TOPIC 25 ANSWERS & MARK SCHEMES
A2 Level
QUESTIONSHEET 1
SINGLE STAGE ALIPHATIC SYNTHESES NOT INVOLVING NITROGEN COMPOUNDS
a) (i)
Reagent
Br2 (1)
Conditions uv light / heat > 400 0C (1)
(ii) Reagent
HBr (1)
Conditions Gas (1) (allow Concentrated HBr(aq) (½) at 100 0C (½))
(iii) Reagent
NaBr / KBr (1) and concentrated H2SO4 / H3PO4 (1)
Or red phosphorus (1) and Br2 (1)
Or concentrated (1) HBr(aq) (1)
Conditions Heat (1)
b) (i)
Reagent
Concentrated (1) H2SO4 (1)
Or concentrated (1) H3PO4 (1)
Or Al2O3 / kaolin / pumice (2)
Conditions 170-180 0C for conc. H2SO4 (1)
Or ∼ 210 0C for conc. H3PO4 (1)
Or ∼ 350 0C for Al2O3 (1)
(ii) Reagent
Aqueous (1) NaBH4 (1)
Or LiAlH4 (1) + dry ether (1)
Conditions Room temperature (1)
(iii) Reagent
Na2Cr2O7 / K2Cr2O7 (1) + dilute H2SO4 (1)
Conditions Add the dichromate to the alcohol (not vice versa) (1)
Distil off CH3CH2CH2CHO as it is formed (1)
c) (i)
CH3CH2COOH + CH3CH2CH2OH → CH3CH2COOCH2CH2CH3 + H2O (1)
(ii) CH3CH2COOH + NaOH → CH3CH2COO-Na+ + H2O (1)
Or 2CH3CH2COOH + Na2CO3 → 2CH3CH2COO-Na+ + H2O + CO2 (1)
A2 Level
TOPIC 25 ANSWERS & MARK SCHEMES
QUESTIONSHEET 2
MULTI-STAGE ALIPHATIC SYNTHESES NOT INVOLVING NITROGEN COMPOUNDS
a) I
Heat (1)
with concentrated (1) H2SO4 (1)
III Boil/heat under reflux (1)
with aqueous(1) sodium hydroxide (1)
IV Boil/heat under reflux (1)
With potassium/sodium dichromate(VI) (1) (conc) H2SO4 (1)
b) (i)
Step 1
Step 2
bromine (in inert solvent) → 1,2-dibromopropane (1)
heat with aq NaOH (1) → propane-1,2-diol (1)
c) Reagents KOH/NaOH (1) Ethanol (1)
Conditions Boil/heat under reflux (1)
A2 Level
TOPIC 25 ANSWERS & MARK SCHEMES
QUESTIONSHEET 3
SINGLE STAGE ALIPHATIC SYNTHESES NOT INVOLVING NITROGEN COMPOUNDS
a) (i)
Heat in a sealed tube (1)
with alcoholic ammonia (1)
then add aqueous sodium hydroxide (1)
(ii) Boil under reflux (1)
with KCN / NaCN (1)
in ethanol / aqueous ethanol (1)
b) (i)
Reduce with hydrogen (1) and Pt / Ni catalyst (1)
Or LiAlH4 (1) in dry ether (1)
Or NaBH4 (1) in water / methanol (1)
(ii) Boil under reflux (1) with dil. H2SO4 / HCl(aq) (1)
Or boil with NaOH(aq) (1) then acidify (1)
c) (i)
Type of compound ammonium salt (1)
Structure of X
CH3CH(CH3)COONH4 (1)
Structure of product CH3CH(CH3)CONH2 (1)
(ii) Types of reaction
neutralization (1) → elimination (1)
d) Recrystallisation (1)
from ethanol (1)
Amides are solids at room temperature (1)
Impurities are likely to remain in solution (1)
Maximum 3 marks
TOPIC 25 ANSWERS & MARK SCHEMES
A2 Level
QUESTIONSHEET 4
MULTI-STAGE ALIPHATIC SYNTHESES INVOLVING NITROGEN COMPOUNDS
a) Step I
NaBr / KBr (1) and concentrated H2SO4 / H3PO4 (1)
Or red phosphorus (1) and bromine (1)
(Allow 1 mark only for PBr3)
Heat / boil under reflux (1)
Or concentrated (1) HBr(aq) (1) Heat (1)
Step II
Na2Cr2O7 / K2Cr2O7 (1) and dilute H2SO4 (1)
Heat / boil under reflux (1)
Step III Alcoholic/ethanolic ammonia (1)
Heat in a sealed tube (1)
then add NaOH(aq) (1)
Step IV Mineral acid (1)
heat (1)
b) I & III
IV
V
Nucleophilic (1) substitution (1)
Hydrolysis (1)
Neutralisation (1)
c) CH3CH2COONH3CH2CH3 (or as ions) (1)
d) Recrystallise (1)
from cold water / ethanol (1)
filter at the pump (1)
and air dry (1)
Maximum 4 marks
Quality of language: two or more sentences with correct spelling, punctuation and grammar in
which the meaning is clear (1)
TOPIC 25 ANSWERS & MARK SCHEMES
A2 Level
QUESTIONSHEET 5
SYNTHESES INVOLVING AN INCREASE IN CHAIN LENGTH
a) C2H4 gas is bubbled through (1)
Br2 solution in a suitable inert solvent (1)
at room temperature (1)
in the dark / shade / diffuse light / absence of u.v. light (1)
No peroxides (1)
Maximum 3 marks
b) 1,2-dibromoethane (1)
c) Boil under reflux (1)
with excess KCN (1)
in alcoholic / aqueous alcoholic solution (1)
d) Boil under reflux with aqueous sodium hydroxide (1)
then acidify with dil. H2SO4 / HCl (1)
Or boil under reflux (1)
with dil. H2SO4 / HCl (1)
e) LiAlH4 (1) in dry ether (1)
f)
CH2=CHCH=CH2
Bubble through Br2 in pentane (1)
Boil under reflux with excess
KCN in aqueous alcohol (1)
COOH
⏐
CH3CHBrCHBrCH3 (1)
Boil under reflux with dil. H2SO4 (1)
CH3-CH-CH-CH3
⏐
COOH
Or boil under reflux with NaOH(aq)
then acidify (1)
CH3CH(CN)CH(CN)CH3 (1)
TOPIC 25 ANSWERS & MARK SCHEMES
A2 Level
QUESTIONSHEET 6
AROMATIC SYNTHESES NOT INVOLVING DIAZONIUM SALTS
a) Reagents & conditions
Name of product
b) Type of reaction
Achievement
Boil under reflux (1) with KMnO4(aq) (1)
made alkaline with Na2CO3 / NaOH (1)
Then acidify with dil. H2SO4 / HCl (1)
Benzenecarboxylic acid / benzoic acid (1)
Reduction (1)
LiAlH4 (1) in dry ether (1)
c) Benzenecarbonyl chloride / benzoyl chloride (1)
COCl (1)
d) Ester (1)
e) The acyl chloride R is more reactive than the carboxylic acid P (1)
because polarisation of both C⎯Cl and C=O bonds increases δ+ on the carbonyl C atom / makes R highly susceptible to
nucleophilic attack (1)
Also, the reaction between an acyl chloride and an alcohol goes to completion / gives a good yield of ester (1)
whereas direct esterification gives an equilibrium mixture / a lower yield (1)
Maximum 3 marks
TOPIC 25 ANSWERS & MARK SCHEMES
A2 Level
QUESTIONSHEET 7
AROMATIC SYNTHESES INVOLVING DIAZONIUM SALTS
a) Ice-cold solutions / ∼ 5 oC (1)
b) Step I
Diazotisation (1)
Step II Coupling/Electrophilic subsitution (1)
Step III Coupling/Electrophilic subsitution (1)
c) C
+
-
-N≡N Cl (2)
HO3S-
+
-
(Allow 1 mark if -N=N-Cl is shown instead of -N≡N Cl)
D
HO3S-
E
Na
+
-
O3S-
-N=N-N=N-
-N(CH3)2 (1)
-OH (2)
(Allow 1 mark if –SO3H is shown instead of –SO3- Na+)
d) Dyes (1)
TOPIC 25 ANSWERS & MARK SCHEMES
A2 Level
QUESTIONSHEET 8
TEST QUESTION I
a) Triiodomethane / iodoform / CHI3 (1)
The compounds contain either CH3CO- (1) or CH3-CH(OH)- (1)
b) C=O i.e. ketone (1)
Carbonyl group because of reaction with 2,4-dinitrophenylhydrazine (1)
but not an aldehyde because no reaction with ammoniacal silver nitrate (1)
c) LiAlH4 in dry ether (1)
[NB Not NaBH4 – will not reduce acids]
d) Carboxyl / carboxylic acid / -COOH (1)
e) J CH3COOH
K CH3COCH2CH3
L CH3CH2OH
M CH3-CH2
C N
CH3
(1)
(1)
(1)
ethanoic acid (1)
butanone
(1)
ethanol
(1)
O2N
N
NO2 (2)
H
N CH3CH(OH)CH2CH3 (1)
f) 3-methylpent-2-ene (1)
CH3CH=C(CH3)CH2CH3 (1)
butan-2-ol (1)
Deduct 1 for each error
TOPIC 25 ANSWERS & MARK SCHEMES
A2 Level
QUESTIONSHEET 9
TEST QUESTION II
a) S contains a benzene ring, since it is highly unsaturated (1)
Therefore, from the formula, there is a methyl side chain (1)
S can be diazotised by treatment with ice-cold NaNO2 & HCl (1)
so must contain an amino group / NH2 group (1)
b) S is produced from R by reduction so R could be
CH3
NO2 (1)
Or CH3
(less likely) (1)
O2N
c) CH3
d) T CH3
N
N
OH (2) Deduct 1 for each error
NH-CO-CH3 (1)
U HOOC
NH-CO-CH3 (1)
V HOOC
NH2 (1)
Name of V
4-aminobenzenecarboxylic acid (1)
e) To block / protect the amino group (1)
from oxidation (1)
TOPIC 25 ANSWERS & MARK SCHEMES
A2 Level
QUESTIONSHEET 10
TEST QUESTION III
a) Bubble gas through a solution of Br2 in an organic solvent / bromine water (1)
which will be decolourised (1)
b) The HCl produced reacts with the KOH present to form KCl (1)
c) CH3CH2CH2CH2Cl → CH3CH2CH=CH2 (2)
CH3
CH3
CH3
C
CH3 → CH3
C
CH2 (2)
Cl
d) 2-chlorobutane (1)
CH3CH2CHClCH3 (1)
1-chloro-2-methylpropane (1)
CH3CH(CH3)CH2Cl (1)
e) (i)
moles gas = 8.0 / (24 x 103) = 3.33 x 10-4 mol (1)
Mr (chlorobutane) = (48 + 9 + 35.5) = 92.5 (1)
∴ n (chlorobutane) = 0.37 / 92.5 = 4.0 x 10-3 mol (1)
∴ % elimination = 100(3.33 x 10-4) / (4.0 x 10-3)
= 8.325 % (8.3%) (1)
(ii) If 8 cm3 ≡ 8.325 %
48 cm3 ≡ (48 / 8)(8.325) = 49.95 i.e. 50% (1)
(iii) Nucleophilic substitution to form an alcohol (1)
(iv) 2-chloro-2-methylpropane is a tertiary haloalkane (1)
which undergoes elimination more readily than primary haloakanes such as 1-chlorobutane (1)
A2 Level
TOPIC 25 ANSWERS & MARK SCHEMES
QUESTIONSHEET 11
TEST QUESTION IV
a) 1/24 mol A →(0.7) / 24 mol B = 0.02917 mol ≡ 4.0 g (1)
∴ Mr (B) = 4.0 / 0.02917 = 137 (1)
If the formula of B is CxHyBr, then CxHy- = (137 – 80) = 57
suggesting four carbons, so A is C4H8 (1)
B is C4H9Br (1)
b) B → D Boil under reflux (1) with (aqueous) alcoholic KCN (1)
D → E Boil under reflux (1) with dil. H2SO4 (1)
c) (i)
Since E is an acid and F is an ester, C must be an alcohol (1)
Since it resists oxidation, it must be a tertiary alcohol (1)
i.e. CH3C(CH3)OHCH3 / 2-methylpropan-2-ol (1)
(ii) B = 2-bromo-2-methylpropane / CH3C(CH3)BrCH3 (1)
D = 2,2-dimethylpropanenitrile / CH3C(CH3)2CN (1)
E = 2,2-dimethylpropanoic acid / CH3C(CH3)2COOH (1)
d) CH3C(CH3)2COOC(CH3)2CH3 (1)
e) Heat (1)
with dil. aqueous NaOH (1)
(minimum heat to reduce elimination)
f) Boil under reflux (1)
with alcoholic KOH / sodium tert-butoxide (1)
TOPIC 25 ANSWERS & MARK SCHEMES
A2 Level
QUESTIONSHEET 12
PREPARATIVE TECHNIQUES I (Preparation of 1-bromobutane)
a) (i)
Because heat is liberated when sulfuric acid is diluted (1)
(ii) To allow the reaction to reach completion (1)
(iii) To avoid loss of material to atmosphere / to condense vapour and hence return material to the flask (1)
(iv) 1-Bromobutane (1) Water (1) Butan-1-ol (1)
Any 2
(v) To remove unreacted butan-1-ol (1)
(vi) HCl dissociates / ionises in butan-1-ol (1)
The ions are water-attracting (1)
(vii) Reason To neutralise traces of HCl (1)
Safety precaution
Pressure must be released frequently (1)
(viii)To dry it (1)
(ix) Because distillation in the presence of hydrated sodium sulfate would give a wet product (1)
b) (i)
CH3(CH2)3OH + NaBr + H2SO4 → CH3(CH2)3Br + NaHSO4 + H2O (1)
Or
(ii)
NaBr + H2SO4 → HBr + NaHSO4
(1)
CH3(CH2)3OH + HBr → CH3(CH2)3 Br + H2O
n (CH3(CH2)3OH) = 6.0/74 = 0.082 mol
n (NaBr)
= 10/103 = 0.097 mol
n (H2SO4)
= 18/98 = 0.184 mol
∴butan-1-ol is most deficient (1)
(iii)
n (C4H9Br) = n (C4H9OH) = 0.082 mol
∴ m (C4H9Br) = 0.082 × 137 = 11.2 g (1)
(iv)
m (C4H9Br) = 1.3 × 5.8 = 7.54 g (1)
∴ yield = (7.54/11.2) × 100 = 67.3 % (1)
(1)
(1)
TOPIC 25 ANSWERS & MARK SCHEMES
A2 Level
QUESTIONSHEET 13
PREPARATIVE TECHNIQUES II (Preparation of phenylamine)
a) Reducing agent / reductant (1)
b) Because the reaction is vigorous / to avoid flooding the condenser (1)
c) To complete the reduction of the nitrobenzene (1)
d) To liberate phenylamine from its ion / from a phenylammonium salt (1)
e) Passing a current of steam through the (heated) mixture in a flask attached to a condenser (1)
f) To separate phenylamine / oily liquids from inorganic compounds (which are not volatile in steam) (1)
g) Because phenylamine is less soluble in salt solution than it is in water (1)
h) Because phenylamine is more soluble in ethoxyethane than in water (1)
i) Because more phenylamine is extracted (1)
j) Add some water (1)
If the liquids are miscible, the layer is aqueous (1)
k) KOH does not react with phenylamine (1)
P4O10 reacts with phenylamine (1) by an acid-base reaction (1)
l) No naked flames / indirect heating / use a hot water bath (1)
Good ventilation / use a fume cupboard (1)
m) Hot phenylamine vapour could crack the condenser if there were cold water in it (1)
n) Liquid boiling below 180 °C is residual ethoxyethane / water (1)
Liquid boiling above 180 °C is unreduced nitrobenzene (1)
Pure phenylamine boils over a narrow range (1)
Any 1
TOPIC 25 ANSWERS & MARK SCHEMES
A2 Level
QUESTIONSHEET 14
PREPARATIVE TECHNIQUES III (Nitration)
a) To avoid introducing more than one nitro group (1)
b) 2H2SO4 + HNO3 à 2HSO4- + NO2+ + H3O+ (species(1) balance (1))
(1)
c) (i) to crystallise the product (1)
(ii) rapid filtration (1)
allows rapid drying (1)
better yield (1)
Maximum 2 marks
(iii) water :
to remove acid residues (1)
methanol : to remove unreacted methyl benzoate (1)
(iv) to minimise solution of the product (1)
d) product is very soluble at high temperature but not on cooling (1)
low boiling point (1)
impurities are far more soluble (1)
e) hydrolysis of reactant/product (1)
absorption during filtration (1)
incomplete reaction (1)
loss during recrystallisation (1)
Maximum 2 marks
f) Actual value compared to expected (1)
The sharpness/range (1)