Each of the parts (a) to (e) below concerns a different pair of isomers. Draw one possible structure for each of the species A to J, using Table 2 on the Data Sheet where appropriate. (a) Compounds A and B have the molecular formula C 5 H 10 A decolourises bromine water but B does not. A B (2) 1 (b) Compounds C and D have the molecular formula C 2 H 4 O 2 Each has an absorption in its infra-red spectrum at about 1700 cm –1 but only D has a broad absorption at 3350 cm –1 C D (2) (c) Compounds E and F are esters with the molecular formula C 5 H 10 O 2 The proton n.m.r. spectrum of E consists of two singlets only whereas that of F consists of two quartets and two triplets. E F (2) Page 1 of 92
Transcript
Each of the parts (a) to (e) below concerns a different pair of isomers.
Draw one possible structure for each of the species A to J, using Table 2 on the Data Sheetwhere appropriate.
(a) Compounds A and B have the molecular formula C5H10
A decolourises bromine water but B does not.
A B
(2)
1
(b) Compounds C and D have the molecular formula C2H4O2
Each has an absorption in its infra-red spectrum at about 1700 cm–1 but only D has a
broad absorption at 3350 cm–1
C D
(2)
(c) Compounds E and F are esters with the molecular formula C5 H10O2
The proton n.m.r. spectrum of E consists of two singlets only whereas that of F consistsof two quartets and two triplets.
E F
(2)
Page 1 of 92
(d) Compounds G and H have the molecular formula C3H6Cl2 G shows optical activity but Hdoes not.
G H
(2)
(e) Compounds I and J have the molecular formula C6H12
Each has an absorption in its infra-red spectrum at about 1650 cm–1 and neithershows geometrical isomerism. The proton n.m.r. spectrum of I consists of a singletonly whereas that of J consists of a singlet, a triplet and a quartet.
I J
(2)
(Total 10 marks)
The alkanes form an homologous series of hydrocarbons. The first four straight-chain alkanesare shown below.
methane CH4
ethane CH3CH3
propane CH3CH2CH3
butane CH3CH2CH2CH3
(a) (i) State what is meant by the term hydrocarbon.
The alkene 3-methylpent-2-ene (CH3CH=C(CH3)CH2CH3) reacts with hydrogen bromide to forma mixture of 3-bromo-3-methylpentane and 2-bromo-3-methylpentane.
(a) The alkene 3-methylpent-2-ene (CH3CH=C(CH3)CH2CH3) exists as E and Zstereoisomers.
Draw the structure of Z-3-methylpent-2-ene.
(1)
4
(b) Name and outline the mechanism for the formation of 3-bromo-3-methylpentane from thisreaction of 3-methylpent-2-ene with hydrogen bromide.
Explain why more 3-bromo-3-methylpentane is formed in this reaction than 2-bromo-3-methylpentane.
(ii) Draw the structure of the chain isomer of but-1-ene.
(2)
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(c) The alkanes and the alkenes are examples of homologous series of compounds.One feature of an homologous series is the gradual change in physical propertiesas the relative molecular mass increases. State two other general features of anhomologous series of compounds.
Which of the following is the general formula of cyclic alkenes such as cyclohexene?
A CnH2n–4
B CnH2n–2
C CnH2n
D CnH2n+2
(Total 1 mark)
6
Compound J, known as leaf alcohol, has the structural formulaCH3CH2CH=CHCH2CH2OH and is produced in small quantities by many green plants.The E isomer of J is responsible for the smell of freshly cut grass.
(a) Give the structure of the E isomer of J.
(1)
7
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(b) Give the skeletal formula of the organic product formed when J is dehydrated usingconcentrated sulfuric acid.
(1)
(c) Another structural isomer of J is shown below.
Explain how the Cahn-Ingold-Prelog (CIP) priority rules can be used to deduce the fullIUPAC name of this compound.
(b) Name and draw the mechanism for the formation of one of the isomers.
Name of mechanism ......................................................................................
Mechanism
(5)
(Total 6 marks)
Dodecane (C12H26) is a hydrocarbon found in the naphtha fraction of crude oil. Dodecane can beused as a starting material to produce a wide variety of useful products. The scheme belowshows how one such product, polymer Y, can be produced from dodecane.
(a) Name the homologous series that both C2H4 and C4H8 belong to.Draw a functional group isomer of C4H8 that does not belong to this homologous series.
Name ...........................................................................................................
(e) Reaction 2 is exothermic. A typical compromise temperature of 200 °C is used industriallyfor this reaction.
Explain the effect of a change of temperature on both the position of equilibrium and therate of reaction, and justify why a compromise temperature is used industrially.
Isooctane (C8H18) is the common name for the branched-chain hydrocarbon that burns smoothlyin car engines. The skeletal formula of isooctane is shown below.10
(b) Deduce the number of peaks in the 13C NMR spectrum of isooctane.
5
6
7
8 (1)
(c) Isooctane can be formed, together with propene and ethene, in a reaction in which onemolecule of an alkane that contains 20 carbon atoms is cracked.
Using molecular formulas, write an equation for this reaction.
(e) Deduce the number of monochloro isomers formed by isooctane.Draw the structure of the monochloro isomer that exists as a pair of optical isomers.
Number of monochloro isomers ....................................................................
Structure
(2)
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(f) An isomer of isooctane reacts with chlorine to form only one monochloro compound.
Draw the skeletal formula of this monochloro compound.
(1)
(g) A sample of a monochlorooctane is obtained from a comet. The chlorine in themonochlorooctane contains the isotopes 35Cl and 37Cl in the ratio 1.5 : 1.0Calculate the Mr of this monochlorooctane.
Mr = ...............................(2)
(h) Isooctane reacts with an excess of chlorine to form a mixture of chlorinated compounds.One of these compounds contains 24.6% carbon and 2.56% hydrogen by mass. Calculatethe molecular formula of this compound.
Molecular formula = ...............................(3)
(Total 12 marks)
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How many isomers have the molecular formula C5H12?
A 2
B 3
C 4
D 5 (Total 1 mark)
11
How many structural isomers have the molecular formula C4H9Br?
A 2
B 3
C 4
D 5 (Total 1 mark)
12
How many secondary amines have the molecular formula C4H11N?
A 2
B 3
C 4
D 5 (Total 1 mark)
13
Central heating fuel, obtained by the fractional distillation of crude oil, contains saturatedhydrocarbons with the molecular formula C16H34
(a) Give the meaning of the terms saturated and hydrocarbon as applied to saturatedhydrocarbons.
(ii) Write an equation for the cracking of one molecule of dodecane into equal amounts oftwo different molecules each containing the same number of carbon atoms.State the empirical formula of the straight-chain alkane that is formed.Name the catalyst used in this reaction.
(iv) When a sample of Y, contaminated with CH3SH, is burned completely in air, a toxicgas is formed.Identify this toxic gas and suggest a compound that could be used to remove thetoxic gas from the products of combustion.
Toxic gas ..............................................................................................
Compound used to remove toxic gas ..................................................
(d) A chemical test can be used to distinguish between separate samples of Isomer 2 andIsomer 3.Identify a suitable reagent for the test.State what you would observe with Isomer 2 and with Isomer 3.
Test reagent ...............................................................................................
Observation with Isomer 2...........................................................................
(iii) Draw the structure of a functional group isomer of pent-1-ene.
(1)
(c) The cracking of one molecule of compound X produces pent-1-ene, ethene and butane in a1:2:1 mol ratio.Deduce the molecular formula of X and state a use for the ethene formed.
Molecular formula of X .................................................................................
(a) Hexane (C6H14) is a hydrocarbon which is a component of LPG (liquid petroleum gas),used as a fuel for heating. When burning fuels in boilers it is important to ensure completecombustion.
(i) Give two reasons why boilers are designed to ensure complete combustion.
(ii) Give one reason why this atom economy of less than 100% is an importantconsideration for the commercial success of this process and predict how a chemicalcompany would maximise profits from this process.
There are seven isomeric carbonyl compounds with the molecular formula C5H10O.The structures and names of some of these isomers are given below.
Structure Name
pentanal
2-methybutanal
2, 2-dimethypropanal
pentan-2-one
(a) (i) Complete the table.
27
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(ii) Two other isomeric carbonyl compounds with the molecular formula C5H10O are notshown in the table. One is an aldehyde and one is a ketone. Draw the structure ofeach.
isomeric aldehyde isomeric ketone
(4)
(b) Pentanal, CH3CH2CH2CH2CHO, can be oxidised to a carboxylic acid.
(i) Write an equation for this reaction. Use [O] to represent the oxidising agent.
(a) The infra-red spectrum of compound A, C3H6O2, is shown below.
Identify the functional groups which cause the absorptions labelled X and Y.
Using this information draw the structures of the three possible structural isomers for A.
Label as A the structure which represents a pair of optical isomers.(6)
28
(b) Draw the structures of the three branched-chain alkenes with molecular formula C5H10
Draw the structures of the three dibromoalkanes, C5H10Br2, formed when these threealkenes react with bromine.
One of these dibromoalkanes has only three peaks in its proton n.m.r. spectrum. Deducethe integration ratio and the splitting patterns of these three peaks.
(10)(Total 16 marks)
Which one of the following can exhibit both geometrical and optical isomerism?
A (CH3)2C=CHCH(CH3)CH2CH3
B CH3CH2CH=CHCH(CH3)CH2CH3
C (CH3)2C=C(CH2CH3)2
D CH3CH2CH(CH3)CH(CH3)C=CH2
(Total 1 mark)
29
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How many different alkenes are formed when 2-bromo-3-methylbutane reacts with ethanolicpotassium hydroxide?
A 2
B 3
C 4
D 5(Total 1 mark)
30
The table below gives some of the names and structures of isomers having the molecularformula C4H9Br
Structure Name
CH3CH2CH2CH2Br
2-bromo - 2-methypropane
1-bromo - 2-methypropane
2-methypropane
Complete the table.(Total 2 marks)
31
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(a) Name and outline a mechanism for the reaction of 2-bromo-2-methylpropane withethanolic potassium hydroxide to form the alkene 2-methylpropene, (CH3)2C=CH2
Name of mechanism ....................................................................................
Mechanism
(4)
32
(b) Two stereoisomers of but-2-ene are formed when 2-bromobutane reacts with ethanolicpotassium hydroxide.
(i) Explain what is meant by the term stereoisomers.
Some alcohols can be oxidised to form aldehydes, which can then be oxidised further to formcarboxylic acids.Some alcohols can be oxidised to form ketones, which resist further oxidation.Other alcohols are resistant to oxidation.
(a) Draw the structures of the two straight-chain isomeric alcohols with molecular formula,C4H10O
(2)
34
(b) Draw the structures of the oxidation products obtained when the two alcohols from part (a)are oxidised separately by acidified potassium dichromate(VI). Write equations for anyreactions which occur, using [O] to represent the oxidising agent.
(6)
(c) Draw the structure and give the name of the alcohol with molecular formula C4H10O whichis resistant to oxidation by acidified potassium dichromate(VI).
(2)(Total 10 marks)
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The compound cis-retinal is shown below.
Which one of the labelled bonds leads to the prefix in the name?(Total 1 mark)
35
Which one of the following is a pair of functional group isomers?
A CH3COOCH2CH3 and CH3CH2COOCH3
B (CH3)2CHCH(CH3)2 and (CH3)3CCH2CH3
C CH3CH2OCH3 and (CH3)2CHOH
D ClCH2CH2CH=CH2 and CH3CH=CHCH2Cl(Total 1 mark)
36
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Mark schemes
(a) A any C5 alkene1
B
1
1
(b) C
or CH3COOH or HCOOCH3
1
D
or HOCH2CHO1
(c) E
1
F
1
Page 50 of 92
(d) G
1
H
1
(e) I
1
J
NOT hex-3-ene1
[10]
(a) (i) Molecule/compound/consists/composed/made up of hydrogen andcarbon only (1)
(ii) CnH2n+2 (1)
(iii) C6H14 only (1)Do not credit structures alone or in addition.
3
2
(b) Chemically similar / react in same way / same chemistryDiffer by CH2
gradation in physical properties OR specified trend e.g. b.p.same functional group
Any 2, 2 marks 1 + 1
Not same molecular formula2
Page 51 of 92
(c) (i) Same molecular formula (1)
NOT same Mr
different structural formula / structures (1)(or atoms arranged in different way)
NOT different spatial arrangementsOnly credit M2 if M1 correct
(ii) 2-methylpentane (1)2,2-dimethylbutane (1)
(iii)
OR correct condensed / structural formula
Penalise “sticks” oncePenalise absence of vertical bonds oncepenalise badly drawn bonds once (vertical between H atoms)
6
(d) (i) M1 % by mass of H = 7.7(0)% (1)M2 mol H = 7.70 / 1 = 7.70 mol C = 92.3 / 12 = 7.69 (1)
M3 (ratio 1:1 ) CH
Correct answer = 3 marks
Credit variations for M2 e.g. 78 × = 6
and = 6
Correct answer 1 mark4
[15]
(ii) (CH has empirical mass of 13 and = 6 ) C6H6 (1)
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(a) M1 curly arrow from lone pair on oxygen of hydroxide ion toH atom on C-H adjacent to C-Br
1
M2 curly arrow from single bond of adjacent C-Hto adjacent single bond C-C
(only credit M2 if M1 is being attempted to correct H atom)1
M3 curly arrow from C-Br bond to side of Br atom
(credit M3 independently)1
3
(b) Ml credit a correct structure for either geometrical E-Z isomer and itsdesignation as either cis or trans.OR credit two correct geometrical E-Z isomer structures(ignore the names)OR credit two correct names for cis pent-2-ene and transpent-2-ene (ignore the structures)
1
M2 credit a second mark if all four parts of the required structures andnames are correct.
(credit “linear” structures)(insist on the alkyl groups being attached clearly by C-C bonds)
1
(c) (i) Ml curly arrow from middle of C = C bond to H atom on H-Br
(penalise M1 if partial negative charge or formal positivecharge on H)(penalise Ml if pent-2-ene is used)
1
M2 curly arrow from H-Br bond to side of Br atom1
M3 correct structure for correct secondary carbocation1
M4 curly arrow from lone pair on bromide ion to the positivecarbon of carbocation, ensuring that bromide ion has anegative charge.
(with the exception of pent-2-ene, if the wrong alkene is used, onlypenalise the structure M3)(penalise the use of two dots in addition to a covalent bond, onceonly)
1
(ii) 1-bromopentane1
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(iii) Ml 2-bromopentane is formed via the secondary (or 2°)carbocation
1
OR 1-bromopentane is formed via the primary (or 1°)carbocationM2 a secondary carbocation is more stable than a primarycarbocation -award this mark only if the quality of language justifiesthe award.
(the argument must involve clear statements about carbocations)1
[12]
(a)
Must show all 4 groups bonded to C=C
Allow CH3− for methyl group; allow C2H5 for ethyl group
Allow correct structure of the style
Allow correct skeletal structure
1
4
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(b) M1 electrophilic addition
NB the arrows here are double-headed1
M2 must show an arrow from the double bond towards the Hatom of the H-Br molecule
1
M3 must show the breaking of the H-Br bond1
M4 is for the structure of the tertiary carbocation1
M5 must show an arrow from the lone pair of electrons on thenegatively charged bromide ion towards the positivelycharged atom (of either a secondary or) of a tertiarycarbocation
1
M6 3-bromo-3-methylpentane is formed from 3y carbocationOR2-bromo-3-methylpentane is formed from 2y carbocation
1
M7 3y carbocation more stable than 2y
1
M2-M5 Penalise one mark from their total if half-headed arrows areused
M2 Ignore partial negative charge on the double bond
M3 Penalise incorrect partial charges on H-Br bond and penaliseformal charges
Penalise M4 if there is a bond drawn to the positive charge
Penalise only once in any part of the mechanism for a line and twodots to show a bond
Max 3 of any 4 marks (M2-5) for wrong organic reactant or wrongorganic product (if shown) or secondary carbocation
Max 2 of any 4 marks in the mechanism for use of bromine
Do not penalise the “correct” use of “sticks”
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For M5, credit attack on a partially positively charged carbocationstructure but penalise M4
M6 is high demand and must refer to product being formedfrom/via correct class of carbocation
M7 is high demand and must be clear answer refers to stabilityof carbocations (intermediates) not products
Candidate that states that products are carbocations wouldlose M6 and M7
M6,7 allow carbonium ion in place of carbocation; or a descriptionof carbocation in terms of alkyl groups/ number of carbon atomsjoined to a positive C
When asked to outline a mechanism, candidates are expected todraw a mechanism with curly arrows (specification 3.3.1.2). On thisoccasion only we would allow a detailed description as shown.
M2 must describe the movement of a pair of electrons / curly arrowfrom the C=C towards the H atom of the H-Br molecule
M3 must describe the breaking of the H-Br bond with the bondingpair of electrons moving to the Br / curly arrow from H-Br bond to Br
M4 is for the structure of the tertiary carbocation (i.e. positive Cbonded to one methyl and two ethyl groups)
M5 must describe the movement of a pair of electrons from the Br−
ion to the positive C atom of the carbocation / curly arrow from thelone pair of electrons on the negatively charged bromide iontowards the positively charged C atom (of either a secondary or) ofa tertiary carbocation
[8]
(a) (i) fractional distillation or fractionation1
(ii) C9H20 only1
(iii) C11H24 + 17O2 → 11CO2 + 12H2O1
(iv) C11H24 + 6O2 → 11C + 12H2O1
5
(b) (i) C10H22 → C3H6 + C7H16
1
(ii) correctly drawn structure of methylpropene
(insist on clearly drawn C-C and C=C bonds)1
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(c) Any two from
o chemically similar or chemically the same or react inthe same way
o same functional group
o same general formula
o differ by CH2
(penalise same molecular formula or same empirical formula)2
[8]
B[1]6
(a)
1
7
(b) 1
(c) Stage 1: consider the groups joined to right hand carbon of the C=C bond
Extended response
Maximum of 5 marks for answers which do not show a sustainedline of reasoning which is coherent, relevant, substantiated andlogically structured.
Consider the atomic number of the atoms attached
M1 can be scored in stage 1 or stage 21
C has a higher atomic number than H, so CH2OH takes priority1
Stage 2: consider the groups joined to LH carbon of the C=C bond
Both groups contain C atoms, so consider atoms one bond further away1
C, (H and H) from ethyl group has higher atomic number than H, (H and H) frommethyl group, so ethyl takes priority
1
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Stage 3: conclusion
The highest priority groups, ethyl and CH2OH are on same side of the C=C bond sothe isomer is Z
Allow M5 for correct ECF conclusion using either or both wrongpriorities deduced in stages 1 and 2
1
The rest of the IUPAC name is 3-methylpent-2-en-1-ol1
AND mass of organic product expected = (8.62 × 10–2) × 98.0 = 8.45 g
Or moles of organic product formed = 6.53 / 98.0 = 6.66 × 10–2
1
% yield = 100 × 6.53 / 8.45
OR = 100 × (6.66 × 10–2) / (8.62 × 10–2)
= 77.294 = 77.3%
AND statement that the student was NOT correct1
[10]
(a) (Compounds with the) same molecular formula but different structural / displayed / skeletalformula
1
8
(b) (basic) elimination1
Mechanism points:
Correct arrow from lone pair on :OH– to H on C adjacent to C–Br1
Correct arrow from C–H bond to C–C1
Correct arrow from C–Br bond to Br1
Structure of chosen product1
Page 58 of 92
OR
[6]
(a) Alkenes19
Correctly drawn molecule of cyclobutane or methyl cyclopropane,need not be displayed formula
1
(b) C6H14 (or correct alkane structure with 6 carbons)
Allow hexane or any other correctly named alkane with 6 carbons1
(c) Poly(but-2-ene)1
(d) High pressure
Allow pressure ≥ MPaMention of catalyst loses the mark
1
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(e) This question is marked using levels of response. Refer to the Mark SchemeInstructions for Examiners for guidance on how to mark this question.
Level 3
All stages are covered and the explanation of each stage is generally correct andvirtually complete.
Answer communicates the whole process coherently and shows a logical progressionfrom stage 1 and stage 2 (in either order) to stage 3.
5–6 marks
Level 2
All stages are covered but the explanation of each stage may be incomplete or maycontain inaccuracies OR two stages are covered and the explanations are generallycorrect and virtually complete.
Answer is mainly coherent and shows progression. Some steps in each stage maybe out of order and incomplete.
3–4 marks
Level 1
Two stages are covered but the explanation of each stage may be incomplete or maycontain inaccuracies, OR only one stage is covered but the explanation is generallycorrect and virtually complete.
Answer includes isolated statements but these are not presented in a logical order orshow confused reasoning.
1–2 marks
Level 0
Insufficient correct chemistry to gain a mark.0 marks
Indicative chemistry content
Stage 1: consider effect of higher temperature on yield
(Or vice versa for lower temperature)
• Le Chatelier’s principle predicts that equilibrium shifts to oppose any increase in temperature
• Exothermic reaction, so equilibrium shifts in endothermic direction / to the left
• So a Higher T will reduce yield
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Stage 2: consider effect of higher temperature on rate
(Or vice versa for lower temperature)
• At higher temperature, more high energy molecules
• more collisions have E>Ea
• So rate of reaction increases / time to reach equilibrium decreases
Stage 3: conclusion
Industrial conditions chosen to achieve (cost-effective) balance ofsuitable yield at reasonable rate
[11]
(a) 2,2,4-trimethylpentane110
(b) 51
(c) C20H42 C8H18 + 2C3H6 + 3C2H4
1
(d) Mainly alkenes formed1
(e) 4 (monochloro isomers)1
1
(f)
1
(g) C8H1735Cl = 96.0 + 17.0 + 35.0 = 148.0
and C8H1737Cl = 96.0 + 17.0 + 37.0 = 150.0
Both required1
Mr of this C8H17Cl = 148.81
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(h) = 2.05 : 2.56 : 2.05
Simplest ratio =
= 1 : 1.25 : 11
Whole number ratio (× 4) = 4 : 5 : 41
MF = C8H10Cl81
[12]
B[1]11
C[1]12
B[1]13
(a) Saturated − single bonds only / no double bonds1
Hydrocarbon − contains carbon and hydrogen (atoms) only1
14
(b) C16H34 + 16.5O2 16CO + 17H2O
Allow multiples1
(c) (On combustion) SO2 produced
Allow equation to produce SO2. Ignore sulfur oxides.1
Which causes acid rain
If formula shown it must be correct
M2 is dependent on M1. But if M1 is sulfur oxides, allow M2.
For M2 allow consequence of acid rain or SO2.
Ignore greenhouse effect and toxic1
(d) (i) C16H34 C8H18 + C2H4 + 2C3H6
Allow multiples1
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(ii) polypropene / propan(-1 or 2-)ol / propane(-1,2-)diol / isopropanol / propanone /propanal
Accept alternative names
Ignore plastic and polymer1
(iii)
1
(e)
Allow any unambiguous representation1
(f) 2,4-dichloro-2,4-dimethylhexane
Only but ignore punctuation1
[10]
(a) Crude oil OR petroleum
Not petrol.1
15
Fractional distillation / fractionation
Not distillation alone.1
(b) C12H26 + 12.5O2 12CO + 13H2O
Allow balanced equations that produce CO2 in addition to CO.
Accept multiples.1
(c) (i) M1 Nitrogen and oxygen (from air) react / combine / allow a correct equation
If nitrogen from petrol / paraffin / impurities CE = 0 / 2.1
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M2 at high temperatures
Allow temperatures above 1000 °C or spark.
Not just heat or hot.
M2 dependent on M1.
But allow 1 mark for nitrogen and oxygen together at hightemperatures.
For M1 penalise Br (or incorrect formula of other correct reagent),but mark on.
M3 Isomer 2: remains orange / red / yellow / brown / the same OR no reaction / no(observable) change OR reference to colour going to the cyclopentane layer
For M1, it must be a whole reagent and / or correct formula.
If oxidation state given in name, it must be correct. If ‘manganate’OR ‘manganate(IV)’ or incorrect formula, penalise M1, but mark on.
Alternatives : potassium manganate(VII)
M1 KMnO4 in acid M2 colourless M3 purple
M1 KMnO4 in alkali / neutral M2 brown solid M3 purple
Credit for the use of iodine
M1 iodine (solution / in KI) M2 colourless M3 (brown) to purple (credit nochange)
Credit for the use of concentrated H2SO4
M1 concentrated H2SO4 M2 brown M3 no change / colourless
Ignore ‘goes clear’.
Ignore ‘nothing (happens)’.
Ignore ‘no observation’.
No credit for combustion observations.3
(c) (i) (Both infrared spectra show an absorption in range) 1620 to 1680 (cm−1)
Ignore reference to other ranges (eg for C–H or C–C).1
(ii) The fingerprint (region) / below 1500 cm−1 will be different or its fingerprintingwill be different
OR
different absorptions / peaks are seen (in the region) below 1500 cm−1 (or aspecified region within the fingerprint range)
Allow the words ‘dip’ OR ‘spike’ OR ‘low transmittance’ asalternatives for absorption.
QoL1
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(d)
All bonds must be drawn.
Ignore bond angles.1
(e) (i) M1 Electrophilic addition
M1 both words needed.
Penalise one mark from their total if half-headed arrows are used.
M2 must show an arrow from the double bond towards the H atom of the H–Brmolecule
M2 Ignore partial negative charge on the double bond.
M3 must show the breaking of the H–Br bond
M3 Penalise incorrect partial charges on H–Br bond and penaliseformal charges.
M4 is for the structure of the tertiary carbocation
Penalise M4 if there is a bond drawn to the positive charge.
Penalise once only in any part of the mechanism for a line and twodots to show a bond.
M5 must show an arrow from the lone pair of electrons on the negativelycharged bromide ion towards the positively charged carbon atom of either asecondary or a tertiary carbocation
For M5, credit attack on a partially positively charged carbocationstructure but penalise M4.
Max 3 of any 4 marks in the mechanism for wrong organicreactant or wrong organic product (if shown) or secondarycarbocation.
Max 2 of any 4 marks in the mechanism for use of bromine.
Do not penalise the correct use of 'sticks”.
NB The arrows here are double-headed5
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(ii) M1 Reaction goes via intermediate carbocations / carbonium ions
M1 is a lower demand mark for knowledge that carbocations areinvolved.
M2 (scores both marks and depends on M1)
Tertiary carbocation / carbonium ion is more stable (than the secondarycarbocation / carbonium ion)
OR
Secondary carbocation / carbonium ion is less stable (than the tertiarycarbocation / carbonium ion)
M2 is of higher demand and requires the idea that the secondarycarbocation is less stable or the tertiary carbocation is more stable.Reference to incorrect chemistry is penalised.
A carbocation may be defined in terms of alkyl groups / number ofcarbon atoms, rather than formally stated.
2
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(f) M1 Elimination
M1 credit ‘base elimination’ but no other qualifying prefix.
Penalise one mark from their total if half-headed arrows are used.
M2 must show an arrow from the lone pair on oxygen of a negatively chargedhydroxide ion to a correct H atom
Penalise M2 if covalent KOH
M3 must show an arrow from a correct C–H bond adjacent to the C–Br bond to acorrect C–C bond. Only award if an arrow is shown attacking the H atom of a correctadjacent C–H bond (in M2)
M4 is independent provided it is from their original molecule BUT penalise M2, M3and M4 if nucleophilic substitution shown
Award full marks for an E1 mechanism in which M2 is on the correct carbocation
NB The arrows here are double-headed
Penalise M4 for formal charge on C or Br of the C–Br bond orincorrect partial charges on C–Br.
Penalise M4 if an additional arrow is drawn from the Br of the C–Brbond to, for example, K+.
Ignore other partial charges.
Penalise once only in any part of the mechanism for a line and twodots to show a bond.
Max 2 of any 3 marks in the mechanism for wrong reactant orwrong organic product (if shown) or a correct mechanism that leadsto the alkene 2-methylbut-2-ene.
Credit the correct use of “sticks” for the molecule except for theC–H being attacked.
M5 hydroxide ion behaves as a base / proton acceptor / electron pair donor / lonepair donor
Penalise M5 if ‘nucleophile’.5
[21]
(a) (i) (nucleophilic) addition-elimination
Not electrophilic addition-elimination
Ignore esterification1
17
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M3 for structure
• If wrong nucleophile used or O–H broken in first step, can only score M2.
• M2 not allowed independent of M1, but allow M1 for correct attack on C+
• + rather than δ+ on C=O loses M2.• If Cl lost with C=O breaking lose M2.
• M3 for correct structure with charges but lone pair on O is part of M4.
• Only allow M4 after correct / very close M3.
• Ignore HCl shown as a product.4
a 20-50 (ppm) or single value or range entirely within this range
If values not specified as a or b then assume first is a.1
b 50-90 (ppm) or single value or range entirely within this range1
Penalise wrong formula for Tollens or missing acid with potassiumdichromate but mark on.
1
J No reaction / no(visible) change /no silver mirror
No reaction / no(visible) change /stays blue / no redppt
No reaction / no(visible) change / staysorange / does not turngreen
Ignore ‘clear’, ‘nothing’.
Penalise wrong starting colour for dichromate.1
K Silver mirror /grey ppt
Red ppt
(allow brick red orred-orange)
(orange) turns green
1
J Two (peaks)
Allow trough, peak, spike.1
K Four (peaks)
Ignore details of splitting.
If values not specified as J or K then assume first is J.1
(c) If all the structures are unlabelled, assume that the first drawn ester is L, the secondester is M; the first drawn acid is N, the second P. The cyclic compound should beobvious.
mark on from incomplete formulae or incorrect oxidation state
M2 (turns) colourless
M3 (stays) purple / no (observed) change / no reaction
In all cases for M3
Ignore “nothing (happens)”
Ignore “no observation”3
(e) (i) Spectrum is for Isomer 1
or named or correctly identified
The explanation marks in (e)(ii) depend on correctly identifyingIsomer 1.
The identification should be unambiguous but candidates shouldnot be penalised for an imperfect or incomplete name. They maysay “the alcohol” or the “alkene” or the “E isomer”
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(ii) If Isomer 1 is correctly identified, award any two from
• (Strong / broad) absorption / peak in the range
3230 to 3550 cm–1 or specified value in this rangeor marked correctly on spectrumand(characteristic absorption / peak for) OH group /alcohol group
• No absorption / peak in range 1680 to 1750 cm–1 orabsence marked correctly on spectrumand(No absorption / peak for a) C=O group / carbonyl group / carbon-oxygendouble bond
• Absorption / peak in the range 1620 to 1680 cm–1
or specified value in this range or marked correctlyon spectrumand
(M2 balanced equation using 2NH3 and leading to NH4Br)
(penalise M1 for use of C4H9NH2 or for incorrect haloalkane, butallow consequent correct balancing of equation with 2 moles ofammonia)
2
(1–)butylamine
(credit 1–aminobutane and butyl–1–amine)
(award QoL mark for correct spelling)1
[13]
(a) 2-bromobutane;133
(b) Elimination;
(penalise “nucleophilic” OR “electrophilic” before the word“elimination”)
1
M1: curly arrow from lone pair on oxygen of hydroxide ion to H atomon correct C-H adjacent to C-Br;
(penalise M1 if KOH shown as covalent with an arrow breaking thebond)
1
M2: curly arrow from single bond of adjacent C-H to adjacentsingle bond C-C;
(only credit M2 if M1 is being attempted to correct H atom)1
M3: curly arrow from C-Br bond to side of Br atom;
(credit M3 independently unless arrows contradict)(Credit possible repeat error from 2(c)(iii) for M3)(If the wrong haloalkane is used OR but-1-ene is produced, awardMAX. 2 marks for the mechanism)(If E1 mechanism is used, give full credit in which M1 and M2 arefor correct curly arrows on the correct carbocation)
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(c) (i) (structural) isomers/hydrocarbons/compounds/they have the samemolecular formula, but different structural formulas/different structures; 1
(penalise statements which are not expressed in good English andwhich do not refer clearly to structural isomers i.e. plural)(penalise statements which refer to “different (spatial)arrangements”)(credit” different displayed formulas”)(Q of L mark)
(ii) Correct structure for but-1-ene;1
[7]
(a) M1: CH3CH2CH2CH2OH;1
M2: CH3CH(OH)CH2CH3;
(penalise incorrect alcohols in part (a), but mark consequentially inpart (b) and in part (c), if relevant)(if three alcohols drawn, award MAX. 1 mark)
1
34
(b) M1, M2 and M3: Correct structures for butanal, butanoneand butanoic acid;
(award these structure marks wherever the structures appear, butinsist that the C=O is shown in each structure and additionally, theC-O in the carboxylic acid
3
M4: balanced equation for the reaction of butan-1-olwith [O] to produce butanal and water;
1
M5: balanced equation for the reaction of butan-1-olwith [O] to produce butanoic acid and water
OR
balanced equation for the reaction of butanal with [O] toproduce butanoic acid;
1
M6: balanced equation for the reaction of butan-2-ol with [O] toproduce butanone and water;
(Credit condensed structures or molecular formulas in eachequation, provided it is obvious to which reaction the equationrefers) (Insist that whatever formula is used in each equation that itis a conventional representation of the compound; for examplepenalise CH3CH2CH2COH for butanal)
1
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(c) M1: Correct structure for 2-methylpropan-2-ol;M2: 2-methylpropan-2-ol
1
OR
methylpropan-2-ol;
(penalise on every occasion in parts (a) and (c), structures for thealcohols that are presented with the alcohol functional group asC-H-O)
