Topic 5.5: Organic chemistry IV (analysis, synthesis and application) Needs Units 2.2, 4.5 and 5.3
Organic analysis – Tests for presence of these
- Tests to distinguish between primary, secondary and tertiary
functional groups:
alcohols
- Tests for the halide group, by alkaline hydrolysis, then
acidification, then testing with silver nitrate(aq)
C C
C Cl
C Br
C know
I
- Must
reactions of
bromine solution,
phosphorus pentachloride,
H
H
2,4-dinitrophenylhydrazine solution,
Fehling’s solution
C OH
C OH
C OH
ammoniacal silver nitrate,
H
sodium or potassium hydrogencarbonate,
iodine in the presence of alkali (or potassium iodide and sodium
O
O
chlorate(I))
solution
C
C O
C
H
O
C
CH3
OH
H
C CH3
OH
functional
group
reagent conditions
result of positive test
-C=C-
bromine in inert solvent
orange bromine
decolourised
-Cl
warm with NaOH(aq)
add HNO3 then AgNO3
then NH3 (aq)
white ppt. of AgCl
soluble in dil. NH3 (aq)
-Br
warm with NaOH(aq)
add HNO3 then AgNO3
then NH3 (aq)
cream ppt. of AgBr
soluble in conc NH3 (aq)
-I
warm with NaOH(aq)
add HNO3 then AgNO3
then NH3 (aq)
yellow ppt. of AgI
Insoluble in conc NH3 (aq)
-OH
add solid PCl5
acrid steamy fumes of HCl
primary
-CH2-OH
warm with acidified
aqueous
potassium dichromate
K2Cr2O7
orange colour changes to
green product tests +ve for
-CHO
secondary
-CH-OH
|
warm with acidified
aqueous conc.
potassium dichromate
K2Cr2O7
orange colour changes to
green
product does not test +ve
for
-CHO
tertiary
|
-C-OH
|
warm with acidified
aqueous conc.
potassium dichromate
K2Cr2O7
no change
-C=O
|
add 2,4dinitrophenylhydrazine
warm with Fehling's
solution
yellow ppt. of hydrazone
no change in blue colour
-CHO
add 2,4dinitrophenylhydrazine
warm with Fehling's
solution
warm with AgNO3(aq) in
NH3(aq)
yellow ppt. of hydrazone
red/brown ppt of Cu2O
forms
silver mirror forms
-COOH
add NaHCO3
-C=O -CHOH
|
|
add iodine then aqueous
CH3 CH3 NaOH
ethanal
ethanol
effervescence CO2 formed
yellow ppt. and
antiseptic smell of
iodoform
ii
iii
Use physical/chemical data to find the structural formula of a compound
a
interpret simple fragmentation patterns from a mass spectrometer
b
interpret simple infra-red spectra
c
interpret simple low-resolution nuclear magnetic resonance spectra limited to proton magnetic resonance
d
interpret simple ultra-violet/visible spectra.
students will not be expected to describe the theory of or the apparatus connected with the production of uv – visible, infra-red or
nuclear magnetic resonance spectra
students will be given tables of data as appropriate.
students will not be expected to recall specific spectral patterns and/or wave numbers, but may be required to inspect given
spectra and tables of data to draw conclusions
Organic synthesis
i
propose practicable pathways for the synthesis of organic molecules
5.5a(iii)(a) interpret simple fragmentation patterns from a mass spectrometer
The large peak on the right is the parent molecular ion and this indicates the relative molecular mass of the compound.
Compound of relative molecular mass 46, each fragment labelled and the structural formula
1-bromopropane
chloride
Ethanoyl
5.5a (iii) (b) interpret simple infra-red spectra
The bonds in organic molecule absorb infra-red radiation.
This happens when the frequency of the radiation matches the natural frequency of vibrations in the bonds.
A spectrometer shines infra-red light at a sample of an organic material and measures how much of the light is absorbed.
A measure of the frequency (wavenumber) is displayed in the spectrum. Each bond has its own frequency (wavenumber) and this
can be used to identify the bonds present in a compound.
bond
wavenumber/cm-1
seen on spectrum
C-H
2840 – 3095
C-C
1610 – 1680
C=O
1680 – 1750
C-O
1000 – 1300
C-Cl
700 – 800
O-H
3233 - 3550
2500 – 3300
N-H
3100 – 3500
Ethanamide
Ethanoic acid
5.5a(iii)(c) Low resolution nuclear magnetic resonance spectra (NMR)
The chemical shift is the difference between the absorption frequencies of the hydrogen nuclei in the compound and those in the
reference compound
Nuclei are placed in a strong magnetic field and then absorb applied radio frequency radiation
The nuclei of hydrogen atoms in different chemical environments within a molecule will have different chemical shifts
The hydrogen nuclei in a CH3 group will have a different chemical shift from those in a CH 2 or in an OH group.
In low resolution NMR, each group will show as a single peak, and the area under the peak is proportional to the number of
hydrogen atoms in the same environment.
Thus ethanol, CH3CH2OH will have three peaks of relative intensities 3:2:1
Methyl propane CH3CH(CH3)CH3 will have two peaks with relative intensities of 9:1
In high resolution NMR spin coupling is observed. This is caused by the interference of the magnetic fields of neighbouring
hydrogen nuclei.
If an adiacent carbon atom has hydrogen atoms bonded to it, they will cause the peaks to split as follows:
1 neighbouring H atom
2 neighbouring H atoms
n neighbouring H atoms
peak splits into 2 lines (a doublet)
peak splits into 3 lines (a triplet)
peak splits into (n + 1) lines
Thus ethanol gives three peaks:
1 peak due to the OH hydrogen, which is a single line (as it is hydrogen bonded)
1 peak due to the CH2 hydrogens, which is split into four lines
by the three H atoms on the neighbouring CH3 group.
1 peak due to the CH3 hydrogens, which is split into three lines
by the two H atoms on the neighbouring CH2
group.
Type of
proton
Chemical
Shift (ppm)
R-CH3
0.9
R-CH2
1.3
R-CH2-O- 4.0
C6H5-
7.5
-O-H
5.0
-CHO
9.5
5.5a (iii) (d) The interpretation of simple ultra-violet/visible spectra.
Some chemical structures absorb electromagnetic radiation in the ultra violet part of the spectrum.
These include conjugated (contain alternate double and single bonds) dienes. E.g. 1,3-butadiene.
The ultraviolet absorption spectrum for 2,5-dimethyl-2,4-hexadiene is shown below.
Ultra-violet wavelengths are from about 200nm to about 400nm.
Visible light has wavelength between 400nm and 800nm. -carotene, which gives carrots their orange colour absorbs at 497nm.
Lycopene, which gives tomatoes their red colour, absorbs at 505nm.
Both of these compounds have 11 conjugated double bonds.
5.5b(i)Pathways for organic synthesis
Compound
Reagent
Alkane
Halogen
bromine
Conditions
UV light
ethane
Product
Haloalkane
bromoethane
Reaction type
Substitution
Alkene
Haloalkane
Br2
Halogen
bromine
Br2
C2H6
ethene
C2H4
Hydrogen halide
Hydrogen bromide
HBr
Alkaline(purple)
potassium manganate(VII)
KMnO4
H2SO4
prop-1-ene
CH3CHCH2
NaOH(aq) or KOH(aq)
bromoethane
C2H5Br
NaOH(ethanol) or KOH(ethanol)
Ethanolic solution
bromoethane
C2H5Br
Heat under reflux
Ethanolic solution
bromoethane
C2H5Br
bromoethane
C2H5Br
Potassium cyanide
KCN(ethanol)
Ammonia
Mg
Heat under reflux with NaOH
Acidify with dilute nitric acid
Add silver nitrate
Alcohol
Combustion
PCl5
Hydrogen halide
Hydrogen bromide
HBr
carboxylic acid
ethanoic acid
CH3COOH
Primary alcohol
Secondary
alcohol
Tertiary alcohol
acid chloride
ethanoyl chloride
CH3COCl
potassium dichromate VI(orange)
dilute sulphuric acid
ethene
C2H4
Dry ether(reflux)
(Ether must be
perfectly dry since
water destroys
resulting Grignard
reagent)
bromoethane
C2H5Br
Chlorides
Bromides
Iodides
dry
ethanol
C2H5OH
ethanol
C2H5OH
concentrated
H2SO4
ethanol
C2H5OH
ethanol
C2H5OH
Heat and distil off
product
ethanol
C2H5OH
concentrated
H2SO4
Heat under reflux
propan-2-ol
CH3CH(OH)CH3
CH3CH2Br
Decolourised from orange to
colourless
Dihaloalkane
1,2-dibromoethane
CH2BrCH2Br
Haloalkane
2-bromopropane
CH3CHBrCH3
Alcohol
ethane-1,2-diol
CH2OHCH2OH
Alcohol
Alcohol
ethanol
C2H5OH
Alkene
ethene
C2H4
Nitrile
propanonitrile
C2H5CN
amine
ethylamine
C2H5NH2
Grignard reagent
C2H5MgBr
ppt of silver halide
white ppt, soluble in dil NH3
cream ppt, souble in conc
NH3
yellow ppt, insoluble in conc
NH3
Carbon dioxide and water
Haloalkane, steamy fumes of
HCl
chloroethane
C2H5Cl + POCl3 + HCl
Haloalkane
bromoethane
C2H5Br
ester
ethyl ethanoate
CH3COOC2H5
(green) aldehyde that will
react with Tollens reagent to
give a silver mirror
ethanal
CH3CHO
(green) ketone will not react
with Tollens reagent
propanone
CH3COCH3
(orange) no reaction
Addition
Electrophilic
addition
Reduction
Electrophilic
addition
Nucleophilic
substitution
Elimination
Nucleophilic
substitution
Grignard reagent
RMgX
Water
Carbon dioxide
C2H5MgBr
Methanal
HCHO
Aldehydes R1CHO
ethanal
CH3CHO
Carboxylic acids
RCOOH
Ketones R1COR2
propan-2-one
(CH3)2CO
Alcohol R1OH
ethanol
C2H5OH
Lithium aluminium hydride
LiAlH4
Phosphorus pentachloride
PCl5
Sodium carbonate/hydrogen
carbonate
Na2CO3 and NaHCO3
Esters RCOOR1
concentrated H2SO4
Heat
concentrated
H2SO4
ethanoic acid
CH3COOH
Dry ether
ethanoic acid
CH3COOH
Dry
ethanoic acid
CH3COOH
ethanoic acid
CH3COOH
ethyl ethanoate
CH3COOC2H5
NaOH(aq)
Aldehydes
RCHO or ketones
RCOR1
Hydrogen
cyanide(HCN(covalent)) and
potassium cyanide
2, 4-dinitrophenylhydrazine
Test for carbonyl(C=O) group
Sodium borohydride NaBH4 or
lithium aluminium hydride
LiAlH4
ethanal
CH3CHO
or
propanone
(CH3)2CO
Dilute sulphuric
acid
ethanal
CH3CHO
or
propanone
(CH3)2CO
Aldehydes
RCHO (not
ketones)
Test for CHO
group
Aldehydes
Ammonical silver nitrate solution
(Tollens reagent)
Fehling’s solution/Benedicts
solution(Blue)
potassium
dichromate(VI)(orange)
acidic conditions
Warm in water bath
ethanal
CH3CHO
Alkane RH
Carboxylic acid RCOOH
propanoic acid
C2H5COOH
Primary alcohol RCH2OH
propan-1-ol
C2H5CH2OH
Secondary alcohol
RCH(OH)R1
butan-2-ol
CH3CH2CH(OH)CH3
Tertiary alcohol RR1R2COH
2-methylpropan-2-ol
(CH3)3OH
Ester RCOOR1
ethyl ethanoate
CH3COOC2H5
Nucleophilic
substitution
Alcohol RCH2OH
ethanol
C2H5OH
Acid chloride RCOCl
ethanoyl chloride
CH3COCl
Sodium salt RCOO-Na+
CO2 gas(gives white ppt with
limewater)
sodium ethanoate
CH3COONa
Alcohol R1OH and acid
RCOOH
ethanol, ethanoic acid
C2H5OH, CH3COOH
Alcohol R1OH and salt
RCOO-Na+
ethanol, sodium ethanoate
C2H5OH, CH3COONa
Cyanohydrin RCH(OH)CN
or RR1C(OH)CN
CH3CH(OH)(CN)
or
CH3C(OH)(CH3)CN
2, 4-dinitrophenylhydrazine
(Orange ppt)
Reduction
Primary alcohol RCH2OH or
secondary alcohol
RCH(OH)R1
Primary alcohol, ethanol
C2H5OH
or
Secondary alcohol propan-2ol
CH3CH(OH)CH3
Silver mirror
Carboxylic acid
ethanoic acid
CH3COOH
Copper(I) oxide ppt (Red)
Nucleophilic
substitution
followed by
elimination
Nucleophilic
substitution
Acid-base
Hydrolysis (equil)
Hydrolysis (equil)
Nucleophilic
substitution
Nucleophilic
substitution
followed by
elimination
Reduction
Reduction of the
silver ion
Reduction of the
copper(II) ion
(green)
Carboxylic acid RCOOH
Oxidation
salt RCOO-X
RCHO
alkaline conditions
Carbonyl
compounds
containing
CH3C=O and
alcohols
containing
CH3CH(OH)
Acid chlorides
ROCl
NaOH + I2
Heat
RCOONa + CHI3
(iodoform/yellow ppt)
Haloform
Water
ethanoyl chloride
CH3COCl
Carboxylic acid
ethanoic acid
CH3COOH
Amide RCONH2
ethanamide
CH3CONH2
Ester RCOOR1
ethyl ethanoate
CH3COOC2H5
N- substituted amide
R1CONHR
CH3CONHC6H5
salt RNH3+ClC2H5NH3+ClN-substituted amide
R1CONHR
CH3CONHC2H5
Nitrile RCN
ethanonitrile
CH3CN
Amine RNH2
methylamine
CH3NH2
Nucleophilic
substitution
Ammonia
NH3
Amines RNH2
Amides RCONH2
Alcohol R1OH
ethanol
CH3CH2OH
Amine R1NH2
phenylamine
C6H5NH2
Aqueous acid
HCl(aq)
Acid chloride R1OCl
ethanoyl chloride
CH3COCl
Phosphorus(V) oxide P4O10
ethylamine
C2H5NH2
ethanamide
CH3CONH2
Bromine followed by NaOH(aq)
Nitriles RCN
HCl(aq)
heat under reflux
ethanonitrile
CH3CN
NaOH(aq)
lithium aluminium hydride
LiAlH4
Dry ether
ethanonitrile
CH3CN
Carboxylic acid
ethanoic acid
CH3COOH
Salt RCOO-Na+
sodium ethanoate
CH3COONa
Amine RCH2NH2
ethylamine
CH3CH2NH2
Salt RCH(NH3+)COOH
Amino acids
RCH(NH3+)COO-
Aqueous acid eg HCl(aq)
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
C6H5NO2 + 6H+ + 6e- C6 H5NH2
+ 2H2O
nitrobenzene
aminobenzene
(phenylamine)
Sodium hydroxide
Sodium phenoxide
Acid-base
Nucleophilic
substitution
Dehydration
Substitution
followed by
rearrangement
and elimination
Hydrolysis
Reduction
Acid-base
Reaction type
C6H5O-Na+(aq) + H2O(l)
C6H2Br3OH(aq) +
substitution
3HBr(aq)
2,4,6-tribromophenol
(TCP)
Phenol
ethanoyl chloride
Dry
ester
C6H5OH
CH3COC
CH3COOC6H5 + HCl
phenylethanoate
Phenylamine
nitrous acid
5oC
C6H5NH2 + HNO2 + HCl
C6H5NH2
HNO2
NaNO2 and dil
 2H2O + C6H5N2+ClHCl situ
Diazonium ion
Diazonium ion
phenol
5oC
Yellow azo dye
C6H5N2+ClC6H5OH
C6H5N2C6H5OH
ii
propose suitable apparatus, conditions and safety precautions for carrying out organic syntheses, given suitable
information
iii
Know practical techniques used in organic chemistry
mixing, heating under reflux,
fractional distillation,
filtration under reduced pressure (filter pump and Buchner funnel),
recrystallisation,
determination of Mt & Bt
heating with a variety of sources, with safety and the specific hazards of the reaction/chemicals
it will be assumed that students wear eye protection during all practical work
iv
demonstrate an understanding of the principles of fractional distillation in terms of the graphs of boiling point against
composition.
students will not be expected to recall experimental procedures for obtaining graphs of boiling point against composition
knowledge of systems that form azeotropes will not be expected
Organic compounds may be hazardous because of
- Flammability - Avoid naked flames. Use electrical heater, water bath.
- Toxicity – fume cupboards
C6H5OH
Phenol
C6H5OH
NaOH
Bromine
Br2
Separating a mixture of immiscible liquids (Separating a mixture of water and hexane)
Water and hexane are immiscible forming 2 separate layers and are separated using a separating funnel
Separating a solvent from solution Simple distillation
Separating a liquid from a mixture of miscible liquids
Fractional distillation Separates mixtures of miscible liquids with different Bt’s, using a fractionating column increasing efficiency of red
with inert material(glass beads) increasing surface area where vapour may condense.
- When mixture is boiled vapours of most volatile component(lowest Bt) rises into the vertical column where they condense to liquids.
- As they descend they are reheated to Bt by the hotter rising vapours of the next component.
- Boiling condensing process occurs repeatedly inside the column so there is a temperature gradient.
- Vapours of the more volatile components reach the top of the column and enter the condenser for collection
Boiling under reflux is necessary when either the reactant has a
low Bt or the reaction is slow at RT
- condenses vapours and returns reagents to flask, prevents loss of
reactants/products, prolonged heating for slow reactions
- For preparation of aldehyde/carboxylic acid from alcohol
(1)Reason for heating the mixture but then taking the flame away
(1)provide Ea, exothermic/prevent reaction getting out of control
Solid can be iden
Solid must be pur
Impurities lower t
Thermometer doe
Recrystallisation
Dissolve the solid
Filter the hot solu
paper.
Allow to cool.
Filter under reduc
Wash with a little
c
Applied organic chemistry
Know organic compounds use in pharmaceuticals, agricultural products and materials. Only need to know:
i
changes to the relative lipid/water solubility of pharmaceuticals by the introduction of non-polar side-chains or ionic
groups
ii
the use of organic compounds such as urea as sources of nitrogen in agriculture and their advantages as compared with
inorganic compounds containing nitrogen
iii
the use of esters, oils and fats(from the viewpoint of saturation) , to include flavourings, margarine, soaps and essential
oils,
Lipid/water solubility of pharrnaceutical .
Those which are ionic which can form hydrogen bonds with water, will tend to be retained in aqueous (non-fatty) tissue,
and excreted
Compounds with no ionic groups and non-polar side chains, will be retained in fatty tissue and stored in the body
Esters - Food flavourings, perfumes, glues, varnishes and spray paints.
Fats - Soap
Oils - Margarine
iv
properties and uses of addition polymers of ethene, propene, chloroethene, tetrafluoroethene and phenylethene, and of the
condensation polymers (polyesters and polyamides).
this should include consideration of the difficulties concerned with the disposal of polymers
no specific reactions will be the subject of recall questions. Students will be expected to give some examples of compounds and
reactions to illustrate their answers.
Polymers(Addition or condensation)
Addition polymer Monomers contain one or more C=C group
Ethene, polyethene
plastic bags, bottles
Propene, polypropene
ropes, sacks, carpets
Chloroethene, PVC
Raincoats, electrical insulator, packaging
Tetrafluroethene
non-stick coating on frying pans
Condensation polymer Both the monomers have 2 functional groups, one at each end.
Polyester
Conveyor belt, safety belt
Polyamide
Parachutes, brushes
Questions
propenal
(a) (i) State what is observed when propenal reacts with 2,4-dinitrophenylhydrazine
Yellow/orange precipitate
(b) Explain why propenal has three peaks in its low-resolution n.m.r. spectrum. Suggest the relative areas under these peaks.
 Hydrogen nuclei  is in 3 different environments
 Ratio 2:1:1
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
1. A chemist has synthesised a compound W believed to be
(a) State and explain what you would see if W is reacted with: (i) sodium carbonate solution
effervescence  COOH present /acidic/contains H+
(ii) bromine water.
containsC=C/unsaturated

 Decolourises  compound
 white ppt so is a phenol
(b) W shows both types of stereoisomerism. (i) How many stereoisomers of W are there? Briefly explain your answer
 Four
 (Two) cis/trans (or geometric), and (two) chiral/optical isomers/ enantiomers
OR
 Two cis-trans/geometric isomers  Two optical isomers/enantiomers
(ii) Explain why W shows optical isomerism
 Molecule has a chiral centre/chiral
carbon/carbon with four different groups
 having non-superimposable mirror
images
(c) Describe how you would show that W contains chlorine.
 NaOH (solution)
 acidify with /add excess HNO3
 add silver nitrate (solution)
 white precipitate
 soluble in dilute/aqueous ammonia
5. Consider the reaction scheme below, which shows how the compound methyl methacrylate, CH 2=C(CH3)COOCH3, is
prepared industrially from propanone
(a) (i) State the type of reaction which occurs in
Step 2.
Elimination/dehydration
(ii) Name the reagent in Step 2.
Concentrated sulphuric acid /
concentrated phosphoric acid / aluminium oxide
(iii) State the type of reaction which occurs in Step 3.
hydrolysis
(iv) State the type of reaction which occurs in Step 4.
eterification
(v) Give the organic reagent required for Step 4.
methanol
(b) (i) Give the mechanism for the reaction in Step 1 between the hydrogen cyanide and propanone.
OR
(ii) The reaction in (b)(i) is carried out at a carefully controlled pH. Given that hydrogen cyanide is a weak acid, suggest why this
reaction occurs more slowly at both high and low concentrations of hydrogen ions.
+
-
 High [H ] insufficient CN (available for nucleophilic attack)
+
+
 Low [H ] insufficient H / HCN for the second stage
+
 High [[H ] surpresses ionisation / shifts equilibrium to left and
+
low [H ] shifts equilibrium to right max
(c) Methyl methacrylate polymerises in a homolytic addition reaction to form the industrially important plastic, Perspex.
(i) Identify the type of species that initiates this polymerisation.
Free radical/peroxide
(ii) Draw a sufficient length of the Perspex polymer chain to make its structure clear.
(iii) Suggest why it is not possible to quote an exact value for the molar mass of Perspex, but only an average value.
The polymer chain lengths are different (due to different termination steps) / different size
molecules/ different numbers of monomer (units)
4. (a) (i) Describe the appearance of the organic product obtained when an aqueous solution of bromine is added to aqueous
phenol.
White ppt
(ii) Give the equation for the reaction in (a)(i).
(iii) Phenol reacts with ethanoyl chloride to form an ester. Complete the structural formula to show the ester produced in this
reaction.
(iv) Suggest, in terms of the bonding in ethanoyl chloride, why the reaction in (a)(iii) proceeds without the need for heat or a
catalyst.
 C (atom) is (very) δ+ because Cl
highly electronegative OR Cl electron withdrawing
 (so C atom) susceptible to
nucleophilic attack OR (so C atom) strongly electrophilic
(b) Phenylamine, C6H5NH2, is formed by the reduction of nitrobenzene, C6H5NH2 Give the reagents which are used
Sn and HCl acid OR Fe and HCl acid
(c) Phenylamine is used to prepare azo dyes. (i) State the reagents needed to convert phenylamine into benzenediazonium
chloride.
•
Sodium nitrite OR NaNO2 OR sodium nitrate(III)
• HCl acid OR dilute
sulphuric acid OR aqueous sulphuric acid
(ii) The reaction in (c)(i) is carried out at a temperature maintained between 0 °C and 5 °C. Explain why this is so.
(iii) Addition of benzenediazonium chloride solution to an alkaline solution of phenol gives a precipitate of the brightly coloured
dye, 4-hydroxyazobenzene. Give the structural formula of 4-hydroxyazobenzene.
 Below 0°C : reaction too slow
 Above 5°C :
product decomposes OR diazonium ion decomposes
(iv) Describe how recrystallisation is used to purify a sample of the solid dye formed in (c)(iii).
 Dissolve in minimum
volume of boiling/hot solvent NOT “small volume”
 Filter hot OR filter through
heated funnel
 Cool or leave to crystallise
 Filter (under suction)
 Wash solid with cold small
volume of solvent (and leave to dry)
(a) (i) State the catalyst that is needed for Step 1
chloride/AlCl /Al Cl / iron(III) chloride/FeCl
3
2
6
aluminium
3
(ii) Suggest a synthetic pathway that would enable you to make ethanoyl chloride from ethanol in two steps. You should give
reagents, conditions and the structure of the intermediate compound. Experimental details and balanced equations are not
required.
 First step
Potassium dichromate +sulphuric acid
OR acidified dichromate
+
OR H + Cr O
2
27
OR (potassium)
manganate(VII)/permanganate + acid/alkali/neutral
 heat / reflux
 Intermediate: CH COOH/CH CO H
 Second step
3
3
2
PCl / PCI / SOCl
5
3
2
(b) Give the reagents and conditions needed for, step 2 & 3
Step 2
 LiAlH
OR
 NaBH
OR
OR
 Na
 H
 ethanol
 Pt OR
Step 3
 KMnO4
 NaOH/
OR
 I2
4
 dry ether /
ethoxyethane (followed by hydrolysis)
4
 aqueous
ethanol/water
2
Ni+heat OR Ni + specified temperature
alkali
 Heat
warm
The IR spectra for compounds A and B are shown.
 NaOH

(i) Using Table 1, give evidence from the spectra which shows that compound A has been reduced, comment on both spectra
shows bond due to OH at 3230-3550cm– 1
 A, spectrum shows bond due to C=O at 1680-1700cm– 1
 B, spectrum
 A has no OH / no bond at 3230-3550 OR B has no C=O bond / no bond at
1680-1700
(ii) Compound B is chiral. The IR spectra of the two optical isomers of B are identical. Suggest why this is so.
 IR spectra due to bonds present
 Same
bonds/functional groups in both isomers
(d) Both compounds A and B will react with iodine in sodium hydroxide solution to give a yellow precipitate of triiodomethane
(iodoform).
(i) B is oxidised to A during the reaction. Suggest the identity of the oxidising agent.
-
Iodine/I /sodium iodate(I) / NaOI /NaIO/iodate(I)/ OI /IO
-
2
(ii) Give the equation for the reaction of A with iodine in sodium hydroxide.
–
–
–
C H COCH + 3I + 4OH C H COO + CHI + 3I + 3H O
6
5
3
2
6
5
3
2
OR
C H COCH + 3I + 4NaOH 
6
5
3
2
C H COONa + CHI + 3NaI + 3H O
6
5
3
2
(iii) Describe a chemical test to show that triiodomethane contains iodine.
 (Hydrolyse with) NaOH / alkali
 acidify / neutralise with HNO3/ excess HNO3
 add silver nitrate (solution)
 yellow ppt
4. (a) The following equation shows the reaction of propane with chlorine to produce 1-chloropropane
CH3CH2CH3 + Cl2 →
CH3CH2CH2Cl + HCl
(i) Name the mechanism of the above reaction
free radical substitution
(ii) State ONE essential condition.
UV
radiation/sunlight/white light/heat
(b) The boiling temperature of 1-chloropropane is 46 °C and that of 1-bromopropane is 71 °C.
Draw a boiling temperature/composition diagram for a mixture of these two substances. Use it to explain how fractional
distillation could be used to separate this mixture
Diagram labelled axes, lozenge and b.pt. values
lines from anywhere except 100%
Explanation
Vapour richer in more volatile/chloropropane
Pure chloropropane distilled off / bromopropane left as residue
At least 2 horizontal + 2 vertical tie
Condense and then reboil
(c) Describe how to distinguish between pure samples of 1-chloropropane and 1-bromopropane using chemical tests.
heat with NaOH
add excess HNO3 OR acidify with HNO3
add AgNO3
chloro gives white and bromo gives cream ppt
white/off white/ pale yellow ppt soluble in dil NH3, cream ppt, slightly/partially
soluble in dil NH3 , (or soluble in conc NH3)
(d) Suggest which technique, mass spectrometry or low resolution n.m.r., would be used to distinguish between 1-chloropropane
and 1-bromopropane.
MS shows
different m/e values for molecular ion
Because
molar masses different / or reason why different
Nmr give
same number/3 peaks with both
OR
Nmr shows
different chemical shifts
Due to
different halides
In MS
molecular ion peak often absent
5. (a) An acidified solution of potassium manganate(VII) contains MnO4– ions, and can oxidise bromide ions, Br–, to bromine.
It was found that 23.90 cm3 of 0.200 mol dm–3 potassium manganate(VII) solution was required to oxidise a solution containing
2.46 g of sodium bromide dissolved in dilute sulphuric acid.
Calculate the ratio of the number of moles of manganate(VII) ions reacting to the number of moles of bromide ions reacting.
Hence write the equation for the oxidation of bromide ions by manganate(VII) ions in acid solution.
Moles manganate = 0.0239 x
0.2 = 0.00478
Moles bromide = 2.46/103 =
0.0239
ratio MnO4− : Br− = 1:5
OR
ratio Br− : MnO4− = 5:1
MnO4- + 5Br- + 8H+ → Mn2+ +
4H2O + 2.5Br2
(b) Acidified potassium manganate(VII) solution can be safely stored in containers made of poly(ethene).
(i) Suggest a property of poly(ethene) which makes it suitable for the storage of this solution.
Not oxidised by manganate(VII)/ does not react with oxidising
agents
OR
Not hydrolysed by acid
(ii) Explain ONE environmental problem which may be caused by the disposal of a poly(ethene) container.
biodegradable therefore fills landfill sites
non-