QUESTIONSHEET 1 SINGLE STAGE ALIPHATIC SYNTHESES NOT INVOLVING NITROGEN COMPOUNDS a) (i) Reagent Br 2 Conditions uv light / heat > 400 0 C (ii) Reagent HBr Conditions Gas (allow Concentrated HBr(aq) (½) at 100 0 C (½)) (iii) Reagent NaBr / KBr and concentrated H 2 / H 3 PO 4 Or red phosphorus and Br 2 Or concentrated HBr(aq) Conditions Heat b) (i) Reagent Concentrated H 2 Or concentrated H 3 PO 4 Or Al 2 O 3 / kaolin / pumice (2) Conditions 170-180 0 C for conc. H 2 Or 210 0 C for conc. H 3 PO 4 Or 350 0 C for Al 2 O 3 (ii) Reagent Aqueous NaBH 4 Or LiAlH 4 + dry ether Conditions Room temperature (iii) Reagent Na 2 Cr 2 O 7 / K 2 Cr 2 O 7 + dilute H 2 Conditions Add the dichromate to the alcohol (not vice versa) Distil off CHO as it is formed c) (i) COOH + OH COO + H 2 O (ii) COOH + NaOH COO - Na + + H 2 O Or 2 COOH + Na 2 CO 3 2 COO - Na + + H 2 O + CO 2
QUESTIONSHEET 2 MULTI-STAGE ALIPHATIC SYNTHESES NOT INVOLVING NITROGEN COMPOUNDS a) I Heat with concentrated H 2 III Boil/heat under reflux with aqueous sodium hydroxide IV Boil/heat under reflux With potassium/sodium dichromate(vi) (conc) H 2 b) (i) Step 1 bromine (in inert solvent) 1,2-dibromopropane Step 2 heat with aq NaOH propane-1,2-diol c) Reagents KOH/NaOH Ethanol Conditions Boil/heat under reflux
QUESTIONSHEET 3 SINGLE STAGE ALIPHATIC SYNTHESES NOT INVOLVING NITROGEN COMPOUNDS a) (i) Heat in a sealed tube with alcoholic ammonia then add aqueous sodium hydroxide (ii) Boil under reflux with KCN / NaCN in ethanol / aqueous ethanol b) (i) Reduce with hydrogen and Pt / Ni catalyst Or LiAlH 4 in dry ether Or NaBH 4 in water / methanol (ii) Boil under reflux with dil. H2SO4 / HCl(aq) Or boil with NaOH(aq) then acidify c) (i) Type of compound ammonium salt Structure of X CH( )COONH 4 Structure of product CH( )CONH 2 (ii) Types of reaction neutralization elimination d) Recrystallisation from ethanol Amides are solids at room temperature Impurities are likely to remain in solution Maximum 3 marks
QUESTIONSHEET 4 MULTI-STAGE ALIPHATIC SYNTHESES INVOLVING NITROGEN COMPOUNDS a) Step I NaBr / KBr and concentrated H 2 / H 3 PO 4 Or red phosphorus and bromine (Allow 1 mark only for PBr 3 ) Heat / boil under reflux Or concentrated HBr(aq) Heat Step II Na 2 Cr 2 O 7 / K 2 Cr 2 O 7 and dilute H 2 Heat / boil under reflux Step III Alcoholic/ethanolic ammonia Heat in a sealed tube then add NaOH(aq) Step IV Mineral acid heat b) I & III Nucleophilic substitution IV Hydrolysis V Neutralisation c) COONH 3 (or as ions) d) Recrystallise from cold water / ethanol filter at the pump and air dry Maximum 4 marks Quality of language: two or more sentences with correct spelling, punctuation and grammar in which the meaning is clear
QUESTIONSHEET 5 SYNTHESES INVOLVING AN INCREASE IN CHAIN LENGTH a) C 2 H 4 gas is bubbled through Br 2 solution in a suitable inert solvent at room temperature in the dark / shade / diffuse light / absence of u.v. light No peroxides Maximum 3 marks b) 1,2-dibromoethane c) Boil under reflux with excess KCN in alcoholic / aqueous alcoholic solution d) Boil under reflux with aqueous sodium hydroxide then acidify with dil. H 2 / HCl Or boil under reflux with dil. H 2 / HCl e) LiAlH 4 in dry ether f) Bubble through Br 2 in pentane =CHCH= CHBrCHBr COOH -CH-CH- COOH Boil under reflux with dil. H 2 Or boil under reflux with NaOH(aq) then acidify Boil under reflux with excess KCN in aqueous alcohol CH(CN)CH(CN)
QUESTIONSHEET 6 AROMATIC SYNTHESES NOT INVOLVING DIAZONIUM SALTS a) Reagents & conditions Boil under reflux with KMnO 4 (aq) made alkaline with Na 2 CO 3 / NaOH Then acidify with dil. H 2 / HCl Name of product Benzenecarboxylic acid / benzoic acid b) Type of reaction Reduction Achievement LiAlH 4 in dry ether c) Benzenecarbonyl chloride / benzoyl chloride COCl d) Ester e) The acyl chloride R is more reactive than the carboxylic acid P because polarisation of both C Cl and C=O bonds increases δ + on the carbonyl C atom / makes R highly susceptible to nucleophilic attack Also, the reaction between an acyl chloride and an alcohol goes to completion / gives a good yield of ester whereas direct esterification gives an equilibrium mixture / a lower yield Maximum 3 marks
QUESTIONSHEET 7 AROMATIC SYNTHESES INVOLVING DIAZONIUM SALTS a) Ice-cold solutions / 5 o C b) Step I Diazotisation Step II Coupling/Electrophilic subsitution Step III Coupling/Electrophilic subsitution c) C + - HO 3 S- -N N Cl (2) + - (Allow 1 mark if -N=N-Cl is shown instead of -N N Cl) D HO 3 S- -N=N- -N( ) 2 E Na + - O 3 S- -N=N- -OH (2) - (Allow 1 mark if SO 3 H is shown instead of SO 3 Na + ) d) Dyes
QUESTIONSHEET 8 TEST QUESTION I a) Triiodomethane / iodoform / CHI 3 The compounds contain either CO- or -CH(OH)- b) C=O i.e. ketone Carbonyl group because of reaction with 2,4-dinitrophenylhydrazine but not an aldehyde because no reaction with ammoniacal silver nitrate c) LiAlH 4 in dry ether [NB Not NaBH 4 will not reduce acids] d) Carboxyl / carboxylic acid / -COOH e) J COOH ethanoic acid K CO butanone L OH ethanol M - O 2 N C N N NO 2 (2) Deduct 1 for each error H N CH(OH) butan-2-ol f) 3-methylpent-2-ene CH=C( )
QUESTIONSHEET 9 TEST QUESTION II a) S contains a benzene ring, since it is highly unsaturated Therefore, from the formula, there is a methyl side chain S can be diazotised by treatment with ice-cold NaNO 2 & HCl so must contain an amino group / NH 2 group b) S is produced from R by reduction so R could be NO 2 Or (less likely) O 2 N c) N N OH (2) Deduct 1 for each error d) T NH-CO- U HOOC NH-CO- V HOOC NH 2 Name of V 4-aminobenzenecarboxylic acid e) To block / protect the amino group from oxidation
QUESTIONSHEET 10 TEST QUESTION III a) Bubble gas through a solution of Br 2 in an organic solvent / bromine water which will be decolourised b) The HCl produced reacts with the KOH present to form KCl c) Cl CH= (2) CH3 C CH2 (2) C Cl d) 2-chlorobutane CHCl 1-chloro-2-methylpropane CH( ) Cl e) (i) moles gas = 8.0 / (24 x 10 3 ) = 3.33 x 10-4 mol M r (chlorobutane) = (48 + 9 + 35.5) = 92.5 n (chlorobutane) = 0.37 / 92.5 = 4.0 x 10-3 mol % elimination = 100(3.33 x 10-4 ) / (4.0 x 10-3 ) = 8.325 % (8.3%) (ii) If 8 cm 3 8.325 % 48 cm 3 (48 / 8)(8.325) = 49.95 i.e. 50% (iii) Nucleophilic substitution to form an alcohol (iv) 2-chloro-2-methylpropane is a tertiary haloalkane which undergoes elimination more readily than primary haloakanes such as 1-chlorobutane
QUESTIONSHEET 11 TEST QUESTION IV a) 1/24 mol A (0.7) / 24 mol B = 0.02917 mol 4.0 g M r (B) = 4.0 / 0.02917 = 137 If the formula of B is C x H y Br, then C x H y - = (137 80) = 57 suggesting four carbons, so A is C 4 H 8 B is C 4 H 9 Br b) B D Boil under reflux with (aqueous) alcoholic KCN D E Boil under reflux with dil. H 2 c) (i) Since E is an acid and F is an ester, C must be an alcohol Since it resists oxidation, it must be a tertiary alcohol i.e. C( )OH / 2-methylpropan-2-ol (ii) B = 2-bromo-2-methylpropane / C( )Br D = 2,2-dimethylpropanenitrile / C( ) 2 CN E = 2,2-dimethylpropanoic acid / C( ) 2 COOH d) C( ) 2 COOC( ) 2 e) Heat with dil. aqueous NaOH (minimum heat to reduce elimination) f) Boil under reflux with alcoholic KOH / sodium tert-butoxide
QUESTIONSHEET 12 PREPARATIVE TECHNIQUES I (Preparation of 1-bromobutane) a) (i) Because heat is liberated when sulfuric acid is diluted (ii) To allow the reaction to reach completion (iii) To avoid loss of material to atmosphere / to condense vapour and hence return material to the flask (iv) 1-Bromobutane Water Butan-1-ol Any 2 (v) To remove unreacted butan-1-ol (vi) HCl dissociates / ionises in butan-1-ol The ions are water-attracting (vii) Reason To neutralise traces of HCl Safety precaution Pressure must be released frequently (viii)to dry it (ix) Because distillation in the presence of hydrated sodium sulfate would give a wet product b) (i) ( ) 3 OH + NaBr + H 2 ( ) 3 Br + NaH + H 2 O Or NaBr + H 2 HBr + NaH ( ) 3 OH + HBr ( ) 3 Br + H 2 O (ii) n ( ( ) 3 OH) = 6.0/74 = 0.082 mol n (NaBr) = 10/103 = 0.097 mol n (H 2 ) = 18/98 = 0.184 mol butan-1-ol is most deficient (iii) n (C 4 H 9 Br) = n (C 4 H 9 OH) = 0.082 mol m (C 4 H 9 Br) = 0.082 137 = 11.2 g (iv) m (C 4 H 9 Br) = 1.3 5.8 = 7.54 g yield = (7.54/11.2) 100 = 67.3 %
QUESTIONSHEET 13 a) Reducing agent / reductant PREPARATIVE TECHNIQUES II (Preparation of phenylamine) b) Because the reaction is vigorous / to avoid flooding the condenser c) To complete the reduction of the nitrobenzene d) To liberate phenylamine from its ion / from a phenylammonium salt e) Passing a current of steam through the (heated) mixture in a flask attached to a condenser f) To separate phenylamine / oily liquids from inorganic compounds (which are not volatile in steam) g) Because phenylamine is less soluble in salt solution than it is in water h) Because phenylamine is more soluble in ethoxyethane than in water i) Because more phenylamine is extracted j) Add some water If the liquids are miscible, the layer is aqueous k) KOH does not react with phenylamine P 4 O 10 reacts with phenylamine by an acid-base reaction l) No naked flames / indirect heating / use a hot water bath Good ventilation / use a fume cupboard m) Hot phenylamine vapour could crack the condenser if there were cold water in it n) Liquid boiling below 180 C is residual ethoxyethane / water Liquid boiling above 180 C is unreduced nitrobenzene Pure phenylamine boils over a narrow range Any 1
QUESTIONSHEET 14 PREPARATIVE TECHNIQUES III (Nitration) a) To avoid introducing more than one nitro group b) 2H 2 + HNO 3 à 2H - + NO 2 + + H 3 O + (species balance ) c) (i) to crystallise the product (ii) rapid filtration allows rapid drying better yield Maximum 2 marks (iii) water : to remove acid residues methanol : to remove unreacted methyl benzoate (iv) to minimise solution of the product d) product is very soluble at high temperature but not on cooling low boiling point impurities are far more soluble e) hydrolysis of reactant/product absorption during filtration incomplete reaction loss during recrystallisation Maximum 2 marks f) Actual value compared to expected The sharpness/range