Unit 5.5

Transcript

Topic 5.5: Organic chemistry IV (analysis, synthesis and application) Needs Units Units 2.2, 4.5 and 5.3  – Tests Tests for presence of these Tests to distinguish between primary, secondary and tertiary Organic analysis –  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 sol ution 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 H C CH3 OH 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 NH 3 (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 CO 2 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 for mula 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. spectrum. Each bond has its own frequency (wavenumber) (wavenumber) and this can be used to identify the bonds present in a compound. bond wavenumber/cm-1 seen on spectrum C-H 2840 –  2840 – 3095 3095 C-C 1610 –  1610 – 1680 1680 C=O 1680 –  1680 – 1750 1750 C-O 1000 –  1000 – 1300 1300 C-Cl 700 –  700 – 800 800 O-H 3233 - 3550 2500 –  2500 – 3300 3300 N-H 3100 –  3100 – 3500 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 CH 3 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, CH 3CH2OH will have three peaks of relative intensities 3:2:1 Methyl propane CH 3CH(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 CH 2 hydrogens, which is split into four lines by the three H atoms on the neighbouring CH3 group. 1 peak due to the CH 3 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. 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 8 00nm. -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) 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 NH 3 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 1 Aldehydes R CHO ethanal CH3CHO 1 Carboxylic acids RCOOH 2 Ketones R COR propan-2-one (CH3)2CO 1 Alcohol R OH ethanol C2H5OH Lithium aluminium hydride LiAlH4 Phosphorus pentachloride PCl5 Sodium carbonate/hydrogen carbonate Na2CO3 and NaHCO3 Esters RCOOR 1 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 1 RCOR Hydrogen cyanide(HCN(covalent)) and potassium cyanide 2, 4-dinitrophenylhydrazine Test for carbonyl(C=O) group Sodium borohydride NaBH 4 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 1 RCH(OH)R butan-2-ol CH3CH2CH(OH)CH3 1 2 Tertiary alcohol RR R COH 2-methylpropan-2-ol (CH3)3OH 1 Ester RCOOR 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 1 Alcohol R OH and acid RCOOH ethanol, ethanoic acid C2H5OH, CH3COOH 1 Alcohol R OH and salt + RCOO Na ethanol, sodium ethanoate C2H5OH, CH3COONa Cyanohydrin RCH(OH)CN 1 or RR C(OH)CN CH3CH(OH)(CN) or CH3C(OH)(CH3)CN 2, 4-dinitrophenylhydrazine (Orange ppt) Reduction Primary alcohol RCH2OH or secondary alcohol 1 RCH(OH)R 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 - RCHO alkaline 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 Amide RCONH2 ethanamide CH3CONH2 1 Ester RCOOR ethyl ethanoate CH3COOC2H5 N- substituted amide 1 R CONHR CH3CONHC6H5 + salt RNH3 Cl + C2H5NH3 Cl N-substituted amide 1 R CONHR CH3CONHC2H5 Nitrile RCN ethanonitrile CH3CN Amine RNH2 methylamine CH3NH2 Nucleophilic substitution Ammonia NH3 1 Amines RNH2 Amides RCONH 2 Alcohol R OH ethanol CH3CH2OH 1 Amine R NH2 phenylamine C6H5NH2 Aqueous acid HCl(aq) 1 Acid chloride R OCl ethanoyl chloride CH3COCl Phosphorus(V) oxide P 4O10 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 o below 60 C Product nitrobenzene C6H5NO2 + H2O arene benzene C6H6 Reagent Nitrating mixture nitric acid HNO3 sulphuric acid H2SO4 Bromine Br2 Catalyst (dry) Anhydrous AlCl 3 arene benzene C6H6 arene benzene C6H6 Chloroalkane Chloroethane C2H5Cl Acid chloride Ethanoyl chloride CH3COCl Catalyst (dry) Anhydrous AlCl 3 Catalyst (dry) Anhydrous AlCl 3 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 C6H5OH Phenol C6H5OH NaOH Bromine Br2 - + 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 o Phenylamine nitrous acid 5C C6H5NH2 + HNO2 + HCl + C6H5NH2 HNO2 NaNO2 and dil  2H2O + C6H5N2 Cl HCl situ Diazonium ion o Diazonium ion phenol 5C Yellow azo dye + C6H5N2 Cl C6H5OH 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 naked flames. Use electrical heater, heater, water bath. Toxicity –  Toxicity – fume 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 re with inert material(glass beads) increasing surface area where vapour may condense. - When mixture is boiled vapours of most volatile component(lowest Bt) r ises 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 components reach the top of the column and enter 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 returns reagents to flask, prevents loss of  reactants/products, prolonged heating for slow reactions - For preparation of aldehyde/carboxylic 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 ide Solid must be pu Impurities lower Thermometer do Recrystallisatio Dissolve the soli Filter the hot sol paper. Allow to cool. Filter under redu Wash with a littl 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 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 follo ws: (a) State the reagent and catalyst needed for step 1. Reagent: chloromethane/CH 3Cl Catalyst: (anhydrous) aluminium chloride/AlCl 3/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 PhCH 2• + Cl•  PhCH2Cl OR 2Cl• 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 –  condenser  – need need not be shown as separate ite ms [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 H 2SO4(aq) OR - NaOH(aq) and boil - Boil/heat (under reflux)/reflux 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 PCl5 /PCl3 /SOCl2 OR Add ethanol PCl5 and ethanol (1) PCl5 in ethanol (0) (d) (i) An isomer, X, of phenylethanoic acid has the molecular formula C 8H8O2. 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) 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 ( ester link) would be hydrolysed OR polymer would react to for m 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: 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  Two cis-trans/geometric isomers  Two optical isomers/enantiomers OR (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 co mpound 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 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 str ucture clear. (iii) Suggest why it is not possible to quote an exact value for the molar mass of Perspex, P erspex, but only an average value. The polymer chain lengths are different (due to different termination steps) / di fferent size molecules/ different numbers of monomer (units) 4. (a) (i) Describe the appearance of t he 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 th is reaction. (iv) Suggest, in terms of the bonding in ethanoyl chloride, why the reaction in (a)(iii) pr oceeds 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, C 6H5NH2 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 co loured 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 t he intermediate compound. Experimental details and balanced equations are not required.  First step + Potassium dichromate +sulphuric acid OR acidified dichromate 2- OR H + Cr O 2 OR (potassium) 7 manganate(VII)/permanganate + acid/alkali/neutral  heat / reflux  Intermediate: CH COOH/CH CO H 3  Second step 3 2 PCl  / PCI  / SOCl 5 3 2 (b) Give the reagents and conditions needed for, step 2 & 3 Step 2  LiAlH  dry ether /  OR  NaBH  aqueous OR OR  Na  H  ethanol  Pt OR Step 3  KMnO4  NaOH/  OR  I2 4 ethoxyethane (followed by hydrolysis) 4 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 co mpound A has been reduced, comment on both spectra  – 1  – 1  – 1  – 1  A, spectrum shows bond due to C=O at 1680-1700cm  B, spectrum shows bond due to OH at 3230-3550cm  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 6 5 3 2  –  C 6  –   –  H COO + CHI + 3I + 3H O 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 CH3CH2CH3 + Cl2 → 4. (a) The following equation shows the reaction of propane with chlorine to produce 1-chloropropane 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 NH 3, cream ppt, slightly/partially soluble in dil NH 3 , (or soluble in conc NH 3) (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 Nmr shows OR different chemical shifts Due to different halides In MS molecular ion peak often absent 5. (a) An acidified solution of potassium manganate(VII) contains MnO 4 – 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 + 2.5Br 2 (b) Acidified potassium manganate(VII) solution can be safely stored in containers made of p oly(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-