Download Inorganic and organic chemistry 1

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
1
C
Sulfur has 16 electrons and its electron configuration is 1s2 2s2 2p6 3s2 2p4. Its outer electrons
are in the p-subshell. A p-block element has its outer electrons in a p-subshell.
2
D
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.
3
S8 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. 
4
Silicon 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.
© Alyn G. McFarland and Nora Henry 2015
Philip Allan for Hodder Education
1
Topic 2 Group 2, the alkaline earth
metals
B
1
2
a
Ba2+ 
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.
e
It 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. 
f
In calcium the delocalised electrons are closer to 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.
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Philip Allan for Hodder Education
2
Topic 3 Group 7, the halogens
Trends in properties
Uses of chlorine and chlorate(I)
Tests for ions
A
1
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.
2
D
3
Dissolve in water 
Sodium chloride — white ppt  which is soluble in dilute ammonia 
Sodium bromide — cream ppt  which is insoluble in dilute ammonia 
4
a
NaCl + H2SO4 → NaHSO4 + HCl 
b
Br− ions are bigger than Cl− ions 
Therefore Br− ions more easily oxidised / lose an electron more easily. 
ci
Purple vapour / gas
(white solid goes to) black or black-grey or black-purple solid
Bad egg smell (any two)  
c ii
The 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 iii
Oxidation 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.
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H2SO4 + 8H+ → H2S + 4H2O
Write down the charge on each side of the equation.
LHS = + 8 RHS = 0
Add 8 electrons to the LHS to balance the charge.
H2SO4 + 8H+ + 8e– → H2S + 4H2O
5
Cl2 + 2I− → 2Cl− + I2 
a
Solution changes from colourless to brown. 
Remember that all group 1 and group 2 halides in solution are colourless. Iodine in solution is
brown.
2NaOH + Cl2 → NaCl + NaClO + H2O 
bi
Products are sodium chloride, sodium chlorate(I) and water 
b ii
Cl2(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
1
B
2
B
3
a
Mg + 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.
b
Mg(OH)2 + 2HCl → MgCl2 + H2O 
Used in antacid / indigestion tablets
ci
TiCl4 + 2Mg → Ti + 2MgCl2 
Titanium is reduced 
The oxidation number of titanium decreases from +4 to 0. A decrease in oxidation
number means reduction. 
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4
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 ii
Mg + 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
a
A
Bromide 
Ag + + Br− → AgBr 
B
Sulfate 
Ba2+ + SO42− → BaSO4 
C
Chloride 
Ag+ + Cl− → AgCl 
b
To remove any carbonate ions which may interfere with the test. 
c
Magnesium / calcium 
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5
Organic chemistry
Topic 1 Introduction to organic
chemistry
Nomenclature
1
a
pentane
b
1,3-dichloropropane
c
2-chloropropane
d
2-methylpropane
e
2-chloro-2,3,3-trimethylpentane
f
3,3-dichloro-2,2,4-trimethylhexane
Note that the longest chain includes the C2H5 and that chloro is alphabetically before methyl
(ignore the tri).
g
2-bromo-1-chloro-3-ethylpentane
h
propan-2-ol
i
propanal
j
pentan-3-one
k
butanoic acid
l
propene
m
but-2-ene
n
3-bromobutanoic acid
o
propan-1-ol
p
2-methylpropene
q
2,2-dimethylpropane
r
buta-1,3-diene
2
a
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b
c
d
e
3
a
CnH2n+1OH 
b
CnH2n+2
c
CnH2n 
a
C5H12 C5H12 
b
C4H8O2
C2H4O 
a
butane

alkane 
b
but-1-ene

alkene 
c
butanoic acid

carboxylic acid
d
butan-1-ol

alcohol

e
pentan-2-one

ketone

f
propan-2-ol
 alcohol

4
5
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
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7
Isomerism
Exam-style questions
B
1
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).
B
2
3-methylbut-1-ene has five carbon atoms.
3
C
4
A
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
a
C=C 
OH 
b

6
c
Lack of rotation about the C=C 
d
Structural isomers are compounds with the same molecular formula but different
structural formulae. 
The right-hand carbon of the double bond has a C attached which has a higher atomic
number than H so 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 H attached 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.
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8
Topic 2 Alkanes
Fractional distillation of crude oil
Modification of alkanes by cracking
1
Allow 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 = 16
y = (2 × 4) + 8 + 18 = 34
C16H34 → 2C2H4 + C4H8 + C8H18
High pressure 
High temperature 
(it is thermal cracking)
2
a
C8H18 → C2H4 + C3H6 + C3H8 
Propane 
Zeolite 
b
Petroleum / crude oil 
Fractional distillation
c

Octane 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. 
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9
Combustion of alkanes
Chlorination of alkanes
1

C
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.
2
B
3
Propane C3H8

Butane C4H10

4
They have the same general formula.
They show similar chemical properties.
They show a gradual change in physical properties from propane to butane.
(Any two) 

5
CnH2n+2
6
They contain carbon and hydrogen atoms only

They only contains single C–C bonds — there are no C=C bonds. 
7
8
Propane 
Because it has a lower boiling point as it has lower Mr

and weaker van der Waals forces between its molecules

C2H6 + 3½O2 → 2CO2 + 3H2O

or
2C2H6 + 7O2 → 4CO2 + 6H2O

Note that complete combustion results in the formation of carbon dioxide and water.
9
C4H10 + 4.5 O2 → 4CO + 5H2O
10
Sulfur dioxide


SO2 + CaCO3 → CaSO3
Neutralisation 
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10
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.
The pressure and high temperature in the combustion engine causes the normally
unreactive nitrogen in the air to react with oxygen / N2 + O2 → 2NO2 
11
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
1
C
2
D
A radical has an unpaired electron. In CH4 all electrons are used in bonding so it cannot have an
unpaired electron.
3
a
C3H8 + Br2 → C3H8Br + HBr
UV light / high temp


b
2-bromopropane 
c
Initiation:

Br2 → 2Br•

Propagation: 
Br• + CH3CH2CH3 → CH3CH2CH2• + HBr
or
CH3CH2CH2• + Br2 →CH3CH2CH2Br + Br•

Termination:

CH3CH2CH2• + CH3CH2CH2• → CH3CH2CH2CH2CH2CH3
or Br• + Br• → Br2
CH3CH2CH2• + Br• → CH3CH2CH2Br
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Be careful with the position of the radical dot — it must be to the side or above of the last carbon
CH3CH2CH2• — if it is beside the CH3 or in the middle of the chain it is incorrect.
4
a
Molecular 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 
c
C10H22 + 10½O2 → 10CO + 11H2O 
d
C8H18 + 25NO → 8CO2 + 12½N2 + 9H2O 
Platinum or rhodium 
e
CH3SH + 3O2 → CO2 + 2H2O + SO2 
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Topic 3 Halogenoalkanes
Nucleophilic substitution and elimination
Ozone depletion
C
1
This is an elimination reaction in which HBr is removed forming an alkene.
B
2
A nucleophile is a lone pair donor. CH4 does not have any lone pairs and cannot act as a
nucleophile.
3
a
1-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.

b
CH3CH2CH2CH2Br + NaOH → CH3CH2CH=CH2 + NaBr + H2O 
But-1-ene

Elimination

4
a

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.
b
It has the same molecular formula but a different structural formula.
c
2-methylbut-1-ene

2-methylbut-2-ene


Two different isomers can be formed here depending on which H bonds to the hydroxide ion.
d
CH3CH2CH2CH2Br + NaOH → CH3CH2CH2CH2OH + NaBr + H2O
Butan-1-ol


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This is a substitution reaction, since it occurs in aqueous solution. The bromine is replaced by
OH.
e
Nucleophilic substitution 
f
The 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
a
The 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
d
To remove acidic impurities

e
Shake in a separating funnel

Invert and release the pressure periodically


Allow to settle and run off bottom layer — discard the aqueous layer.
f
To remove water


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Topic 4 Alkenes
Structure, bonding and reactivity
Addition reactions of alkenes
Addition polymers
C
1
Hydrocarbons which contain a double bond, are alkenes and have the general formula C nH2n.
2

A
3
a

But-1-ene


butane

b
The double bond is a centre of high electron density  and can undergo attack by
electrophiles. 
c
Bromine water 
Orange to colourless 
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di
CH3CH=CHCH3 + Br2 → CH3CHBrCHBrCH3
d ii
Electrophilic addition 
 2,3-dibromobutane


e
Major 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
2-bromobutane can also exist as two enantiomers as there are four different groups on one of its
carbon atoms.
f
Poly(but-2-ene) 

Exam-style questions
1
C
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 CH 2. 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.
2
D
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
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16
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
a
A hydrolysis reaction is a reaction in which bonds are broken (C–X) by water
molecules. 
b
Equal amounts of each halogenoalkane
Use halogenoalkanes with the same chain length.
Equal amounts of ethanol
Same temperature
(Any two)
c
d


Silver chloride
white
Silver bromide
cream 
Silver iodide
yellow 
The 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.
e
Iodoalkane

The C–I bond is weakest. 
f
The 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
a
Dichlorodifluoromethane 
There is one carbon so it is meth, the side groups are chloro and fluoro. Remember to place
these in alphabetical order.
b
UV light 
c
CF2Cl2 → CF2Cl• + Cl• 
d
Cl• + O3 → O2 + ClO• 
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ClO• + O3 → 2O2 + Cl• 
The chlorine radical is not destroyed and can act as a catalyst.
5
a
CH2=CHCl + NaCN → CH2=CHCN + NaCl
b
Molecular formula C3H3N. 

Empirical formula C3H3N. 
ci

c ii
It contains a C=C which can break and allow molecules to add on. 
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Topic 5 Alcohols
Alcohol production
1
a
C6H12O6(aq) → 2CH3CH2OH(aq) + 2CO2(g) 
b
Yeast 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). 
ci
Fractional distillation. 
c ii
C2H5OH + 3O2 → 2CO2 + 3H2O 
di
A carbon-neutral activity is one which has no net annual emissions of carbon
dioxide to the atmosphere.

d ii
Mention 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 CO 2. 
d iii
It is a renewable resource. 
d iv
It uses up land which could be used for crops for food./Production of crops is slow.

ei
CH2=CH2 + H2O ⇌ C2H5OH 
In this reaction you must include equilibrium arrows.
Concentrated phosphoric acid / concentrated sulfuric acid
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19
e ii
Electrophilic addition 

e iii
Excess 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
a
butan-2-ol: secondary 
propan-1-ol: primary 
2-methylbutan-2-ol: tertiary 
b
c
orange to green
butan-2-one
orange to green
propanoic acid
stays orange
no oxidation
C4H9OH → C4H8 +H2O
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20
Topic 6 Organic analysis
Identification of functional groups by testtube reactions
Mass spectrometry
Infrared spectroscopy
1
a
Warm 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.
b
Bromine water 
Propene changes it from orange to colourless. 
Propene is an alkene and is unsaturated and will react with bromine water.
c
Warm with Fehling’s solution / Tollens’ reagent.
Butanal — silver mirror / red ppt


Add sodium carbonate — effervescence with the acid 
Butanal is an aldehyde and can be oxidised by Fehling’s solution or Tollens’ reagent (or even
potassium dichromate(VI)) but butanone, a ketone cannot. The butanoic acid reacts with solid
sodium carbonate producing bubbles of carbon dioxide.
2
(2 × N) and (1 × O): 2(14.00307) + 15.99491 = 44.00105 
propane is C3H8 = 44.0 
carbon dioxide is CO2 = 44.0 
dinitrogen oxide is N2O = 44.0 
They have the same Mr (to 1 decimal place) and would give the same molecular ion
peak. 
The molecular ion value gives the relative molecular mass — it is 44.00105. To show that this is
N2O simply add together the precise relative atomic mass values of (2 × N) and (1 × O).
2(14.00307) + 15.99491 = 44.00105. You then need to calculate the relative atomic mass of each
of these molecules using the relative atomic masses given to 1 decimal place in your periodic
table:
© Alyn G. McFarland and Nora Henry 2015
Philip Allan for Hodder Education
21
propane is C3H8
= (3 × 12.0) + (8 × 1.0)
= 44.0
carbon dioxide is CO2
= (1 × 12.0) + (2 × 16.0)
= 44.0
dinitrogen oxide is N2O = (2 × 14.0) + (1 × 16.0)
= 44.0
They have the same Mr (to 1 decimal place) and would give the same molecular ion peak.
A is ethanol as it has a broad OH absorption between 3230 and 3550. 
3
B is ethanoic acid as it has an OH (acid) between 2500 and 3000 and a C=O between
1680 and 1750. 
C is propanone as there is no OH absorption but a C=O absorption at 1680–1750. 
Exam-style questions
C
1
The alcohol has four carbons so it is named but-. The OH group is on the second carbon so it is
butan-2-ol and a secondary alcohol which is oxidised to a ketone.
B
2
To answer this question draw out each structure. The alcohol which has the methyl group and
the OH on the same carbon is the one which is a tertiary alcohol — it has three alkyl groups
bonded to the carbon that is bonded to the OH group.
3
D
4
The broad 3350 peak indicates an alcohol. 
moles C = 70.6/12.0 = 5.88 
moles H = 13.7/1.0 = 13.7 
moles O =15.7/16.0 = 0.98 
Ratio
5.88/0.98 = 6.0; 13.7/0.98 = 14.0; 0.98/0.98 = 1.0 so C6H14O 
m/z = Mr = 102 = (C6H14O) n = 102n. n = 1 so the molecular formula = C6H14O 
hexan-1-ol 
hexan-2-ol 
hexan-3-ol 
5
a
Repeated heating to evaporate a liquid and cooling to condense a vapour. 
Vapour escapes from the liquid mixture, condenses and returns to the liquid mixture
where it is in contact with the oxidation agent again. 
This ensures complete oxidation of propanol and any propanal formed. 
b
CH3CH2OH + H2O → CH3COOH + 4H+ + 4e− 
c
Acidified potassium dichromate(VI) 
© Alyn G. McFarland and Nora Henry 2015
Philip Allan for Hodder Education
22
Orange to green 
d
Change the apparatus for distillation. 
Use a still head containing a thermometer with the condenser horizontal. 
Collect sample at the boiling point of propanal. 
e
Propanal: add Tollens’ reagent and warm  silver mirror 
Or add Fehling’s solution orange  red ppt 
Absence of propanoic acid — add sodium hydrogencarbonate or sodium carbonate

No effervescence observed 
f
Alcohol OH absorption in different place (3230−3550 cm−1 ) 
from acid OH absorption (2500−3000 cm−1) 
or different fingerprint region  (any two)
© Alyn G. McFarland and Nora Henry 2015
Philip Allan for Hodder Education
23