A2 23 Ans - Frankly...
Transcript of A2 23 Ans - Frankly...
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
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
Z-pent-2-ene or (cis pent-2-ene) (1) E-pent-2-ene or (trans pent-2-ene) (1)
b) Suggestion Alicyclic ring prevents rotation / locks the molecule in position (1)
Structures
c) CH2=CHCH
2CH
3No (1)
(CH3)
2C=CH
2No (1)
CHCl=CHCl Yes (1)CHCl=CHCH
3Yes (1)
CHCl=CClCH3
Yes (1)CH
3CH=CClCH
2CH
3Yes (1)
Structural feature Both C atoms of the C=C bond must be joined to different atoms or groups (1)
C CC CCH
3CH
3CH
2
H H
CH3CH
2
CH3
H
H(1)
(1)
(1)
Cl Cl
Cl
Cl
(1)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 2
OPTICAL ISOMERISM
a) (i) Structures
(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)
This leads to (±)-CH3CH(OH)CN / dl-CH
3CH(OH)CN (1)
Or to (±)-lactic acid / dl-lactic acid (1)Maximum 4 marks
C
CH3
C2H
5
OHHC
CH3
C2H
5
HO H
(1) (1)
C
CH3
CN
O -
HC
CH3
NC
- O H(1)
c)
H C HC C C C
H
H
HH
H H
O
H
C
H
HH
CH3
CH3
CH2
CH C
O
O H
NH2
C
H
C N H
OH
H H
CH3
HO
HO
CH2C
CH3
CH3CH
2
CH2
C CH3
H
NH2 H C C
CH2OH
N
CH2Cl
CH3CH3
C O CH2
CH2
CH CH3
O CH3
CHCl
CH2
C O
CH2
CH2
CH2
* *
* * *
*
1 mark per correctly placed * and - ½ per wrongly placed *.
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)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 3
TEST QUESTION ON ISOMERISM
a) (i) Skeletal isomerism CH3CH
2CH
2CH
2OH / CH
3CH(OH)CH
2CH
3 (1)
and (CH3)
2CHCH
2OH / (CH
3)
3COH (1)
Positional isomerism CH3CH
2CH
2CH
2OH (1) and CH
3CH(OH)CH
2CH
3 (1)
Or (CH3)
2CHCH
2OH (1) and (CH
3)
3COH (1)
Functional group isomerism Accept any of the C4 alcohols (1) and any of the C
4 ethers (1)
Stereoisomerism
(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 rotationof atoms (1)∴there is no possibility of geometrical isomerism / existence of cis and trans isomers (1)
b) (i) Geometrical / cis-trans isomerism (1)
(ii)
C
CH3
C2H
5
OHHC
CH3
C2H
5
HO H(1) and (1)
(iii) Melting point / solubility (1)
CH3 CH2
C=N
CH3
.N
H
C6H5 (1)
CH3 CH2
C=N
CH3
..
NH
C6H5
(1).
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 4
STRUCTURE OF BENZENE
a) (i) Platinum catalyst, room temperature / nickel catalyst, heat (1)
(ii)
(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 additionreaction with hydrogen (1)
(ii)
X
X
(1)
(iii) Benzene does not undergo electrophilic addition reactions (1)
(1)
and
X
X(1)Benzene does not produce disubstituted isomers of the form
Electron-diffraction measurements show that all the carbon-carbon bonds are of equal length / intermediatebetween 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
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 5
STRUCTURES OF CYLOALKENES
a) Alkene I Cyclohexene (1)Alkene II 1,3-Cyclohexadiene (1)Alkene III 1,4-Cyclohexadiene (1)
b) + H
2 → (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 -239.2 kJ mol-1 (1)
Reasons π-bonds are too far apart to overlap (1)No delocalisation of π-electrons (1)∴ no increase in stability (1)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
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)
⎯⎯
⎯⎯
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 7
(ii) Aluminium chloride / iron (III) chloride (1)
(iii) Electrophilic (½) substitution (½)
(iv) Heterolytic fission / heterolysis (1)
c) Example CH3CH
2Br CH
3CH
2OH (1)
Nucleophilic substitution (1)
CH3CH
2Br CH
2 = CH
2 (1)
Elimination (1)
REACTION CONDITIONS AND PRODUCT
a) (i) Free radical (½) substitution (½)
(ii) Homolytic fission / homolysis (1)
(iii) Stage 1
Ethanolic / alcoholic NaOH(1)
Aqueous NaOH (1)
Cl Cl.. 2Cl
.(1)
CH3
+ Cl.
+ HCl (1)
Stage 2
.CH
2
+ Cl2
CH2Cl
+ HCl (1)
.CH
2
Stage 3
+ Cl.
(1)
.CH
2 CH2Cl
Or 2Cl.
Cl2
Or .CH
2 CH2 CH
2(1)
b) (i) CH3
+ Cl2
CH3
+ HCl (1)Cl
Or
CH3
+ Cl2
CH3
+ HCl (1)
Cl
(1)
2
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 8
NITRATION OF ARENES
a) Reagents Concentrated nitric acid (1)and concentrated sulfuric acid (1)
Conditions 50 - 60 °C (1)
Equation + HNO
3 → + H
2O (1)
Or C6H
6 + HNO
3 → C
6H
5NO
2 + H
2O (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
(1)
Use High explosive (1)
c) (i) C6H
5NO
2 + 6[H] → C
6H
5NH
2 + 2H
2O (1)
Or + 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)
NO2
CH3
NO2
NO2
O2N
NO2
NH2
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 9
HALOGENATION OF ARENES
a) (i) Conditions Br2 in solution (in a named solvent) (1)
Room temperature / 20 °C (1)
Equation + Br2 → (1)
Or with molecular formulae in place of the structural formulae
(ii) (Electrophilic) addition (1)
b) (i) Conditions Br2 + AlBr
3 / Al catalyst (1) (Allow AlCl
3)
Room temperature / 20 °C (1)
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)
BrBr
Br
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 10
ALKYLATION AND ACYLATION OF ARENES
a) (i) Reagent Chloromethane / bromomethane / iodomethane (1)
Conditions AlCl3 catalyst (1)
Room temperature / 20 °C (1)
Equation + CH3Cl → + HCl (1)
Or C6H
6 + CH
3Cl → C
6H
5CH
3 + HCl (1)
(ii) It is difficult to prevent further alkylation / a dimethylbenzene or trimethylbenzene is formed readily (1)
(iii) Type of reaction Oxidation (1)
Reagent KMnO4 (½) + NaOH(aq) (½)
Or K2Cr
2O
7 (½) + dil. H
2SO
4 (½)
Or dilute HNO3 (1)
b) (i) Conditions AlCl3 catalyst (1)
50 ºC / heat (1)
Equation + CH3COCl → + HCl (1)
Or C6H
6 + CH
3COCl → C
6H
5COCH
3 + HCl (1)
Name of product Phenylethanone / acetophenone (1)
(ii) Type of reaction Reduction (1)
Reagent NaBH4 / Zn + dil. H
2SO
4 / other named combination of metal and protic solvent (1)
CH3
COCH3
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 11
PHENOL
a) Identity of phenols I, II, & IV (1) (All must be correct)Basis of identification A phenol must have at least one –OH group attached directly to a benzene ring (1)
b) (i) Identity of A Sodium phenoxide / sodium phenate / C6H
5O Na (1)
Identity of B Mainly phenol / (saturated) solution of water in phenol (1)(Allow just ‘phenol’)
Equations
Or C6H
5OH + NaOH → C
6H
5O Na + H
2O (1)
Or C6H
5O Na + HCl → C
6H
5OH + NaCl (1)
(ii) Reason 1 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 The phenoxide ion / C6H
5O is stabilised (1)
by delocalisation of –ve charge (1)over O and the benzene ring (1)This encourages dissociation / ionisation / formation of C
6H
5O (1)
No scope for delocalisation in an alkoxide ion / corresponding anion from an alcohol (1)Maximum 3 marks
c) (i) White (½) precipitate (½)
(ii)
+ 3Br2 → + 3HBr (1)
(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)
+
-
OH
+ NaOH → + H2O (1)
O Na
-
+
-
+
-
O Na
+
+ HCl → + NaCl (1)
OH
-
+
-
-
OH OH
Br
BrBr
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 12
ALDEHYDES AND KETONES
a) (i) C (½) Butanone / butan-2-one (½)
(ii) A (½) Propanal (½) andB (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 NaBH
4 (1)
(ii) In dry ether (1) for LiALH4
or in solution in water or methanol for NaBH4 (1)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 13
USE OF 2,4 - DINITROPHENYLHYDRAZINE
a) C H OMoles 69.8/12 11.6/1 18.6/16
5.817 11.6 1.163 mol % (1)Ratio 5 10 1 (1)
Empirical formula is C5H
10O ; relative formula mass = 86 = relative molecular mass (1)
therefore molecular formula is also C5H10O (1)
b) Aldehydes: CH3CH
2CH
2CH
2CHO
CH3CH
2CH(CH
3)CHO
CH3CH(CH
3)CH
2CHO
CH3C(CH
3)
2CHO
Ketones: CH3COCH
2CH
2CH
3
CH3CH
2COCH
2CH
3
CH3COCH(CH
3)
2 (1 mark each)
c)
+ H2N ⎯ NHC O
NO2
NO2
N ⎯ NHC
NO2
NO2
(2) Delete 1 mark for each error+ H2O
d) Test 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 No silver mirror (1)Or no red / brown precipitate / solution remains dark blue for Fehling’s solution (1)
e) (i) 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)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 14
THE TRIIODOMETHANE REACTION
a) (i) Yellow (½) precipitate (½)
(ii) CH3CHO + 3I
2 + 3NaOH → CI
3CHO + 3NaI + 3H
2O (1)
CI3CHO + NaOH → CHI
3 + HCOO- Na+ (1)
(iii) CHI3 is hydrolysed / decomposed by the NaOH (1)
(iv) NaClO / ClO- oxidises (1)KI / I- to I
2 (1)
b) (i) I2 & NaOH oxidises (1)
CH3CHOH to CH
3CO (1)
(ii) less reactive/ (1)
c) A, E, F, G (1 each)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 15
GRIGNARD REAGENTS
a) Preparation 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
Equation CH3I + Mg → CH
3MgI (1)
b) (i) Reagent: methanal (1)
Intermediate: CH3−CH
2OMgl (1)
Conditions for second stage: Acidic hydolysis (1)
(ii) Reagent: Solid (1) carbon dioxide (1) (Allow ‘Drikold’ or ‘dry ice’ for both marks)
Intermediate:
Conditions for second stage Acidic hydolysis (1)
c) A = propanalB = propanoneC = butan-2-olD = 2-methylpropan-2-ol(Award 1 mark for name or formula in each case )
Peak at m/z 29 due to CH3CH
2+ / C
2H
5+ (1)
Peak at m/z 30 due to CH(OH)+ (1) Not CH2O+
(Deduct 1 mark if +ve charges are missing)
CH3C
OMgI
O(1)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 16
CARBOXYLIC ACIDS I
a) (i) Product: Potassium ethanoate (1)
Type of reaction: Acid-base / neutralisation (1)
Equation: CH3COOH + KOH → CH
3COOK + H
2O (1)
(ii) Product: Magnesium ethanoate (1)
Type of reaction: Acid-base / neutralisation (1)
Equation: CH3COOH + MgO → CH
3COO)
2Mg + H
2O (1)
b) (i) Reagent: Na2CO
3(aq) / NaHCO
3(aq) (1)
Observation: Effervescence / gas evolved which turns limewater milky (1)
Equation: 2RCOOH + Na2CO
3 → 2RCOO− Na+ + H
2O + CO
2 (1)
Or RCOOH + NaHCO3 → RCOO− Na+ + H
2O + CO
2 (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 + CH
3CH
2CH
2CH
2OH ¾ CH
3COOCH
2CH
2CH
2CH
3 + H
2O (1)
Type of reaction Esterification / addition-elimination / condensation (1)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
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 CCl
3COO- is stabilised (1)
by delocalisation of –ve charge (1)Maximum 4 marks
b) (i) Reagent PCl5 / PCl
3 / SOCl
2 (1)
Equation CH3CH
2COOH + PCl
5 → CH
3CH
2COCl + POCl
3 + HCl (1)
Or 3CH3CH
2COOH + PCl
3 → 3CH
3CH
2COCl + H
3PO
3 (1)
Or 3CH3CH
2COOH + SOCl
2 → CH
3CH
2COCl + SO
2 + HCl (1)
(ii) Reagent Conc. NH3(aq) (1) Accept just ‘NH
3’
Equation CH3CH
2COCl + NH
3 → CH
3CH
2CONH
2 + HCl (1)
c) (i) Description Effervescence (1)Gas turns limewater milky (1)at room temperature (1)Maximum 2 marks
Formula of product
(ii) Description Reaction on heating (1)with conc. H
2SO
4 (1)
to give a fruity smell (1)Maximum 2 marks
Formula of product
COO Na
COO Na
COOCH3
COOCH3
(1)
(1) Ionic charges must be shown
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)
+
+
-
-
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
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 (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’)
c) (i) 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 CH3COCl + C
2H
5NH
2 → CH
3CONHC
2H
5 + HCl (1)
Name N-ethylethanamide (1)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 19
ESTERS
a) O O OH – C H – C CH
3 – C
OCH2CH
2CH
3 OCH(CH
3)
2 OCH
2CH
3
Propyl methanoate Isopropyl methanoate Ethyl ethanoateor 1-methylethyl methanoate
OCH
3CH
2– C
OCH3
Methyl propanoate
Award (1) for each formula and (1) for each name if correctly related to formulaMaximum 6 marks
b) (i) Compound II is CH2OH
CH2OH (1)
Compound III is CH3CH
2CH
2COOH (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
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
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)∴ M
r (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 + H
2O → CH
3COO-Na+ + NH
3 (1)
(ii) CH3COO-Na+ + H
2SO
4 → CH
3COOH + NaHSO
4 (1)
d) Mr (CH
3CN) = (15 + 12 + 14) = 41 and M
r (CH
3COOH) = 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 / H
2 / Zn + dil H
2SO
4 (1)
Conditions For LiAlH4, dry ether (1)
For H2, Pt / Ni catalyst (1)
For Zn + dil. H2SO
4, heat / boil (1)
Equation CH3C≡N + 4[H] → CH
3CH
2NH
2 (1)
Or CH3C≡N + 2H
2 → CH
3CH
2NH
2 (1) only if H
2 is quoted as the reducing agent
Or CH3C≡N + 2Zn + 2H
2SO
4 → CH
3CH
2NH
2 + 2ZnSO
4 (1)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 21
AMIDES
a) Heat / boil (1)with bromine (1)and alkali / NaOH(aq) (1)
b) (i) Ethanenitrile (1)CH
3CN (1)
(ii) Heat / distil (1)with P
4O
10 (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’)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 22
AMINES
a) (i) CH3NH
2 + H
2O ž CH
3NH
3 + 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 CH
3NH
2, which is a gas (1)
Maximum 2 marks
b) Name of A Phenylammonium chloride (1)
Name of B Phenylamine (1)
Equation 1 C6H
5NH
2 + HCl → C
6H
5NH
3 Cl (1)
Equation 2 C6H
5NH
3 Cl + NaOH → C
6H
5NH
2 + NaCl (1)
c) (i) Strength compared with ammonia Stronger (1)
Reasons Electron releasing effect / +I effect of C2H
5 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 Weaker (1)
Reasons 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 + C
6H
5NH
2 → CH
3CONHC
6H
5 + HCl (1)
Accept C6H
5NHCOCH
3
+
-
-
+
-
+
-
+
+
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
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 C6H
5NH
2 + NaNO
2 + 2HCl → C
6H
5N≡N Cl + NaCl + 2H
2O (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 Slow / low rate of reaction (1)
b) (i) Conditions Cold / ∼ 5 0C (1)In alkaline solution / NaOH(aq) (1)
Observation Red (½) precipitate (½)
Equation
(ii) Coupling (1)
c) A would give a red precipitate (1)but B would not (1)
+
-
OH+
N≡N ClOH
N=N
+ HCl (1)
-
+
(1) (1)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 24
AMINO ACIDS
a) R
H2N –C – COOH (1)
H
b) Glycine forms zwitterions (1)
NH3 – CH
2 – 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 H
3N – R – COOH (1)
Equation (high pH) H2N – R – COOH + OH ž H
2N – R – COO + H
2O (1)
(ii) Amino acids exists as zwitterions (1)which are attracted equally strongly to both electrodes (1)
d) (i) CH3
CH2OH CH
3 CH
2OH
H2N – CH – COOH + H
2N – CH – COOH → H
2N – CH – CONH – CH – COOH (1)
CH2OH CH
3 CH
2OH CH
3
H2N – CH – COOH + H
2N – CH – COOH → H
2N – 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)
+
-
+
+
-
-
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
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)
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 26
TEST QUESTION 1CARBONYL COMPOUNDS
a) C19
H28
O2
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)CH
3O
CH3
O
1 mark for the C = O group and 1 mark for the rest of the molecule.
e) (1)
f) (i) B (1)
(ii) A (1)
O
OH
TOPIC 23 ANSWERS & MARK SCHEMESA2 Level
QUESTIONSHEET 27
TEST QUESTION 1IPHENOLS
a) Ketone (1)
b) C14
H8O
4 (1)
c) C7H
4O
2 (1)
d) (i) O− Na+
O
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 watermolecules (1)
e)
H O CH
Br
Br Br Br
CH2
1 mark for the addition of Br and 1 mark for the substitution of Br