Download Inorganic and organic chemistry 2

WORKBOOK ANSWERS
AQA A-level Chemistry
Inorganic and organic chemistry 2
This Answers document provides suggestions for some of the possible answers that might be
given for the questions asked in the workbook. They are not exhaustive and other answers
may be acceptable, but they are intended as a guide to give teachers and students feedback.
Inorganic chemistry
Topic 4 Properties of period 3 elements
and their oxides
1a
4Na + O2 → 2Na2O 
2Mg + O2 → 2MgO 
4Al + 3O2 → 2Al2O3 
Si + O2 → SiO2 
P4 + 5O2 → P4O10 
S + O2 → SO2 
Remember that it is the oxide formed on reaction of the element with oxygen. Phosphorus and
sulfur form two oxides. It is P4O10 and SO2 that form on reaction with oxygen.
1b
SO2 / sulfur dioxide 
The ionic (Na2O, MgO, Al2O3) oxides are solids as is the giant covalent oxide SiO2. P4O10 is a
solid as well due to the van der Waals forces between the molecules, but SO2 is a gas as it is a
small molecule with weak intermolecular forces between the molecules.
2a
Basic 
2b
Bonding: ionic 
Structure: ionic lattice 
2c
Na2O + H2O → 2NaOH 
2d
Na2O + H2SO4 → Na2SO4 + H2O 
Sodium, magnesium and aluminium form ionic compounds with oxygen. The oxides of sodium
and magnesium are basic oxides. Aluminium oxide is amphoteric.
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1
MgO + H2O → Mg(OH)2 
3a
3b
i
MgO + 2HCl → MgCl2 + H2O 
ii
MgO + H2SO4 → MgSO4 + H2O 
4a
Al2O3 + 6H+ → 2Al3+ + 3H2O 
4b
Al2O3 + 2OH− + 3H2O → 2Al(OH)4− 
4c
Aluminate(III) or aluminate ion 
5a
Bonding: covalent 
Structure: giant/macromolecular 
5b
2NaOH + SiO2 → Na2SiO3 + H2O 
Sodium silicate 
6a
Bonding: covalent 
Structure: simple/molecular 
6b
P4O10 + 6H2O → 4H3PO4 
Phosphoric(v) acid 
6c
P4O10 + 12NaOH → 4Na3PO4 + 6H2O 
Sodium phosphate(V) 
6d
P4O10 + 6MgO → 2Mg3(PO4)2 
The oxidation state is important in the IUPAC name of an acid or its salt at this level. Remember
to include it when needed and also to recognise it.
7a
SO2 + H2O → H2SO3 
Sulfuric(IV) acid 
SO3 + H2O → H2SO4 
Sulfuric(VI) acid 
7b
SO2 + CaO → CaSO3 
7c
SO3 + 2NaOH → Na2SO4 + H2O 
The difference between sulfuric(IV) acid and sulfuric(VI) acid is important, as is the difference
between sulfate(IV) and sulfate(VI) ions. Make sure you look for the oxidation state as this is
important.
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2
Topic 5 and 6 Transition metals and
reactions of ions in aqueous solution
1
Answer is C 
The coordination number is the number of coordinate bonds between the ligand(s) and the
central metal atom or ion. EDTA4− is a hexadentate ligand, so one ion forms six coordinate
bonds. The other ligands are monodentate so each forms a single coordinate bond with the
central metal atom or ion.
2
Answer is A 
All contain 24 electrons as iron has an atomic number of 26, so the number of electrons in a 2+
ion should be 24. This is an important way to check this. Add up the superscript numbers.
Transition metal atoms lose their 4s electrons first. The most obvious distractor here is C as this
would imply that iron atoms would lose their 3d electrons to form the 2+ ion.
3a
Answer is A 
3b
Answer is D 
3c
Answer is E 
3d
Answer is E 
3e
Answer is C 
The shapes and coordination number of complexes are important. Create a list of complexes
with shape, coordination number, oxidation state of the transition metal atom or ion and colour
of any complex you are supposed to remember.
4a
A = [Al(H2O)6]3+ 
B = [Cu(H2O)6]2+ 
C = [Fe(H2O)6]2+ 
4b
[Al(H2O)6]3+ + 3OH− → Al(OH)3(H2O)3 + 3H2O 
4c
[Cu(NH3)4(H2O)2]2+ 
4d
[Fe(H2O)6]2+ + CO32− → FeCO3 + 6H2O 
4e
Carbon dioxide 
The reactions of ions in aqueous solution can be used to identify the complex. Make sure you
know the identity of any precipitates and be able to write equations for the reactions.
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Exam-style question
Oxidation state is +3 
1a
Coordination number is 6 
Shape is octahedral 
Brown precipitate 
1b
Does not redissolve in excess sodium hydroxide solution 
1c
[Fe(H2O)6]3+ + 3OH− → Fe(OH)3(H2O)3 + 3H2O 
1d
Brown precipitate 
Bubbles of gas 
2[Fe(H2O)6]3+ + 3CO32− → 2Fe(OH)3(H2O)3 + 3CO2 + 3H2O 
1e
1f
i
 = 410 × 10−9 m 
ΔE =
ii
hc
l
=
6.63 ´10-34 ´ 3 ´108
1.989 ´10-25

=
= 4.85 × 10−19 J 
410 ´10-9
410 ´10-9
For yellow solution it is bigger 
Smaller  means larger ν and larger ΔE 
iii
Any two from the list below: 
identity of the metal/type of ligands/coordination number/oxidation state of the metal
or charge on the ion/shape of the complex
This type of question on one particular complex tests your knowledge of many different parts of
this section. The observations and equations for the reaction of [Fe(H2O)6]2+, [Fe(H2O)6]3+,
[Cu(H2O)]2+ and [Al(H2O)6]3+ with sodium hydroxide solution, ammonia solution and sodium
carbonate solution are important. Remember that 3+ ions form the hydroxide precipitates with
sodium carbonate solution. The answer to (f)(i) using the other wavelength given (480 nm) would
be 4.14 × 10−19 J. Try the calculation backwards as well to make sure you can work out the
wavelength from the given value of ΔE.
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Variable oxidation states and catalysts
1
Answer is A 
The substance that could carry out a reduction of vanadium must be on the right-hand side of
the half-equations, so it cannot be iodine.
Sulfur dioxide being oxidised to sulfate(VI) ions has a potential of −0.17 V. Adding this to each of
the potentials for the vanadium half-equations gives +0.83 V (so +5 to +4 is feasible), +0.17 V (so
+4 to +3 is feasible) and −0.42 V (so +3 to +2 is not feasible), so sulfur dioxide reduces vanadium
from +5 to +3 but not to +2.
Zinc being oxidised from Zn to Zn2+ has a potential of +0.76 V and this means that zinc will
reduce vanadium from +5 to +2 as all of the reactions are feasible.
Iodide ions being oxidised to iodine has a potential of −0.54 V, so adding to each of the
potentials for the vanadium half-equations gives +0.46 V (so +5 to +4 is feasible), 0.20 V (so +4 to
+3 is not feasible) and −0.80 V, so +3 to +2 is also not feasible. Iodide ions would reduce
vanadium from +5 to +4 but no further.
2a
Colourless to pink 
This titration is self-indicating, so the last drop of the purple potassium manganate(VII) solution
added does not decolourise but changes the colourless solution to pink. Mn 2+, Fe2+ and Fe3+ ions
have a very pale colour in solution compared to the intense colour of manganate( VII).
2b
5Fe2+ + MnO4− + 8H+ → 5Fe3+ + Mn2+ + 4H2O 
ratio of Fe2+ : MnO4− = 5:1 
2c
Moles of MnO4− =
14.25 ´ 0.0248
= 3.534 × 10−4 mol 
1000
Moles of Fe2+ in 25 cm3 = 3.534 × 10−4 × 5 = 1.767 × 10−3 mol 
Moles of Fe2+ in 1 dm3 = 1.767 × 10−3 × 40 = 0.0707 mol 
Mr of FeSO4.7H2O = 277.9 
Mass of FeSO4.7H2O = 0.0707 × 277.9 = 19.6 g 
Using given ratio of 4:1 answers are 3.534 × 10−4 mol , 1.414 × 10−3 , 0.0565 mol ,
277.9  and 15.7 g 
Often a ratio or numerical answer will be given to enable you to carry on with the rest of the
question and make sure you do this. The ×5 is from the ratio and ×40 is to multiply from 25 cm 3
to 1 dm3. Remember to give your answer to three significant figures. This question could have
asked for the degree of hydration of hydrated iron(II) sulfate using the initial mass (19.6 g) and
the final titre. Check that you would get 7 (or 6.96 which is what is obtained using the 19.6 g). It
is only the last couple of steps in the calculation that are different. Check it out below.
Moles of MnO4− =
= 3.534 × 10−4 mol 
Moles of Fe2+ in 25 cm3 = 3.534 × 10−4 × 5 = 1.767 × 10−3 mol 
Moles of Fe2+ in 1 dm3 = 1.767 × 10−3 × 40 = 0.0707 mol 
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5
Mr of FeSO4.xH2O =
19.6
0.0707
= 277.2 
Mr of xH2O = 277.2 − (55.8 + 32.1 + 4 × 16.0) = 125.3
x = 125.3 / 18.0 = 6.96 (i.e. 7) 
3a
S2O82− + 2I− → 2SO42− + I2 
3b
2Fe2+ + S2O82− → 2Fe3+ + 2SO42− 
2I− + 2Fe3+ → I2 + 2Fe2+ 
Reactions can occur in either order 
3c
Homogeneous catalysts are in the same state as the reactants 
All ions here in solution/aqueous 
This reaction is commonly used as an example of homogeneous catalysis. Make sure you know
the ionic equations for the main reaction and the catalysed reactions.
4
moles of MnO4− =
9.25 ´ 0.214
= 1.98 × 10–3 mol 
1000
Ratio of FeC2O4 : MnO4− = 5:3 
moles of FeC2O4 in 25 cm3 =
1.98 ´10-3
× 5 = 3.3 × 10-3 
3
moles of FeC2O4 in 250 cm3 = 3.3 × 10-3 × 10 = 0.033 mol 
Mr of FeC2O4.xH2O =
5.93
= 179.7 
0.033
Mr of xH2O = 179.7 – (55.8 + (2 × 12.0) + (4 × 16.0)) = 35.9 
x = 35.9/18.0 = 1.99 so x = 2 
The most difficult part of this calculation is the ratio of FeC2O4 : MnO4– = 5:3. It is important you
remember this or can work it out and sometimes the question will take you through the halfequations and the ionic equations and working out the ratio. If you remember the overall ratio
then you know where you are going with the question. Both the Fe 2+ and C2O42– ions react with
MnO4−. Try repeating the calculation with an average titre of 8.40 cm3. This should give x = 3.
This is not a true value but allows you another chance to do the calculation.
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Exam-style questions
1a
i
5C2O42− + 2MnO4− + 16H+ → 10CO2 + 2Mn2+ + 8H2O 
This equation can be worked out from the two half-equations by multiplying the top one by 2 and
the bottom one by 5 to get 10e− in each. The equations when added together form the ionic
equation.
ii
C oxidation state changes from + 3 to + 4 
C is oxidised 
Mn oxidation state changes from + 7 to +2 
Mn is reduced 
Make sure you can clearly identify the element being oxidised and the element being reduced
using oxidation states. This was first met at AS but often forms parts of the A-level papers.
iii
Bubbles of gas 
Purple solution 
Changes to colourless 
You can use an equation to determine the observations expected during a reaction even if you
have never seen the reaction. In this case carbon dioxide is produced so bubbles should be
seen in the solution. Also manganate(VII) ions are purple in solution and Mn2+ ions are virtually
colourless. You should know this from manganate(VII) titrations.
1b
i
Two negative ions repel each other/high activation energy 
ii
Production of Mn2+ 
Autocatalysis by Mn2+ 
4Mn2+ + MnO4− + 8H+ → 5Mn3+ + 4H2O 
2Mn3+ + C2O42− → 2Mn2+ + 2CO2 
iii
Manganate(VII) ions are coloured/purple 
Colorimetry 
Using solutions of known concentration/calibration curve 
iv
Ions are used up 
This reaction is often used as an example of autocatalysis, as it shows the Mn2+ ions being
formed that catalyse the reaction. This is why the reaction is slow initially as it has no catalyst
present. As the Mn2+ ions are formed in the reaction the reaction rate increases. You may be
asked to sketch this shape of graph based on your knowledge of the reaction.
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Organic chemistry
Topic 7 Optical isomerism
1
Answer is C 
Optical isomers have a chiral centre which is a carbon with four different atoms or groups
attached.
2
Answer is C 
It its good practice to quickly sketch out the isomers of C4H10O butan-1-ol, butan-2-ol, 2methylpropan-1-ol and 2-methylpropan-2-ol. Butan-2-ol has a chiral centre, and therefore
stereoisomers, hence the total is 5.
3
a, c, d and e have optical isomers 
b and f do not 
First decide if the structures have chiral centres. You may need to draw the structure out.
4a
Rotates the plane of plane polarised light 
4b


Remember to find the chiral centre, with four different groups attached, then draw two threedimensional structures, arrange the four groups on the structure and draw a mirror image.
4c
Mixing equal amounts  of the same concentration of two enantiomers 
No effect on plane polarised light 
One isomer rotates plane polarised light in one direction, the other rotates plane
polarised light in the opposite direction and the effect cancels 
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Topic 8 Aldehydes and ketones
1a
Propanone 
1b
Ethanal 
1c
Pentan-3-one 
1d
Butanal 
1e
3-methylpentanal 
1f
Pentan-2-one 
1g
Propanal 
1h
2-methylpropanal 
1i
4-methylpentanal 
1j
Methanal 
1k
2-bromo-4-methylhexanal 
1l
Hexan-2-one 
2a
(a) and (g)  (e) and (l)/(i)  (c) and (f) 
2b
C7H13OBr 
3
Answer is C 
4a
CH3CH2CHO + 2[H] → CH3CH2CH2OH 
Reduction of an aldehyde produces a primary alcohol, in this case propan-1-ol
4b
Heat under reflux  propan-1-ol 
4c
Hydride ion 
4d

4e


CH3COCH3 or
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
An isomer has the same molecular formula (C3H6O) and a different structure hence propanone is
the isomer.
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Topic 9 Carboxylic acids and derivatives
Carboxylic acids and esters
1a
Ethyl ethanoate 
1b
Pentanoic acid 
1c
Butanoic acid 
1d
3-hydroxypropanoic acid 
1e
Methyl butanoate 
1f
Propanoic acid 
1g
Propyl butanoate 
1h
Propyl methanoate 
2
Answer is D
There cannot be a straight-chain ketone with two carbon atoms as the C=O has to be in the
middle of a molecule, so the simplest straight-chain ketone is propanone C3H6O.
3a
2CH3CH2COOH + Mg → (CH3CH2COO)2Mg + H2 
3b
2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2 
3c
CH3CH2CH2COOH + NaOH → CH3CH2COONa + H2O
3d
CH3COOH + CH3CH2CH2OH ⇌ CH3COO CH2CH2CH3 H2O 
3e
CH3CH2COOH + CH3OH ⇌ CH3CH2COOCH3 + H2O 
Remember that acid + metal gives salt and hydrogen, acid and base gives salt and water and
that acid and carbonate gives salt and water and carbon dioxide. The salts are -anoates.
Carboxylic acids react with alcohols to give esters. Esters are named as alkyl carboxylates
where the alkyl name comes from the alcohol and the carboxylate name comes from the
carboxylic acid from which they are formed.
4a
Name: propane-1, 2, 3-triol 

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4b

4c

4d
Heat with methanol 


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Acylation
1
Answer is D 
An acid chloride reacts with an alcohol at room temperature. In this case the ester formed is
propyl ethanoate. It is not a reversible reaction.
2a
Nucleophilic addition–elimination 

2b


N-methylpropanamide 
Exam-style questions
1
Answer is A 
Always make sure you number from one side; this eliminates answers B and C. The smallest
combination of numbers is chosen, hence D is incorrect.
2
Answer is C 
An optically active compound has a chiral centre; a carbon with four different atoms or groups
attached. A and D do not have chiral centres. Tollens’ reagent is only reduced by aldehydes. C is
an aldehyde with a chiral centre.
3
Answer is C 
The structure is an acyl chloride and it reacts with alcohols to form esters, in a reaction where
HCl is removed.
4a
CH2O 
The empirical formula is the simplest ratio. C16H32O2 is the molecular formula so the empirical
formula is CH2O
4b


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
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
5a
CH3CHO + HCN → CH3CHOHCN 
5b
2-hydroxypropanenitrile 
5c
Nucleophilic addition



5d
i

ii

CH3CH=CHCHO
Remember that en means that a double bond is present and that it is on position 2
6a
CH3OH + CH3CH2COOH ⇌ CH3CH2COOCH3 + H2O
6b
Catalyst 
6c
H bonds between molecules of propanoic acid 
These are stronger than the dipole–dipole attractions between methyl propanoate

When explaining the difference in boiling points in molecular covalent substances, always refer
to intermolecular forces.
6d
Goes to completion 
No heat or catalyst needed 
6e
C2H5COONH3  C2H5CH2OH  C2H5COONa 
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Topic 10 Aromatic chemistry
1a
C6H5COOC2H5 + HNO3 → C6H4NO2COOC2H5 + H2O 
1b
Conc sulfuric acid 
Conc nitric acid 
1c
Moles of ethyl 3-nitrobenzoate = 5.85/195.0 = 0.03 
For 70% yield this (0.03 × 100)/70.0 = 0.043 moles 
Ratio 1:1
0.043 moles ethyl benzoate = 0.043 × 150.0 = 6.45 g 
1d
Substitution of NO2 at different positions forms different isomers / multiple substitution
or nitration can also occur 
1e
To remove impurities 
Dissolve in minimum amount of hot ethanol / methanol 
Filter while hot and allow (filtrate) to cool and crystals to form 
Suction filtration to separate crystals 
1f
Between filter papers / cool oven / desiccator 
1g
Sharp m.pt  compare to 42°C 
2a
Benzene is more stable than cyclohexatriene 
Expected enthalpy of hydrogenation of C6H6 is 3(−120) = −360 kJ mol−1 
Actual value is less because of delocalisation, which makes benzene more stable 
2b
Cyclohexatriene orange → colourless 
Benzene stays orange 
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Topic 11 Amines
1a
Ethylamine/aminoethane 
1b
N-methylpropylamine 
The prefix N is used to show that the alkyl groups are attached to the main chain via the nitrogen
atom.

1c
Phenylamine
1d
N-methylethylamine 
1e
2-aminopropane 
1f
2,4-diaminopentane 
The prefix diamino is used if a compound contains two amino groups.
1g
1,3-diaminopropane 
1h
Diethylamine 
1i
2-aminopropanoic acid 
1j
N,N-diethylpropylamine 
2a
i
CH3CH2NH2 + HCl → CH3CH2NH3Cl 
ii
2CH3NH2 + H2SO4 → (CH3NH3)2SO4 
iii
C6H5NH2 + HNO3 → C6H5NH3NO3 
iv
CH3CH2NH2 + H2O → CH3CH2NH3+ + OH− 
v
CH3CH2NH3Cl + NaOH → CH3CH2NH2 + NaCl + H2O 
i
Ethylamine as the electron-donating alkyl group releases electrons to the N,
meaning the lone pair is more able to accept a proton 
ii
Ammonia as in phenylamine the lone pair on N becomes delocalised into the pi
system so the electron density on the N is less and the lone pair is less available to
accept a proton 
iii
Butylamine as it has a larger electron-donating group 
2b
Remember that aliphatic amines are more basic than ammonia, which in turn is more basic than
aromatic amines.
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Exam-style questions
Answer is A 
1
First calculate the number of moles of 4-hydroxyphenylamine (10.9/109.0 = 0.1).
Moles of paracetamol produced = 0.1. Multiply this by the formula mass, 151.0, to get 15.1 g if the
yield is 100%. The yield is 80% so the answer is 15.1 × 0.8 = 12.1 g.
1,3-dinitrobenzene 
2a
The NO2 group is nitro. There are two of them, so dinitro is used in the name. The position of the
nitro groups must be given; remember to place commas between numbers and a hyphen
between a number and word.
1,2-dinitrobenzene or 1,4-dinitrobenzene 
2b
The position of the NO2 group can change to give an isomer, a different structure with the same
molecular formula.
2c
i
NO2+ 
HNO3 + 2H2SO4 → NO2+ + 2HSO4− + H3O+ 
ii
Electrophilic substitution 



Be careful that the + is on the N and that the arrow comes from the ring. The horseshoe must not
extend beyond C2 to C6.
iii
2d
Manufacture of explosives / formation of amines / organic synthesis 
H2/Ni or H2/Pt or Sn/HCl or Fe/HCl 
3a

Phenylethanone 
+
3b
CH3COCl+ AlCl3 ® CH3 C O+ AlCl4- 
3c
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
4a


A nucleophile is a lone pair donor.
Amines have a lone pair on the nitrogen and so can act as nucleophiles. 
4b
CH3CH2CH2Br + 2NH3 → CH3CH2CH2NH2 + NH4Br 
4c
Quaternary ammonium salt 
Topic 12 Polymers
1
Answer is C 
A and B would both form condensation polymers; A would form a polyamide polymer and B
would form a polyester polymer. D would form an addition polymer. The alcohol groups in
HO(CH2)2OH and the amino groups in H2N(CH2)4NH2 cannot react to form a polymer.
2a
 or other appropriate
Any proper repeating unit is correct such as all the ones shown below. They are shown in
different colours to show you that the repeating unit can start from any part in the polymer
structure.
There are six possible repeating groups based on where you start from in the structure of the
polymer.
2b
Benzene-1,4-dicarboxylic acid 
Benzene-1,4-diamine 
Kevlar 
Clothing / bullet-proof vests 
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Topic 13 Amino acids, proteins and DNA
1a
2-amino-4-methylpentanoic acid 
1b
2,6-diaminohexanoic acid 
1c


1d

At low pH the NH2 groups and COOH groups are protonated. All NH2 groups are protonated.
2a

i
ii
Adenine 
© Alyn G. McFarland and Nora Henry 2016
Hodder Education
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2b
iii
Thymine 
iv
Condensation 
Cisplatin is [Pt(NH3)2Cl2] 
Coordinate bonds between nitrogen atom in guanine and cisplatin 
DNA cannot be replicated/copied 
Cells which are replicating are killed 
Topic 14 Organic synthesis
1
Route A uses KCN 
In ethanol (or aqueous) 
Intermediate is butanenitrile or CH3CH2CH2CN 
Lithal/LiAlH4 or H2 
In dry ether for lithal or Ni/Pt/Pd for H2 
Route B uses NH3 
Excess NH3 
It is important to be able to process the organic reactions you have met. Make sure you think
through the reactions. In Route A the carbon chain is lengthened so the reaction of the
halogenoalkane with KCN provides an extra carbon atom in the chain. The nitrile produced is
then reduced using either lithal or hydrogen to form the amine. The longer chain halogenoalkane
in Route B reacts directly with ammonia to form the amine. The conditions required in each step
are necessary. These questions are a great way to revise your organic chemistry.
Topic 15 Nuclear magnetic resonance
spectroscopy
1
Answer is C 
As the molecule is symmetrical, the 1H nuclei in the two CH3 groups are chemically equivalent
and so occur at the same chemical shift. As there are no 1H nuclei bonded to the middle carbon
atom, there is no spin‒spin splitting, so there is only one singlet in the spectrum.
2
Answer is C 
The first thing to consider here is the number of chemical environments. A and B have four
chemical environments whereas the spectrum shows only three sets of peaks which would
indicate three environments. C and D have three environments of 1H nuclei. In order of
increasing chemical shift, C would produce a triplet, a singlet and then a quartet as shown in the
spectrum. D would produce a triplet, a quartet and the singlet would be at the highest  value.
© Alyn G. McFarland and Nora Henry 2016
Hodder Education
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3
L is 2-methylbutanal

3-methylbutanal is not optically active and the only one which has four different groups bonded
to the same carbon atom is 2-methylbutanal. Be able to work out that C2H5, CH3 CHO and H are
the different groups.
M is 2-methylbut-2-en-1-ol or 2-methylbut-1-en-1-ol or 3-methylbut-1-en-1-ol.
Or any other appropriate structure 
Any correct E‒Z isomer would be accepted but these are the answers accepted.
N is pentan-3-one.

This is the only ketone that has three peaks in its 13C NMR spectrum.
Topic 16 Chromatography
1a
For spot 1 Rf = 2.5/8 = 0.3125 
For spot 2 Rf = 5/8 = 0.625 
For spot 3 Rf = 6.1/8 = 0.7625 
Rf values are calculated by dividing the distance moved by the spot by the distance moved by
the solvent.
1b
The most polar substance is 1 
Slowest movement in non-polar solvent and held by polar stationery phase 
The silica gel support is the stationary phase and is polar and so polar substances are held by
the stationary phase.
© Alyn G. McFarland and Nora Henry 2016
Hodder Education
21