
Rate = k[R-Br] R=the alkyl group
This is an SN1 mechanism with the following
steps:
(CH3)3C-Br
------> (CH3)3C+ + Br-
slow
(CH3)3C+ + OH -
------>
(CH3)3COH
fast
The reaction is thus first order with respect to R-Br but zero order with respect to OH-. First order kinetics is good evidence that the rate determining step is unimolecular. The reaction is therefore given the symbol SN1. Since the nucleophile is not involved in the rate determining step, the mechanism must involve at least two steps.
The reaction between bromomethane and hydroxide ion is
another nucleophilic substitution reaction.
CH3Br + OH - -----> CH3OH
+ NaBr
The rate equation is Rate = k[CH3Br][OH-]
In this case the reaction is first order with respect
to both [CH3Br] and [OH-]. It is thus second
order overall. This is a bimolecular reaction. The reaction is thus given the symbol SN2.
It is thought to proceed in a single step involving a transition state.
HO - + CH3------Br --------> [
HO - - - CH3 - - - Br] -
[ HO - - - CH3 - - - Br] ----------->
HO-----CH3 + Br -
The cyanide ion CN- can also take part in nucleophilic
substitution
See an animation of hetrolytic nucleophilic substitution
5.3(b)(v).1
Describe an SN1 mechanism and give a rate equation for the reaction
of 2-methyl-2-iodopropane with aqueous potassium
hydroxide.
5.3(b)(v).1 Describe
an SN2 mechanism and give a rate equation for the reaction of bromoethane
with aqueous hydrogen cyanide.
5.3(b)(v).1 Describe
a mechanism for the reaction of 2-iodopropane with
aqueous sodium hydroxide if the rate equation is
rate = k[2-iodopropane].
5.3 (b) (vi) Heterolytic, nucleophilic addition
HCN made in situ from KCN and HCl. Alkali is added
to ensure a reasonable concentration of CN- by shifting the
following equilibrium to the left:
CN- + H2O <-----> HCN
+ OH-
CH3CHO + HCN ------> CH3CH2CH(OH)CN
ethanal
2-hydroxypropanonitrile (a cyanohydrin)
This is a nucleophilic addition reaction with a two step
mechanism.
CH3
CH3
|
|
N=C-
C=O ------> N=C-C-O-
|
|
H
H
CH3
CH3
|
|
N=C-C-O-
H+ ------>
N=C-C-O-H
|
|
H
H
Task5.3(b)(vi)
Describe mechanisms for the reaction of HCN with propanal
and propanone.