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CCEA A2CHEMISTRY EXAM PRACTICE
Alyn G McFarlandNora Henry
MARK SCHEMEThis mark scheme is for use in conjunction with the book Chemistry Exam Practice for CCEA A2, ISBN: 978 1 78073 254 1
Colourpoint EducationalAn imprint of Colourpoint Creative LtdColourpoint HouseJubilee Business Park21 Jubilee RoadNewtownardsCounty DownNorthern IrelandBT23 4YH
This material is provided solely for use with the accompanying textbook. Unauthorised reproduction, in whole or part, without prior written permission from the publisher is prohibited.
Publisher’s Note: These answers have been written to help students preparing for the A2 Chemistry specification from CCEA. While Colourpoint Educational and the author have taken every care in its production, we are not able to guarantee that it is completely error-free. Additionally, while the mark scheme has been written to closely match the CCEA specification, it is the responsibility of each candidate to satisfy themselves that they have fully met the requirements of the CCEA specification prior to sitting an exam set by that body. For this reason, and because specifications change with time, we strongly advise every candidate to avail of a qualified teacher and to check the contents of the most recent specification for themselves prior to the exam. Colourpoint Creative Ltd therefore cannot be held responsible for any errors or omissions in this mark scheme or any consequences thereof.
v2
2
Unit A2 1:
Further Physical and Inorganic Chemistry
Answers
11
4.1 LATTICE ENTHALPY
4.1 Lattice Enthalpy
1 C [1]
2 B [1]
Enthalpy of solution = lattice enthalpy + hydration enthalpy of calcium ion + 2 × hydration enthalpy of chloride ion = + 2237 + (–1650) + 2(–364) = –141 kJ mol–1. Most common error would be forgetting the 2 × for the enthalpy of hydration of the chloride ion. This would give answer C.
3 (a) the energy required to convert one mole of gaseous atoms into gaseous ions with a single positive charge. [2]
(b) enthalpy change when one mole of a compound is formed from its elements under standard conditions. [2]
(c) enthalpy change when one mole of gaseous atoms is formed from the element in its standard state. [2]
(d) enthalpy change when one mole of an ionic compound is converted to gaseous ions. [2]
All other reactions produce a gas which would increase entropy as gases are disordered. Reaction B only contains gases but the number of moles of gas decreases so entropy decreases.
2 C [1]
Reaction is exothermic and there is an increase in entropy.
8 (a) using known concentrations of iodine solution [1]plot calibration curve/absorbance against concentration of iodine [1]set up reaction with large excess of other reactants and record absorbance against time [1]convert absorbance to concentration (using calibration curve) [1]plot concentration of iodine against time [1]take tangents and measure the gradient [1]plot graph of rate against concentration and determine order from shape of graph [1] max [6]
(b) (i) zero order [1]constant gradient/straight line [1] [2]
(ii) rate = 1.00
400 = 2.50 × 10–3 mol dm–3 s–1 [1]
(iii) horizontal line [1]at 2.50 (× 10–3) on graph [1] [2]
(c) (i) rate = k[CH3COCH
3][H+] [1]
(ii) k = 0.104 [1] mol–1 dm3 s–1 [1] [2]
99
4.3 RATES OF REACTION
9 (a)
00.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
10 20 30 40 50 60 70 80 90 100
[A] /
mol
dm
–3
time / s
correctly drawn tangent at t = 0 s [1](b) gradient of tangent calculated from drawing
rate = 0.80
34 = 0.024 [1] mol dm–3 s–1 [1] [2]
(c) 0.40 = 20 s0.20 = 40 s [1]
(d) order = 1 [1](e) rate = k[A] [1]
(f) k = 0.024
0.80 = 0.03 [1] s–1 [1] [2]
(g) Step 1 as A is a reactant and present in rate equation [1]
1010
4.4 EQUILIBRIUM
4.4 Equilibrium
1 Answer is C [1]
moles at equilibrium: CH
3COOH = 0.35
CH3CH
2OOCCH
3 = H
2O = 0.65
Kc =
[CH3CH
2OOCCH
3][H
2O]
[CH3CH
2OH][CH
3COOH]
= (0.65)2
(0.35)2 = 3.4
2 Answer is D [1]
Kc = (2.40)2 = 5.76
3 (a) CH4 + 2H
2O s CO
2 + 4H2 [1]
(b) forward reaction is endothermic as K
c increases as temperature increases [1]
(c) (i) Kc =
[CO][H2]3
[CH4][H
2O]
[1]
(ii) CH4 = 0.09 [1]
H2O = 0.245 [1]
CO = 0.155 [1]H
2 = 0.465 [1] [4]
(iii) Kc =
[CO][H2]3
[CH4][H
2O]
= (0.155)(0.465)3
(0.09)(0.245) = 0.707 mol2 dm–6 [3]
4 (a) (i) Kc =
[HI]2
[H2][I
2]
[1]
(ii) concentrations cancel as (mol dm–3)2 / (mol dm–3)2 [1](iii) equilibrium moles:
H2 = I
2 = 0.4
HI = 3.00
Kc =
(3.00)2
(0.4)(0.4) = 56.3 [3]
(b) forward reaction exothermic so at lower temperatureK
c increases [1]
position of equilibrium moves to the right [1] [2](c) no effect on position of equilibrium or K
c [1]
same number of moles of gas on both sides [1]K
c not affected by changes in pressure [1] [3]
1111
4.4 EQUILIBRIUM
5 equilibrium molesmoles of CH
3COOH = 0.010 – 9.0 × 10–3 = 1.0 × 10–3
moles of C5H
10 = 0.020 – 9.0 × 10–3 = 0.011
moles of CH3COOC
5H
11 = 9.0 × 10–3
equilibrium concentrationsCH
3COOH = 1.67 × 10–3
C5H
10 = 0.0183
CH3COOC
5H
11 = 0.015
Kc =
[CH3COOC
5H
11]
[CH3COOH][C
5H
10]
Kc =
(0.015)
(1.67×10−3)(0.0183) = 490 mol–1 dm3 [4]
Answer is given to 2 significant figures as this is appropriate based on the data in the question.
6 (a) Kc =
1
8.00 = 0.354 mol dm–3 [2]
(b) (i) moles of H2 reacting = 3.00 – 0.30 = 2.70
equilibrium molesN
2 = x – 0.90
H2 = 0.30
NH3 = 1.80
equilibrium concentrationsN
2 = 5(x – 0.90)
H2 = 1.50
NH3 = 9.00 [3]
(ii) Kc =
[NH3]2
[N2][H
2]3 =
(9.00)2
(5(x − 0.90))(1.5)3 = 8.00
5(x – 0.90) = 81
8.00 × 3.375 = 3.00
x – 0.90 = 0.60x = 1.5 [4]
(iii) no effect on Kc as only affected by changes in temperature [1]
lower pressure so position of equilibrium moves to the left [1]4 moles of gas on left and 2 moles of gas on right [1] [3]
In this buffer moles of acid = moles of salt formed so the [H+] = Ka.
pH = – log10
Ka.
17 (a) A solution which resist changes in pH on addition of small amounts of acid or alkali [1](b) lactate ion reacts with H+ and removes them maintaining pH [1]
CH3CH(OH)COO– + H+ CH
3CH(OH)COOH [1]
(explanation in terms of CH3CH(OH)COOH s CH
3CH(OH)COO– + H+ position of equilibrium
moving to the left to remove H+ ions will be accepted) [2](c) lactic acid reacts with OH– ions and removes them so maintaining pH [1]
CH3CH(OH)COOH + OH– CH
3CH(OH)COO– + H
2O [1]
(explanation in terms of CH3CH(OH)COOH s CH
3CH(OH)COO– + H+ position of equilibrium
moving to the right to replace the H+ ions which have reacted with OH– ions, i.e. H+ + OH– H2O
In this buffer calculation the solid sodium lactate added does not dilute the acid so the concentration of the acid is unchanged. The concentration, in mol dm–3, of the salt must be calculated and then used to calculate [H+].
The pH of 4.06 is obtained using the rounded [H+] (8.61 × 10–3 mol dm–3) however using 8.613 × 10–5 mol dm–3 or the value in the calculator throughout gives 4.07 and both 4.06 and 4.07 would be accepted. The moles of the acid remaining and the moles of the salt may be used to calculate [H+]. Ideally you should convert these moles to concentration, in mol dm–3, by dividing by the total volume (50.0 cm3 here) and multiplying by 1000 but it gives the same value for [H+] but you must convert both of them but using the moles if faster. In 17(d) you had to calculate the concentration, in mol dm–3, of the sodium lactate as you were using the concentration of the acid in mol dm–3
Whilst this type of calculation in (b) is beyond the specification, you should expect the unexpected at A2. However, you would be guided through it as shown here.
(c) C6H
5COOH + OH– C
6H
5COO– + H
2O [1]
benzoate ions react with OH– ions and remove them so maintaining pH [1](explanation in terms of C
6H
5COOH s C
6H
5COO– + H+ position of equilibrium moving to the right
to replace the H+ ions which have reacted with OH– ions, i.e. H+ + OH– H2O will be accepted) [2]
19 (a) nitrite ion reacts with H+ and removes them maintaining pH [1]NO
2
– + H+ HNO2 [1]
(explanation in terms of HNO2 s NO
2
– + H+ position of equilibrium moving to the left to remove H+ ions will be accepted) [2]
3 reacts with added H+/position of equilibrium moves to the left [1]
added H+ removed and pH maintained [1] [2]
2121
4.5 ACID-BASE EQUILIBRIA
Titration curves and salts21 Answer is A [1]
A salt of a strong acid and a weak base (A) will have an acidic pH. A salt of a strong acid and a strong base (C) will have a neutral pH. A salt of a weak acid and a strong base (D) will have an alkaline pH. A salt of a weak acid and a weak base (B) could be acidic, alkaline or neutral depending on the parent acid and base but it would not be as acidic as A or as alkaline as D, if the concentrations are the same.
22 Answer is D [1]
Ethanoic acid is the weaker weak acid (higher pKa so lower K
a). A weaker weak acid will have a higher
concentration of the undissociated acid so when the anion is added to water it reacts to a greater extent giving a higher pH (A– + H
2O HA + OH–). The sodium ion does not affect the pH but the ammonium ion
will cause some acidity in the solution counteracting the effect of the anion.
23 (a) HCl + NaOH NaCl + H2O [1]
(b) (i) [H+] = 0.270 mol dm–3
pH = – log10
[H+] = – log10
(0.270) = 0.57 [2]
(ii) moles of HCl = 25.0 × 0.270
1000 = 6.75 × 10–3
moles of NaOH = 6.75 × 10–3
volume of NaOH required = 6.75 × 10−3 × 1000
0.320 = 21.1 cm3 [3]
(iii)
0.00
2
4
6
8
10
12
14
1
3
5
7
9
11
13
5.0 10.0 15.0 20.0 25.0 30.0
pH
Volume of sodium hydroxide solution / cm3
starts around pH 0.57/0.6 [1]vertical region between pH 3 and 10 approximately [1]vertical region at approximately 21 cm3 [1] [3]
2222
4.5 ACID-BASE EQUILIBRIA
24 (a) [H+] = 2 × 0.145 = 0.290 mol dm–3
pH = – log10
[H+] = – log10
(0.290) = 0.54 [2](b) [OH–] = 0.500 mol dm–3
[H+] = 1.00 × 10−14
0.500 = 2.00 × 10–14 mol dm–3
pH = – log10
[H+] = – log10
(2.00 × 10–14) = 13.70 [2]
(c) moles of H2SO
4 =
25.0 × 0.145
1000 = 3.625 × 10–3
moles of KOH = 3.625 × 10–3 × 2 = 7.25 × 10–3
volume of KOH = 7.25 × 10−3 × 1000
0.500 = 14.5 cm3 [3]
(d) moles of H2SO
4 = 3.625 × 10–3
moles of H+ = 7.25 × 10–3
moles of OH– = 14.0 × 0.500
1000 = 7.00 × 10–3
moles of H+ in excess = 7.25 × 10–3 – 7.00 × 10–3 = 2.50 × 10–4
total volume = 39.0 cm3
new [H+] = 2.50 × 10−4
39.0 × 1000 = 6.41 × 10–3 mol dm–3
pH = – log10
[H+] = – log10
(6.41 × 10–3) = 2.19 [5](e) moles of H
2SO
4 = 3.625 × 10–3
moles of H+ = 7.25 × 10–3
moles of OH– = 15.0 × 0.500
1000 = 7.50 × 10–3
moles of OH– in excess = 7.50 × 10–3 – 7.25 × 10–3 = 2.50 × 10–4
total volume = 40.0 cm3
new [OH–] = 2.50 × 10−4
40.0 × 1000 = 6.25 × 10–3 mol dm–3
[H+] = 1.00 × 10−14
6.25 × 10−3 = 1.60 × 10–12 mol dm–3
pH = – log10
[H+] = – log10
(1.60 × 10–12) = 11.80 [5]
2323
4.5 ACID-BASE EQUILIBRIA
(f)
0.00
2
4
6
8
10
12
14
1
3
5
7
9
11
13
5.0 10.0 15.0 20.0 25.0 30.0
pH
Volume of potasium hydroxide solution added / cm3
starts around pH 0.54/0.5 [1]vertical region between pH 3 and 10 approximately [1]vertical region at approximately 14.5 cm3 [1] [3]
(g) phenolphthalein or methyl orange [1]indicator pH range colour change in pH range of vertical region [1] [2]
25 Answer is B [1]
Vertical region for this titration between 3 and 8/9 so only indicator with a colour change in this range would be bromophenol blue
2424
4.5 ACID-BASE EQUILIBRIA
26 (a) 2.65 (accept 2.6 to 2.7) [1](b) 20 cm3 [1](c) at 10 cm3, pH = pK
a = 4.00 [1]
Ka = 10–4.00 = 1 × 10–4 mol dm–3 [1] [2]
(d) Ka =
[ClC6H
4COO−][H+]
[ClC6H
4COOH]
[1]
(e) [H+] = 10–2.65 = 2.24 × 10–3 mol dm–3
[acid] = [H+]2
Ka
= (2.24 × 10−3)2
1 × 10−4 = 0.0502 mol dm–3 [3]
(f) moles of ClC6H
5COOH =
25.0 × 0.0502
1000 = 1.255 × 10–3
moles of NaOH = 1.255 × 10–3
concentration of NaOH = 1.255 × 10−3 × 1000
20 = 0.0628 mol dm–3 [3]
(g) adding acid:4-chlorobenzoate ion reacts with H+ and removes them maintaining pH [1]ClC
6H
4COO– + H+ ClC
6H
4COOH [1]
(explanation in terms of ClC6H
4COOH s ClC
6H
4COO– + H+ position of equilibrium moving to the
left to remove H+ ions will be accepted)
adding alkali:4-chlorobenzoic acid reacts with OH– ions and removes them so maintaining pH [1]ClC
6H
4COOH + OH– ClC
6H
4COO– + H
2O [1]
(explanation in terms of ClC6H
4COOH s ClC
6H
4COO– + H+ position of equilibrium moving to
the right to replace the H+ ions which have reacted with OH– ions, i.e. H+ + OH– H2O will be
accepted) [4]
27 (a)
N NNCH
3
CH3
all N atoms indicated [1](b) (i) A CH
3COOH + NH
3 CH
3COONH
4 [1]
B CH3COOH + NaOH CH
3COONa + H
2O [1]
C HNO3 + NH
3 NH
4NO
3 [1]
D HNO3 + NaOH NaNO
3 + H
2O [1] [4]
(ii) C and D [1]indicator pH range colour change in pH range of vertical region / titrations involving a strong acid [1] [2]
(iii) CH3COO– + H
2O CH
3COOH + OH– [1]
OH– ions cause pH > 7 [1]sodium ion does not react with water [1] [3]
(iv) NH4
+ + H2O NH
3 + H
3O+ [1]
H3O+ ions cause pH < 7 [1]
nitrate ion does not react with water [1] [3](v) salt of a strong acid and a strong base [1]
both ions in the salt do not react with water [1] [2]
2525
4.6 ISOMERISM
4.6 Isomerism
Optical isomers
1 molecules which have the same molecular formula but a different structural formula. [2]
2
CH
H
H
C
H
O
C
H
H
CH
H
H
C
O
C H
H
H
propanal propanone [2]
3 (a)
CH
H
H
O C
O
C H
H
H
CH
H
H
C
H
H
O
O
C H
methyl ethanoate ethyl methanoate [2]
(b)
CH
H
H
C
OH
O
C
H
H
propanoic acid [1]
4 An atom which has four different atoms or groups attached [1]
5 (a) no [1](b) no [1](c) yes [1](d) no [1](e) yes [1](f) yes [1](g) yes [1]
6 Answer is D [1]
7 Answer is A [1]
8 (a)
HCOH
CH2
COOH
COOH
[1]
(b) molecules which exist as non-superimposable mirror images [1](c) hydroxyl carboxyl [1]
2626
4.6 ISOMERISM
9 (a)
HC
Cl
C2H
5
CH3
H3C
C
Cl
H5C
2
H
[2](b) pass plane polarised light through the sample [1]
each enantiomer will rotate the light in a different direction [1] [2](c) 1-chlorobutane/2-chloro-2-methylpropane /1-chloro-2-methylpropane [1]
does not contain a chiral centre/no atom with 4 different groups or atoms attached [1] [2]
10 (a) It contains a carbon which has 4 different atoms/ groups attached/ has a chiral centre. [1](b)
H3C
C
H
COOH
OH HOC
H
HOOC
CH3
[2] bond angle = 109.5° [1] [3]
(c) CH3COCOOH [1]
(d) pass plane polarised light through samples of each [1]it will be rotated by the single enantiomer [1]it will be unaffected by the racemate [1]the racemate contains equal amounts of each isomer and so the two opposite rotations cancel out [1] [4]
11 Answer is B [1]
12 (a) A sample which rotates the plane of plane polarised light. [1](b)
H3C
C
H
CH2CHO
OH HOC
H
OHCH2C
CH3
[2](c) a 50:50 mixture of the 2 optical isomers [1]
one isomer rotates plane polarised light one direction and the other isomer rotates it an equal amount the other direction [1] [2]
(d) the receptor sites are stereospecific [1]
13 (a) CH3CH
2CH(OH)CH
3 [1]
(b)
H3C
C
H
CH2CH
3
OH HOC
H
H3CH
2C
CH3
[2](c) light which vibrates in one plane [1](d) pass plane polarised light into a solution of each of the isomers [1]
one isomer rotates plane polarised light to the right and the other to the left [1] [2](e)
2 they can both form a hydrogen bond with water [1]between lone pair of the O of carbonyl and δ+ H of water [1] [2]
3 2-methylpropanal is branched [1]intermolecular forces between the molecules are weaker [1] [2]
4 cool the collection vessel in ice [1]
5 all have similar van der Waals’ forces between their molecules, as there is a similar Mr/similar number of electrons [1]the boiling point of the aldehyde is much higher than the boiling point of the alkane due to permanent dipole attractions between the carbonyl groups on neighbouring aldehyde molecules [1]the boiling point of the alcohol is much higher than the boiling point of the aldehyde due to it also having hydrogen bonds between the hydroxyl groups on neighbouring alcohol molecules [1] [3]
6
Hydrogen bondH C
H
O
C
H
H
O
H
H
δ+
δ+
δ+
δ–
δ–
[2]
Oxidation and reduction of aldehydes and ketones7 (a) E [1]
highest priority groups are on opposite sides of the double bond [1] [2](b) warm with Tollens’ reagent [1] silver mirror [1]
warm with Fehling’s solution [1] red ppt [1] [2](c) O
H + [O]
O
OH
[1]acidified potassium dichromate(VI) [1] [2]
2828
4.7 ALDEHYDES AND KETONES
8 Answer is D [1]
9 Answer is B [1]
10 (a)
CH
H
H
C OH
H
H [1] ethanol [1]
(b)
CC
H
OH
H
H
H
C H
H
H [1] propan-2-ol [1]
(c)
H C
OH
O
C
H
H [1] ethanoic acid [1]
(d)
CH
H
H
C
H
HO
C H
[1] propanone [1]
11 (a) A carbonyl (aldehyde) and hydroxyl [1]B carbonyl (ketone) and hydroxyl [1] [2]
[1] per row [2](b) freshly prepare Fehling’s solution by mixing 1 cm3 of Fehling’s solution A with 1 cm3 of Fehling’s
solution B [1]add a few drops of the liquid to 1 cm3 of Fehling’s solution reagent in a test tube [1] warm in a hot water bath [1]remains blue [1] [4]
Nucleophilic addition reactions
14 (a)
C
H
H H
CCH
H
H
C
HO
H C
H
H H
CCH
H
H CN
C
HOH
H+ HCN
[1] 2-hydroxy-2-methylbutanenitrile [1] [2]
(b) nucleophilic addition [1]
O CH3CH
2CCH
3
H+
CN
O–
CH3CH
2CCH
3
CN
OH
C
H3C
CN–
H3CH
2C
[4](c) the carbonyl group is planar CN– can attack from either side [1]
equal amounts of each enantiomer forms (racemate/racemic mixture) [1]rotates plane polarised light equally in opposite directions and effect cancels [1]
2-hydroxy-2-methylpentanenitrile [1] [2](d) product is a racemic mixture/equal amounts of enantiomers [1]
one enantiomer rotates plane polarised light one direction and the other rotates it equally the opposite direct so no overall rotation [1]product from B has no chiral centre/optically inactive [1] [3]
(e) blue solution remains [1]
17 (a) C Oδ+ δ– [1]
(b) an ion or molecule, with a lone pair of electrons, that attacks regions of low electron density [2](c) the carbon δ+ is susceptible to attack by nucleophiles seeking such an electron
• place (5cm3) of 2,4-dinitrophenylhydrazine solution in a test tube/suitable container and add some drops of propanone • cool (the mixture in iced water) • filter off the crystals using suction filtration. • dry by sucking air over the crystals in the Büchner or in a low temperature oven/desiccator • determine the melting point• compare to data book to identify
(d) dissolve the impure crystals in the minimum volume of hot solvent [1]filter when hot by gravity filtration, using a hot funnel, or fluted filter paper (to remove insoluble impurities) [1]allow filtrate to cool and crystallise filter off the crystals using suction filtration [1]solid must be soluble in hot solvent, but less soluble in cold solvent [1] [4]
(e) moles of product C9H
10O
4 = 10 ÷ 238 = 0.0420
mass of propanone = 0.0420 × 58 = 2.436 gvolume of propanone = 2.436 ÷ 0.790 = 3.084 cm3
for 80% yield 3.084 × 100 ÷ 80 = 3.855cm3 = 3.9 cm3 [4](f) place some solid in a capillary tube sealed at one end/melting point tube [1]
heat slowly (using melting point apparatus) [1]record the temperature at which the solid starts and finishes melting [1]repeat and average and compare the temperature(s) with known values in a data book [1] [4]
3232
4.7 ALDEHYDES AND KETONES
19 (a)
C
CH3CH
2
CH3CH
2
H3C
H2OC
H3C
O +
+
H2N N
NO2
H
NO2
N N
NO2
H
NO2
[3](b)
C
H
H
H2OC
O +
+
H2N N
NO2
H
NO2
N N
NO2
H
NO2
[3]
3333
4.8 CARBOXYLIC ACIDS
4.8 Carboxylic Acids
Nomenclature and physical properties1 (a) 2-chlorobutanoic acid [1]
[1](b) 3-hydroxypropanoic acid [1](c) 4-methylpent-2-enoic acid [1](d) it can form hydrogen bonds [1]
between the lone pair on O and the δ+ H in water [1] [2]
3434
4.8 CARBOXYLIC ACIDS
4 Hydrogen bonds
H C
O
O
C
H
H
H
HH
O
OH
H
δ+ δ+
δ+
δ+
δ–
δ–
δ+
[2]
5 Same relative molecular mass, and so similar van der Waals’ forces between the molecules. [1]there are 2 hydrogen bonds between 2 molecules of ethanoic acid and only one between 2 molecules of propan-1-ol/hydrogen bonding occurs between two molecules of the acid which doubles the size of the molecule and increases the van der Waals’ forces [1] [2]
Preparation of acids
6 Answer is B [1]
7 (a) CH3CH
2CHO + [O] CH
3CH
2COOH [1]
(b) CH3CH
2CH
2OH+ 2[O] CH
3CH
2COOH + H
2O [1]
(c) CH3CH
2COOC
2H
5 + H
2O CH
3CH
2COOH + C
2H
5OH [1]
(d) CH3CH
2CN + 2H
2O + HCl CH
3CH
2COOH + NH
4Cl [1]
(e) CH3CH
2CN + H
2O + NaOH CH
3CH
2COONa + NH
3 [1]
CH3CH
2COONa + HCl CH
3CH
2COOH + NaCl [1] [2]
8 (a) CH3CH
2CH
2CH
2OH + 2[O] CH
3CH
2CH
2COOH + H
2O [1]
(b) indicative content:• add concentrated sulfuric acid to water in a pear shaped/round bottomed flask • swirl the solution and cool the flask to dissipate the heat and to prevent spitting • add potassium dichromate(VI) solution and swirl the mixture• add anti-bumping granules and attach a vertical condenser• add the alcohol slowly, down the condenser, to the acidified potassium dichromate(VI) solution, • cooling the reaction flask in a water bath.• heat the mixture under reflux• distil off the acid
3 (a) hydrogen bonds form [1] between the lone pair on O of the ester and the δ+ H of water [1] [2]
(b) as the length of the hydrocarbon chain increases the boiling point decreases [1]hydrocarbon chain is non-polar [1] [2]
4 (a) 2CH3CH
2COOH + CaCO
3 (CH
3CH
2COO)
2Ca + H
2O + CO2 [1]
(b) (i) CH3COOCH
3 methyl ethanoate [1]
HCOOC2H
5 ethyl methanoate [1] [2]
(ii) between propanoic acid molecules there are strong hydrogen bonds [1]between ester molecules there are dipole-dipole attractions which are weaker [1] [2]
[1](c) (i) prevents the formation of large bubbles/promotes smooth boiling [1]
(ii) catalyst [1] dehydrating agent/removes water and promotes forward reaction [1] [2]
(iii) repeated boiling and condensing of a (reaction) mixture [1](iv) sand bath/electric heater [1]
no naked flames as the organic compounds are flammable [1] [2](d) (i) distillation [1]
mixture heated in a flask with still head containing a thermometer [1]water cooled condenser connected to the still head and suitable named collecting vessel [1] [3]
(ii) stopper and shake ester with portion of sodium carbonate solution in separating funnel [1]invert and release pressure [1]allow to settle and run off layers. Discard aqueous layer/keep organic layer [1] [3]
(iii) add a spatula of anhydrous magnesium sulfate/anhydrous sodium sulfate/anhydrous calcium chloride and swirl [1] add more of the drying agent until the liquid is clear/no longer cloudy [1]decant/filter off the liquid [1] [3]
(iv) distill [1] sharp boiling point/compare to data book [1] [2]
Reaction 3: mineral acid/HCl [1] [2](e) phosphorus pentachloride/phosphorus(V) chloride [1]
14 (a) add sodium carbonate [1]with methyl ethanoate no reaction occurs [1] with propanoic acid bubbles of gas [1] [3]
(b) warm with acidified potassium dichromate(VI) solution [1]butan-2-ol orange to green [1]2 methylpropan-2-ol shows no reaction/stays orange [1] [3]
Fats and oils
15 (a) propane-1, 2, 3-triol [1]
OH
OH
OH
C
C
H
CH
H
H
H
[1] [2](b)
+ +
(CH2)
16CH
3
3CH3(CH
2)
16COOH
H2C C
O
(CH2)
16CH
3HC C
O
O
O
(CH2)
16CH
3H
2C
3H2O
CO
H2C
HC
OH
OH
H2C OH
O
[2]
3939
4.9 DERIVATES OF CARBOXYLIC ACIDS
(c)
++
(CH2)
14CH
3
3CH3(CH
2)
14COONa
H2C C
O
(CH2)
14CH
3HC C
O
O
O
(CH2)
14CH
3H
2C
3NaOH
CO
H2C
HC
OH
OH
H2C OH
O
[2](d) heat with methanol [1]
acid catalyst [1]
++
(CH2)
14CH
3
3CH3(CH
2)
14COOCH
3
H2C C
O
(CH2)
14CH
3HC C
O
O
O
(CH2)
14CH
3H
2C
3CH3OH
CO
H2C
HC
OH
OH
H2C OH
O
[2](e) a fuel, similar to diesel, which is made from vegetable sources [1](f) a reaction where the alkyl group of an ester is exchanged with the alkyl group of an alcohol [2]
16 (a)
OH
OH
OH
C
C
H
CH
H
H
H
[1] glycerol [1] [2]
4040
4.9 DERIVATES OF CARBOXYLIC ACIDS
(b)
(CH2)
7CH=CHCH
2CH=CH(CH
2)
4CH
3
(CH2)
7CH=CHCH
2CH=CH(CH
2)
4CH
3
(CH2)
7CH=CHCH
2CH=CH(CH
2)
4CH
3
H2C
H2C
HC
C
O
C
O
O
O
CO
O
[1](c) breaking up molecules by reaction with water [1]
aqueous sodium hydroxide [1] [2]
17 (a) (i)
++
(CH2)
16CH
3
3CH3(CH
2)
16COOK
C
C
C
H
H
H
H
C
OH
(CH2)
16CH
3C
O
O
O
(CH2)
16CH
3
3KOH
CO
C
C
C
H
H
H
H
H
OH
OH
OH
O
[2](ii) propane-1,2,3-triol [1]
(b) (i) C19
H38
O2 [1]
(ii) 2C19
H38
O2 +55O
2 38CO
2 + 38H
2O [2]
18 (a)
++
CH2OCOC
15H
31
CH2OCOC
15H
31
CHOCOC15
H31
CH2OH
CH2OH
CHOH 3C15
H31
COOH 3H2O
[1](b) It has no carbon-carbon double bonds/alkyl group is C
20 (a) can form hydrogen bonds between the δ+ H atom and lone pair of O of water [1]the long hydrocarbon chain in dodecanoic acid is non-polar [1] [2]
(b) solid disappears and fizzing [1](c) place some solid in a capillary tube sealed at one end [1]
heat slowly (using melting point apparatus) [1]record the temperature at which the solid starts and finishes melting [1]repeat and average the temperatures [1]if pure melting point is sharp/melting point should be same as that in data book [1] [4]
(e) each carbon also has one electron in a p orbital. The p orbitals overlap sideways to form a ring of charge above and below the plane [1] the 6 pi electrons spread across this ring and are delocalised [1] [2]
(f) CH3
[1](g) 12 [1] (h) pi electrons are spread over several atoms [1]
2 (a) bromine water [1](b) with benzene it remains orange; with cyclohexene it changes from orange to colourless [1]
3 Answer is C [1]
4 indicative content
bonding • each C has three (covalent) sigma bonds • one electron per carbon in a p orbital• p orbitals overalp overlap • delocalisation of electrons to form a pi ring
shape• planar • hexagon/6 carbon ring
type of reaction • addition causes delocalised ring to break • substitution maintains the stability of the ring
theoretical yield = 9.05 ÷ 181 = 0.05 moles0.05 moles of methyl benzoate minimum mass of methyl benzoate = 0.05 × 136 = 6.8 g [3]
(iii) concentrated nitric acid and concentrated sulfuric acid [1]HNO
3 + 2H
2SO
4 NO
2
+ + 2HSO4
– + H3O+ [1] [2]
(iv) nitronium [1](v) COOCH
3 COOCH3
NO2
+ ++
COOCH3
NO2
H H+
NO2
[3] electrophilic substitution [1] [4]
(d) (i) to remove impurities/purify the product [1] to ensure the hot solution would be saturated/crystals would form on cooling/minimise loss of product [1] [2]
(ii) Büchner funnel Büchner flask suction pump 3 correct = [2] marks, 2 correct = [1] mark [2]
4444
4.10 AROMATIC CHEMISTRY
(iii) to remove soluble impurities [1](iv) suck air over crystals in the Büchner funnel/low temperature oven/desiccator [1](v) sharp melting point [1]
matches the data book value [1] [2](vi) water would cause the melting point to be lower [1]
12 (a) O
Cl
[1](b) (i) hydrogen atom on the benzene is replaced by an acyl group [1]
secondary [1] [2](d) C as it has a chiral centre attached to the benzene ring [1]
17 Answer is B [1]
4545
Unit A2 2:
Analytical, Transition Metals, Electrochemistry and Organic
Nitrogen Chemistry
Answers
4646
5.1 MASS SPECTROMETRY
5.1 Mass Spectrometry
1 (a) ethene and carbon monoxide [1](b) M
r C
2H
4 = (2 × 12.0000) + (4 × 1.0078) = 28.0312 [1]
Mr CO = (12.0000 + 15.9949) = 27.9949 [1]
C2H
4 value matches the molecular ion peak [1] [3]
2 Answer is D [1]
(12.0000 × 10) + (14 × 1.00788 ) + 15.9949
3 (a) (relative) abundance [1](b) mass/charge ratio [1](c) 46 [1](d) a peak produced by an ion formed by the removal of one electron from a molecule [2](e) peak of greatest abundance in a mass spectrum [1]
31 [1] [2]
4 Answer is D [1]
5 Answer is C [1]
6 (a) 29 [1](b) M+1 peak [1]
due to presence of one 13C atom [1] [2](c) (i) a positively charged ion produced when the molecular ion breaks apart [1]
(ii) 31 = CH3O+ [1]
57 = CH3CH
2CO+ [1] [2]
7 any two possible fragments, e.g. CH
3+ at m/z 15
C2H
5+ at m/z 29
C3H
7+ at m/z 43
C4H
9+ at m/z 57
OH+ = 17
CH2OH+ 31 or CH
3CH
2CH
2CH
2O+ at m/z = 73 [2]
8 from IR absorption, A contains O–H/alcohol [1]C H O 70.59/12 13.72 15.69/16 5.8825 13.72 0.9806 [1]empirical formula = C
6H
14O [1]
Mr = 102 empirical formula is the molecular formulae C
6H
14O = 102 [1]
OH
OHOH
[2] for 3 structures [1] for 2 structures [6]
4747
5.1 MASS SPECTROMETRY
9 (a) butanedioic acid [1](b) (i) 56 [1]
(ii) a peak produced by a molecular ion with an increased mass due to the presence of one carbon-13 atom [1]
(iii) 119 [1](iv) 45 COOH+ [1]
73 HOOCCH2CH
2+ [1] [2]
10 (a) 46 [1](b) a peak produced by a molecular ion with an increased mass due to the presence of one
carbon-13 atom [1](c) CH
2OH+ [1]
(d) 15 CH3+ [1]
17 OH+ [1]
29 CH3CH
2+ [1] [3]
(e) C2H
5OH+ [1]
11 (a) 31 CH2OH+ [1]
57 C4H
9+ [1] [2]
(b) 59 (CH3)
2 COH+ [1]
74 C4H
9OH+ [1] [2]
(c) spectrum A [1]C
4H
9+ peak present/CH
2OH+ peak present [1] [2]
12 (a)
butan-1-ol ethoxyethane
CH
H
H
C
HH
HH
C
H
H
C OH CH
H
H
C
H
H
O
H
H
C
H
H
C H
(b) CH2OH+ [1]
(c) 29 C2H
5+ [1]
45 C2H
5+ [1]
57 C2H
5OCH
2+ [1] [3]
(d) spectrum 2 [1]ethoxyethane cannot have a peak at 31 [1] [2]
4848
5.1 MASS SPECTROMETRY
13
Compound m/z Fragment ion
butanone 29 C2H
5+
butanone 43 CH3CO+
ethyl methanoate 29 C2H
5+ or HCO+
ethyl methanoate 45 C2H
5O+ or HCOO+
ethyl methanoate 74 HCOOC2H
5+
CH3CH
2CONH
257 CH
3CH
2CO+
CH3CH
2CONH
244 CONH
2+
[1] per row [7]
14 (a) due to the isotopes of Cl-35 and Cl-37 [1](b) 35 35Cl+ [1]
it has two hydrogens on adjacent carbon (n+ 1) [1] [2](d) a high resolution spectrum is one which shows the spin-spin splitting pattern and a low
resolution nmr spectrum does not [1]
3 (a) 2 [1](b) triplet [1]
as it has two hydrogens on adjacent carbon [1] [2](c) area under the peaks which is proportional to the relative number of hydrogens atoms
in each environment [1](d) 3:2 [1]
4 (a) CDCl3/CD
2Cl
2/C
6D
6/CCl
4/tetramethylsilane [1]
(b) CH3CH
2CH
2CH
2OH 5 [1]
(CH3)
3COH 2 [1] [2]
(c)
CH3
Si
CH3
CH3
CH3
[1]tetramethylsilane [1] [2]
(d) any two from:sharp strong signal as all 12 hydrogens are in the same environment [1]unreactive/volatile liquid easily removed and recovered [1]produces a peak to the extreme right away from most other peaks [1] [2]
5 two peaks as 2 chemically equivalent environments [1] no splitting/singlets as no adjacent protons [1]integration is 1:1 same ratio of each type of hydrogen [1] [3]
6
chemical shift / ppm5 3 14 2
1
3
0
2 singlets [1]integration 1:3 [1] [2]
7 Answer is C [1]
8 Answer is A [1]
9 Answer is C [1]
5050
5.2 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY
10 (a)
CH
H
H
C
O
O
C
H
H C H
H
H[1]
CH
H
H O
O
C
C H
H
H
C
H
H[1] [2]
(b)
CH3C
CH3
CH3
C
[1]
O
O CH3
CH3C
CH3
CH3
C
[1]
O
C CH3
[2]
11 (a) 3 peaks [1] quartet triplet singlet [1]integration 3:2:1 [1] [3]
12 indicative pointsinfrared • absorption at 1000-1300 shows C–O present • absorption at 1650-1800 shows C=O present • absorption at 2850-3000 shows C–H presentnmr• there are 3 peaks which indicates 3 different environments of chemically equivalent hydrogen atoms• the integration ratio = 0.8:1.2:1.2 – simplest whole number ratio is 2:3:3 • the quartet at chemical shift 4 must be due to a CH
2 group with an adjacent CH
3
• the singlet at chemical shift 2 must be due to a CH3 with no H atoms bonded to adjacent
carbon atoms• the quartet at chemical shift 1.2 must be due to a CH
14 (a) due to TMS standard [1](b) proton(s) adjacent to CH
2/2H atoms bonded to adjacent carbon [1]
split into n+1 (2+1=) 3 peaks [1] [2] (c) proton(s) adjacent to a methyl/CH
3 group/3H atoms bonded to adjacent carbon [1]
split into n+1 (3+1 =) 4 peaks [1] [2](d) protons adjacent to nitrogen which is deshielding/withdraws electrons [1](e) an extra peak [1]
integration would show 3:2:1:3 [1] some/all the peaks would change their chemical shift [1] max [2]
(f) a signal which appears as 3 lines in the approximate intensity ratio 1:2:1 [2]
5252
5.3 VOLUMETRIC ANALYSIS
5.3 Volumetric Analysis
Iodine-thiosulfate titrations1 Answer is C [1]
2 (a) blue-black to colourless [1](b) H
2O
2 + 2H+ + 2I– I
2 + 2H
2O [2]
(c) using titrations 2 and 4 [1] concordant results/within 0.1 cm3 [1]given to 3 significant figures [1] [3]
(d) moles of thiosulfate = 17.9 × 0.245
100 = 4.3855 × 10–3
moles of I2 =
4.3855 × 10−3
2 = 2.19275 × 10–3
moles of H2O
2 = 2.19275 × 10–3
concentration of H2O
2 = 2.19275 × 10–3 × 40 = 0.0877 (mol dm–3) [4]
(e) ratio H2O
2:H
2SO
4 = 1:1
moles of H2SO
4 required = 2.19275 ×10–3
volume of 1.50 mol dm–3 H2SO
4 required =
2.19275 × 10−3 × 10001.50
= 1.46 cm3 [3]
(f) react with H2O
2 and required to form triiodide ion or /allow iodine to remain in solution [1]
3 (a) indicative points• weigh out accurately 2.89 g of solid KIO
3
• dissolve in suitable volume of deionised water• transfer to a 250 cm3 volumetric flask• wash all apparatus with deionised water with washings going into volumetric flask• add deionised water until the bottom of the meniscus is on the line/mark• stopper and invert flask to mix
(iii) it forms an irreversible complex with the iodine when large amount of iodine present [1]
(c) moles of IO3– =
25.0 × 0.05401000
= 1.35 × 10–3
moles of S2O
32– required = 1.35 × 10–3 × 6 = 8.10 × 10–3
volume of Na2S
2O
3 required =
8.10 × 10−3 × 10000.445
= 18.2 cm3 [3]
5353
5.3 VOLUMETRIC ANALYSIS
4 (a) (i) colourless to brown [1]iodine produced in reaction between iodate(V) ions and iodide ions [1]IO
3– + 6H+ + 5I– 3I
2 + 3H
2O [1] [3]
(ii) chloride in HCl could be oxidised [1]sulfur in H
2SO
4 in highest oxidation state [1] [2]
(b) (i) starch [1]blue-black to colourless [1] [2]
(ii) 2S2O
32– + I
2 S
4O
62– + 2I– [1]
(iii) moles of IO3– =
25.0 × 0.01501000
= 3.75 × 10–4
moles of S2O
32– = 3.75 × 10–6 × 6 = 2.25 × 10–3
moles of S2O
32– in 1.00 dm3 =
2.25 × 10−3
22.5 × 1000 = 0.1
RFM of Na2S
2O
3.xH
2O =
24.80.1
= 248
xH2O = 248 – 158 = 90
x = 9018
= 5 [5]
5 Answer is C [1]
moles of S2O
32– =
12.5 × 0.1001000
= 1.25 × 10–3
moles of H2O
2 =
1.25 × 10−3
2 = 6.25 × 10–4
molarity of diluted H2O
2 solution = 6.25 × 10–4 × 40 = 0.0250 M
molarity of original solution = 0.025 × 25 = 0.625 M
6 (a)
Rough titration
Titration 1 Titration 2 Titration 3
Initial burette reading /cm3
Final burette reading /cm3 24.20 42.60
Titre /cm3 12.70 11.85
(b) estimate the end point/faster subsequent accurate titrations [1](c) lower concentration of sodium thiosulfate [1]
higher titre [1] [2](d) 11.55 cm3 [1]
titres from titrations 1 and 3 as they are concordant [1] [2](e) 3
26 [1]
5454
5.3 VOLUMETRIC ANALYSIS
(f) moles of S2O
32– =
11.55 × 0.6401000
= 7.392 × 10–3
moles of IO3– in 25.0 cm3 =
7.392 × 10−3
6 = 1.232 × 10–3
moles of IO3– in 250.0 cm3 = 1.232 × 10–3 × 10 = 0.01232
RFM of MIO3 =
2.440.01232
= 198
RAM of M = 198 – 175 = 23
Identity of M = Na/sodium [5]
7 (a) Cu + 4H+ + 2NO3– Cu2+ + 2NO
2 + 2H
2O [1]
(b) 2NO3– + 4H+ + 2I– I
2 + 2NO
2 + 2H
2O [1]
(c) 2Cu2+ + 4I– 2CuI + I2 [1]
(d) creamy white colour is due to precipitate of CuI present [1]
(e) moles of S2O
32– =
14.2 × 0.1251000
= 1.775 × 10–3
moles of I2 =
1.775 × 10−3
2 = 8.875 × 10–4
moles of Cu2+ = moles of Cu = 8.875 × 10–4 × 2 = 1.775 × 10–3
mass of Cu = 1.775 × 10–3 × 64 = 0.1136 g
% Cu = 0.11360.245
× 100 = 46.4 % [5]
Manganate(VII) titrations8 Answer is B [1]
9 (a) graduated pipette with pi pump [1]rinse pipette with iron(II) sulfate solution [1]pipette 10.0 cm3 of the solution into a 250 cm3 volumetric flask [1]add deionised water until the bottom of the meniscus is on the line or mark [1]stopper flask and invert to mix [1] [5]
(b)
Initial burette reading / cm3
Final burette reading / cm3 Titre / cm3
Rough titration 0.0 22.5 22.5
Accurate titration 1 22.5 44.1 21.6
Accurate titration 2 20.7 42.4 21.7
[1](ii) 21.7 cm3 (allow 21.65 cm3) [1]
5555
5.3 VOLUMETRIC ANALYSIS
(iii) moles of = 21.7 × 0.0175
1000 = 3.7975 × 10–4
moles of Fe2+ in 25.0 cm3 = 3.7975 × 10–4 × 5 = 1.899 × 10–3
moles of Fe2+ in 250.0 cm3 = 0.01899
moles of Fe2+ in original 10.0 cm3 = 0.01899
molarity of FeSO4 solution = 0.01899 × 100 = 1.90 M [5]
(iv) concentration in g dm–3 = 1.90 × 152 = 289 g dm–3
solubility in g/100 g water = 28.9 g/100 g water [2]
10 (a) 5Fe2+ + MnO4– + 8H+ 5Fe3+ + Mn2+ + 4H
2O [1]
(b) colourless to pink [1]
(c) moles of MnO4– =
17.85 × 0.01501000
= 2.6775 × 10–4
moles of Fe2+ = 2.6775 × 10–4 × 5 = 1.339 × 10–3
moles of Fe2+ in 250 cm3 = 0.01339
mass of FeSO4 = 0.01339 × 152 = 2.035 g
% Fe2SO
4 =
2.0352.75
× 100 = 74 % [5]
11 (a) 5Fe2+ + MnO4– + 8H+ 5Fe3+ + Mn2+ + 4H
2O [1]
(b) 5C2O
42– + 2MnO
4– + 16H+ 10CO
2 + 2Mn2+ + 8H
2O [1]
(c) 5FeC2O
4 + 3MnO
4– + 24H+ 5Fe3+ + 10CO
2 + 3Mn2+ + 12H
2O [2]
(d) (i) purple colour of [1]last drop of MnO
4– added not decolourised [1] [2]
(ii) moles of MnO4– =
17.7 × 0.001201000
= 2.124 × 10–5
moles of FeC2O
4 in 25 cm3 =
2.124 × 10−5
3 × 5 = 3.54 × 10–5
[FeC2O
4] = 0.001416 mol dm–3
Concentration of FeC2O
4 = 0.001416 × 144
= 0.204 g dm–3 = 204 mg dm–3 [4]
12 (a) moles of MnO4– =
20.8 × 0.01301000
= 2.704 × 10–4
moles of Fe2+ in 25 cm3 = 2.704 × 10–4 × 5 = 1.352 × 10–3
moles of Fe2+ in 1 dm3 = 1.352 × 10–3 × 40 = 0.05408 mol dm–3
moles of (NH4)
2 Fe(SO
4)
2.xH
2O = 0.05408
RFM = 19.6
0.05408 = 362.4
RFM of (NH4)
2Fe(SO
4)
2 = 284
RFM due to xH2O = 362.4 – 284 = 78.4
x = 78.418
= 4.4 [5]
(b) average degree of hydration across the sample [1]
5656
5.3 VOLUMETRIC ANALYSIS
(c) moles of (NH4)
2Fe(SO
4)
2.6H
2O =
19.6392
= 0.05
moles of Fe2+ in 25.0 cm3 = 0.0540
= 1.25 × 10–3
moles of MnO4– required =
1.25 × 10−3
5 = 2.5 × 10–4
volume of MnO4– required =
2.5 × 10−4 × 10000.0130
= 19.2 cm3 [4]
Back titrations13 (a) colourless to pink [1]
(b) Mg + 2HCl MgCl2 + H
2 [1]
2Al + 6HCl 2AlCl3 + 3H
2 [1]
Ca + 2HCl CaCl2 + H
2 [1]
Zn + 2HCl ZnCl2 + H
2 [1] [4]
(c)
Metal Mean titre / cm3 Percentage of metal in sample / %
Magnesium 19.9 78.4 [1]
Aluminium 12.0 80.1 [1]
Calcium 29.1 [1] 75.2
Zinc 32.8 [1] 86.7
Method for Mg
moles of NaOH = moles of HCl in 25 cm3 = 19.9 × 0.075
1000 = 1.4925 × 10–3
moles of HCl in 1 dm3 = 1.4925 × 10–3 × 40 = 0.0597
moles of HCl added initially = 50.0 × 2.50
1000 = 0.125
moles of HCl which reacted with the metal = 0.125 – 0.0597 = 0.0653
moles of metal = 0.0653
2 = 0.03265
mass of metal = 0.03265 × 24 = 0.7836 g
% Mg = 0.7836
1.00 × 100 = 78.4 %
Method for Al
moles of NaOH = moles of HCl in 25 cm3 = 12.0 × 0.075
1000 = 9.00 × 10–4
moles of HCl in 1 dm3 = 1.4925 × 10–4 × 40 = 0.0360
moles of HCl added initially = 50.0 × 2.50
1000 = 0.125
moles of HCl which reacted with the metal = 0.125 – 0.0360 = 0.0890
moles of metal = 0.0890
3 = 0.02967
mass of metal = 0.02967 × 27 = 0.801 g
% Mg = 0.8011.00
× 100 = 80.1 %
5757
5.3 VOLUMETRIC ANALYSIS
Method for Camass of metal = 0.752 g
moles of metal = 0.752
40 = 0.0188
moles of HCl which reacted with metal = 0.188 × 2 = 0.0376
moles of HCl added initially = 50.0 × 2.50
1000 = 0.125
moles of HCl left over = 0.125 – 0.0376 = 0.0874
moles of HCl in 25 cm3 = moles of NaOH required = 0.0874
40 = 2.185 × 10–3
volume of NaOH = 2.185 × 10−3 × 1000
0.0750 = 29.1 cm3
Method for Znmass of metal = 0.867 g
moles of metal = 0.867
65 = 0.0133
moles of HCl which reacted with metal = 0.133 × 2 = 0.0266
moles of HCl added initially = 50.0 × 2.50
1000 = 0.125
moles of HCl left over = 0.125 – 0.0266 = 0.0984
moles of HCl in 25 cm3 = moles of NaOH required = 0.0984
40 = 2.46 × 10–3
volume of NaOH = 2.46 × 10−3 × 1000
0.0750 = 32.8 cm3
(d) calcium also reacts with water [1]
14 (a) CaCO3 + 2HCl CaCl
2 + CO
2 + H
2O [1]
NaOH + HCl NaCl + H2O [1] [2]
(b) colourless to pink [1](c) 13.9 cm3 [1] (allow 13.85)
titrations 1 and 3 are concordant [1] [2]
(d) moles NaOH = moles of HCl in 25 cm3 = 13.9 × 0.350
moles of HCl which reacted with CaCO3 = 0.075 – 0.04865 = 0.02635
moles of CaCO3 =
0.026352
= 0.013175
mass of CaCO3 = 0.013175 × 100 = 1.3175 g
mass of CaCO3 in 1 tablet =
1.31755
= 0.2635 g = 264 mg [5]
5858
5.3 VOLUMETRIC ANALYSIS
15 moles NaOH = moles of HCl in 25 cm3 = 15.6 × 0.100
1000 = 1.56 × 10–3
moles HCl in 250 cm3 = 1.56 × 10–3 × 10 = 0.0156
moles of HCl added initially = 50.0 × 1.00
1000 = 0.0500
moles of HCl which reacted with CaCO3 = 0.0500 – 0.0156 = 0.0344
moles of CaCO3 =
0.03442
= 0.0172
mass of CaCO3 = 0.0172 × 100 = 1.72 g
mass of CaCO3 in 1 tablet =
1.722.00
= 86 % [5]
16 (a) Ba(OH)2 + 2HCl BaCl
2 + 2H
2O [1]
(b) moles NaOH = moles of HCl in 25 cm3 = 27.95 × 0.500
1000 = 0.013975
moles of HCl added initially = 50.0 × 0.750
1000 = 0.0375
moles of HCl which reacted with Ba(OH)2= 0.0375 – 0.013975 = 0.023525
moles of Ba(OH)2 =
0.0235252
= 0.0117625
mass of Ba(OH)2 = 0.0117625 × 171 = 2.011 g
mass of Ba(OH)2 in 1 tablet =
2.0112.25
= 89.4 % [5]
17 (a) 2Cr + 3O2 Cr
2O
3 [1]
(b) Cr2O
3 + 6HCl 2CrCl
3 + 3H
2O [1]
(c) colourless to green [1](d) green colour diluted so it does not interfere with the colour change of the indicator [1](e) mass of Cr which reacts = 4.046 g
moles of Cr which reacts = 4.046
52 = 0.0778
moles of Cr2O
3 formed =
0.077812
= 0.0389
moles of HCl which react = 0.0389 × 6 = 0.2334
moles of HCl added initially = 100.0 × 3.50
1000 = 0.350
moles of HCl remaining = 0.350 – 0.2334 = 0.1166
moles of HCl in 25 cm3 = moles of NaOH required = 0.1166
80 = 1.4575 × 10–3
volume of NaOH required = 1.4575 × 10−3 × 1000
0.145 = 10.1 cm3 [5]
5959
5.3 VOLUMETRIC ANALYSIS
(f) moles of NaOH = moles of HCl in 25 cm3 = 17.4 × 0.145
1000 = 2.523 × 10–3
moles of HCl in 2 dm3 = 2.523 × 10–3 × 80 = 0.20184
moles of HCl added initially = 100.0 × 3.50
1000 = 0.350
moles of HCl which reacted with Cr2O
3 = 0.350 – 0.20184 = 0.14816
moles of Cr2O
3 =
0.148166
= 0.02469
moles of Cr = 0.02469 × 2 = 0.04938
mass of Cr which reacted = 0.04938 × 52 = 2.568 g
% of Cr which reacted = 2.5684.25
× 100 = 60.4 % [5]
6060
5.4 CHROMATOGRAPHY
5.4 Chromatography
1 Answer is D [1]
2 (a)
SpotDistance moved by
spot in solvent 1 / mmDistance moved by
spot in solvent 2 / mmRf value solvent 1 Rf value solvent 2
W 30 1630
122 = 0.246
1687
= 0.184
X 72 2072
122 = 0.590
2087
= 0.230
Y 90 4390
122 = 0.738
4387
= 0.494
Z 40 8040
122 = 0.328
8087
= 0.920
[2] per column [8]
(b) indicative points• draw two base lines in pencil close to edges of paper• spot sample onto the intersection of the lines using a capillary tube• allow to dry and repeat to make spot concentrated• place paper in tank containing a small amount of solvent and cover with lid• allow solvent to run up over the spots until almost at top of the paper• remove and mark position of solvent front; allow to dry• rotate the paper through 90 ° and repeat run in second solvent
(c) solvent 1 more polar than solvent 2 [1]Y greater R
f than Z in solvent 1 / Z greater R
f than Y in solvent 2 [1] [2]
3
Chromatography Mobile phase Stationary phase
Gas-liquid chromatographyinert carrier gas/
nitrogen/helium etc [1]film of liquid on solid
support
TLC solvent [1] silica/alumina [1]
Paper chromatography solvent [1]water bonded to
cellulose on the paper [4]
6161
5.4 CHROMATOGRAPHY
4 (a) pencil does not run in the solvent/does not interfere with the results [1](b)
Dis
tanc
e / m
m A
Solvent front
D
B
C
0
10
20
30
40
50
60
70
solvent front marked at 65 mmspot A at 40 mmspot B at 12.5 mmspot C at 57.5 mmspot D at 20 mm [3]
(c) dye C [1]runs furthest in non-polar solvent [1] [2]
5 (a) avoid contamination of TLC plate from hands [1](b) not to dissolve the sample/remain below the sample [1](c) provide better separation of components [1](d) toxic solvent [1](e) compounds are colourless so position viewed under UV [1]
(f) (i) Rf =
682
= 0.0732 [2]
(ii) moves the least [1]more attracted to polar stationary phase [1] [2]
(iii) provide better separation of the components/prevent spots overlapping [1]
6 (a) stationary phase: liquid/solvent on solid support [1]mobile phase: inert carrier gas [1]sample enters as a gas [1]components partition between liquid and gas phase [1] [4]
(b) some components may have the same retention time under these conditions [1](c) each component analysed by mass spectrometry [1]
identified as pure substance/mixture by comparison with database of known spectra [1] [2](d) total integration = 8.3
(b) amino acid are colourless and ninhydrin stains them [1](c) UV/iodine [1](d) measure the distance moved by the spot to the centre of the spot [1]
measure the distance moved by the solvent [1]both from base line/origin [1]
Rf =
distance moved by spotdistance moved by solvent
[1] [4]
(e)
Solvent front
Origin
lys
cysval
tyrphe
lys at approximately 10 from base linecys at approximately 23 from base lineval at approximately 27 from base linetyr at approximately 34 from base linephe at approximately 39 from base line [3]
(f) lysine most polar and phenylalanine least polar [1] solvent relatively non-polar and silica stationary phase polar [1]lysine binds to stationary phase/less soluble in solvent but phenylalanine more soluble in solvent [1] [3]
6363
5.5 TRANSITION METALS
5.5 Transition Metals
General properties of transition metals and complexes1 Answer is B [1]
2 Answer is A [1]
3 Answer is C [1]
4 Answer is B [1]
5 (a)
Ni
OH2
OH2
2+OH2
OH2
H2O
H2O
[1]octahedral [1] [2]
(b) [Ni(H2NCH
2CH
2NH
2)
3]2+ + edta4– [Ni(edta)]2– + 3H
2NCH
2CH
2NH
2 [1]
(c) 1,2-diaminoethane [1](d) 6 [1](e) +3 [1](f) (i) same number and type of co-ordinate/covalent bonds broken and formed [1]
Ni—N co-ordinate/covalent bonds [1] [2](ii) increase in entropy [1]
4 species in solution on left and 7 on right [1] [2](g) (i) two chloro ligands beside each other [1]
(ii)
Co
NH3
Cl
+NH3
NH3
Cl
H3N
[1]
6 (a) co-ordination number changes from 6 to 4 [1]chloro ligands too large to have 6 around the central ion [1] [2]
(b) (i) green solution formed [1](ii) NiSO
4 + 6H
2O [Ni(H
2O)
6]2+ + SO
42– [1]
(iii) Ag+(aq) + Cl–(aq) AgCl(s) [2]
(iv) moles of NiSO4.6H
2O = moles of [Ni(H
2O)
6]2+ in 100 cm3 =
3.30263
= 0.01255
moles of [Ni(H2O)
6]2+ in 10 cm3 = 1.255 × 10–3
moles of HCl added = moles of Cl– added = 10 × 1.5
1000 = 0.015
moles of Cl– which reacted = 1.255 × 10–3 × 4 = 0.00502
(v) white (precipitate) [1]moles of AgCl formed = 9.98 × 10–3
mass of AgCl formed = 9.98 × 10–3 × 143.5 = 1.432 g = 1400 mg [2] [3]
6464
5.5 TRANSITION METALS
7 (a) a ligand which uses two lone pairs of electrons to form two co-ordinate bonds with a central metal atom or ion in a complex [2]
(b) aminoethanoate (ion) [1](c) [Cu(H
2NCH
2COO)
3]– [1]
(d) –
OFe
NH2
O
NH2 NH
2
CH2
H2
O
CH2
C
C
C
CO
O
O
[1]
8 (a) [Fe(H2O)
6]3+ + 2C
2O
42– [Fe(H
2O)
2(C
2O
4)
2]– + 4H
2O [1]
(b) ethanedioate (ion) [1](c) large ion so three do not fit around the central ion [1](d) same number and type of co-ordinate/covalent bonds broken and formed [1]
Fe—O co-ordinate/covalent bonds [1]increase in entropy [1]3 species in solution on left and 5 on right [1] [4]
(e) 6 [1]
9 (a) hexaaquacopper(II) (ion) [1](b) N less electronegative than O [1]
lone pair more readily donated from N/more stable complex formed [1] [2](c) high conc of Cl– [1] pushes equilibrium to right hand side [1] [2](d) blue solution [1] changes to a deep/dark blue solution [1] [2](e)
Complex Shape Bond angle / ° Co-ordination number
[Cu(H2O)
6]2+ octahedral 90 6
[CuCl4]2– tetrahedral 109.5 4
[1] per row [2]
10 (a) element which forms at least one stable ion with a partially filled d-subshell [1](b) a central metal atom or ion with ligands bonded by co-ordinate bonds [1](c) (i) oxidation state of Fe changes from +2 to +3 = oxidation [1]
oxidation state of Ir changes from +4 to +3 = reduction [1]oxidation and reduction occur in the same reaction = redox [1] [3]
(ii) K2IrCl
6 [1]
(iii) hexachloroiridate(III) (ion) [1]
6565
5.5 TRANSITION METALS
(d) (i) [AuCl2]– = linear [1]
[AuCl4]– = tetrahedral [1] [2]
(ii) disproportionation = oxidation and reduction of the same element in the same reaction [1]oxidation state of Au changes from +1 in [AuCl
2]– to 0 in Au = reduction [1]
oxidation state of Au changes from +1 in [AuCl2]– to +3 in [AuCl
4]– = oxidation [1] [3]
11 (a) 2KCl + PdCl2 K
2[PdCl
4] [1]
(b)
PdCl
2– 2–
90°
ClCl
ClNi
ClCl
109.5°
[1] [1]
Cl
Cl
[2](c) (i) [NiCl
4]2– + 3H
2NCH
2CH
2NH
2 [Ni(H
2NCH
2CH
2NH
2)
3]2+ + 4Cl– [2]
(ii) [NiCl4]2– + edta4– [Ni(edta)]2– + 4Cl– [2]
(iii)
Ligand Monodentate Bidentate HexadentateCN–
H2O
Cl–
1,2-diaminoethane
edta4–
all correct [2]; one error [1]; more than one error [0] [2]
Colour of transition metal ions and their identification12 Answer is B [1]
13 Answer is D [1]
14 (a) CrCl3.6H
2O [1]
(b) dissolve in water and add dilute nitric acid/dissolve in dilute nitric acid [1]add silver nitrate solution [1]white precipitate [1]add aqueous ammonia and white precipitate is soluble forming a colourless solution [1] [4]
(c) indicative points• make solutions of both solids• add sodium hydroxide solution until in excess to a sample of the solution• Cr3+ – green-blue precipitate which is soluble in excess• Ni2+ – green precipitate which is insoluble in excess• add aqueous ammonia until in excess to a sample of the solution• Cr3+ – green-blue precipitate• Ni2+ – green precipitate which is soluble in excess forming a blue solution• Cr3+ + 3OH– Cr(OH)
(d) green precipitate [1]which remains on addition of excess sodium hydroxide solution [1] [2]
16 Answer is C [1]
17 (a)
Transition metal ion
Observation on addition of a few drops of ammonia solution
Observation on addition of excess ammonia solution
Mn2+
Fe2+ ppt is insoluble [1]
Fe3+ brown ppt [1] ppt is insoluble [1]
Cr3+ green-blue ppt [1] ppt is insoluble [1]
Ni2+ green ppt [1]ppt is soluble forming
a blue solution [1]
Co2+ blue ppt [1]ppt is soluble forming a yellow solution [1]
Cu2+
each error [–1] [4](b) Cu2+ + 2OH– Cu(OH)
2 [1]
(c) Cu(OH)2 + 4NH
3 + 2H
2O [Cu(NH
3)
4(H
2O)
2]2+ + 2OH– [2]
(d) 2Mn(OH)2 + ½O
2 Mn
2O
3.2H
2O [2]
6767
5.5 TRANSITION METALS
(e) (i) green solution to yellow [1]yellow solution to orange [1] [2]
(ii) Cr(OH)3 + H
2O CrO
42– + 5H+ + 3e–
(or Cr3+ + 4H2O CrO
42– + 8H+ + 3e–) [2]
(iii) 2Cr(OH)3 + 3H
2O
2 2CrO
42– + 4H+ + 4H
2O
(or 2Cr3+ + 2H2O + 3H
2O
2 2CrO
42– + 10H+) [2]
(iv) 2CrO42– + 2H+ Cr
2O
72– + H
2O [1]
Vanadium chemistry18 Answer is A [1]
Use a table to answer this style of question
Oxidation Ag Ag+ Fe Fe2+ Sn Sn2+ SO2 SO
42–
Oxidation potential / V –0.80 V +0.44 –0.14 –0.17
Vanadium reduction
+5 +4 + 1.00 V
+4 +3 + 0.32 V ×
+3 +2 – 0.26 V × × ×
Once all the information is in the table add together the oxidation potential value (negative of the reduction electrode potential given) and the vanadium reduction electrode potential and if you get a positive value then it is feasible () and if it is negative then it is not feasible (×). Note that iron would reduce vanadium from +5 to +2 whereas both tin and sulfur dioxide would reduce vanadium from +5 to +3.
19 (a)
Vanadium compound Oxidation state of vanadium Colour of compound
NH4VO
3+5 yellow
VSO4
+2 violet
VCl3
+3 green
VO(NO3)
2+4 blue
[1] per row [4](b) (i) VO
3– + 2H+ VO
2+ + H
2O [1]
(ii) vanadium oxidation state remains at +5 [1](c) (i) 2VO
2Cl + 8HCl + 3Zn 3ZnCl
2 + 2VCl
2 + 4H
2O [2]
(ii) initial colour = yellowfinal colour = violet [1]
6868
5.5 TRANSITION METALS
20 (a) SO2 + ½O
2 SO3 [1]
(b) yellow [1](c) vanadium(V) oxide is solid and reactants/SO
2 + O
2 are gases [1]
heterogeneous catalyst in different state to reactants [1] [2](d) (i) V
2O
5 + 2HNO
3 2VO
2NO
3 + H
2O [2]
(ii) V2O
5 + 6NaOH 2Na
3VO
4 + 3H
2O [2]
(iii) VO43– + 6H+ + e– VO2+ + 3H
2O [1]
(iv) intermediate between the yellow and blue [1]
(v) 2Na3VO
4 + 3Fe + 8H
2SO
4 3Na
2SO
4 + 3FeSO
4 + 2VSO
4 + 8H
2O [1]
(vi) VO2Cl = yellow
VOSO4 = blue
VSO4 = violet
V2(SO
4)
3 = green
all correct [2]; 1 error [1]; more than 1 error [0] [2]
21 (a) (i) Pb2+ [1]E Pb2+/Pb > E V3+/V2+ /EMF = +0.13 V [1]E Pb2+/Pb < E VO2+/V3+/EMF = – 0.45 V [1] [3]
Reduction I2 2I– Br
2 2Br– Ag+ Ag Pb2+ Pb
Reduction potential / V +0.54 +1.09 +0.80 -0.13
Vanadium oxidation
+2 +3 + 0.26 V
+3 +4 – 0.32 V ×
+4 +5 – 1.00 V × × ×
(ii) violet to green [1](iii) 2V2+ + Pb2+ 2V3+ + Pb [1]
(b) (i) 3Cl2 + 2V2+ + 4H
2O 2VO
2+ + 6Cl– + 8H+ [2]
(ii) yellow [1](iii) +2 +3 EMF = +1.36 – (–0.26) = +1.62 V [1]
complete circuit with salt bridge at 298 K [1] and external circuit with voltmeter [1]left hand cell with H
2 gas at 100 kPa [1]
into a solution of H+ at 1 mol dm–3 (e.g. 1 mol dm–3 hydrochloric acid) [1]platinum electrodes in both solutions [1]right hand cell with Fe2+ and Fe3+ both 1 mol dm–3 [1] [6]
(b) (i) 2Fe + 3Cl2 2FeCl
3 [2]
(ii) E (Cl2/Cl–) > E (Fe2+/Fe)/EMF = + 1.80 V [1]
E (Cl2/Cl–) > E (Fe3+/Fe2+)/EMF = + 0.59 V [1] [2]
(c) (i) oxidation state of O changes from –2 in H2O to 0 in O
2 [1]
oxidation state of Cl changes from 0 in Cl2 to –1 in HCl [1]
redox is oxidation and reduction occurring in the same reaction [1] [3](ii) Pt|Cl
2|Cl–||H
2O|O
2|H+|Pt
EMF = + 1.36 – (+1.23) = +0.13 V [3](iii) chlorine removed so potentially harmful bacteria no longer killed [1]
7070
5.6 ELECTRODE POTENTIALS
5 (a) salt bridge [1]completes the circuit without introducing a metal/electrical connection between the two half cells or electrodes/maintains the balance of anions and cations [1] [2]
(b) chloride ions would form a complex with Cu2+ ions [1](c) on left: Cu Cu2+ + 2e– [1]
lower concentration of Cu2+ [1]on right Cu2+ + 2e– Cu [1]higher concentration of Cu2+ [1] [4]
(d) Cu|Cu2+||Cu2+|Cu [2](e) the concentrations of Cu2+ in the two half cells would be equal [1](f) (i) 0 V/zero [1]
(ii) 0.03 = 0.34 – xx = +0.31 V [2]
6 (a) salt bridge [1]completes the circuit without introducing a metal/electrical connection between the two half cells or electrodes/maintains the balance of anions and cations [1] [2]
(b) B = copper [1]C = platinum [1] [2]
(c) D = Cu2+ ions (e.g. copper(II) sulfate/copper(II) nitrate) [1]1 mol dm–3 [1] E = Fe2+ and Fe3+ ions (e.g. iron(II) nitrate and iron(III) nitrate) [1] both 1 mol dm–3 [1] [4]
2 (a) primary amine = only one carbon atom directly bonded to the nitrogen atom and therefore has the (–NH
2) group [1]
secondary amine = two carbon atoms directly bonded to the nitrogen atom, i.e. –NH [1] [2](b)
Amine1°
(primary)2°
(secondary)3°
(tertiary)
methylamine
ethylamine
dimethylamine
phenylamine
triethylamine
[1] per correct [5]
3 (a) NH3C
CH3
H [1]N dimethylamine/N-methylmethamine [1] [2]
(b) for the secondary amine the nitrogen in the middle of the chain [1](makes the dipole less) and the hydrogen bonds between molecules are less strong and so less energy is needed to break them [1] [2]
(c) hydrogen bonds form [1]between the lone pair on N of amine and δ+ hydrogen of water/between lone pair on O of water and δ+ hydrogen of amine [1] [2]
4 (a) propylamine/1-aminopropane/propan-1-amine [1](b) N
tertiary
H3C
CH3
CH3
CH
primary
H3C
NH2
CH3
N
secondary
H3C
H
CH2CH
3
[3](c) A can form strong hydrogen bonds between molecules [1]
tertiary amines cannot form hydrogen bonds between molecules [1] [2]
tin and concentrated hydrochloric acid [1] [2](ii) phenylammonium chloride/C
6H
5NH
3Cl produced which is soluble [1]
sodium hydroxide added to liberate the amine [1] [2]
10 (a) methylbenzene [1](b)
NO2
CH3
accept any position of NO2 [1]
(c) reducing agent: tin and concentrated hydrochloric acid [1]
6[H]+NO
2
CH3
2H2O+
NH2
CH3
[2](d) ammonia/NH
3 [1]
7474
5.7 AMINES
Amines as bases11 (a) CH
3CH
2NH
2 + HCl CH
2CH
2NH
3Cl [1]
ethylammonium chloride [1] [2]
(b) 2CH3NH
2 + H
2SO
4 (CH
3NH
3)
2SO
4 [1]
methylammonium sulfate [1] [2]
(c) C6H
5NH
2 + HNO
3 C
6H
5NH
3NO
3 [1]
phenylammonium nitrate [1] [2]
(d) CH3CH
2NH
3Cl + NaOH CH
3CH
2NH
2 + NaCl + H
2O [1]
ethylamine/aminoethane/ethanamine [1] [2]
(e) 2C6H
5NH
2 + H
2SO
4 (C
6H
5NH
3)
2SO
4 [1]
phenylammonium sulfate [1] [2]
12 (a) ethylamine [1]the electron donating alkyl group releases electrons to the N atom so the lone pair on N atom is more able to accept a proton [1] [2]
(b) ammonia [1]in phenylamine the lone pair of the N atom becomes delocalised into the pi system so the electron density on the N atom and the lone pair is less available to accept a proton [1] [2]
(c) butylamine [1]larger electron donating alkyl group releases electrons to the N atom so the lone pair is more able to accept a proton [1] [2]
13 propylamine strongest then ammonia then phenylamine weakest [1]phenylamine lone pair on N (less available) delocalised into ring [1]lone pair in propylamine is more available [1]propyl group increases electron density (on N)/is an electron donating group [1] [4]
14 Answer is D [1]
15 lone pair on the nitrogen on side chain is more available/more able to be donated [1]lone pair on nitrogen in ring is delocalised into ring [1] [2]
16 (a) tin and concentrated HCl [1] sodium hydroxide [1] [2]
(b) 1,3-diaminobenzene ethane-1,2-diamine
[1] [1]
NH2
NH2
CH2N
H
H
C NH2
H
H
[2](c) both have a lone pair of electrons on the nitrogen [1]
that can bond to a hydrogen ion/proton [1] [2](d) aromatic ring is electron withdrawing/lone pair delocalised into ring [1]
the lone pair of electrons on nitrogen less available [1] [2]
7575
5.7 AMINES
Reactions of amines17 (a) C
6H
5NH
2 + CH
3COCl CH
3CONHC
6H
5 + HCl [1]
N-phenylethanamide [1] [2]
(b) CH3CH
2NH
2 + CH
3CH
2COCl CH
3CH
2CONHCH
2CH
3 + HCl [1]
N-ethylpropanamide [1] [2]
(c) C6H
5NH
2 + HNO
2 + HCl C
6H
5N
2Cl + 2H
2O [1]
benzene diazonium chloride [1] [2]
(d) CH3NH
2 + HNO
2 CH
3OH + N
2 + H
2O [1]
methanol [1] [2]
(e) C6H
5N
2Cl + C
6H
5OH C
6H
5N=NC
6H
5OH + HCl [1]
4-hydroxyazobenzene/azo dye [1] [2]
18 (a)
NH2
HClHNO2
+ N N Cl+ –
2H2O++
[2]
(b) dissolve phenylamine in (dilute) hydrochloric acid [1]below 10 °C [1]add sodium nitrite [1] [3]
(c) a reaction in which two benzene rings are linked together through an azo(–N=N–) group [1](d)
N
N N
OH NaOHN Cl+ – + +
OH H2O+ NaCl+
[2]
(e) < 10 °C [1]yellow ppt [1] [2]
19 (a) NO
2N N
+
[1]
(b) C12
H9N
3O
4 [1]
259 [1] [2]
(c) moles of G = 28
259 = 0.10811
75% 100% 0.10811 ÷ 75 × 100 = 0.1441 (1:1) number of moles of 4-nitrophenylamine = 0.1441 mass of 4-nitrophenylamine = 0.1441 × 138 = 20 g [3]
energy levels close together (so absorb in visible region) [1]less energy required for electron transitions [1]electrons move to higher energy level removing a colour [1] [4]
(f) sodium nitrite and hydrochloric acid [1]<10 °C [1] [2]
20 (a) hydrogen bonds form [1]between lone pair on N atom and δ+ hydrogen atom in water or lone pair of O atom of water with δ+ hydrogen atom of NH
2 [1] [2]
(b) H2N(CH
2)
4NH
2 + 2CH
3COCl H
3CCONH(CH
2)
4NHCOCH
3 + 2HCl [2]
(c) find the melting point of the solid [1]compare with data book/actual melting point [1] [2]
2 hydrogen bonds form [1]between the lone pair of one molecule and the δ+ H atom of the other [1] [2]
3 both have similar van der Waals’ forces between molecules [1]propanamide has stronger hydrogen bonding between molecules which takes more energy to break [1] [2]
4 (a) breaking up molecules by reaction with water [1](b) CH
3CONH
2 + HCl + H
2O CH
3COOH + NH
4Cl [1]
ethanoic acid [1] [2]
(c) CH3CONH
2 + NaOH CH
3COONa + NH
3 [1]
sodium ethanoate [1] [2]
(d) 2CH3CH
2CONH
2 + H
2SO
4 + 2H
2O 2CH
3CH
2COOH + (NH
4)
2SO
4 [1]
propanoic acid [1] [2]
5 (a) a reaction which involves the elimination of water from the amide [1](b) phosphorus pentoxide/phosphorus(V) oxide [1](c) CH
bubbles/fizzing/effervescence [1] [2](i) zwitterion forms ionic interactions with polar water molecule [1]
5 (a) sequence of amino acids joined by peptide links in the chain [1](b) (i) A = alpha helix [1]
B = beta pleated sheet [1] [2](ii) hydrogen bonds [1]
form between a lone pair of electrons on an O atom and a δ+ H [1] [2](c) the bending/folding of the secondary structure to give a precise 3D shape held together by
hydrogen bonding/disulfide bridges/ionic interactions/van der Waals’ forces [2]
8383
5.9 AMINO ACIDS
(d) (i) the substrate induces a change of shape of the active site of the enzyme [1]the reaction occurs in active site with a path of lower activation energy [1] [2]
(ii) enzymes lose their activity at (low) or high temperature [1]enzymes lose their activity at low or high pH [1]enzyme structure is denatured [1]with heat hydrogen bonds of tertiary structure break/pH change breaks the ionic attractions of tertiary structure [1] [4]
(iii) enzymes break down proteins/carbohydrates/ stains on clothes [1]reaction can occur at lower temperature/ less agitation required [1] [2]
(iv) the site on the surface of the enzyme into which the substrate fits [1]
6
CHH3N
CH2
OH
C
O
O
(Na )– +
CHH2N
CH2
CH2
C
C
O
O
[1] [2]
O
(Na )– +
O(Na )–+
[3]
7 (a) 4 [1](b)
R
H
C COOHH2N
where R = H, CH3, CH
2OH or CH
2C
6H
5 [1]
(c)
N
H H H
H
CH2
OH
C
O
C
H2C
N C
H H
HO
C N C
H CH3
HO
C
O
CN C
any one of circles shown [1]
(d) hydrogen bonding [1](e) hydrolysis [1]
dilute acid/alkali e.g. hydrochloric acid or sodium hydroxide solution [1] [2]
8 (a) HOOCCH2CH
2CH(NH
2)COOH + Na
2CO
3 NaOOCCH
2CH
2CH(NH
2)COONa + H
2O + CO
2 [2]
(b) the NH2 and COOH group attached to the same carbon [1]
(c) 2-aminopentanedioic acid [1]
9 Answer is C [1]
8484
5.10 POLYMERS
5.10 Polymers
1 (a) (i)
(CH2)
4(CH
2)
6
O
C
O H H H H
C N N (CH2)
4(CH
2)
6
O
C
O
C N N
[2](ii) polyamide [1](iii) polymers formed by the elimination of small molecules such as water or hydrogen chloride
when monomers bond together [2](iv)
(CH2)
4C
Cl
OO
Cl
C
[1]
(b) (i)
CHO
H
H
H
H
C OH
[1]
(ii) ethane-1,2-diol [1] condensation [1] [2]
(c) (i) 2 [1](ii)
C
OH
OO
HO
C H2N NH
2
[2](iii) Kevlar as there are hydrogen bonds between molecules which require more energy
to break [1] Terylene only has weaker permanent dipole–dipole attractions/van der Waals’ forces [1] [2]
(d) (i) 3-methylpent-2-ene [1](ii) addition [1](iii) Terylene is biodegradable [1]
(d) (i) the process by which a double stranded DNA molecule is copied to produce two identical DNA molecules [2]
(ii) [Pt(NH3)
2Cl
2] + H
2O [Pt(NH
3)
2(H
2O)Cl]+ + Cl– [2]
(iii) C5H
5N
5O [1]
(e) K2PtCl
4/K
2[PtCl
4] [1]
7 (a)
CN
CH2COONaNaOOCH
2C
NaOOCH2C CH
2COONa
H
H
C
H
H
N
CH2N
H
H
C
H
H
NH2
++ 4H2O
+ 4NH3
+4NaCN 4HCHO
[2](b) (i) sequesters Ca2+ [1]
prevent blood from clotting [1] [2](ii) C
10H
13O
8N
2K
3 [1]
(iii) mass of tripotassium edta = 1.5 × 4 = 6 mg
moles of tripotassium edta = 0.006406
= 1.48 × 10–5 mol [2]
(iv) A ligand which uses many lone pairs of electrons to form more than two co-ordinate bonds with a central metal atom or ion in a complex [2]
(c) (i)
C
O
C
OH
CHO
O
OH
C
O
OH
H2C
H2C [1]
(ii) C6H
8O
7 + 3NaOH C
6H
5O
7Na
3 + 3H
2O [2]
8989
5.11 CHEMISTRY IN MEDICINE
8 (a) 4 [1](b) haem bonds to protein/globin by co-ordinate bond [1]
haem and 4 protein molecules form haemoglobin complex [1]O
2 binds reversibly to the iron/iron(II) ion in haem [1] [3]
(c) CO binds irreversibly to iron in haemoglobin [1]forming carboxyhaemoglobin [1]lower O
2 carrying capacity [1] [3]
9 (a) C
O
O
O
C CH3
OH
[1](b)
C
+
C
C ++
O
O
O
C 2NaOHCH3
OH O
OH H2OH
3C
ONaO
ONa
[2](c) (i) more soluble in ethanol than water [1]
(ii) some impurities in the tablets insoluble in ethanol [1](iii) increase rate of hydrolysis [1]
(iv) moles of aspirin = 0.9180
= 5 × 10–3
moles of NaOH added = 25.0 × 0.5
1000 = 0.0125
moles of NaOH which reacted = 5 × 10–3 × 2 = 0.01moles of NaOH left over = 0.0125 – 0.01 = 2.5 × 103
moles of NaOH left over in 25.0 cm3 = mole of HCl = 2.5 × 10−3
10 = 2.5 ×10–4
volume of HCl required = 2.5 × 10−4 × 1000
0.0125 = 20.0 cm3 [5]
(v) phenolphthalein [1] or methyl orange [1]pink to colourless [1] or yellow to red [1] [2]
(vi) hydrolysis not complete so more NaOH present than expected [1]
10 (a) reacts with water/becomes aquated [1]ligands bind to nitrogen atoms in guanine [1]inter or intra strand links [1]prevent DNA replication [1]cell death [1] max [4]
(b) fewer side effects [1](c) [Pt(NH
3)
2Cl
2] [1]
(d) same group in Periodic Table/directly above Pt in Periodic Table/similar electronic configuration [1]
9090
5.11 CHEMISTRY IN MEDICINE
11 (a) (i) use up all 2-hydroxybenzoic acid [1]excess ethanoic anhydride can be hydrolysed [1] [2]
(ii) moles of 2-hydroxybenzoic acid = 0.500138
= 3.623 × 10–3
moles of ethanoic anhydride required = 3.623 × 10–3
mass of ethanoic anhydride required = 3.623 × 10–3 × 102 = 0.370 g
volume of ethanoic anhydride = 0.3701.08
= 0.342 cm3 [3]
(iii) moles of aspirin = 3.623 × 10–3
mass of aspirin = 3.623 × 10–3 ×180 = 0.652 g = 652 mg [3]
(iv) % yield = 485652
× 100 = 74.4 % [1]
(iv) loss by mechanical transfer/not all recrystallised/side reactions/reaction not complete [1]
(b) % atom economy = 180240
× 100 = 75 % [1]
(c) more soluble in water [1]
12 (a) remove an insoluble substances which may react [1](b) titre values within 0.1 cm3 (allow 0.2 cm3) [1](c) CaCO
3 + 2HCl CaCl
2 + CO
2 + H
2O [1]
NaOH + HCl NaCl + H2O [1] [2]
(d) moles of NaOH = moles of HCl in 25.0 cm3 = 13.1 × 0.250
1000 = 3.275 × 10–3
moles of HCl in 250 cm3 = 3.275 × 10–3 × 10 = 0.03275
moles of HCl added initially = 50.0 × 1.50
1000 = 0.075
moles of HCl which reacted with CaCO3 = 0.075 – 0.03275 = 0.04225
moles of CaCO3 =
0.042252
= 0.021125
mass of CaCO3 = 0.021125 ×100 = 2.1125 g
mass of CaCO3 in 1 tablet =
2.11255
= 0.4225 g = 423 mg [5]
13 (a) gas-liquid chromatography attached to a mass spectrometer [1](b) mobile phase = inert carrier gas [1]
stationary phase = liquid on solid support [1]mixture injected with carrier gas [1]components separate based on solubility in liquid [1]components identified by mass spectrometry by comparison with known spectra [1] [5]
9191
5.11 CHEMISTRY IN MEDICINE
14 (a) (i)
C C
+ +
O OH O
O
O
C
C
O
O
C O
CH3
OH
OH
H3C
H3C
C
HO
O
H3C
[2](ii) ethanoyl chloride too reactive [1]
ethanoic acid reaction not complete [1] [2](iii) 2-hydroxybenzoic acid [1](iv) H
3PO
4 [1]
catalyst [1] [2](v) hydrolyse any unreacted ethanoic anhydride [1](vi) remove any soluble impurities [1](vii) remove any acid [1](viii) mass of ethanoic anhydride = 1.08 × 5 = 5.4 g
moles of ethanoic anhydride = 5.4102
= 0.0529
moles of salicylic acid = 2.0138
= 0.0145
salicylic acid limitingmoles of aspirin = 0.0145mass of aspirin = 0.0145 ×180 = 2.61 g
% yield = 2.242.61
× 100 = 85.8 % (85.9 % with no rounding) [5]
(b) (i) draw pencil line and × about 1 cm from the bottom of the plate [1]using a capillary tube spot the mixture onto the line [1]allow to dry and repeat [1] [3]
(ii) place in a developing tank with ethyl ethanoate to a depth of < 1 cm andcover with a lid [1]allow to run and remove plate when solvent near the top [1]mark solvent front [1] spray with potassium manganate(VII) solution/view under UV [1]
calculate Rf values using R
f =
distance moved by spotdistance moved by solvent
[1]
(iii) compare to known values for pure samples [1]no spot for salicylic acid [1] [2]