Unit 5.3 Organic chemistry III (reaction mechanisms and aromatic compounds)
Needs Unit 2.2/4.5 nomenclature, isomerism, bond polarity, bond enthalpy, reagents and reaction conditions
Structure of benzene
Explain using the different types of covalent bonds, and bond enthalpy, the structure and stability of the benzene ring
Show benzene as the electrophilic substitution in aromatic systems
Recall, reagents and reaction conditions, reaction of:
benzene with a nitrating mixture; bromine and chloroalkanes and acid chlorides in the presence of anhydrous aluminium chloride
aromatic compounds with carbon-containing side chains with alkaline potassium manganate(VII) solution, resulting in the
oxidation of the side-chains
phenol with sodium hydroxide, bromine and ethanoyl chloride
aromatic nitro-compounds with tin and conc HCl acid reducing them to amines
phenylamine with nitrous acid; and the subsequent reaction of benzenediazonium ions with phenol
to represent movement of an electron pair or single electron,
recall the following reaction mechanisms with reagents and conditions, and an overall equation
homolytic, free radical substitution (alkanes with chlorine)
homolytic, free radical addition (polymerisation of ethene)
iii
heterolytic, electrophilic addition (symmetrical and unsymmetrical alkenes with halogens and hydrogen halides)
explanations of the orientation of addition should be in the context of the relative stability of the intermediate
carbocation.
Markovnikov’s Rule will not be examined
iv
heterolytic, electrophilic substitution showing generation of the electrophile, eg NO 2+ (benzene with nitrating mixture,
bromine, chloroalkanes, acid chlorides)
the orientation of substitution in benzene derivatives will not be examined
v
heterolytic, nucleophilic substitution (halogenoalkanes with hydroxide ions and cyanide ions) S N1 and SN2.
students are not expected to carry out reactions involving cyanide
vi
heterolytic, nucleophilic addition (carbonyl compounds with hydrogen cyanide).
Benzene C6H6
3 C-C single bonds and 3 C-C double bonds. All C-C bonds are the same
1 electron left over for each C atom. These 6 electrons are delocalised across all C atoms
Compound
arene
benzene C6H6
Conditions
heat under reflux
below 60oC
Product
nitrobenzene
C6H5NO2 + H2O
arene
benzene C6H6
Reagent
Nitrating mixture
nitric acid HNO3
sulphuric acid H2SO4
Bromine
Br2
Catalyst (dry)
Anhydrous AlCl3
arene
benzene C6H6
arene
benzene C6H6
Chloroalkane
Chloroethane
C2H5Cl
Acid chloride
Ethanoyl chloride CH3COCl
Catalyst (dry)
Anhydrous AlCl3
Catalyst (dry)
Anhydrous AlCl3
arene
methylbenzene
C6H5CH3
Aromatic nitro
compounds
Potassium manganate VII
KMnO4
alkaline
conditions
heat under reflux
heated under reflux
with tin in conc.
HCl as reducing
agent
halogenoarene
bromobenzene
C6H5Br(l) + HBr(g)
ethylbenzene
C6H5C2H5(l) + HCl(g)
Ketone
phenylethanone
C6H5COCH3(l) + HCl(g)
Carboxylic acid
benzoic acid
C6H5COOH + H2O
Amines
Phenol
C6H5OH
Phenol
C6H5OH
C6H5NO2 + 6H+ + 6e- C6 H5NH2
+ 2H2O
nitrobenzene
aminobenzene
(phenylamine)
Sodium hydroxide
NaOH
Bromine
Br2
Phenol
C6H5OH
ethanoyl chloride
CH3COC
Dry
Phenylamine
nitrous acid
5oC
Sodium phenoxide
C6H5O-Na+(aq) + H2O(l)
C6H2Br3OH(aq) +
3HBr(aq)
2,4,6-tribromophenol
(TCP)
ester
CH3COOC6H5 + HCl
phenylethanoate
C6H5NH2 + HNO2 + HCl
Reaction type
substitution
C6H5NH2
HNO2
Diazonium ion
C6H5N2+Cl-
phenol
C6H5OH
NaNO2 and dil
HCl situ
5oC
 2H2O + C6H5N2+ClDiazonium ion
Yellow azo dye
C6H5N2C6H5OH
Homolytic, free radical substitution (alkanes with chlorine)
The reaction is exothermic but energy in the form of UV light must be supplied to initiate the reaction.
Chlorine absorbs the UV light, energy supplied is greater than the bond strength of the Chlorine molecule which then splits into Cl
atoms.
Homolytic fission – 2 electrons in the bond are given one by one to the two species, making 2 free radicals
Free radical - An atom or group of atoms which posses an unpaired electron
Free radical substitution has 3 steps
step 1: Initiation
covalent bond between the atoms
Cl---Cl ----> 2Cl
step 2: Propagation
methane giving a methyl radical.
H3C---H +
.
Each atom retains one electron from the
.
.
Cl -----> CH3 + HCl
.
H3C + Cl---Cl ----> CH3Cl + Cl
molecule of Chlorine to form chloromethane and a Cl atom
.
step 3: Termination
Each chlorine atom reacts with a molecule of
.
The methyl radical reacts with a
.
H3 C +
Cl ---> CH3Cl
Radicals combine
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
.
Propagation C6H5 + H2C=CH2 ---> C6H5CH2CH2
.
.
(C6H5COO)2 ----> 2C6H5COO -----> 2C6H5 + 2CO2
.
.
.
C6H5CH2CH2 + H2C=CH2 ---> C6H5CH2CH2CH2CH2
Termination Termination results in chains hundreds of monomer units long
.
.
C6H5(CH2CH2)nCH2CH2 + C6H5(CH2CH2)mCH2CH2 ---> C6H5CH2CH2(CH2CH2)n+mCH2CH2H5C6
Heterolytic, electrophilic addition (alkenes)
Heterolytic fission – 2 shared electrons in the bond are split unequally between the two atoms. One of the atoms keeps both
electrons, giving ions.
Electrophile – ‘electron seeking’ A species which attacks a carbon atom by accepting an electron pair. It is thus a Lewis acid
Carbonium ions – Where the carbon atom bears the positive charge eg C2H5 +
Addition reaction
Alkene provides 2 electrons for the new bond
movement of a pair of electrons
CH2+
CH2
||
d+
d-
+ Br -Br --------->
CH2
Ethene Bromine
|
CH2Br
Br/Cl is electrophilic
CH2+
+ Br
-
|
CH2Br
CH2Br
-
+ Br ---------> |
CH2Br
1,2-dibromoethane
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 which combine
to form the product.
Heterolytic, electrophilic substitution (benzene)
Electrophile takes 2 of the delocalised [pi] electrons on the benzene ring
An unstable [pi] complex containing both an electrophile and a leaving group is formed as an intermediate.
Nitration carried out under reflux at 55-60oC using a nitrating mixture (equal amounts of conc. nitric & sulphuric acid which react
to generate the nitryl cation NO2+)
HNO3 + H2SO4 -------> NO2+ + H2O + HSO4-
step 1
----->
+
H+ step 2
Aluminium chloride is used as a catalyst which helps form the electrophile
Br-Br + AlCl3 ---> Br+---Br---AlCl3CH3COCl + AlCl3 ---> CH3C+=O + Cl-AlCl3-
CH3-Br + AlCl3 ---> +CH3---Br---AlCl3-
Heterolytic, nucleophilic substitution (SN1 and SN2)
Nucleophile - Attacks a carbon atom with a partial positive charge by donating an electron pair.
SN1 mechanism:
(CH3)3C-Br
------> (CH3)3C+ + Br-
(CH3)3C+ + OH - ----
--> (CH3)3COH
2-methyl-2-bromopropane
nucleophile
2-
methylpropan-2-ol
Rate = k[R-Br] R=the alkyl group
(CH3)3CBr + NaOH ------> (CH3)3COH + NaBr
Unimolecular reaction. Since the nucleophile is not involved in the rate determining step, the mechanism must involve at least
two steps.
SN2 mechanism: HO - + CH3------Br --------> [ HO - - - CH3 - - - Br] -> HO-----CH3 + Br -
[ HO - - - CH3 - - - Br] ----------
Rate = k[CH3Br][OH-]
CH3Br + OH - -----> CH3OH + NaBr
Bimolecular reaction. It is thought to proceed in a single step involving a transition state. The cyanide ion CN - can also take part
in nucleophilic substitution
Heterolytic, nucleophilic addition
This is a nucleophilic addition reaction with a two step mechanism
CH3CHO + HCN ------> CH3CH2CH(OH)CN
ethanal
2-hydroxypropanonitrile (a
cyanohydrin)
CH3
CH3
|
|
N=C- C=O ------> N=C-C-O|
|
H
H
CH3
|
N=C-C-O- H+
|
H
CH3
|
------> N=C-C-O-H
|
H
2. (a) Equations for the hydrogenation of three compounds are given below, together with the corresponding enthalpy changes.
Explain, in terms of the bonding in benzene, why the enthalpy change of hydrogenation of benzene is not –360 kJ mol-1
 Delocalisation / π-system
 due to overlap of six p-orbitals OR
due to overlap of p-orbitals around the ring
 Confers stability/ benzene at a lower energy level / more energy needed to break bonds compared with having three separate π /
double bonds / cyclohexatriene, Kekule structure
(b) Benzene can be converted into phenylamine, C6H5NH2 in two stages. Give the reagents needed for each step and identify the
intermediate compound formed
st
 1 step: sulphuric and nitric
acid
 concentrated
 Intermediate: Nitrobenzene /
C6H5NO2
nd
 2 Step: Tin / iron and conc
HCl (followed by addition of alkali)
(c) Benzene, C6H6, reacts with bromoethane, CH3CH2Br, in the presence of a catalyst, to form ethylbenzene, C6H5CH2CH3, and
HBr
(i) Give the formula of a catalyst for this reaction.
AlBr /FeBr / AlCl / Al Cl / FeCl / Fe Cl
3
3
3
2
6
3
2
6
(ii) Give the mechanism for the reaction between benzene and bromoethane, including the formation of the species that reacts with
the benzene molecule.
+
−
AlBr3 + CH3CH2Br CH3CH2 + AlBr4
(iii) Name the type of mechanism involved in this reaction
Electrophilic substitution
(d) A mixture of ethylbenzene (boiling point 136°C) and
benzene (boiling point 80°C) can be separated by fractional
distillation.
A labelled boiling point/composition diagram for this mixture is
shown below.
 At least two horizontal and two vertical tie-lines drawn
from 60% ethylbenzene
 Vapour condensed and then reboiled
 Vapour (from 60% ethylbenzene liquid) gets richer in the
more volatile component (benzene) / residue gets richer in
ethylbenzene
 Pure benzene distilled off / ethylbenzene left as residue
Use the diagram to explain what happens when a mixture
containing 60% ethylbenzene and 40 % benzene is fractionally
distilled
1. (a) Pent-1-ene, CH3CH2CH2CH=CH2, polymerises in a similar manner to ethene.
(i) Draw enough of the chain of poly(pent-1-ene) to make the structure of the polymer clear.
(ii) Give the mechanism for the polymerisation of pent-1-ene, using a peroxide initiator RO–OR that produces RO. radicals. Show
only the initiation and two propagation steps. Include the use of an appropriate type of arrow to show the movement of an
electron.
Initiation
Propagation step 1
Propagation step 2
(b) Pent-1-ene reacts with hydrogen bromide to give 2-bromopentane as the major product. (i) Give the mechanism for this
reaction.
(ii) By considering the nature of the intermediates in this reaction, explain why the major product is 2-bromopentane rather than 1bromopentane
 Secondary cation is more stable than primary CONDITIONAL on reference to cations
 Structures of the 2 intermediate carbocations / intermediate cation giving 2-bromopentane is
secondary and primary for 1-bromopentane
(c) Molecules of 2-bromopentane are chiral. If a single isomer of 2-bromopentane is reacted with hydroxide ions, the SN1
reaction that results gives pentan-2-ol, but the product mixture shows no optical activity. (i) How would you test for optical
activity?
 Sample in polarimeter / use of crossed
polaroids / pass polarised light through sample
 Rotates the plane of (polarisation of plane)polarised (monochromatic) light
(ii) Explain, in terms of the reaction mechanism, why the product mixture does not show optical activity.

intermediate (carbocation) planar

equal (probability of) attack from either side

(leads to) racemic/ 50:50 / equimolar mixture
5. (a) Give the structural formula of the organic product when phenol is reacted with:
(i) sodium hydroxide solution
–
+
C6H5O Na / C6H5ONa/ C6H5O
–
(ii) aqueous bromine
OR
(iii) ethanoyl chloride.
(b) An azo dye can be made from benzenediazonium chloride.
(i) State the reagents and conditions needed to make benzenediazonium chloride from phenylamine.
NaNO / sodium nitrite / nitrate(III)
2
conc aq / dil HCl / hydrochloric acid, NOT HCI
o
o
Any temperature between 0 – 10 C OR range between 0-10 C
(ii) Write an equation, using structural formulae, to show the reaction between benzenediazonium ions and phenol to give the azo
dye.
(iii) What condition is required for the reaction in (ii) above?
Alkaline/alkali/sodium hydroxide/ NaOH /KOH/potassium hydroxide/
sodium carbonate/sodium hydrogencarbonate
3. This question concerns the following reaction scheme starting from benzene, C 6H6
(a) Explain why the low resolution n.m.r. spectrum of benzene contains only a single peak
nuclei/atoms/protons in same (chemical) environment
(b) (i) Identify the reagent and the catalyst needed for reaction 1.
ethanoyl chloride
aluminium chloride / AlCl3/Al2Cl6
(ii) What type of reaction is reaction 1?
substitution
(iii) Draw the mechanism for reaction 1.
OR
(c) (i) Identify the reagents needed for reaction 2.
HCN + KCN
OR
KCN + Acid
OR
HCN + Base/alkali
OR
HCN/KCN pH 5 - 9
(ii) What type of reaction is reaction 2?
Nucleophilic addition
(iii) Draw the mechanism for reaction 2.
OR
All hydrogen
Reagent =
Catalyst (anhydrous)
Electrophilic
(d) The product of reaction 2 is a mixture of two isomers.
dimensional shape clear.
(i) Draw the structures of these two isomers, making their three-
(ii) What would be the effect of this mixture on monochromatic plane-polarised light? Give reasons for answer with reference to
your mechanism in (c)(iii).
 (No effect) as ketone planar
 Attack possible from top or bottom
 Producing racemic/50:50 mixture (of
enantiomers) / rotations cancel out
(e) How could infra-red spectroscopy be used to show that the product of reaction 2 did not contain traces of the reactant
phenylethanone?
 No absorption corresponding to C=O / carbonyl
OR
No absorption around 1700 cm-1
propenal
(a)Give the structural formula of the compound formed in the reaction when propenal reacts with 2,4-dinitrophenylhydrazine
(c) Propenal reacts with hydrogen cyanide as shown by the following equation
(i) Write the mechanism for the reaction.
CH2—CHCHO + HCN  CH2—CHCH(OH)CN
OR
(ii) Name the type of mechanism involved in this reaction.
Nucleophilic addition
(d) Propenal reacts with hydrogen bromide as shown by the following equation CH 2—CHCHO + HBr  CH3CHBrCHO
(i) Write the mechanism for the reaction.
(ii) Name the type of mechanism involved in this reaction.
Electrophilic addition
(e) C=O and C=C bonds have the same electronic structure but their reactions occur by different mechanisms. Explain why this is
so.
• C = O is a
polar bond OR O more electronegative than C
• C = C has
high electron density OR C = C is electron rich
• Cδ+ can be attacked by a nucleophile OR (C in) C = O can be attacked by
nucleophile OR C = C attacked by electrophile
4. Phenylethanoic acid occurs naturally in honey as its ethyl ester: it is the main cause of the honey’s smell. The acid has the
structure
Phenylethanoic acid can be synthesised from benzene as follows:
(a) State the reagent and catalyst needed for step 1.
Reagent: chloromethane/CH3Cl
Catalyst:
(anhydrous) aluminium chloride/AlCl3/Al2Cl6
(b) (i) What type of reaction is step 2?
Free radical substitution
(ii) Suggest a mechanism for step 2. The initiation step, the two propagation steps and a termination step. You may use Ph to
represent the phenyl group, C6H5.
- Cl2  2Cl•
- PhCH3 + Cl•  PhCH2• + HCl
-
PhCH2• + Cl2  PhCH2Cl + Cl•
2PhCH2•  PhCH2CH2Ph
OR
PhCH2• + Cl•  PhCH2Cl
OR
2Cl•  Cl2
(iii) Draw an apparatus which would enable you to carry out step 2, in which chlorine is bubbled through boiling methylbenzene,
safely. Do not show the uv light source.
- flask and vertical condenser – need not be shown as separate items [Ignore direction of water flow; penalise sealed
condenser]
- gas entry into liquid in flask [allow tube to go through the side of the flask, but tube must not be blocked by flask wall]
- heating from a electric heater/heating mantle/sandbath/water bath/oil bath
(c) (i) Give the structural formula of compound A.
(ii) Give the reagent and the conditions needed to convert compound A into phenylethanoic acid in step 4.
HCl (aq) OR dilute H2SO4(aq)
- Boil/heat (under reflux)/reflux
OR
- NaOH(aq) and boil
- Acidify
(iii) Suggest how you would convert phenylethanoic acid into its ethyl ester.
- ethanol and (conc) sulphuric acid
- heat/warm/boil/reflux conditional on presence of ethanol
OR
PCl5 /PCl3/SOCl2
Add ethanol PCl5 and ethanol (1) PCl5 in ethanol (0)
(d) (i) An isomer, X, of phenylethanoic acid has the molecular formula C8H8O2. This isomer has a mass spectrum with a large
peak at m/e 105 and a molecular ion peak at m/e 136. The ring in X is monosubstituted. Suggest the formula of the ion at m/e
105 and hence the formula of X.
X is
OR
(ii) Another isomer, Y, of phenylethanoic acid is boiled with alkaline potassium manganate(VII) solution and the mixture is then
acidified. The substance produced is benzene-1,4-dicarboxylic acid:
Suggest with a reason the structure of Y.
Side-chain(s) oxidised to COOH
(e) Benzene-1,4-dicarboxylic acid can be converted into its acid chloride, the structural formula of which is
This will react with ethane-1,2-diol to give the polyester known as PET.
(i) What reagent could be used to convert benzene-1,4-dicarboxylic acid into its acid chloride?
PCl5 /Phosphorus pentachloride/phosphorus(V) chloride
OR PCl3/ Phosphorus trichloride/phosphorus(III) chloride
OR SOCl2/Thionyl chloride/sulphur oxide dichloride
(ii) Give the structure of the repeating unit of PET.
(iii) Suggest, with a reason, a type of chemical substance which should not be stored in a bottle made of PET.
(concentrated) acid/alkali (ester link) would be hydrolysed OR polymer would react to form the monomers/alcohol and acid