· Web viewVariable 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,

WORKBOOK ANSWERSAQA A-level ChemistryInorganic 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.

© Alyn G. McFarland and Nora Henry 2016 Hodder Education 1

3a MgO + H2O → Mg(OH)2

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.


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.


Exam-style question1a Oxidation state is +3

Coordination number is 6

Shape is octahedral

1b Brown precipitate

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

1e 2[Fe(H2O)6]3+ + 3CO32− → 2Fe(OH)3(H2O)3 + 3CO2 + 3H2O

1f

i = 410 × 10−9 m

ΔE = = = = 4.85 × 10−19 J

ii 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.


Variable oxidation states and catalysts1 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. Mn2+, 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− 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

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 cm3 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


Mr of FeSO4.xH2O = = 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− → 2SO4

2− + I2

3b 2Fe2+ + S2O82− → 2Fe3+ + 2SO4

2−

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− = = 1.98 × 10–3 mol

Ratio of FeC2O4 : MnO4− = 5:3

moles of FeC2O4 in 25 cm3 = × 5 = 3.3 × 10-3

moles of FeC2O4 in 250 cm3 = 3.3 × 10-3 × 10 = 0.033 mol

Mr of FeC2O4.xH2O = = 179.7

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 half-equations 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 Fe2+ and C2O4

2– 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.


Exam-style questions1a

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.


Organic chemistryTopic 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, 2-methylpropan-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 three-dimensional 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


Topic 8 Aldehydes and ketones1a 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


An isomer has the same molecular formula (C3H6O) and a different structure hence propanone is the isomer.


Topic 9 Carboxylic acids and derivatives

Carboxylic acids and esters1a 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


4b

4c

4d Heat with methanol

Acid catalyst


Acylation1 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 questions1 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


5a

5b CH3CHO + HCN → CH3CHOHCN

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 molecules

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


Topic 10 Aromatic chemistry1a 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


Topic 11 Amines1a 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

2b

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

Remember that aliphatic amines are more basic than ammonia, which in turn is more basic than aromatic amines.


Exam-style questions1 Answer is A

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.

2a 1,3-dinitrobenzene

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.

2b 1,2-dinitrobenzene or 1,4-dinitrobenzene

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 Manufacture of explosives / formation of amines / organic synthesis

2d H2/Ni or H2/Pt or Sn/HCl or Fe/HCl

3a

Phenylethanone

3b

3c


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 Polymers1 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


Topic 13 Amino acids, proteins and DNA1a 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

iii Thymine


iv Condensation

2b 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 synthesis1 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.


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 Chromatography1a 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.


Documents

· Web viewVariable 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,