WORKBOOK ANSWERS
AQA AS/A-level Chemistry
Inorganic and organic chemistry 1
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 1 Periodicity
Classification
Physical properties of period 3 elements
1C
Sulfur has 16 electrons and its electron configuration is 1s22s22p63s22p4. Its outer electrons are in the p-subshell. A p-block element has its outer electrons in a p-subshell.
2D
Across the third period the atomic radius decreases, the ionisation energy increases, the melting point increases to Si and then decreases. The correct trend is D — the nuclear charge is increasing and so there is similar shielding as the electron is being removed from the same shell. Hence there is a smaller atomic radius, as the outermost electron is held closer to the nucleus by the greater nuclear charge because there is an increase in nuclear attraction.
3S8 molecules have a larger Mr than Cl2 molecules.
Hence there are stronger van der Waals forces between the molecules and so they are held more tightly together and need more energy to break.
4Silicon is a giant covalent structure and more energy is needed to break the many strong covalent bonds than is needed to break the weaker metallic bonds in aluminium.
Notice that in this answer it is important that you mention the type of bonding in each element, and compare their strength.
Topic 2 Group 2, the alkaline earth metals
1B
2
aBa2+
CO32−
The first ionisation energy, melting point and solubility of sulfates decrease down the group.
b Barium is a very reactive element.
c Ba(s)+ 2H2O(l) → Ba(OH)2(aq) + H2(g)
Always include brackets in Ba(OH)2,BaOH2 is incorrect. Remember it is an aqueous solution.
d MgSO4
The solubility of sulfates decreases down the group. Hence, the most soluble sulfate is magnesium sulfate. Remember to work out the formula correctly:the magnesium ion is Mg2+ and the sulfate ion is SO42−so the formula is MgSO4.
eIt is taken as a ‘barium meal’ as it absorbs X-rays and allows the gut to be seen.
It is insoluble and so, although toxic, safe to use.
fIn calcium the delocalised electronsare closerto cations,as the cation has fewer electron shells and is smaller.
Hence there is a stronger attraction between the cations and the delocalised electrons and stronger metallic bonding.
In this case the converse argument is also accepted — barium has more electron shells and so the attraction between the delocalised electrons and the cation is smaller and so it takes less energy to pull atoms apart and for the solid to melt.
Topic 3 Group 7, the halogens
Trends in properties
Uses of chlorine and chlorate(i)
Tests for ions
1A
The question says dilute sulfuric acid: do not get confused with concentrated sulfuric acid which reacts differently from dilute sulfuric acid. The dilute sulfuric acid will only react with the carbonate.
2D
3Dissolve in water
Sodium chloride — white ppt which is soluble in dilute ammonia
Sodium bromide — cream ppt which is insoluble in dilute ammonia
4
aNaCl+H2SO4→NaHSO4+HCl
bBr− ions are bigger than Cl−ions
Therefore Br−ions more easily oxidised / lose an electron more easily.
c iPurple vapour/gas
(white solid goes to) black or black-grey or black-purple solid
Bad egg smell (any two)
c iiThe iodide ion(s) lose (an) electron(s)
2I−→ I2 + 2e–
Note that this question needs answered in terms of electrons; oxidation numbers are not acceptable. The I−changes to I by losing an electron.
c iiiOxidation state of S changes from +6 to −2
H2SO4 + 8H+ + 8e–→H2S + 4H2O
H2SO4→H2S
Balance the oxygen by adding H2O.
H2SO4→H2S + 4H2O
Balance the hydrogen by adding H+ to the left-hand side.
H2SO4 + 8H+→H2S + 4H2O
Write down the charge on each side of the equation.
LHS = + 8RHS = 0
Add 8 electrons to the LHS to balance the charge.
H2SO4 + 8H+ + 8e–→ H2S + 4H2O
5
aCl2 + 2I−→ 2Cl−+ I2
Solution changes from colourless to brown.
Remember that all group1 and group 2 halides in solution are colourless. Iodine in solution is brown.
bi2NaOH+Cl2→NaCl+NaClO+H2O
Products are sodium chloride, sodium chlorate(i) and water
b iiCl2(g) +H2O(l)⇌HCl+HClO
Products are hydrochloric acid and chloric(i) acid
The oxidation numbers are essential in the names and the equilibrium arrows are essential in the second equation. To work out the oxidation number of chlorine in NaClO, it is a compound so the sum of the oxidation numbers is zero, Na has oxidation number +1 and O is −2.
Hence 0 = oxidation no. Na(+1) + oxidation no. Cl + oxidation no. O (−2)
0 = +1 + oxidation no. Cl −2
0 + 2 −1 = + 1 = oxidation no. Cl
Exam-style questions
1B
2B
3
aMg + 2H2O →Mg(OH)2 + H2
Magnesium reacts with cold water to give magnesium hydroxide — note that magnesium oxide is only produced if the magnesium reacts with steam.
bMg(OH)2+ 2HCl → MgCl2 + H2O
Used in antacid /indigestion tablets
c iTiCl4 +2Mg → Ti + 2MgCl2
Titanium is reduced
The oxidation number of titanium decreases from +4 to 0. A decrease in oxidation number means reduction.
When working out the oxidation number of titanium, remember that the sum of the oxidation numbers in a compound equals zero and the oxidation number of chlorine is −1.
Hence 0 = oxidation no. of titanium+ 4 ×oxidation no. of Cl (−1)
0 = Ti − 4
Ti = +4
You must include the signs for oxidation numbers.
c iiMg + H2SO4→ MgSO4 + H2
The magnesium sulfate is soluble and the Ti solid can be filtered off.
This reaction is an acid and metal reaction, producing a salt and hydrogen. Hydrogen is diatomic.
4
aA BromideAg ++ Br−→ AgBr
B SulfateBa2++ SO42−→ BaSO4
C ChlorideAg++ Cl−→ AgCl
bTo remove any carbonate ions which may interfere with the test.
cMagnesium/calcium
Organic chemistry
Topic 1 Introduction to organic chemistry
Nomenclature
1
apentane
b1,3-dichloropropane
c 2-chloropropane
d2-methylpropane
e2-chloro-2,3,3-trimethylpentane
f 3,3-dichloro-2,2,4-trimethylhexane
Note that the longest chain includes the C2H5and that chloro is alphabetically before methyl (ignore the tri).
g 2-bromo-1-chloro-3-ethylpentane
h propan-2-ol
ipropanal
j pentan-3-one
k butanoic acid
l propene
mbut-2-ene
n 3-bromobutanoic acid
o propan-1-ol
p 2-methylpropene
q 2,2-dimethylpropane
r buta-1,3-diene
2
a
b
c
d
e
3
aCnH2n+1OH
bCnH2n+2
cCnH2n
4
aC5H12C5H12
bC4H8O2C2H4O
5
abutanealkane
bbut-1-enealkene
cbutanoic acidcarboxylic acid
dbutan-1-olalcohol
epentan-2-oneketone
fpropan-2-ol alcohol
Isomerism
Exam-style questions
1B
The criteria for E–Z isomers are that a carbon=carbon double bond must be present and each carbon in the double bond must be attached to two different groups. B has a C=C and on each carbon there is an H and a CH3(i.e. 2 different groups).
2B
3-methylbut-1-ene has five carbon atoms.
3C
4A
There are 5 carbons in the longest chain — pent. The methyl group is at position two and the double bond between carbon 1 and 2. Remember to choose the smallest combination of numbers and to count both the side group and double bond from the same end of the chain.
5
aC=COH
b
cLack of rotation about the C=C
dStructural isomers are compounds with the same molecular formula but different structural formulae.
6The right-hand carbon of the double bond has a C attached which has a higher atomic number than Hso the CH2Br has higher priority.
The left-hand carbon of the double bond has a carbon attached in each group, so consider the atoms one bond further away. C H H from the propyl group are attached to one carbon and have higher atomic number than the H H Hattached to the carbon of the methyl group. The propyl group has priority.
The highest priority group –CH2CH2CH3 and –CH2Br are on the same side so it is a Z isomer.
There are 6 carbons in the longest chain — hex, with a bromo on carbon one, a methyl on carbon 3 and the double bond on carbon 2.
(Z)-1-bromo-3-methylhex-3-ene.
Topic 2 Alkanes
Fractional distillation of crude oil
Modification of alkanes by cracking
1Allow CxHy to be the hydrocarbon and then write down the formula of the products using the information in the question:
CxHy→C2H4 + C4H8 + C8H18
Then insert the number of moles given in the question:
CxHy→2C2H4 + C4H8 + C8H18
Remember that the equation must balance so the number of the carbons and hydrogens on both sides must equal.
x= (2 × 2) + 4+ 8 = 16y= (2 × 4) + 8 + 18 = 34
C16H34→2C2H4 + C4H8 +C8H18
High pressure
High temperature
(it is thermal cracking)
2
aC8H18 → C2H4 + C3H6 + C3H8
Propane
Zeolite
bPetroleum/crude oil
Fractional distillation
cOctane and propane are both molecular covalent structures, in which the molecules are held together by van der Waals forces. Boiling these substances means breaking the van der Waals forces. Octane has a higher Mr and so more electrons and stronger van der Waals forces.
Octane has a higher Mr and more electrons.
Hence it has stronger van der Waals forces between the molecules.
Combustion of alkanes
Chlorination of alkanes
1C
To answer this question you need to realise that alkanes combust completely to produce carbon dioxide and water. Then write the balanced symbol equation. Butane has four carbons and its formula is C4H10.
C4H10+ 6.5O2→ 4CO2 + 5H2O
To balance the equation 6.5 moles of oxygen are required.
2B
3Propane C3H8
Butane C4H10
4They have the same general formula.
They show similar chemical properties.
They show a gradual change in physical properties from propane to butane.
(Any two)
5CnH2n+2
6They contain carbon and hydrogen atoms only
They only contains single C–C bonds — there are no C=C bonds.
7Propane
Because it has a lower boiling point as it has lower Mr
and weaker van der Waals forces between its molecules
8C2H6 + 3½O2 → 2CO2 + 3H2O
or
2C2H6 + 7O2 → 4CO2 + 6H2O
Note that complete combustion results in the formation of carbon dioxide and water.
9C4H10 + 4.5 O2→ 4CO + 5H2O
10Sulfur dioxide
SO2 + CaCO3→ CaSO3
Neutralisation
The major impurity present in most fuels is sulfur. When sulfur burns it produces sulfur dioxide S + O2→ SO2 which is a toxic acidic gas. On passing through calcium carbonate the sulfur dioxide is neutralised.
11The pressure and high temperature in the combustion engine causes the normally unreactive nitrogen in the air to react with oxygen/ N2 + O2→ 2NO2
Harm caused — React with unburned hydrocarbons to produce smog / Dissolve in water to form acid rain.
Incomplete combustion in very limited oxygen can produce carbon particles.
Carbon particles exacerbate asthma.
Note that nitrogen(iv) oxide is NO2 where the oxidation number of nitrogen is +4.
Exam-style questions
1C
2D
A radical has an unpaired electron. In CH4 all electrons are used in bonding so it cannot have an unpaired electron.
3
aC3H8 + Br2→ C3H8Br + HBr
UV light/high temp
b
2-bromopropane
cInitiation:Br2 → 2Br•
Propagation:
Br• + CH3CH2CH3→ CH3CH2CH2• + HBror
CH3CH2CH2• + Br2→CH3CH2CH2Br + Br•
Termination:
CH3CH2CH2•+ CH3CH2CH2•→ CH3CH2CH2CH2CH2CH3
orBr• + Br•→ Br2
CH3CH2CH2•+ Br•→ CH3CH2CH2Br
Be careful with the position of the radical dot — it must be to the side or above of the last carbonCH3CH2CH2• — if it is beside the CH3 or in the middle of the chain it is incorrect.
4
aMolecular formula C10H22
Empirical formula C5H11
Alkanes have the general formula CnH2n+2.n is the number of carbon atoms so n = 10.
Hence 2n + 2 = 22 so decane is C10H22. The empirical formula is the simplest ratio of atoms so it is C5H11.
b Fractional distillation
cC10H22 + 10½O2 → 10CO + 11H2O
dC8H18 + 25NO → 8CO2 + 12½N2 + 9H2O
Platinum or rhodium
eCH3SH + 3O2→ CO2 + 2H2O + SO2
Topic 3 Halogenoalkanes
Nucleophilic substitution and elimination
Ozone depletion
1C
This is an elimination reaction in which HBr is removed forming an alkene.
2B
A nucleophile is a lone pair donor. CH4 does not have any lone pairs and cannot act as a nucleophile.
3
a1-iodobutane
The C–I bond has the lowest bond enthalpy/strongest bond.
Bond enthalpy decreases down the group with increasing size of the halogen atom.
bCH3CH2CH2CH2Br + NaOH → CH3CH2CH=CH2 + NaBr + H2O
But-1-ene
4
aElimination
The curly arrows show movement of electrons and must come from the lone pair on the oxygen of the OH− to the hydrogen and from the middle of the C–H bond to the C–C bond.
bIt has the same molecular formula but a different structural formula.
c2-methylbut-1-ene
2-methylbut-2-ene
Two different isomers can be formed here depending on which H bonds to the hydroxide ion.
dCH3CH2CH2CH2Br+ NaOH → CH3CH2CH2CH2OH + NaBr + H2O
Butan-1-ol
This is a substitution reaction, since it occurs in aqueous solution. The bromine is replaced by OH.
eNucleophilic substitution
fThe halogen is more electronegative than the carbon.
The electrons in the covalent bond are attracted more to the halogen and form a partial negative charge on the halogen.
5
aThe reaction is exothermic and adding conc acid slowly with stirring controls the reaction and prevents dangerous spitting.
b Bromine
The solid potassium bromide reacts with the conc sulfuric acid to form hydrogen bromide which is further oxidised by the conc sulfuric acid to form bromine.
This is a reaction of a solid halide with conc sulfuric acid.
NaBr+H2SO4→NaHSO4+HBr
2HBr+H2SO4→Br2+SO2+2H2O
c Repeated boiling of a liquid and condensing of the vapour
dTo remove acidic impurities
eShake in a separating funnel
Invert and release the pressure periodically
Allow to settle and run off bottom layer — discard the aqueous layer.
fTo remove water
Topic 4 Alkenes
Structure, bonding and reactivity
Addition reactions of alkenes
Addition polymers
1C
Hydrocarbons which contain a double bond, are alkenes and have the general formula CnH2n.
2 A
3
aBut-1-ene
butane
bThe double bond is a centre of high electron densityand can undergo attack by electrophiles.
cBromine water
Orange to colourless
diCH3CH=CHCH3 + Br2→ CH3CHBrCHBrCH32,3-dibromobutane
d iiElectrophilic addition
eMajor product is 2-bromobutane.
Minor product is 1-bromobutane.
But-1-ene is an unsymmetrical alkene and as a result two different carbocations may form in the mechanism depending on which carbon the hydrogen from the HBr adds on to. The major product is formed from the most stable carbocation — the secondary carbocation is more stable than the primary carbocation. If you have studied the complete A level, you will realise that 2bromobutane can also exist as two enantiomers as there are four different groups on one of its carbon atoms.
fPoly(but-2-ene)
Exam-style questions
1C
A hydrocarbon contains the elements carbon and hydrogen only. Hence there is 84.7 g of carbon present and 100 − 84.7 = 15.3 g of hydrogen. To find the empirical formula it is best to first calculate the number of moles of carbon and of hydrogen present.
Moles C = 84.7/12.0 = 7.1 moles H = 15.3/1.0 = 15.3
Hence the ratio of moles of carbon to hydrogen is 1 : 2 and the empirical formula is CH2. This rules out answers A and B and D — in D the empirical formula works out to be C2H5.
Finally, the Mr of C4H8 in C works out to be 56.0 which answers the question.
2D
When hydrogen bromide reacts with ethene, an addition reaction occurs and the product is bromoethane, which does not contain any double bonds and is saturated. The mechanism for this reaction is electrophilic addition, which has a carbocation present in the transition state, which is therefore not neutral. The hydrogen bromide splits into two ions. There is only one position for the bromine atom. Hence there are no isomers formed.
3
aA hydrolysis reaction is a reaction in which bonds are broken (C–X) by water molecules.
bEqual amounts of each halogenoalkane
Use halogenoalkanes with the same chain length.
Equal amounts of ethanol
Same temperature
(Any two)
cSilver chloridewhite
Silver bromidecream
Silver iodideyellow
dThe halogenoalkane is hydrolysed and an alcohol is formed.
Halide ions are released.
The halide ions react with the silver ions in silver nitrate.
An insoluble ppt is formed.
Ag+ + X−→ AgX where X = Cl, Br or I
The halogen in the halogenoalkane is bonded covalently and does not react with silver nitrate. As the halogenoalkane hydrolyses a halide ion is released and this then reacts with the silver nitrate to form a precipitate.
eIodoalkane
The C–I bond is weakest.
fThe ethanol is the solvent for the halogenoalkane.
The ethanol is the solvent, making sure the halogenoalkane(insoluble in water) and the aqueous silver nitrate mix together and react.
4
aDichlorodifluoromethane
There is one carbon so it is meth, the side groups are chloro and fluoro. Remember to place these in alphabetical order.
bUV light
cCF2Cl2→ CF2Cl•+ Cl•
dCl•+ O3→ O2 + ClO•
ClO•+ O3→ 2O2+ Cl•
The chlorine radical is not destroyed and can act as a catalyst.
5
a CH2=CHCl + NaCN→ CH2=CHCN+ NaCl
bMolecular formula C3H3N.
Empirical formula C3H3N.
ci
c iiIt contains a C=C which can break and allow molecules to add on.
Topic 5 Alcohols
Alcohol production
1
aC6H12O6(aq) → 2CH3CH2OH(aq) + 2CO2(g)
bYeast produces an enzyme which converts glucose to ethanol and carbon dioxide.
A temperature of 35°C. The enzyme in yeast works best around this temperature. Above this the enzyme is denatured, below this the reaction is too slow.
Air is kept out to prevent the oxidation of the ethanol formed to ethanoic acid (vinegar).
ciFractional distillation.
c iiC2H5OH + 3O2→ 2CO2 + 3H2O
diA carbon-neutral activity is one which has no net annual emissions of carbon dioxide to the atmosphere.
d iiMention any process: for example, harvesting, planting, transport of materials, distilling the ethanol solution, producing fertilisers for crops.
The specified process burns (fossil) fuel that releases CO2.
d iiiIt is a renewable resource.
d ivIt uses up land which could be used for crops for food./Production of crops is slow.
eiCH2=CH2+ H2O⇌C2H5OH
In this reaction you must include equilibrium arrows.
Concentrated phosphoric acid /concentrated sulfuric acid
e iiElectrophilic addition
e iiiExcess ethene OR Excess steam / water OR remove the ethanol as it forms OR recycle the ethane
60 atm pressure
600 K temperature
It is an equilibrium reaction so using excess ethene or steam causes the reaction to move to the right producing ethanol. If the ethanol is removed as it is formed, again the reaction moves to the right to replace it.
Oxidation of alcohols
Elimination
2
abutan-2-ol: secondary
propan-1-ol: primary
2-methylbutan-2-ol: tertiary
borange to greenbutan-2-one
orange to greenpropanoic acid
stays orangeno oxidation
cC4H9OH → C4H8 +H2O
Topic 6 Organic analysis
Identification of functional groups by test-tube reactions
Mass spectrometry
Infrared spectroscopy
1
aWarm with acidified potassium dichromate(vi).
Pentan-1-ol changes from orange to green.
In this example you must realise that pentan-1-ol is a primary alcohol and 2-methylpentan-2-ol is tertiary and cannot be oxidised. Hence there is no colour change.
bBromine water
Propene changes it from orange to colourless.