Download questionsheet 1 e/z (cis/trans) isomerism

A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 1
E/Z (CIS/TRANS) ISOMERISM
a) Explanation
Restricted rotation (1)
of atoms or groups about a C=C bond / when C atoms are joined together by a double bond (1)
(Do not allow restricted rotation of molecules.)
Due to the π- bond locking atoms in position (1)
Structures and names
CH 3CH 2
C C
H
CH 3CH 2
CH 3 (1)
H
H
Z-pent-2-ene or (cis pent-2-ene) (1)
b) Suggestion
H
C C
(1)
CH 3
E-pent-2-ene or (trans pent-2-ene) (1)
Alicyclic ring prevents rotation / locks the molecule in position (1)
Structures
(1)
Cl Cl
c) CH2=CHCH2CH3
(CH3)2C=CH2
CHCl=CHCl
CHCl=CHCH3
CHCl=CClCH3
CH3CH=CClCH2CH3
Structural feature
Cl
(1)
Cl
No (1)
No (1)
Yes (1)
Yes (1)
Yes (1)
Yes (1)
Both C atoms of the C=C bond must be joined to different atoms or groups (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 2
OPTICAL ISOMERISM
a) (i) Structures
CH 3
CH 3
C
C
OH
H
(1)
C 2H 5
H
HO
C 2H 5
(1)
(Tapered bonds (or similar) to show tetrahedral distribution about C are essential)
Means of distinguish between them
Equal rotation (1)
of the plane of polarised light (1)
but in opposite directions (1)
(ii) Butan-2-ol has a chiral centre / asymmetric carbon atom (1)
but butanone does not (1)
b) (i)
Isomers whose molecules bear an object to mirror image relationship (1)
(ii) No effect on the plane of polarised light / optically inactive (1)
(iii) Racemic mixture / racemate (1)
(iv) Ethanal is trigonal planar about the carbonyl C atom (1)
There is an equal probability / 50:50 chance of CN- attacking from above and below the plane (1)
so that equal amounts of the mirror image ions are formed (1)
CH 3
CH 3
C
O
H
C
-
-
CN
c)
(1)
H
O
NC
This leads to (±)-CH3CH(OH)CN / dl-CH3CH(OH)CN (1)
Or to (±)-lactic acid / dl-lactic acid (1)
Maximum 4 marks
H
O
H O H H
CH 3
*CH
2
CH
C
O
H
H
*C
C
H
NH 2
H
H
C
C
C
H
H
CH 3
C
H
CH 2
CH 3CH 2
H
H
NH 2
CH 2
*C
CH 3
HO
H
C
C
OH H
CH 3
O
CH 3
*C
HO
H
O
CH 2 CH 2
CH
CH 3
N
CH 3
H
H
d) Only one of the isomers is beneficial (1)
So half the dose needed if single isomer used (1)
The other isomer may have unpleasant side effects (1)
*
CH2Cl
*CHCl
CH 2
C
CH 2
CH 2
CH 2
1 mark per correctly placed * and - ½ per wrongly placed *.
CH 2OH
C C N
O
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 3
TEST QUESTION ON ISOMERISM
a) (i)
Skeletal isomerism
CH3CH2CH2CH2OH / CH3CH(OH)CH2CH3 (1)
and (CH3)2CHCH2OH / (CH3)3COH (1)
Positional isomerism
CH3CH2CH2CH2OH (1) and CH3CH(OH)CH2CH3 (1)
Or (CH3)2CHCH2OH (1) and (CH3)3COH (1)
Functional group isomerism
Accept any of the C4 alcohols (1) and any of the C4 ethers (1)
Stereoisomerism
CH 3
CH 3
(1)
C
H
C
and
HO
OH
(1)
H
C 2H 5
C 2H 5
(ii) Butan-2-ol has a chiral centre / asymmetric C atom (1)
∴ can exhibit optical isomerism / exist in two non-superimposable forms (1)
but none of the compounds has a C = C bond (or other structural feature) which can cause restricted rotation
of atoms (1)
∴there is no possibility of geometrical isomerism / existence of cis and trans isomers (1)
b) (i) Geometrical / cis-trans isomerism (1)
(ii)
CH 3
H
CH 2
CH 3
N
C=N. .
C 6H 5
CH 3
(iii) Melting point / solubility (1)
CH 2
..
C=N
(1)
CH 3
H
N
C 6H 5
(1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 4
STRUCTURE OF BENZENE
a) (i)
Platinum catalyst, room temperature / nickel catalyst, heat (1)
(ii)
(1)
(iii) Since three double bonds are hydrogenated,
∆H = 3(-120) = -360 kJ mol-1 (1)
(iv) Benzene is more stable than expected (1)
by (360 – 208) = 152 kJ mol-1 (1)
This is because benzene contains a delocalised system of electrons (1)
b) (i)
Delocalised
Electrons are not restricted to “2-electron, 2-centre” bonds (1)
but are dispersed / spread / mobile over a greater number of nuclear centres (1)
Unsaturated
One or more (carbon-carbon) bonds are multiple bonds / the compound will undergo an addition
reaction with hydrogen (1)
(ii)
(1)
(iii) Benzene does not undergo electrophilic addition reactions (1)
X
X
Benzene does not produce disubstituted isomers of the form
X
and
X
Electron-diffraction measurements show that all the carbon-carbon bonds are of equal length / intermediate
between C-C and C=C (1)
The enthalpy of hydrogenation of benzene is less exothermic than expected (1)
The enthalpy of formation of benzene is less endothermic than expected (1)
Maximum 3 marks
(1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 5
STRUCTURES OF CYLOALKENES
a) Alkene I
Alkene II
Alkene III
Cyclohexene (1)
1,3-Cyclohexadiene (1)
1,4-Cyclohexadiene (1)
b)
+ H2 →
(1)
+ 2H2 →
(1)
c) If the π-orbitals of II were completely localised / by comparison with cyclohexene (1)
enthalpy of hydrogenation of II could be calculated as –119.6 x 2 = -239.2 kJ mol-1 (1)
This is 7.5 kJ mol-1 different from the experimental value (1)
∴II is energetically more stable than if it had two completely localised π-bonds (1)
A slight degree of delocalisation of π-electrons must be occurring (1)
Maximum 4 marks
d) Shortest bonds will be a and c (1)
Longest bonds will be d, e and f (1)
b will be slightly shorter than d, e, and f (1)
because of the slight degree of delocalisation (1)
e) Estimate
Reasons
-239.2 kJ mol-1 (1)
π-bonds are too far apart to overlap (1)
No delocalisation of π-electrons (1)
∴ no increase in stability (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 6
BOND TYPE AND REACTIVITY
a) (i)
Electrophiles are attracted by electrons of the π-bond (1)
The π-bond is weak / easily broken (1)
(ii) C⎯H bonds are little polarised / C and H have similar electronegativity (1)
For an alkane to react, strong σ-bonds would have to be broken (1)
b) (i)
Polarisation of the bond / Cδ+ ⎯Xδ- (1)
due to the high electronegativity of halogens compared with carbon (1)
(ii) RCl < RBr < RI (1)
(iii) Bond strength decreases from C⎯Cl to C⎯I (1)
hence a halide ion can leave more readily (1)
This outweighs the decreasing strength of the –I effect / the lower electronegativity of the halogens (1)
Maximum 2 marks
c) (i)
Nucleophilic reagents / nucleophiles (1)
(ii) Bond polarisation Cδ+ ⎯
⎯ Oδ- / oxygen is much more electronegative than carbon (1)
The π-bond is easily polarised (1)
(iii) The π-bond is weak / easily broken (1)
but the σ-bond is strong (1)
(iv) They are repelled by the π-electrons of the C ⎯
⎯ C bond (1)
(v) HCN would have to undergo heterolytic fission / dissociation (to H+ and CN-) (1)
This is difficult because HCN is a very weak acid / because of the strength of the H⎯CN bond (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 7
REACTION CONDITIONS AND PRODUCT
a) (i)
Free radical (½) substitution (½)
(ii) Homolytic fission / homolysis (1)
(iii) Stage 1
.
.
Cl . Cl
2Cl (1)
Stage 2
.
CH 3
+ Cl
.
CH2
.
+ HCl
CH2
(1)
CH2Cl
+ Cl2
+ HCl
(1)
Stage 3
.
CH2
CH2Cl
+ Cl
b) (i)
(1)
.
Or
Or
.
Cl2
2Cl
2
.
CH 3
(1)
CH2
CH 2
CH 3
+ Cl2
CH 2
(1)
Cl
+ HCl
(1)
+ HCl
(1)
Or
CH 3
CH 3
+ Cl2
Cl
(ii) Aluminium chloride / iron (III) chloride (1)
(iii) Electrophilic (½) substitution (½)
(iv) Heterolytic fission / heterolysis (1)
c) Example
Aqueous NaOH (1)
CH3CH2Br
Nucleophilic substitution (1)
CH3CH2OH (1)
Ethanolic / alcoholic NaOH(1)
CH3CH2Br
Elimination (1)
CH2 = CH2 (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 8
NITRATION OF ARENES
a) Reagents
Concentrated nitric acid (1)
and concentrated sulfuric acid (1)
Conditions
50 - 60 °C (1)
NO2
Equation
+ HNO3 →
+ H2O (1)
Or C6H6 + HNO3 → C6H5NO2 + H2O (1)
b) (i)
Lower temperature / 30 - 40 °C (1)
(ii) Use fuming nitric acid / more concentrated nitric acid (1)
and a higher temperature (1)
(iii) Formula
O2N
CH3
NO2
(1)
NO2
Use
c) (i)
High explosive (1)
C6H5NO2 + 6[H] → C6H5NH2 + 2H2O (1)
NO2
Or
NH2
+ 6[H] →
+ 2H2O (1)
(ii) Reduction(1)
(iii) Tin (1) and concentrated hydrochloric acid (1)
(Accept HCl (aq) but not just ‘HCl’.)
(iv) Primary aromatic amines form the basis of the azo dyestuffs industry (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 9
HALOGENATION OF ARENES
a) (i)
Conditions
Equation
Br2 in solution (in a named solvent) (1)
Room temperature / 20 °C (1)
Br
Br
+ Br2 →
(1)
Or with molecular formulae in place of the structural formulae
(ii) (Electrophilic) addition (1)
b) (i)
Conditions
Br2 + AlBr3 / Al catalyst (1) (Allow AlCl3)
Room temperature / 20 °C (1)
Br
Equation
+ Br2 →
+ HBr (1)
Or with molecular formulae in place of the structural formulae
(ii) (Electrophilic) substitution (1)
c) π-electrons in cyclohexene are localised between two adjacent C atoms (1)
π-electrons in benzene are delocalised over the six-membered ring (1)
Therefore π-electrons in benzene attract bromine / the reagent / the electrophile less strongly (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 10
ALKYLATION AND ACYLATION OF ARENES
a) (i)
Reagent
Chloromethane / bromomethane / iodomethane (1)
Conditions
AlCl3 catalyst (1)
Room temperature / 20 °C (1)
CH3
+ CH3Cl →
Equation
+ HCl (1)
Or C6H6 + CH3Cl → C6H5CH3 + HCl (1)
(ii) It is difficult to prevent further alkylation / a dimethylbenzene or trimethylbenzene is formed readily (1)
(iii) Type of reaction
b) (i)
Oxidation (1)
Reagent
KMnO4 (½) + NaOH(aq) (½)
Or K2Cr2O7 (½) + dil. H2SO4 (½)
Or dilute HNO3 (1)
Conditions
AlCl3 catalyst (1)
50 ºC / heat (1)
COCH3
+ CH3COCl →
Equation
Or
Name of product
(ii) Type of reaction
Reagent
+ HCl (1)
C6H6 + CH3COCl → C6H5COCH3 + HCl (1)
Phenylethanone / acetophenone (1)
Reduction (1)
NaBH4 / Zn + dil. H2SO4 / other named combination of metal and protic solvent (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 11
PHENOL
a) Identity of phenols
Basis of identification
b) (i)
Identity of A
Identity of B
I, II, & IV (1) (All must be correct)
A phenol must have at least one –OH group attached directly to a benzene ring (1)
-
Sodium phenoxide / sodium phenate / C6H5O Na+ (1)
Mainly phenol / (saturated) solution of water in phenol (1)
(Allow just ‘phenol’)
-
OH
+
O Na
Equations
+
NaOH
→
+ H2O (1)
-
+
Or C6H5OH + NaOH → C6H5O Na
-
+
O Na
OH
+ HCl →
-
Or C6H5O Na
(ii) Reason 1
(ii)
+
+ NaCl (1)
→ C6H5OH + NaCl (1)
+ HCl
Lone pair of electrons on O (1)
is drawn towards the benzene ring (1)
O⎯H bond is weakened / increasingly polarised (1)
so that H+ is lost relatively easily (1)
Maximum 3 marks
-
Reason 2
c) (i)
+ H2O (1)
The phenoxide ion / C6H5O is stabilised (1)
by delocalisation of –ve charge (1)
over O and the benzene ring (1)
This encourages dissociation / ionisation / formation of C6H5O (1)
No scope for delocalisation in an alkoxide ion / corresponding anion from an alcohol (1)
Maximum 3 marks
White (½) precipitate (½)
OH
OH
Br
Br
+ 3Br2 →
+ 3HBr (1)
Br
(iii) Lone pair of electrons on O (1)
is drawn towards the benzene ring / occupies a p-orbital which overlaps the delocalised π-orbital of the ring (1)
Hence electron density on the ring is increased (1)
and the ring is activated towards electrophiles / electrophiles are more strongly attracted (1)
Maximum 3 marks
d) Antiseptics / disinfectants (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 12
ALDEHYDES AND KETONES
a) (i)
C (½) Butanone / butan-2-one (½)
(ii) A (½) Propanal (½) and
B (½) 2-methylpropanal (½)
(iii) C (½) Butanone (½)
Allow if named in (i)
b) (i)
Silver mirror (1)
(ii) propanoic acid (1)
(iii) Fehlings (1) Red precipitate (1)
or Potassium dichromate (VI) (1) orange to green (1)
or potassium manganate(VII) (1) purple to colourless (1)
c) (i)
LiALH4 or NaBH4 (1)
(ii) In dry ether (1) for LiALH4 or in solution in water or methanol for NaBH4 (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 13
USE OF 2,4 - DINITROPHENYLHYDRAZINE
a)
C
69.8/12
5.817
5
Moles
Ratio
H
11.6/1
11.6
10
O
18.6/16
1.163 mol %
1
(1)
(1)
Empirical formula is C5H10O ; relative formula mass = 86 = relative molecular mass (1)
therefore molecular formula is also C5H10O (1)
b) Aldehydes: CH3CH2CH2CH2CHO
CH3CH2CH(CH3)CHO
CH3CH(CH3)CH2CHO
CH3C(CH3)2CHO
Ketones:
CH3COCH2CH2CH3
CH3CH2COCH2CH3
CH3COCH(CH3)2
(1 mark each)
c)
C
O + H2N ⎯ NH
C
N ⎯ NH
NO2
NO2
d) Test
NO2 + H2O
(2) Delete 1 mark for each error
NO2
Tollens’ reagent / ammoniacal silver nitrate (1)
Or Fehling’s solution (1)
Conditions Warm (not boil) for Tollens’ reagent (1)
Or boil / heat for Fehling’s solution (1)
Observation with aldehydes Silver mirror (1)
Or red / brown precipitate for Fehling’s solution (1)
Observation with ketones
e) (i)
No silver mirror (1)
Or no red / brown precipitate / solution remains dark blue for Fehling’s solution (1)
Compare the melting point with that obtained from a data source (1)
(ii) Recrystallise (1) to make the compound pure (1),
because impurities change/ lower the melting point. (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 14
THE TRIIODOMETHANE REACTION
a) (i)
Yellow (½) precipitate (½)
(ii) CH3CHO + 3I2 + 3NaOH → CI3CHO + 3NaI + 3H2O (1)
CI3CHO + NaOH → CHI3 + HCOO- Na+ (1)
(iii) CHI3 is hydrolysed / decomposed by the NaOH (1)
(iv) NaClO / ClO- oxidises (1)
KI / I- to I2 (1)
b) (i)
I2 & NaOH oxidises (1)
CH3CHOH to CH3CO (1)
(ii) less reactive/ (1)
c) A, E, F, G (1 each)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 15
GRIGNARD REAGENTS
a) Preparation
Equation
b) (i)
Magnesium (½) and iodomethane / methyl iodide (½)
in dry (½) ether (½) under reflux condenser (½)
with iodine (½) as a catalyst (½)
Maximum 3 marks
CH3I + Mg → CH3MgI (1)
Reagent:
methanal (1)
Intermediate: CH3−CH2OMgl (1)
Conditions for second stage:
Acidic hydolysis (1)
(ii) Reagent: Solid (1) carbon dioxide (1) (Allow ‘Drikold’ or ‘dry ice’ for both marks)
O
(1)
Intermediate: CH3C
OMgI
Conditions for second stage
Acidic hydolysis (1)
c) A = propanal
B = propanone
C = butan-2-ol
D = 2-methylpropan-2-ol
(Award 1 mark for name or formula in each case )
Peak at m/z 29 due to CH3CH2+ / C2H5+ (1)
Peak at m/z 30 due to CH(OH)+ (1) Not CH2O+
(Deduct 1 mark if +ve charges are missing)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 16
CARBOXYLIC ACIDS I
a) (i)
Product:
Potassium ethanoate (1)
Type of reaction: Acid-base / neutralisation (1)
Equation:
(ii) Product:
CH3COOH + KOH → CH3COOK + H2O (1)
Magnesium ethanoate (1)
Type of reaction: Acid-base / neutralisation (1)
Equation:
b) (i)
CH3COOH + MgO → CH3COO)2Mg + H2O (1)
Reagent:
Na2CO3(aq) / NaHCO3(aq) (1)
Observation:
Effervescence / gas evolved which turns limewater milky (1)
Equation:
2RCOOH + Na2CO3 → 2RCOO− Na+ + H2O + CO2 (1)
Or RCOOH + NaHCO3 → RCOO− Na+ + H2O + CO2 (1)
Or similar alternatives
c) Reagents
Ethanoic acid (1)
Butan-1-ol (1)
Concentrated sulfuric acid (1)
Conditions
Heat / boil under reflux (1)
Equation
CH3COOH + CH3CH2CH2CH2OH ¾ CH3COOCH2CH2CH2CH3 + H2O (1)
Type of reaction
Esterification / addition-elimination / condensation (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 17
CARBOXYLIC ACIDS II
a) (i)
Polarisation of the C=O bond (1)
increases polarisation of the O⎯H bond / weakens the O⎯H bond (1)
so that there is a greater degree of dissociation / [H+] is increased (1)
(ii) -I effect / electron - withdrawing effect / high electronegativity of Cl atoms (1)
further increases polarisation of / weakens the O⎯H bond (1)
so that degree of dissociation / [H+] is increased (1)
Also, the anion CCl3COO- is stabilised (1)
by delocalisation of –ve charge (1)
Maximum 4 marks
b) (i)
Reagent
PCl5 / PCl3 / SOCl2 (1)
Equation CH3CH2COOH + PCl5 → CH3CH2COCl + POCl3 + HCl (1)
Or 3CH3CH2COOH + PCl3 → 3CH3CH2COCl + H3PO3 (1)
Or 3CH3CH2COOH + SOCl2 → CH3CH2COCl + SO2 + HCl (1)
(ii) Reagent
Conc. NH3(aq) (1) Accept just ‘NH3’
Equation CH3CH2COCl + NH3 → CH3CH2CONH2 + HCl (1)
c) (i)
Description Effervescence (1)
Gas turns limewater milky (1)
at room temperature (1)
Maximum 2 marks
Formula of product
-
+
COO Na
(1) Ionic charges must be shown
-
COO Na
+
(ii) Description Reaction on heating (1)
with conc. H2SO4 (1)
to give a fruity smell (1)
Maximum 2 marks
Formula of product
COOCH3
(1)
COOCH3
d) Benzene-1,4-dicarboxylic acid (1)
Hydrogen bonding can lead to the formation of “long-chains” of molecules (1)
but in benzene-1,2-dicarboxylic acid hydrogen bonding is largely intramolecular (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 18
ACYL CHLORIDES
a) Name
Propanoyl chloride (1)
Two advantages
Better yield / reaction goes to completion / an equilibrium mixture is not obtained (1)
Faster reaction (1)
Two disadvantages
Acyl chlorides are relatively expensive (1)
Fumes of hydrogen chloride / toxic fumes (1)
b) (i)
Acyl chloride
Benzenecarbonyl chloride / benzoyl chloride (1)
Reagent
Aqueous ammonia (1)
(ii) Addition-elimination / condensation (1)
(iii) Problem
c) (i)
(White) smoke of ammonium chloride (1)
Origin
Reaction between HCl (formed in the main reaction) and unreacted NH3 (1)
Precaution
Carry out the experiment in a fume cupboard (1)
(Allow ‘limited ammonia’)
Mix / react ethanoyl chloride and the amine at room temperature (1)
Recrystallise the product (1)
Determine its melting point (1)
Compare this m.p. with that in a data book / textbook (1)
Maximum 3 marks
(ii) Equation
Name
CH3COCl + C2H5NH2 → CH3CONHC2H5 + HCl (1)
N-ethylethanamide (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 19
ESTERS
a)
O
O
H–C
OCH2CH2CH3
Propyl methanoate
O
CH3 – C
H–C
OCH(CH3)2
Isopropyl methanoate
or 1-methylethyl methanoate
OCH2CH3
Ethyl ethanoate
O
CH3CH2– C
OCH3
Methyl propanoate
Award (1) for each formula and (1) for each name if correctly related to formula
Maximum 6 marks
b) (i)
Compound II is CH2OH
CH2OH (1)
Compound III is CH3CH2CH2COOH (1)
(ii) Compound I
Ester (1)
Compound II
Alcohol / diol (1)
Compound III
Carboxylic acid (1)
(iii) Hydrolysis / saponification (1)
(iv) Alkaline hydrolysis goes to completion (1)
Acidic hydrolysis is reversible / results in an equilibrium mixture (1)
∴ yield is higher from alkaline hydrolysis (1)
Maximum 2 marks
c) Perfumes (1)
Solvents / thinners (1)
Food flavourings (1)
Plasticisers (1)
Maximum 3 marks
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 20
NITRILES
a) n (NaOH) = 0.1 × 16.65/1000 = 1.665 × 10-3 mol (1)
∴ n (H) = 1.665 × 10-3 mol in 25.0 cm3
= 1.665 × 10-2 mol in 250 cm3 (1)
∴ Mr (H) = 1/1.665 × 10-2 = 60.06 / 60 (1)
b) If formula of H is R-COOH, Mr of (R-) is 60.06 – (12 + 32 + 1) = 15.06 (1)
∴ R- is CH3- (1)
∴ H is ethanoic acid / CH3COOH (1)
and G must be ethanenitrile / CH3CN (1)
c) (i)
CH3CN + NaOH + H2O → CH3COO-Na+ + NH3 (1)
(ii) CH3COO-Na+ + H2SO4 → CH3COOH + NaHSO4 (1)
d) Mr (CH3CN) = (15 + 12 + 14) = 41 and Mr (CH3COOH) = 60
4.0 g CH3CN ≡ 4.0 / 41 mol (1)
∴ theoretical yield of H is 60(4.0) / 41 = 5.85 g (1)
% yield is 4.5(100) / 5.85 = 76.9% / 77% (1)
e) Reducing agent LiAlH4 / H2 / Zn + dil H2SO4 (1)
Conditions For LiAlH4, dry ether (1)
For H2, Pt / Ni catalyst (1)
For Zn + dil. H2SO4, heat / boil (1)
Equation
CH3C≡N + 4[H] → CH3CH2NH2 (1)
Or CH3C≡N + 2H2 → CH3CH2NH2 (1) only if H2 is quoted as the reducing agent
Or CH3C≡N + 2Zn + 2H2SO4 → CH3CH2NH2 + 2ZnSO4 (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 21
AMIDES
a) Heat / boil (1)
with bromine (1)
and alkali / NaOH(aq) (1)
b) (i)
Ethanenitrile (1)
CH3CN (1)
(ii) Heat / distil (1)
with P4O10 (1)
(iii) Elimination (of water) / dehydration (1)
c) Smelly compound in Solution A Ethanoic acid (1)
Compound in Solution B
Sodium ethanoate (1)
Gas C
Ammonia (1)
(Accept formulae in lieu of names)
d) (i)
Peptide link (1)
(ii) Nylon (1)
(iii) Protein (1) (Allow ‘polypeptide’)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 22
AMINES
a) (i)
-
+
CH3NH2 + H2O ž CH3NH3 + OH (1)
-
(ii) The high concentration of OH ions from the NaOH (1)
disturbs equilibrium to the left hand side (1)
to give molecules of CH3NH2, which is a gas (1)
Maximum 2 marks
b) Name of A Phenylammonium chloride (1)
Name of B Phenylamine (1)
+
-
Equation 1 C6H5NH2 + HCl → C6H5NH3 Cl (1)
+
Equation 2 C6H5NH3 Cl
c) (i)
-
+ NaOH → C6H5NH2 + NaCl (1)
Strength compared with ammonia
Reasons
Stronger (1)
Electron releasing effect / +I effect of C2H5 group (1)
increases electron availability on the N atom (1)
so that a lone pair of electrons on N is donated more readily (1)
+
and a proton / H is more strongly accepted (1)
Maximum 3 marks
(ii) Strength compared with ammonia
Reasons
Weaker (1)
Lone pair of electrons on N is drawn towards the benzene ring (1)
so that electron availability on the N is reduced (1)
Donation of the lone pair occurs less readily (1)
+
and a proton / H is less strongly attracted (1)
Maximum 3 marks
d) (i)
Substitution / replacement of H on the N atom (1)
by an acyl group / RCO (1)
(ii) Ethanoyl chloride / ethanoic anhydride / benzenecarbonyl chloride (1)
(iii) N-substituted amide / substituted amide / amide (1)
(iv) Mark consequentially from (ii), e.g.
CH3COCl + C6H5NH2 → CH3CONHC6H5 + HCl (1)
Accept C6H5NHCOCH3
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 23
DIAZOTISATION
a) (i)
Treatment of a primary aromatic amine with nitrous acid to form a diazonium salt (1)
(ii) Compounds needed for diazotisation
Sodium nitrite (1) and hydrochloric acid (1)
Name of product
Benzenediazonium chloride (1)
Equation
C6H5NH2 + NaNO2 + 2HCl → C6H5N≡N Cl + NaCl + 2H2O (1)
+
(iii) Optimum temperature
-
0-10 0C (1)
Problem if the temperature were too high
Hydrolysis of benzenediazonium chloride (1)
to give phenol (1)
or decomposition of benzenediazonium chloride (1)
to give chlorobenzene (1)
Problem if the temperature were too low
b) (i)
Slow / low rate of reaction (1)
Conditions
Cold / ∼ 5 0C (1)
In alkaline solution / NaOH(aq) (1)
Observation
Red (½) precipitate (½)
Equation
+
OH
+
(1)
(ii) Coupling (1)
c) A would give a red precipitate (1)
but B would not (1)
N≡N Cl
N=N
-
OH
+
(1)
HCl (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 24
AMINO ACIDS
a)
R
H2N –C – COOH (1)
H
b) Glycine forms zwitterions (1)
+
-
NH3 – CH2 – COO (1)
Attraction between zwitterions is stronger (1)
than hydrogen bonding between molecules of glycollic acid (1)
c) (i)
+
+
Equation (low pH)
H2N – R – COOH + H ž H3N – R – COOH (1)
Equation (high pH)
H2N – R – COOH + OH
-
-
ž H2N – R – COO + H2O (1)
(ii) Amino acids exists as zwitterions (1)
which are attracted equally strongly to both electrodes (1)
d) (i)
CH3
CH2OH
CH3
CH2OH
H2N – CH – COOH + H2N – CH – COOH → H2N – CH – CONH – CH – COOH (1)
CH2OH
CH3
CH2OH
CH3
H2N – CH – COOH + H2N – CH – COOH → H2N – CH – CONH – CH – COOH (1)
(ii) They are amino acids because they both contain –NH2 and –COOH groups (1)
but they are not α-amino acids because these groups are attached to different carbon atoms (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 25
PROTEINS
a) (i)
Name
Peptide link (1)
O H
Formula
– C – N – (1)
(ii) Condensation polymerisation (1)
(iii) Amino acid residues can be in different sequences (1)
(iv) Protein molecules contain > 30 – 40 amino acid residues / polypeptides contain < 30 – 40 (1)
Protein molecules are always hydrated / polypeptides are not hydrated (1)
Maximum 1 mark
b) Hydrogen bonding holds a protein chain in a coil or spiral / α-helix (1)
Water molecules are held to a protein chain by hydrogen bonding (1)
c) (i)
Hydrolysis (1)
Heat / boil under reflux (1)
with concentrated aqueous acid / hydrochloric acid (1)
(ii) Chromatography (1)
d) They undergo hydrolysis (1)
due to the catalytic action of digestive enzymes (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 26
TEST QUESTION 1
CARBONYL COMPOUNDS
a) C 19 H 28 O 2
b) (i)
Red/yellow/orange precipitate (1)
(ii) Orange to green (1)
(iii) Orange/brown to colourless or decolourised (1)
c)
CH3 OH
CH3
HO
1 mark for the HO group and 1 mark for the rest of the molecule.
d)
CH3 O
CH3
O
1 mark for the C = O group and 1 mark for the rest of the molecule.
e)
OH
(1)
O
f) (i)
B (1)
(ii) A (1)
A2 Level
TOPIC 23 ANSWERS & MARK SCHEMES
QUESTIONSHEET 27
TEST QUESTION 1I
PHENOLS
a) Ketone (1)
b) C14H8O 4 (1)
c) C7H4O2 (1)
d) (i)
O
O− Na+
O
O− Na+
1 mark for both Na+ ions and 1 mark for the rest of the compound
(ii) The compound is ionic (1) and attracts water molecules (1), but quinizarin is non-polar and does not attract water
molecules (1)
Br
e)
H
O
Br
Br
CH
CH2
Br
1 mark for the addition of Br and 1 mark for the substitution of Br