Topic 5.5: Organic chemistry IV (analysis, synthesis and application) Needs Units 2.2, 4.5 and 5.3Organic analysis – Tests for presence of these functional groups:

- Tests to distinguish between primary, secondary and tertiary alcohols

- Tests for the halide group, by alkaline hydrolysis, then acidification, then testing with silver nitrate(aq)

- Must know reactions of bromine solution, phosphorus pentachloride, 2,4-dinitrophenylhydrazine solution, Fehling’s solution ammoniacal silver nitrate, sodium or potassium hydrogencarbonate, iodine in the presence of alkali (or potassium iodide and sodium

chlorate(I)) solution

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 AgBrsoluble in conc NH3 (aq) 

-I warm with NaOH(aq)add HNO3 then AgNO3 then NH3 (aq) 

yellow ppt. of AgIInsoluble 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 2,4-dinitrophenylhydrazine warm with Fehling's solution 

yellow ppt. of hydrazone no change in blue colour 

-CHO 

add 2,4-dinitrophenylhydrazine 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

C C C Cl C Br

C OHH

HC OHH

C OH

C O CO

HCO

OH

CO

CH3C CH3OH

H

C I

-COOH  add NaHCO3 effervescence CO2 formed 

-C=O   -CH-OH  |           | CH3      CH3  

ethanal ethanol

add iodine then aqueous NaOH

yellow ppt. and antiseptic smell of iodoform

ii Use physical/chemical data to find the structural formula of a compoundiii a interpret simple fragmentation patterns from a mass spectrometer

b interpret simple infra-red spectrac interpret simple low-resolution nuclear magnetic resonance spectra limited to proton magnetic resonanced 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 spectrastudents 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 synthesisi propose practicable pathways for the synthesis of organic molecules

5.5a(iii)(a) interpret simple fragmentation patterns from a mass spectrometerThe 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 Ethanoyl chloride

5.5a (iii) (b) interpret simple infra-red spectraThe 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 – 1680C=O 1680 – 1750

C-O 1000 – 1300C-Cl 700 – 800

O-H 3233 - 35502500 – 3300

N-H 3100 – 3500Ethanamide 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 radiationThe nuclei of hydrogen atoms in different chemical environments within a molecule will have different chemical shiftsThe hydrogen nuclei in a CH3 group will have a different chemical shift from those in a CH2 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 peak splits into 2 lines (a doublet)2 neighbouring H atoms peak splits into 3 lines (a triplet)n neighbouring H atoms 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 1.3R-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 synthesisCompound Reagent Conditions Product Reaction typeAlkane Halogen

bromine  UV lightethane 

Haloalkanebromoethane 

Substitution

Br2 C2H6 CH3CH2BrAlkene Halogen

bromine  Br2

ethene  C2H4

Decolourised from orange to colourlessDihaloalkane1,2-dibromoethane  CH2BrCH2Br

Addition

Hydrogen halideHydrogen bromideHBr

prop-1-ene  CH3CHCH2

Haloalkane2-bromopropane  CH3CHBrCH3

Electrophilic addition

Alkaline(purple)potassium manganate(VII)KMnO4

ethene  C2H4

Alcohol ethane-1,2-diol  CH2OHCH2OH

Reduction

H2SO4 Alcohol Electrophilic addition

Haloalkane NaOH(aq) or KOH(aq) bromoethane  C2H5Br

Alcoholethanol  C2H5OH

Nucleophilic substitution

NaOH(ethanol) or KOH(ethanol) Ethanolic solutionbromoethane  C2H5Br

Alkeneethene  C2H4

Elimination

Potassium cyanideKCN(ethanol)

Heat under refluxEthanolic solutionbromoethane  C2H5Br

Nitrilepropanonitrile  C2H5CN

Nucleophilic substitution

Ammonia bromoethane  C2H5Br

amine  ethylamine C2H5NH2  

Mg Dry ether(reflux)(Ether must be perfectly dry since water destroys resulting Grignard reagent)bromoethane  C2H5Br

Grignard reagentC2H5MgBr

Heat under reflux with NaOHAcidify with dilute nitric acidAdd silver nitrate

ChloridesBromidesIodides

ppt of silver halidewhite ppt, soluble in dil NH3

cream ppt, souble in conc NH3

yellow ppt, insoluble in conc NH3

Alcohol Combustion Carbon dioxide and waterPCl5 dry

ethanol  C2H5OH

Haloalkane, steamy fumes of HClchloroethane  C2H5Cl + POCl3 + HCl

Hydrogen halideHydrogen bromideHBr

ethanol  C2H5OH

Haloalkanebromoethane  C2H5Br

carboxylic acid  ethanoic acid  CH3COOH

concentrated H2SO4

ethanol  C2H5OH

esterethyl ethanoate  CH3COOC2H5

acid chloride  ethanoyl chloride  CH3COCl

ethanol  C2H5OH

Primary alcohol potassium dichromate VI(orange)dilute sulphuric acid

Heat and distil off productethanol  C2H5OH

(green) aldehyde that will react with Tollens reagent to give a silver mirrorethanal  CH3CHO

Secondary alcohol

concentrated H2SO4

Heat under refluxpropan-2-ol  CH3CH(OH)CH3

(green) ketone will not react with Tollens reagentpropanone  CH3COCH3

Tertiary alcohol (orange) no reaction

Grignard reagentRMgX

Water Alkane RH Nucleophilic substitutionCarbon dioxide C2H5MgBr Carboxylic acid RCOOH

propanoic acid  C2H5COOH

Methanal HCHO

Primary alcohol RCH2OHpropan-1-ol  C2H5CH2OH

Aldehydes R1CHOethanal  CH3CHO

Secondary alcohol RCH(OH)R1

butan-2-ol  CH3CH2CH(OH)CH3

Ketones R1COR2

propan-2-one  (CH3)2CO

Tertiary alcohol RR1R2COH2-methylpropan-2-ol  (CH3)3OH

Carboxylic acids RCOOH

Alcohol R1OHethanol  C2H5OH

Heat concentrated H2SO4

ethanoic acid  CH3COOH

Ester RCOOR1

ethyl ethanoate  CH3COOC2H5

Nucleophilic substitution followed by elimination

Lithium aluminium hydrideLiAlH4

Dry etherethanoic acid  CH3COOH

Alcohol RCH2OHethanol  C2H5OH

Reduction

Phosphorus pentachloridePCl5

Dryethanoic acid  CH3COOH

Acid chloride RCOClethanoyl chloride  CH3COCl

Nucleophilic substitution

Sodium carbonate/hydrogen carbonateNa2CO3 and NaHCO3

ethanoic acid  CH3COOH

Sodium salt RCOO-Na+

CO2 gas(gives white ppt with limewater)sodium ethanoateCH3COONa

Acid-base

Esters RCOOR1 concentrated H2SO4 ethyl ethanoate  CH3COOC2H5

Alcohol R1OH and acid RCOOHethanol, ethanoic acid  C2H5OH, CH3COOH

Hydrolysis (equil)

NaOH(aq) Alcohol R1OH and salt RCOO-Na+

ethanol, sodium ethanoateC2H5OH, CH3COONa

Hydrolysis (equil)

Aldehydes RCHO or ketones RCOR1

Hydrogen cyanide(HCN(covalent)) and potassium cyanide

ethanal  CH3CHOorpropanone  (CH3)2CO

Cyanohydrin RCH(OH)CN or RR1C(OH)CNCH3CH(OH)(CN)orCH3C(OH)(CH3)CN

Nucleophilic substitution

2, 4-dinitrophenylhydrazineTest for carbonyl(C=O) group

Dilute sulphuric acid

2, 4-dinitrophenylhydrazine(Orange ppt)

Nucleophilic substitution followed by elimination

Sodium borohydride NaBH4 or lithium aluminium hydride LiAlH4

ethanal  CH3CHO

orpropanone  (CH3)2CO

Primary alcohol RCH2OH or secondary alcohol RCH(OH)R1

Primary alcohol, ethanol  C2H5OHorSecondary alcohol propan-2-ol  CH3CH(OH)CH3

Reduction

Aldehydes RCHO (not ketones)Test for CHO group

Ammonical silver nitrate solution (Tollens reagent)

Warm in water bathethanal  CH3CHO

Silver mirrorCarboxylic acidethanoic acid  CH3COOH

Reduction of the silver ion

Fehling’s solution/Benedicts solution(Blue)

Copper(I) oxide ppt (Red) Reduction of the copper(II) ion

potassium dichromate(VI)(orange)

(green)

Aldehydes RCHO

acidic conditions Carboxylic acid RCOOH Oxidationalkaline conditions salt RCOO-X

Carbonyl compounds containing CH3C=O and alcohols containing CH3CH(OH)

NaOH + I2 Heat RCOONa + CHI3 (iodoform/yellow ppt)

Haloform

Acid chlorides ROCl

Water ethanoyl chloride  CH3COCl

Carboxylic acid ethanoic acid  CH3COOH

Nucleophilic substitution

AmmoniaNH3

Amide RCONH2

ethanamide  CH3CONH2

Alcohol R1OHethanol CH3CH2OH

Ester RCOOR1

ethyl ethanoate  CH3COOC2H5

Amine R1NH2

phenylamine  C6H5NH2

N- substituted amide R1CONHRCH3CONHC6H5

Amines RNH2 Aqueous acid HCl(aq)

ethylamine  C2H5NH2

salt  RNH3+Cl-

C2H5NH3+Cl-

Acid-base

Acid chloride R1OClethanoyl chlorideCH3COCl

N-substituted amide R1CONHRCH3CONHC2H5

Nucleophilic substitution

Amides RCONH2 Phosphorus(V) oxide P4O10 ethanamide  CH3CONH2

Nitrile RCNethanonitrile  CH3CN

Dehydration

Bromine followed by NaOH(aq) Amine RNH2

methylamine  CH3NH2

Substitution followed by rearrangement and elimination

Nitriles RCN HCl(aq) heat under refluxethanonitrile  CH3CN

Carboxylic acid ethanoic acid  CH3COOH

Hydrolysis

NaOH(aq) Salt RCOO-Na+

sodium ethanoate  CH3COONa

lithium aluminium hydride LiAlH4

Dry ether ethanonitrile  CH3CN

Amine RCH2NH2

ethylamine  CH3CH2NH2

Reduction

Amino acids RCH(NH3

+)COO-Aqueous acid eg HCl(aq) Salt RCH(NH3

+)COOH Acid-base

Compound Reagent Conditions Product Reaction typearene  benzene  C6H6

Nitrating mixturenitric acid  HNO3

sulphuric acid H2SO4

heat under reflux  below 60oC 

nitrobenzeneC6H5NO2 + H2O

arene  benzene  C6H6

BromineBr2

Catalyst (dry) Anhydrous AlCl3

halogenoarenebromobenzeneC6H5Br(l)   + HBr(g)

arene  benzene  C6H6

ChloroalkaneChloroethane C2H5Cl

Catalyst (dry) Anhydrous AlCl3

ethylbenzeneC6H5C2H5(l) + HCl(g)

arene  benzene  C6H6

Acid chloride Ethanoyl chloride CH3COCl

Catalyst (dry) Anhydrous AlCl3

KetonephenylethanoneC6H5COCH3(l) + HCl(g)

arene methylbenzene  C6H5CH3

Potassium manganate VII KMnO4

alkaline conditions 

heat under reflux

Carboxylic acidbenzoic acid  C6H5COOH + H2O

Aromatic nitro compounds

C6H5NO2 + 6H+ + 6e- C6 H5NH2 + 2H2O nitrobenzene                         aminobenzene

(phenylamine)

heated under reflux with tin in conc. HCl as reducing agent

Amines

PhenolC6H5OH

Sodium hydroxideNaOH

Sodium phenoxideC6H5O-Na+(aq) + H2O(l)

PhenolC6H5OH

BromineBr2

C6H2Br3OH(aq) + 3HBr(aq)2,4,6-tribromophenol (TCP)

substitution

PhenolC6H5OH

ethanoyl chloride  CH3COC

Dry esterCH3COOC6H5 + HClphenylethanoate

PhenylamineC6H5NH2

nitrous acid  HNO2

5oCNaNO2 and dil HCl situ

C6H5NH2 + HNO2 + HCl 2H2O  +  C6H5N2

+Cl-

Diazonium ionDiazonium ionC6H5N2

+Cl-phenol  C6H5OH

5oC Yellow azo dyeC6H5N2C6H5OH

ii propose suitable apparatus, conditions and safety precautions for carrying out organic syntheses, given suitable informationiii 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 & Btheating with a variety of sources, with safety and the specific hazards of the reaction/chemicalsit 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 compositionknowledge 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

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 redistillation process, packed 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 identified from its Mt and BtSolid must be purified by recrystallisation to have accurate Mt Impurities lower the melting point.Thermometer doesn’t come into contact with the glass

RecrystallisationDissolve the solid in a minimum of hot solvent.Filter the hot solution through a preheated funnel using fluted filter paper. Allow to cool. Filter under reduced pressure (Buchner funnel),Wash with a little cold solvent and allow to dry.

c Applied organic chemistryKnow 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 groupsii the use of organic compounds such as urea as sources of nitrogen in agriculture and their advantages as compared with inorganic compounds containing nitrogeniii 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 polymersno 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 groupEthene, polyethene plastic bags, bottlesPropene, polypropene ropes, sacks, carpetsChloroethene, 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 beltPolyamide Parachutes, brushes

Questions

propenal

(a) (i) State what is observed when propenal reacts with 2,4-dinitrophenylhydrazineYellow/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 2:1:14. 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) chlorideOR PCl3/ Phosphorus trichloride/phosphorus(III) chlorideOR 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

acid1. 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. Decolourises compound containsC=C/unsaturated 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 ammonia5. Consider the reaction scheme below, which shows how the compound methyl methacrylate, CH2=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 aluminium chloride/AlCl3/Al2Cl6 / iron(III) chloride/FeCl3

(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+ + Cr2O72- OR (potassium)

manganate(VII)/permanganate + acid/alkali/neutral heat / reflux Intermediate: CH3COOH/CH3CO2H Second step PCl5 / PCI3 / SOCl2

(b) Give the reagents and conditions needed for, step 2 & 3 Step 2 LiAlH4 dry ether / ethoxyethane (followed by hydrolysis)

OR NaBH4 aqueous ethanol/water

OR Na ethanol OR H2 Pt OR

Ni+heat OR Ni + specified temperatureStep 3 KMnO4 NaOH/

alkali HeatOR I2 NaOH

warm The IR spectra for compounds A and B are shown.

(i) Using Table 1, give evidence from the spectra which shows that compound A has been reduced, comment on both spectra

A, spectrum shows bond due to C=O at 1680-1700cm– 1 B, spectrum shows bond due to OH at 3230-3550cm– 1

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/I2/sodium iodate(I) / NaOI /NaIO/iodate(I)/ OI- /IO-

(ii) Give the equation for the reaction of A with iodine in sodium hydroxide.

C6H5COCH3 + 3I2 + 4OH – C6H5COO – + CHI3 + 3I – + 3H2O OR C6H5COCH3 + 3I2 + 4NaOH C6H5COONa + CHI3 + 3NaI + 3H2O(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

Diagramlabelled axes, lozenge and b.pt. values At least 2 horizontal + 2 vertical tie lines from anywhere except 100%

Explanation Vapour richer in more volatile/chloropropane Condense and then reboil Pure chloropropane distilled off / bromopropane left as residue

(c) Describe how to distinguish between pure samples of 1-chloropropane and 1-bromopropane using chemical tests.heat with NaOHadd excess HNO3 OR acidify with HNO3

add AgNO3

chloro gives white and bromo gives cream pptwhite/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. non-biodegradable therefore fills landfill sites