docshare02.docshare.tipsdocshare02.docshare.tips/files/23041/230415604.pdf1. (i) when calcium oxide...

1. (i) When calcium oxide and coke are heated in an electric furnace, the products are carbon monoxide and calcium dicarbide (CaC2). Write the equation for this reaction.

CaO + 3C CaC2 + CO (1)

(ii) Addition of water to calcium dicarbide leads to the formation of calcium hydroxide and ethyne (ethyne has the formula C2H2 and is the starting point for the manufacture of a variety of substances, including rayon and PVC). Write the equation for the production of ethyne.

CaC2 + 2H2O Ca(OH)2/CaO2H2 + C2H2 (1)

(iii) Until recently most ethyne was made in industry by the method in (ii). Today, ethyne is produced in industry by heating methane (CH4) alone at a high temperature for 0.01 second. Write an equation for this reaction.

2CH4 C2H2 + 3H2 (1)

(iv) Draw a diagram showing the electron arrangement in the dicarbide ion C22–.

Triple bond with correct total number of electrons (2)

(5)

2. (a) The enthalpy change for the reaction

CH2 = CH2 + H2 CH3CH3

is - 120 kJ mol-1, whereas that for the reduction of benzene, C6H6, to cyclohexane, C6H12,

is -208 kJ mol-1. What may be deduced from the fact that this value is not three times the first one?

delocalised bonding in benzene (1) benzene does not have 3 (localised) C=C bonds (1) benzene ring has high stability (1) less susceptible to addition (1)

(2)

(b) (i) State the conditions under which benzene may e nitrated to form mononitrobenzene. less than 60 C (1)

(ii) Both of the reagents that are used to nitrate benzene are usually regarded as acids. However, in this instance, one of them behaves as a base. Show how this is so. HNO3 + 2H2SO4 NO2 + 2HSO3- + H3O+ (2) OR HNO3 + H2SO4 H2NO3+ + HSO4- OR HNO3 + H2SO4 NO2+ + HSO4- + H2O (1) sulphuric acid donates H+ to HNO3 (1)

Khalid Mazhar Qureshi www.edexcel-cie.com

(iii) Give the mechanism for the nitration of benzene

NO2

HNO2 NO2

++H

(1) allow this mark also for H at end

(1) for plus charge incomplete ring delocalisationand

+

OR equivalent using Kekule forms - intermediate should have two double bonds only

(iv) Explain why benzene tends to undergo substitution rather than addition reactions. addition destroys delocalisation OR substitution does not (1)

(8) (Total 10 marks)

3. (i) State the conditions under which benzene may e nitrated to form mononitrobenzene. less than 60 °C (1)

(ii) Both of the reagents that are used to nitrate benzene are usually regarded as acids. However, in this instance, one of them behaves as a base. Show how this is so.

HNO3 + 2H2S04 N02 + 2HSO3- + H30+ (2) OR HNO3 + H2SO4 H2N03+ + HS04- OR HNO3 + H2SO4 N02+ + HSO4- + H20 (1) sulphuric acid donates H+ to HNO3 (1)

(iii) Give the mechanism for the nitration of benzene

NO2

HNO2 NO2

++H

(1) allow this mark also for H at end

(1) for plus charge incomplete ring delocalisationand

+

OR equivalent using Kekule forms - intermediate should have two double bonds only

(iv) Explain why benzene tends to undergo substitution rather than addition reactions. addition destroys delocalisation OR substitution does not (1)

4. Low density (poly(ethene) is used for packaging and plastic bags. The exothermic reaction by which poly(ethane) is made is shown by the following equation:

nC2H4 (C2H4)n

(a) Write the structural formulae, showing all the bonds, of


H H

H H

C C C C C C C C

H H H H H H

H H H H H H(1)(1)

n

(ii)(i)

(2)

(b) Typical conditions used in the process are a high pressure of 2000 atmospheres and a relatively low temperature of 200 C.

Suggest reasons for these conditions. Pressure: formation of poly(ethene) occurs with volume decrease (1)

favoured by pressure increase / suitable reason (1) If rate award (2)

Temperature: forward reaction, being exothermic (1) favoured by low temperature / suitable reason (1) Allow economic reason with justification Allow Le Chatelier

(4)

(c) The reaction proceeds via a free radial mechanism. The production of poly(ethane) may be initiated by the reaction.

R + CH = CH R–CH –CH2 22 2

Suggest an equation to show how

(i) a subsequent stage occurs; RCH2CH2

• + H2C=CH2 RCH2CH2CH2CH2• (1)

(ii) the polymerisation might terminate. R(CH2CH2)nCH2CH2

• + R(CH2CH2)nCH2CH2•

R(CH2CH2)nCH=CH2 + R(CH2CH2)nCH2CH3 (2) or R(CH2CH2)n+CH2CH2CH2CH2(CH2CH2)nR

For 2 plausible radicals reacting (2) For R• + R• R2 (1) for H• (0) (1) for single-headed curly arrows in (i) or (ii)

(4)

(d) High density poly(ethane), used for articles such as buckets and crates, is made under other conditions, using a catalyst.

(i) Suggest why this form of poly(ethene) has a higher density. polymer chains more closely packed (1) allow 'crystallites' not less branched or cross-linked

(2)

(ii) Other than density, suggest ONE physical property which would be different for high density poly(ethene). more rigid, higher softening point (1) allow tensile strength, Tm, hard, elasticity not boiling point (ignore comparative remarks)


(e) Draw a representative length of the molecule of poly(2-methylpropene), showing three

repeating units.

C C C C C C

H H H

H H H

CH

CH

CH

CH

CH

CH

3

3

3

3

3

3 (1) (1)

(f) (i) Write the structural formula, showing all covalent bonds, for the product obtained by reacting 2-methylpropene with bromine.

C

C

C

C

C

C

C

C

C

C

H

H

H

H

H

H

H

H

H

H

H

H

H

H

Br

OH

Br

OH

Br

OH

Br

OH

(1)

(1)

H–C–H

H–C–H

CH

CH

(ii)

accept

accept

3

3

3

3

H C

H C

H

H

consequential from (i)


5. Phenylethene (styrene),

CH CH2

is used in large quantities to make poly(phenylethene). A possible synthetic route from benzene to phenylethene is shown below.

CH CH2

+ CH Br AStep 1

Step 6

Step 2 Step 3

Step 4 Cr O /H

Step 5

Cl , u.v.23

CH Cl CH OH

CHO

2 2

B

Conc H SOheat

2 4 (1) CH MgBr

(2) H3

+

2 72– +

phenylethene


(a) (i) Give the name of a suitable catalyst for step 1 and give the structure of compound A.

aluminium chloride / aluminium bromide / iron(III) bromide (or any that is used) (1)

CH3

(1)

(ii) Why is ultraviolet light necessary in step 2? to split the chlorine or homolysis (1) generation of free radicals (1)

(iii) Suggest a suitable reagent and conditions for step 3. aqueous KOH / NaOH (1) heat (1) (correct reagent but no heat (1) heat but incorrect reagent (0))

(iv) The oxidation step 4 has to be done with care to obtain a\ good yield of aldehyde. Suggest how this is achieved. avoid excess oxidising agent (1) add oxidising agent to organic compound and not the other way round (1) (recognition of avoiding the second stage if neither first point scored (1))

(v) How would you prepare the Grignard reagent, methylmagnesium bromide, CH3MgBr used in step 5? Mg + CH3Br (1) in dry ether (1) OR with iodine catalyst (1) (max 2)

(vi) Give the structural formula of B. 3CHCH

OH all hydrogen atoms must be shown except ring (1)

(10)

(b) (i) What type of polymerisation does phenylethene undergo? addition (1) OR free radical (1)

(ii) Draw the structure of the polymer chain showing at least TWO monomer units.

C C C C

H C H C HH

H H H H n

6 65 5

(open ends but no [ ]n (1)) (2)

(4)


(c) Commercially, pheynylethene is made by the series of reactions:

AlCl3

900 K

Fe O2 3

CH CH2 3

2CH CH2

CH CH2

2+ + H

Suggest two reasons why this method id better than the route given earlier. better yield / shorter cheaper / useful by product H2 / no Grignard reagent not cheaper on its own


6. The following is adapted for a textbook of practical chemistry which gives practical details for the preparation of the azo dye phenylazo-2-naphthol.

Dissolve 2.5 g of phenylamine in a mixture of 8 cm3 of concentrated hydrochloric acid and 8 cm3 of water in a small beaker. Place in an ice bath; ignore any crystals that may appear. When the temperature is between 0 C and 5 C, add drop by drop a solution of 2 g of sodium nitrite dissolved in 10 cm3 of water, not allowing the temperature to rise above 5 C. Addition of the sodium nitrite solution should continue until, after a wait of 3-4 minutes, a drop of the reaction mixture given an immediate blue colouration with starch-iodine paper.

Prepare a solution of 3.9 g of 2-naphthol in 10% aqueous sodium hydroxide solution in a 250 cm3 beaker, and cool in ice bath to below 5 C; add 10-15 g of crushed ice to this solution. Stir the mixture, and add the diazonium salt prepared as above very slowly; red crystals of the azo compound will separate. When addition is complete, allow the mixture to stand in ice for 10 minutes and then filter the product, using gentle suction on a Buchner funnel. Wash with water; the product may be recrystallised from glacial ethanoic acid. it has a melting point of 131 C.

(a) Phenylamine is toxic by inhalation and skin absorption; concentrated hydrochloric acid is corrosive and gives harmful fumes. What specific precautions would you therefore take when doing this experiment?

fume cupboard (1) gloves (1) (safety requirements must be specific to question)

(2)

(b) Write the equation for the reaction between phenylamine and hydrochloric acid. C6H5NH2 + HCI C6H5NH3+CI– must have correct charges ignore CI– if H+ on the L.H.S.

(1)

(c) (i) Why should the reaction mixture be kept between 0 C and 5C? < 0 C too slow (1) > 5 C product decomposes (1)

(ii) Is the diazotisation reaction exothermic or endothermic? How do you know? exothermic (1) have to keep temperature below 5 C (1)

(iii) Why should you wait before testing with starch-iodide paper for the presence of excess nitrite ions? reaction slow so NO2

– needs time to be used (1)

(5)


(d) Give the equation, using structural formulae, for the diazo coupling reaction between

benzene diazonium chloride and 2-naphthol.

N Cl + N=N

OH

ignore HCl

+ –2

reactants (1) products (1)

+ HCl

allow –NN or –NN

but not ––N

=N or –N=N

CI– not essential

(2)

(e) The aqueous solution of the diazo compound prepared as in the first paragraph will, if allowed to warm up, turn from very pale yellow to a turbid orange-red, and will evolve nitrogen. Why is this?

diazo compound gives phenol (1) which reacts with diazo compound to give dye (1)


7. (a) (i) What are isotopes? atoms with same Z different A (1) not mass no. / atomic no.

(ii) The oxygen isotope 19O is formed by bombardment of 18O with another particle; the only other product is -radiation. Write the equation for this reaction.

O + n188

10

198O + (1)

(iii) 19O is a -emitter; give the equation for its decay. O + F + (1)19

8199

0–1

0–1

e OK, enot –

(iv) The table below gives the count-rate of a sample of 19O measured at various times. Show that the decay of 19O is first order and determine graphically or otherwise the half-life of this nuclide.

time/s

counts/s

0

2000

10

1550

20

1200

30

930

40

720

50

550

60

430

70

330

80

260–1

plot either raw data vs. t showing constant t½ and relating this to first order kinetics; or plot ln (counts) vs. t. showing straight line and relating this to first order.

In either case: points (1) labelled axes right way round (1) line (1)

{Value of t½ = 27 s (1)

{Statement recognising graph as arising from 1st order process (1)

One or other showing a mean determination of t½ for both marks


(v) The rate of decay of the radioactive species 40K is given by

rate of decay = N

where N is the number of atoms of 40K present at time t and is the decay constant. The human body contains about 4 moles of potassium, of which 0.012% is the radioactive isotope 40K. If the decay constant for this nuclide is 1.6 × 10–17 s–1, how many potassium atoms decay per second in the average human? (Avogadro constant = 6.0 × 1023 mol–1)

no. of 40K atoms N = 4 × 6.0 × 1023 × )1(1000

012.0

= 2.88 × 1020 (1)

rate = = 1.6 × 10–17 × 2.89 × 1020 (1)

= 4600 (s–1) (1)

(12)

(b) The rate equation describing the reaction of some halogenoalkanes with nucleophiles is also first order; eg:

(CH3)3CBr + OH– (CH3)3COH + BrWrite the rate equation for this reaction and show how it is related to the mechanism for the reaction.

rate = k [(CH3)3CBr] (1)

(CH ) C–Br (CH ) C + :Br

:OH do not allow if arrow is from negative charge– no need for lone pair

(1)

3 3 3 3–r.d.s.

(1)

+

fast (1)

3 3(CH ) COHfor process leadingto carbonium ion(1)

mech:

‘OH– fast’ can score if somehow the non-involvement of OH– in r.d.s. is made explicit

(5)

(c) An estimate of the relative rates of reaction of halogenoalkanes RX (where X is Cl, Br or I) with aqueous hydroxide ions can be obtained by reacting a sample of each separately with silver nitrate dissolved in aqueous ethanol.

(i) What would you see as the reaction proceeds? Give reasons for your answer. precipitate (1) of silver halide (1) ionic equation + states for pptn of AgX (1)

(ii) why do you think aqueous ethanol is used as the solvent for the silver nitrate in the reaction with halogenoalkanes? improves miscibility of reagents (1) or solvent for both reagents (or words to that effect)

(iii) Write the halogenoalkanes Rbr, RI and RCI in order of rate of reaction, fastest first. iodide > bromide > chloride (1)

(iv) Give your reasons for this order. C–I bond weakest (1)

(6)


(d) By considering the importance of the shapes of both reactants and intermediates in SN1

and SN2 reactions, suggest why the compound 1-bromotypticene (below) is virtually inert to nucleophilic attack.

Br

C

C

H

SN1 needs coplanar carbocation; not achievable (1) SN2 needs backside approach blocked by rings (1) or repulsion of OH– by electrons in benzene rings


8. An estimate of the relative rates of reaction of halogenoalkanes RX (where X is Cl, Br or I) with aqueous hydroxide ions can be obtained by reacting a sample of each separately with silver nitrate dissolved in aqueous ethanol.

(i) What would you see as the reaction proceeds? Give reasons for your answer. precipitate (1) of silver halide (1) ionic equation + states for pptn of AgX (1)

(ii) why do you think aqueous ethanol is used as the solvent for the silver nitrate in the reaction with halogenoalkanes?

improves miscibility of reagents (1) or solvent for both reagents (or words to that effect)

(iii) Write the halogenoalkanes Rbr, RI and RCI in order of rate of reaction, fastest first. iodide > bromide > chloride (1)

(iv) Give your reasons for this order. C–I bond weakest (1)

(Total 6 marks)

9. Propranolol is a chiral compound used in some 40 pharmaceutical preparations for the treatment of high blood pressure and cardiac pain. It is a base and is usually used as its hydrochloride salt, which is a white powder soluble to the extent of 50 g dm–3 in cold water and much more so in hot.

Propranolol is manufactured from blycidyl butanoate.

CH CH CH C O CH CH CH

O

O

3 2 22 2

which is a chiral ester. This is made from glycidol, which has a boiling point of 56 C, and butanoyl chloride, the latter being made from butan-1-ol via butanoic acid.


The esterification gives a racemic mixture of the ester but proporanolol requires only one of the

optical isomers in this mixture.

Butan-1-ol is made commercially from natural gas; it can also be made from an aldehyde and a Grignard reagent, but this is not economic. The alcohol is oxidised to butanoic acid, which is then converted to the acid chloride and then the ester.

(a) (i) What is meant by the term chiral? compound non-superimposable on mirror image (1) (not bonded to 4 different groups...)

(ii) Draw the two stereoisomers of glycidol, CH CHCH OH

O

2 2 .

H H

C C

CH2 CH2

CH2CH OH &

CH OH

2

2CH2

O O

O Oor

3D structure (1) mirror image (1)

(iii) How is chirality detected experimentally? rotate plane of plane-polarised light (1) not bend, nor reflect, nor refract

(5)

(b) Write the equations, stating briefly the necessary conditions, to show how you would bring about the following:

(i) conversion of butan-1-ol to butanoic acid;

CH CH CH CH OH CH CH CH COOHCr O / H

3 2 2 2 3 2 2

2 7 2– +

heat reagents (1) Heat (1) equations (1)

(ii) conversion of butanoic acid to butanoyl chloride;

CH CH CH COOH + SOCl CH CH CH COCl + SO + HCl3 2 2 2 3 2 2 2r.t.

PClPCl

H POPOCl + 2HCl

3 3 3

5 3

34 equation with appropriate stoichiometry (1) conditions (1)

(iii) reaction of butanoyl chloride with glycidol.

CH –CH–CH OH + ClOCCH CH CH

CH –CH–CH O CCH CH CH + HCl

2

2

2

2

2

2

2

2

3

3

O

O

O

formulae of reactants (1) products (1) (full structures not necessary) consequential on acid chloride

(6)


(c) (i) Suggest how butan-1-ol could be prepared using a Grignard reagent and

an aldehyde.

CH CH CH MgBr + HCHO (1)3 2 2

2 23CH CH CH

C

H H

OMgBr(1)

+(1) H

butan –1 –ol

(ii) State how the Grignard reagent you have suggested can be prepared from a halogenoalkane.

CH CH CH Br + Mg3 2 2 2 23CH CH CH MgBrdry

ether(1) (1)

or (1)I catalyst 2

(5)

(d) Suggest why only one of the optical isomers of propranolol is effective as a drug. enzymes chiral reagents so reagent and ester only fit if stereochemistry correct (1) or metabolism stereochemically sensitive

(1)

(e) Give experimental details of how you could purify propranolol hydrochloride by recrystallisation. How would you assess its purity?

dissolve in min vol (1) boiling water (1) filter hot (1) cool and filter (1) wash with small vol cold water (1) dry (in desiccator) (1) determine melting temperature and compare with data book value or mixed m.p. (1)

(7)

(f) How would you liberate the base propranolol from its hydrochloride salt? add sodium hydroxide solution (1)


10. Consider the following reaction scheme, then answer the question which follow.

CCH CO H HOCH CH CO H ClCH CH CO H CH CHCO H

H

O A B C D

step 1 step 2 step 3

2 2 2 2 2 2 2 2 22


(a) From your knowledge of the characteristic reactions of functional groups, how would you

show the presence of:

(i) The –CHO group in A; eg Fehling’s solutions / Benedict’s / ammoniacal silver nitrate (1) not 2,4-DNP unless an oxidation is also carried out warm (1) red / orange / brown precipitate (1) or correct result for reagent

(ii) The –CO2H group in B? Add Na2CO3 (aq) / NaHCO3 or a named carbonate or Mg (1) CO2 (H2 if Mg used) evolved / effervesces / fizzes (1) neutral iron(III) (1) red / brown colour (1) allow named indicator (1) and correct colour (1) but indicator and no colour (0)

or alcohol (1) + conc H2SO4 and warm (1) produce sweet smell (1) but (max 2)

(5)

(b) Give the reagents and conditions required to carry out:

(i) Step 1: eg NaBH4 / LiBH4 (1) if NaBH4 (aq) (2)

followed by acid / in solution (1) or any reducing agents that work eg lithium aluminium hydride (1) in dry ether (1) hydrogen (1) with platinum (1) sodium (1) in ethanol (1)

(ii) Step 2: ethanolic (1) KOH / NaOH (1) not OH– ions heat/followed by acidification (1)

(4)

(c) Name the type of reaction occurring in:

(i) Step 1: reduction / nucleophilic addition-elimination / hydrogenation / addition (1).

(ii) Step 3: elimination / dehydrohalogenation (1)

(2)

(d) Give the essential structural features necessary for geometrical isomerism to occur in sample organic molecules. Hence deduce whether or note this type of isomerism exists in compound D.

C=C / C = N double bond (1) this can be implied in a diagram correct orientation of groups required (1) or something about groups on a given carbon not being the same does not occur in D (1)



11. Suggest the conditions under which ammonia could be used to convert CH3CHCICO2H to

CH3CH(NH2)CO2H. concentrated (ammonia) / sealed tube / metal bomb / high pressure / alcoholic (ammonia) (1) not gaseous heat/warm / heat under reflux (1) reflux only (0)

(Total 2 marks)

12. (a) Give the reagents and conditions necessary to convert nitrobenzene to phenylamine eg tin / iron / zinc (1) + concentrated HCI / ethanoic acid (with iron only) (1) HCI only (½) heat / reflux (1) if LiAIH4 (1) and dry ether (1) (max 2)

(3)

(b) In order to prepare a diazonium salt, phenylamine is dissolved in excess hydrochloric acid. A solution of sodium nitrite is added in small quantities to this solution until in excess, maintaining the temperature between 0 and 10 C, preferably around 5C. The diazonium salt is formed in solution according to the equation:

C H NH + 2HCl + NaNO C H N NCl + NaCl + 2H O6 5 2 2 6 5+ –

2

(i) what structural feature of the phenylamine molecule allows it to dissolve readily in hydrochloric acid? Give the formulae of the species formed and hence explain, in energetic terms, why phenylamine is soluble in hydrochloric acid but not very soluble in water despite its ability to form hydrogen bonds. lone pair of electrons on N/NH2 group (allows phenylamine to act as a base (1) C6H5NH4+ (and CI–) (1) ionic water soluble (1) hydration energy of ions > lattice energy (1) large hydrophobic phenyl group makes phenylamine less soluble in water (1)

(ii) The function of the sodium nitrite is to react wit excess hydrochloric acid to form nitrous acid. Write an ionic equation for this reaction. Identify the acid/base conjugate pairs. H3O+ + NO2

– H2O + HNO2 (1) must be an ionic equation A1 B2 B1 A2 1 mark for each conjugate pair (2) if acid and bases are indicated but not paired then (1)

if H+ + NO2– HNO2 get equation mark only

(iii) Suggest why the sodium nitrate solution is added in small quantities. exothermic (1) temperature may rise about 5 C/10 C / other reactions may occur (1)

(10)


(c) If the solution of the diazonium salt is warmed to about 50 C, it decomposes as follows:

C H N NCl + H O C H OH + N + HCl6 5

+ –2 6 5 2

Describe in detail an experiment you would perform in order to measure the rate of this reaction. Include in your account how the measurements taken could be used to determine the rate at a particular time during the decomposition.

diagram (1) suitable container (1) suitable method of collecting gas (1) apparatus leak proof (1) (thistle funnel negates) thermometer in reacting mixture (1) start cold (1) warm up to 50 C (1) maintain constant temperature (1) start clock at correct point (1) (max 8) note gas volume at various time intervals (1) plot graph: Volume N2 vs time (1) shape of graph shown (1) find gradient at given time (1)

OR pH method (max 3) plus (4) marks for: measure pH plot [H+] shape gradient


13. Give the reagents and conditions necessary to convert nitrobenzene to phenylamine eg tin / iron / zinc (1) + concentrated HCI / ethanoic acid (with iron only) (1) HCI only (½) heat / reflux (1) if LiAIH4 (1) and dry ether (1) (max 2)

(Total 3 marks)

14. In order to prepare a diazonium salt, phenylamine is dissolved in excess hydrochloric acid. A solution of sodium nitrite is added in small quantities to this solution until in excess, maintaining the temperature between 0 and 10 C, preferably around 5C. The diazonium salt is formed in solution according to the equation:

C H NH + 2HCl + NaNO C H N NCl + NaCl + 2H O6 5 2 2 6 5+ –

2

(a) what structural feature of the phenylamine molecule allows it to dissolve readily in hydrochloric acid? Give the formulae of the species formed and hence explain, in energetic terms, why phenylamine is soluble in hydrochloric acid but not very soluble in water despite its ability to form hydrogen bonds.

lone pair of electrons on N/NH2 group (allows phenylamine to act as a base (1) C6H5NH4+ (and CI–) (1) ionic water soluble (1) hydration energy of ions > lattice energy (1) large hydrophobic phenyl group makes phenylamine less soluble in water (1)


(b) The function of the sodium nitrite is to react wit excess hydrochloric acid to form nitrous

acid. Write an ionic equation for this reaction. Identify the acid/base conjugate pairs. H3O+ + NO2

– H2O + HNO2 (1) must be an ionic equation A1 B2 B1 A2 1 mark for each conjugate pair (2) if acid and bases are indicated but not paired then (1)

if H+ + NO2– HNO2 get equation mark only

(c) Suggest why the sodium nitrate solution is added in small quantities. exothermic (1) temperature may rise about 5 C/10 C / other reactions may occur (1)

(Total 10 marks)

15. (a) Explain why light is essential for the reaction between methane and chlorine to form chloromethane at room temperature. Explain also why this method is not regarded as a good way of synthesising pure chloromethane.

light a form of energy / hv (1) required to break the CI-Cl bond / split the chlorine molecule (1)

into free radicals (OR equation CI2 2CI.) (1) (Cl2 2Cl. (2)) homolytic fission earns the 2nd and 3rd marks ie (2) accept if the rest of the answer demonstrates a lack of understanding

mechanism: CI + CH4 CH3 + HCl (1)

.CH3 + CI2 CH3CI + CI. (1)

or the equivalent in words for the above equation repetition of process gives dichloromethane etc. (or chain reaction goes further) / or termination step showing product other than CH3CI (1) hence product not pure (1)

(7)

(b) Explain why the first ionisation energy of phosphorus is greater than that of both silicon and sulphur.

(implied) definition of first ionisation energy (1) (P > Si) since nuclear charge P > Si (1) and shielding by inner shells same / outer electrons in same shell/orbital (1) (P > S) since S has paired electrons in 3p shell (1) greater repulsion between these makes them easier to remove / or given a half filled ‘p shell’ which conveys some stability (1)

(5)

(c) An acidic solution of potassium manganate (VII) will liberate chlorine from dilute sodium chloride solution but an acidic solution of potassium dichromate (VI) will not. Solid potassium dichromate (VI) will liberate chlorine gas from concentrated hydrochloric acid.

Explain these observations with reference to the data below.

E / V

MnO4– (aq) + 8H+(aq) + 5e– Mn2+ (aq) + 4H2O + 1.52

Cl2(g) + 2e- 2Cl– (aq) + 1.36

Cr2O72–(aq) + 14H+(aq) + 7e– 2Cr3+(aq) + 7H2O + 1.33


liberation of CI2 is oxidation (implied) (1) 2CI– – 2e– CI2 (1)

oxidant having a more positive E values will do this (1) hence manganate(VII) will and dichromate(VI) will not (1) but E values refer to molar solutions /standard conditions (1) concentrated HCI not 1 mol dm-3 (1) difference in concentration makes up for small difference in E values (1)

(7)

(d) By application of the concept of bond energy, predict a value for H for the reaction:

C2H5OH + CH3CO2H CH3CO2C2H5+H2O

Hence predict the effect of an increase in temperature on the equilibrium constant for this reaction.

same bonds broken as are formed (1) 1 C – O and 1 H-O bonds are broken and formed (1) total H = 0 (1) marked consequentially on bonds deducing Kc = constant from H = 0 (2) justification (1) eg temperature change has no effect on the equilibrium position or

Kc = [ ][ ]

[ ][ ]

ester water

acid alcohol implied

or le Chatelier argument if H = +ve or –ve, mark Kc and justification consequentially for full (3)


16. Explain why light is essential for the reaction between methane and chlorine to form chloromethane at room temperature. Explain also why this method is not regarded as a good way of synthesising pure chloromethane.

light a form of energy / hv (1) required to break the CI–Cl bond / split the chlorine molecule (1)

into free radicals (OR equation CI2 2CI.) (1) (Cl2 2Cl. (2)) homolytic fission earns the 2nd and 3rd marks ie (2) accept if the rest of the answer demonstrates a lack of understanding

mechanism: CI. + CH4 .CH3 + HCl (1)

.CH3 + CI2 CH3CI + CI. (1)

or the equivalent in words for the above equation repetition of process gives dichloromethane etc. (or chain reaction goes further) / or termination step showing product other than CH3CI (1) hence product not pure (1)

(7)


17. Consider the following series of reactions and answer the questions which follow.

NaOH (aq)

NaOH(aq)

ethanolic KOH

HCl(aq)

C H O6 12 2 C H O4 10 CH CO Na3 2– +

C H4 8

C H4 10

2H /Pt

C H Br4 9 CH CO H3 2

+A B C

DEF

G

(a) (i) What type of reaction is represented by the conversion of E to F? elimination (1)

ignore nucleophilic or electrophilic

(1)

(ii) The product F exists in two steroisomeric forms. Draw them, and state the feature of the molecule which makes this isomerism possible.

H H H

CH CH3 3 CH H

CH

3

3

C C C C allow stick for H’s

(1) (1) restricted/no/stops rotation about C=C (1) this mark is conditional on two correct cis/trans structures being given

(3)

(iii) Give a simple chemical test, stating what you would se, for the functional group present in F. bromine (water) (1) yellow / brown / orange colourless (not clear) (1) just decolourise without stating original colour (½) or any suitable equivalent: eg potassium manganate(VII) purple brown/ colourless/green is OK

(2)

(iv) F is more reactive than G. Suggest in terms of the bonding of these compounds why this is so. alkenes have accessible / high electron density / description of and bonding (1) or C=C has bond which is weaker than bond so alkenes react (2) if just ‘alkene has a bond’ is given then (1) only if ‘double bonds add on and single bonds undergo substitution’ is given then (1) only

(2)


(b) Compound E displays optical isomerism.

(i) State what this means. rotates (plane of polarisation of) plane-polarised (monochromatic) light (not bends/ twists) or non-superimposable mirror image (1) if ‘contains chiral carbon’ is given this must be explained or indicated on the diagram for (1)

(1)

(ii) Sketch the optical isomers of E.

CH CH CH CH

H H

C CBr Br

2 23 3

CH3 CH3 structure (1) correct mirror image (1) must be 3D drawing for (2) if it is drawn flat (1) only if a correct mirror image of an incorrect but chiral compound (1) – the chiral compound must be feasible to get this mark

(2)

(c) The reaction of E to give B is a nucleophilic substitution.

(i) What is meant by the term nucleophile? species with lone pair (½) to donate (½) ‘seeks positive centres’ (½) only not ‘seeks protons in nucleus’ or ‘nucleus loving’ not has a negative charge’

(1)

(ii) Give the structural formula for B.

H C C C C H

H H H H

H H OH H

(1)

consequential alcohol on (b)(ii) the structure must be sufficient to show butan-2-ol CH3CH2CH(OH)CH3 is OK

(1)

(iii) Give a simple chemical test for the functional group in B, and say what you would see. +PCI5 (1) (steamy) (acid) fumes / white gas (1) if evolution of HCI given (½) if PCI5(aq) then (0) any other suitable test and result acceptable e.g.

sodium (1) Bubbles (1) if ‘H2 formed’ is given (½)

or potassium dichromate (VI (1)green (1)

or potassium manganate(VII) (1)brown/colourless (1)

or add carboxylic acid and conc. sweet smell (1) if ‘ester formed’ given then (½) sulphuric acid) (1)

or ZnCI2/HCI (1)white ppt (1)

or I2/NaOH or KI/NaCIO (1)yellow ppt (1) if ‘forms CH3’ (½) consequential on answer given to (c)(iii)

(2) (d) The type of reaction exemplified by A B + C is important in the manufacture of soap.


(i) What type of reaction is this? hydrolysis / saponification (1) not substitution

(1)

(ii) A could be reacted with aqueous acid to give B and D. If the same quantity of A was treated with aqueous acid, instead of aqueous alkali, how would the yield of B differ? Explain your answer. yield falls; (½) reaction with acid is an equilibrium (½)

(1)

(e) Consider the reaction C D.

(i) Name D. acetic/ethanoic acid (1)

(1)

(ii) Identify the acid-base conjugate pair in this reaction. CH3COOH / CH3COO– (1) CH3COONa given (½) if C and D given (½) only

(1)

(iii) Explain why the reaction occurs. HCI/H3O+ is a stronger acid (1) than ethanoic acid (1) or CH3COO– strong base (1) whereas CI– weaker base (1)

(2)

(iv) A quantity of A was reacted with NaOH, acidified and the volatile acid D distilled out. D was then added to a solution containing 0.0250 mol (an excess) of NaOH. The excess NaOH from this reaction required 28.7 cm3 of 0.100 mol dm–3 HCl for complete reaction. What mass of a was reacted with NaOH in the first place?

NaOH left mol 28 7

100001 1 2 87 103.. ( ) .

NaOH used = 0.025 – 2.87 × 10-3 (1) = 0.0221 mol

0.0221 mol of A

mass of A = 0.0221 × 116 (1) = 2.56/7 g (1)


18. Polymers may be naturally occurring or synthetic. Name

(a) a synthetic polyalkene any suitable (1) must contain C and H only eg polythene

(1)


(b) a synthetic polyamide

nylon (1) not acrylic

(1)

(c) a natural polyalkene protein or a natural protein such as keratin / gelatine / silk / (1) not hair / wool / fingernails

(1)

(d) a synthetic polymer containing no hydrogen PTFE / Teflon (1) must be names

(1) (Total 4 marks)

19. Terylene is a polymer made from ethane 1,2-diol and benzene– 1,4-dicarboxylic acid. It is a condensation polymer.

(a) Draw the structural formulae of the two compounds which are used to make Terylene.

COOH

COOH

(1)

HOCH2CH2OH (1)

(2)

(b) Give a structural formula for the polymer.

( OCH CH OCO2 2

OC )n correct structure (1) minor fault (–½) but ester link must be present line through bracket and n (1)

(2) 2nd mark conditional on 1st

(2)

(c) Explain the meaning of the term condensation in this context. 2 molecules joined together by loss of water (1)

(1)

(d) Suggest why polyesters are not suitable for use under strongly alkaline conditions. hydrolyses (easily) / saponification (1)

(1) (Total 6 marks)


20. Carvone

C

C

C

O

CH

CH

CH3

CH2 CH2

CH2H C3

O

CH3

CH2H C3

which may alsobe shown as

is the main flavouring material in spearmint oil.

(a) Draw the structure of the product obtained when carvone reacts with

CH

(2)

3

3

N–NH 2

2

NO

2NO

any N–N link at carbonyl (1)

CH BrBr

Br

Br

CH

O

(2) (1)

(2)

for each double bond that has been reactedif use bromine water if get addition to give ignore relative positions of these

(i)

(ii)

Br OH

(4)

(b) Explain why carvone does not react with ammoniacal silver nitrate. ketone so not easily oxidised (and therefore no reaction)/

no reducing group present / no aldehyde group present (1)

(1)


(c) Carvone is chiral and shows optical activity.

(i) Explain the meaning of chiral. mirror-image and original non superimposable / two non-superimposable mirror images / molecule can be left handed or right handed / molecule with a carbon atom attached to 4 different groups (1)

(1)

(ii) Circle on the structure of carvone the features of the molecule which makes it chiral.

CH3

O

(1)

(1)

(iii) What is optical activity? ability to rotate (plane of polarisation) (1) (of plane–) polarised (monochromatic) light (1)

(2)

(d) Suggest a synthetic method, including reference to reagents and conditions, by which carvone could be converted to

Cl

CH3

CH2H C3

C

route (1) (ie state the C=O group is reduced or give the intermediate)

C=O to –CHOH: reagent (1) conditions (1)

–CHOH to –CH–Cl: reagent (1) conditions (1)

eg reduce C=O to –CHOH: LiAIH4 (1) (dry) ether (1) convert –CHOH to –CH–CI: PCI5 (1) heat/warm/room temperature (1)



21. Alkanes and alkenes differ strikingly in their reactivity, for example with chlorine; this is a

consequence of their structure.

(a) Using a charge-cloud representation of the – and conds show the structures of ethane and ethene.

ignore in C–H concentrate on C–C ignore absence of charge cloud here mark for correct representation of

(2)

(b) (i) Give the equation for the reaction of one mole of ethane with one mole of chlorine.

CH CH + Cl CH CH Cl + HCl allow C H / C H Cl etc(1)3 3 2 3 2 2 6 2 5 (1)

(ii) Give the mechanism for this reaction.

Cl 2Cl (1)CH CH + Cl CH CH + HCl (1)CH CH + Cl CH CH Cl + Cl (1)

2

3 3 3 2

3 2 2 3 2 (3)

(iii) Butane is a by-product by this reaction. How does it arise?

2CH CH butane (1)if start with methane then ( for the Cl 2Cl and(max 2) (1)

(1) for a consequentially correct 2nd propagation step)

3 2

2

if start with methane then (max 2) (1) for the CI2 2 CI• and (1) for a consequentially correct 2nd propagation step)

(1)

(c) (i) Give the mechanism for the reaction of chlorine with ethene.

C C

Cl–Cl :Cl–

C CC C

Cl Cl Cl

each arrow (½)(1) for intermediate (1) for arrow from chlorine – the lone pair does not have to be drawn in but the arrow must not come from the negative charge

(3)

(ii) Explain in terms of their structures why the reactions of ethane and of ethene with chlorine are so different. ethene has more accessible electron density (1) which polarises CI2 to form electrophile (1)

in ethane strong bonds need to be broken / bond less strong than (1) (or explain in terms of saturation of ethane)

(3)


(iii) Why are alkenes not used as fuels?

too valuable as synthetic intermediates / used to make polymers (1) not H values as these are about the same

(1)

(d) Alkynes such as ethyne

H—CC—H

are unsaturated compounds and undergo addition reactions in a similar fashion to alkenes. Under suitable conditions ethyne will react with water, and the reaction might be expected to be

H—CC—H + H2O CH2==CH—OH ethenol

(i) Using the bond enthalpies given below, suggest why the product of this reaction is not in fact ethenol, but its structural isomer ethanal, CH3CHO.

H/kJ mol–1: C—H412, C==C612, C—O 360, C==O 743, O—H 463

C—C 348, C837. sum of bond enthalpies of ethenol = 3 × 412 + 612 + 360 + 463 = 2671 kJ mol–1 (1) sum of bond enthalpies of ethanal = 4 × 412 + 348 + 743 = 2739 kJ mol–1 (1)

production of ethanal more exothermic (1) so thermodynamically favoured / product more stable (1) if answers incorrect but in correct order or calculation not attempted last 2 marks can still be scored

(4)

(ii) The catalyst used for the reaction is mercury(II) sulphate. The process was responsible for a serious pollution incident in Japan in the 1950’s, when mercury(II) sulphate effluent was converted by bacteria in river muds into methylmercury, (CH3)2Hg. This then entered the food chain via fish caught locally. HgSO4 is a typical inorganic compound, (CH3)2Hg is typically organic and is very much more toxic. Suggest a reason why? Me2Hg covalent and fat-soluble (. retained in body) (1) allow similar argument based on solubility of HgSO4

(1)

(e) Alkanes are obtained by the fractional distillation of crude oil; this system is complex, but it does not form any azeotropes. The boiling-point/composition diagram shown is for a pair of liquids A and B which doe form an azeotrope:

0 1

Temperature

Mole fraction of AX

(i) State and explain what would happen if a mixture of composition × was fractionally


distilled. distillate richer in B / residue richer in A (1) pure B distils off first (1) until azeotropic composition reached (1) when azeotrope distils unchanged (1)

(4)

(ii) Does this system show a positive or negative deviation from Raoult’s Law? State how this deviation arises. negative (1) interactions between molecules/intermolecular forces in mixture greater than in either of pure substances (1)


22. (a) Give the equation for the reaction of one mole of ethane with one mole of chlorine.

CH CH + Cl CH CH Cl + HCl allow C H / C H Cl etc(1)3 3 2 3 2 2 6 2 5 (1)

(b) Give the mechanism for this reaction.

Cl 2Cl (1)CH CH + Cl CH CH + HCl (1)CH CH + Cl CH CH Cl + Cl (1)

2

3 3 3 2

3 2 2 3 2 (3)

(c) Butane is a by-product by this reaction. How does it arise?

2CH CH butane (1)if start with methane then ( for the Cl 2Cl and(max 2) (1)

(1) for a consequentially correct 2nd propagation step)

3 2

2

if start with methane then (max 2) (1) for the CI2 2 CI• and (1) for a consequentially correct 2nd propagation step)

(1) (Total 5 marks)


23. Phenylamine, C6H5NH2, can be made in several ways from benzene. The flowchart shows two

of these in outline.

NO2

2

Step 1

Step 2

Step 3

NH

C H8 10

C H O7 6 2

C H OCl7 5

C H NO7 7

A

B

C

D

MnO /OH4– –

phenylamine

AlCl /C H Cl3 2 5

(a) (i) Give the reagents and conditions needed for step 1. concentrated nitric + (concentrated) sulphuric (1) 40 – 60 °C/heat/warm to max 60 °C (1)

(2)

(ii) Write the mechanism for this reaction.

HNO + 2H SO 2HSO + H O + NO

or HNO + H SO HSO + H O + NO (1)

(1)3 2 4 4 3 2

3 2 4 4 2 2– +

NO2

HNO2 NO2

++H

(4)

(iii) Give the reagents and conditions for step 2. tin / iron (1) + concentrated HCI (1) allow ethanoic acid for iron heat (1) not reflux not LiAIH4

(3)


(b) (i) In the alternative pathway identify the compounds A and D inclusive.

CH CH

COCl

COOH

CONH

2

2

3A

C

B

D

(1) each = (4) allow correct unambiguous names ie A – ethylbenzene / phenylethane B – benzenecarboxylic acid/benzoic acid C – benzenecarbonyl chloride/benzoyl chloride D – benenecarboxamide/benzamide

(4)

(ii) State the reagents and conditions needed for step 3. bromine (1) alkali / NaOH (1) heat (1)

(3)

(iii) Why is the first pathway commercially preferable for the manufacture of phenylamine? quicker formation of product/simplest path best because yields higher / more economic (1)

(1)

(c) Phenylamine reacts with nitrous acid in the presence of concentrated hydrochloric acid at 0–5 C to produce benzenediazonium chloride. If the temperature of the solution rises, this compound reacts with the solvent water to give phenol, C6H5OH, and nitrogen gas. The rate of the reaction can be followed by measuring the volume of nitrogen produced at various times, pressure and temperature remaining constant. The amount of benzenediazonium chloride remaining in the solution is proportional to V – Vt, where Vt is the volume of gas at time t and V is that at the end of the reaction.

(i) Give the equation for the reaction between benzenediazonium chloride and water.

N Cl2 OH

2 2+ H O + N + HCl

+ –

allow C6H5 or kekule if no charges (1)

(2)


(ii) Some results of this experiment are shown below.

250

200

150

100

50

00 500 1000

t/s

( –

)/

cmV

V

3

t

gets half-life from graph at least two readings (2) half-life constant (1) characteristic of first order reactions (1) if say not constant but then say not 1st order award the last mark

Deduce the order of reaction with respect to benzenediazonium chloride from the graph.

(4)

(iii) Is the reaction necessarily of this order overall? Explain your answer. the other reagent/water present (1) which may / may not contribute to the overall order (1)



24. (a) Phenylamine, C6H5NH2, can be made in several ways from benzene. The flowchart

shows two of these in outline.

NO2

2

Step 1

Step 2

Step 3

NH

C H8 10

C H O7 6 2

C H OCl7 5

C H NO7 7

A

B

C

D

MnO /OH4– –

phenylamine

AlCl /C H Cl3 2 5

(i) Give the reagents and conditions needed for step 1. concentrated nitric + (concentrated) sulphuric (1) 40 – 60 °C/heat/warm to max 60 °C (1)

(2)

(ii) Write the mechanism for this reaction.

HNO + 2H SO 2HSO + H O + NO

or HNO + H SO HSO + H O + NO (1)

(1)3 2 4 4 3 2

3 2 4 4 2 2– +

NO2

HNO2 NO2

++H

(4)

(iii) Give the reagents and conditions for step 2. tin / iron (1) + concentrated HCI (1) allow ethanoic acid for iron heat (1) not reflux not LiAIH4

(3)


(b) (i) In the alternative pathway identify the compounds A and D inclusive.

CH CH

COCl

COOH

CONH

2

2

3A

C

B

D

(1) each = (4) allow correct unambiguous names ie A – ethylbenzene / phenylethane B – benzenecarboxylic acid/benzoic acid C – benzenecarbonyl chloride/benzoyl chloride D – benenecarboxamide/benzamide

(4)

(ii) State the reagents and conditions needed for step 3. bromine (1) alkali / NaOH (1) heat (1)

(3)

(iii) Why is the first pathway commercially preferable for the manufacture of phenylamine? quicker formation of product/simplest path best because yields higher / more economic (1)


25. An organic compound X contains 82.75% carbon and 17.25% hydrogen by mass.

(a) (i) Calculate the empirical formula of X. 82.75

12

17.25

1

=6.9

6.9 =

17.25

6.9

= 1 = 2.5 (1) (ie for correct ratio)

empirical formula = C2H5 (1) if just C2H5 then (0)

(2)


(ii) Deduce from the mass spectrum below the relative molecular mass of X, giving a

reason for your choice. Hence show that the molecular formula of X is C4H10.

20

40

60

80

100

605040302010

58

43

29

15

Relativeintensity

m/e 58

highest m/e molecular ion / parent ion / molecule / parent peak (1)

which corresponds to C4 H10 ( = 48 + 10) = 58 (1)

(2)

(b) There are two possible structures A and B for this molecule:

C C C C

H H H H

H H H H

H H C C

C

C

H H H

H H H

H H

3

A B

(i) Identify the species responsible for the peaks in the mass spectrum at 43, 29 and 15. 43: +C3 H7 (1)

29: +C2 H5 (1)

15: +CH3 (1)

penalise (–1) once if any +’s are omitted

(3)

(ii) Hence deduce which of the structures A or B is present, giving a reason for your answer. A + reason pinpointing +C2H5 (2)

A + reason which is vague in terms of species (1)

(2)

(c) Complete combustion of X in oxygen gives carbon dioxide and water only.

(i) Write an equation for this combustion reaction. C4H10 + 6½O2 4CO2 + 5H2O (or doubled) (1)

(1)


(ii) Calculate the total volume of the gaseous mixture produced when 2 cm³ of gaseous

C4H10 is mixed with 15 cm³ of oxygen and completely burned. All volumes are measured at room temperature and pressure. 2 vols 13 vols 8 vols (1) which can be implied

hence 8 cm3 CO2 (1)

+ 2 cm3 O2 in excess 10 cm3 in total (1)

if calculation includes 10 cm3 gas for H2O to get 20 cm3 as answer then (2) only

marked consequentially on equation

allow correct method using 24000 or 22400 cm3 mol–1


26. (a) Give the formulae of the three structural isomers of C4H8 which are non-cyclic. CH3CH2CH=CH2 (1)

CH3CH=CHCH3 (1)

(CH3)2C=CH2 (1)

(3)

(b) One of these isomers show a type of stereoisomerism.

(i) Give the structure of stereoisomers and name them. clear structures of both isomers (½) each

names: cis and trans (½) but-2-ene (½)

(2)

(ii) Suggest how these stereoisomers might be distinguished. different melting / boiling points or difference in i.r. spectra

or difference in n.m.r. or difference in dipoles (1)

(1)

(c) One of the isomers of C4H8 in (a) reacts with HBr to give two different products, the major one of which is a chiral molecule.

(i) Identify this isomer of C4H8. but-1-ene (1) allow name or correct formula

(1)


(ii) Give the mechanism for the reaction of this isomer with HBr.

CH CH CH=CH CH CH CH–CH CH CH CHCH3 2 2 3 2 2 3 2 3

Br–H H Br(½)

(½) :Br –

(1)+

the arrow from the double bond can point to the H

the arrow from the Br– must not point to the positive charge or come from the –ve sign.

the arrows must not be straight or v-shaped

allow triangular structured intermediate

the intermediate must have a + charge not +

(3)

(iii) Why is the major product chiral but the minor one not so? the product has a carbon atom with four different groups attached/ has a non-superimposable minor image (1)

not just ‘assymetryric carbon’

allow inverse argument (ie the minor product does not have....)

(1)

(iv) Why is the chiral product the major one, and the non-chiral product the minor one? intermediate secondary carbocation more stable than primary (1)

allow inverse argument (ie in minor product intermediate....)

statement of Markownikof rule alone (0)

(1)

(v) The major product of this addition reaction is found to be optically inactive. Explain why this is so. both d– and I– forms / optical isomers are present (1)

in equal (molar) quantities / racemic mixture formed (1)


27. The characteristic reactions of benzene can be classified as electrophilic substitution reactions.

(a) Select a reaction of benzene which illustrates this type of reaction. Give the reagents and the equation for this reaction, the conditions under which it occurs and the name of the organic product.

I II III

reagent conc (½)

HNO3/H2SO4 (½)

Br2/R–Hal/RCOHal (½) + AlCI3 (½) allow iron bromide

conc/fuming (½)H2SO4 (½)

equation correct equation (1) correct equation (1) correct equation (1)

conditions warm/heat (½) to max of 60°C (½) or 40–60°C as specific temperature range (1)

anhydrous heat (1) heat (1)

name of organic product

nitrobenzene (1) any unambiguous name (1)

benzene sulphonic acid (1)

(4) (b) For the reaction selected in (a):


(i) identify the electrophile; electrophile for reaction in (a)

eg No2+ / RCO+ / R+ / Br+ (1) name or formula

(1)

(ii) give an equation to show its formation; equation for reaction in (a)

eg HNO3 + 2H2SO4 2HSO4– + H3O+ + NO2

+ (2)

if only HNO3 + H2SO4 HSO4– + H2O then H2O + NO2+ then (2)

but if written in 2 stages ie H2NO3+ then H2O + NO2+ then (2)

eg R–Hal + AICI3 R+ + AICI4– (2)

penalise (–1) for each error

(2)

(iii) give the mechanism for the substitution reaction. arrow to electrophile (1)

must be from ring or inside the ring

not to +ve sign or to oxygen

intermediate with incomplete ring and one positive charge

or correct position of one positive charge if Kekule (1)

loss of proton ie arrow on C–H bond (1)

(+ve charge can be outside bracket but if show TWO positive charges on

intermediate (max 1))

eg

NO2+

+ +NO

NO2

2H

H+

(3)

(c) Give two specific safety precautions you would need to take in carrying out the reaction in (a).

specific safety precautions 2 × (1) – any two of:

gloves

fume cupboard or ventilation

indirect heating

benzene carcinogenic and talk about risk assessment or replacement

(ignore other suggestions such as safety goggles or lab coats as these

are not specific to these systems

however if the suggestions are patently absurd in these cases (–1))

(2)


(d) Give one reaction of the functional group which was introduced into the benzene ring as a

result of the reaction performed in (a). give the reagents, conditions and equation for this reaction.

the answers in this section must relate to the functional group not to the benzene ring

Reagent (1) name or formula of compound(s) eg for reduction of –NO2 – Sn/HCI (not LiAIH4) conditions (1) applicable for the reagent stated eg heat (under reflux) not just reflux equation (1) will accept [H} or H2 but must be balanced


28. (a) When excess methane reacts with chlorine, the main product is chloromethane.

(i) Under what conditions does this reaction occur? sunlight / UC (1)

(1)

(ii) Give the mechanism for this reaction. (initiation) CI–CI CI• + CI• (1)

(propagation) CI• + HCH3 CI–H + •CH3 (1) CI–CI + •CH3 CI• + CI–CH3 (1)

(termination) eg CI• + CI• CI2 (1)

if correct use of half arrow these take place of •

• does not have to be on the C atom ie CH3• is acceptable.

(4)

(b) Heptane and octane are liquids at room temperature with boiling points of 98C and 126C respectively.

(i) What type of intermolecular forces of attraction are present in such liquids? van der Waals’/dispersion / induced dipole-induced dipole /

instantaneous dipole (1)

(1)


(ii) On the axes below, sketch a graph of how the boiling points of various mixtures of

these two liquids vary with composition. Show clearly the composition of the liquid and vapour phases.

0

Temperature

mole fraction of heptane 1

lines correct shape (½) and labels (½)

there must be 2 curves without a max or min in the diagram

the liquid line must not be straight

correct orientation according to axes (1)

(2)

(iii) By what process would a mixture containing a mole fraction of 0.5 heptane be separated into pure heptane and pure octane? fractional distillation (1)

(1)

(c) Liquid alkanes such as heptane and octane occur in petrol used as fuel to drive cards.

(i) Give two reasons why liquid fuels are generally preferred to gaseous ones. any TWO from:

liquids have smaller volume can carry more fuel OWTTE (1)

(ie some reference to volume and consequence)

gases would need high pressures and therefore more likely escape (1)

(ie some reference to pressure and reason)

liquids easier to transfer (1) (max 2)

(2)

(ii) The combustion characteristics of fuels for internal combustion engines in cars can be considerably improved by adding branched chain alkanes, cycloalkanes, aromatic hydrocarbons or tetraethyl lead(IV) (lead tetraethyl). Two of these are now considered to be hazardous to health. Select these two and identify the health hazard with which each is associated. aromatics (½) – carcinogenic (½)

lead compounds (½) – lead poisoning etc (½)

hazard is consequential on compound chosen



29. This question relates to the following reaction scheme

step 1 step 2 step 3

C2H5Br C2H5CN C2H5CH2NH2 C2H5CH2OH

(a) Give the reagents, the conditions required and the equation for step 1 and step 2.

(i) Step 1 KCN / NaCN (1) or name

not HCN

ethanolic / aqueous ethanol/alcoholic (½)

boil/heat (under reflux) (½)

equation (1) molecular or ionic

(3)

(ii) Step 2 Na in EtOh / LiAIH4 / hydrogen (1) not sodium borohydride

correct condition for reagent used ie room temp or warm (for Na in EtOH) / dry ether (for LiAIH4) / Pt or Ni catalyst (for hydrogen) (1)

equation (using [H] or H2 acceptable but if using hydrogen gas must show as H2) (1)

(3)

(b) (i) give a mechanism for step 1. mechanism – SN2

(1) :CN–

C – C – Br

(1)

C CC C

CN CN

Br Br

–

+.. -

the arrow from CN– must not come from the lone pair on the nitrogen

arrows must not be straight or v–shaped

do not need to show the lone pair

the intermediate must show an overall negative charge

if no Br– on right hand side then penalise (–½)

(3)

(ii) What type of mechanism is this? nucleophilic substitution (1)

(1)


(c) The conversion of C2H5Br to C2H+5+CH2OH by this method is ineffective, not least

because step 3 gives a very poor yield of only 7%. Outline an alternative synthetic route, stating clearly the reagents and conditions of the steps you suggest.

There are several alternatives for this part of the question. The two most likely are listed below. If a candidate suggests an alternative route which is feasible the schemes below should act as a guide as how the marks should be apportioned.

Mg (1) dry ether (1) (C2H5Br + Mg) C2H5MgBr (1) ie mark for product – name or formula methanal (1) name or formula dil HCI or water (1) hydrolyse(1)

OR Step 1 of question the same (1) dil HCI / Na OH (1) boil/heat (under reflux) (1) some reference to the need to remove water or purify the acid (1) LiAIH4 (1) dry ether (1)


30. The following are some common ions of chromium together with their characteristic colours.

[Cr(H2O)63+ Green CrO42– Yellow

[Cr(H2O)6]2+ Blue Cr2O72– Orange

A green solution of chromium(III) sulphate on treatment with dilute sodium hydroxide gave a green precipitate A, which dissolved in excess sodium hydroxide to give a dark green solution B. Treatment of solution B with hydrogen peroxide. H2O2, gave a yellow solution C which on acidification with dilute sulphuric acid, gave an orange solution D.

(a) (i) Write equations for the formation of A and B. FORMATION OF A:

[Cr(H2O)6]3+ + 3OH– Cr(OH)3(H2O)3 + 3H2O

species (1)

balancing (1) – conditional on correct species

if products given as Cr(OH)3 + 6H2O then (1) only

if Cr3+(aq) + 3OH– (aq) Cr(OH)3(s)2 (1) only

molecular equation equivalent to: Cr2(SO4)3 + 6NaOH 2Cr(OH)3 + 3Na2SO4 (1) only


FORMATION OF B:

[Cr(OH)4(H2O)2]– (1)

also allow [Cr(OH)5(H2O)]2– or [Cr(OH)6]3–

not CrO3 –(aq)

if give Cr(OH)3 + OH– [Cr(OH)4]– (1)

[Cr(OH5]2– (1)

[Cr(OH)6]3– (1)

allow molecular formula if molecular equation given for A

(3)

(ii) What type of reaction is shown in both equations? deprotonation / acid base (1)

(1)

(iii) What is the function of sodium hydroxide? base / proton acceptor (1)

(1)

(b) (i) What ion is responsible for the colour of solution C? CrO42– / chromate(VI) (1) not Cr6+ or Cr(VI)

(1)

(ii) What type of reaction is the conversion of B to C? redox / oxidation (1)

(1)

(iii) What is the function of the hydrogen peroxide in this reaction? oxidant / oxidising agent (1)

(1)

(c) (i) Write an equation for the conversion of C to D 2CrO42– + 2H+ Cr2O72– + H2O (2)

if correct species but not balanced then (1) only

allow molecular equation

if give 2CrO42– + H+ Cr2O72– + OH– then (1) only

(2)

(ii) Is this a redox reaction? Give a reason for your answer. no (1) conditional on correct reason

no change in oxidation number/state (of chromium) (1)

(2)

(d) Acidified potassium dichromate(VI) is frequently used as an oxidising agent in organic chemistry.

(i) Identify the organic product of the reaction between acidified potassium dichromate(VI) and butan-2-ol. butanone (1)

OR clear structural formula

CH CH CCH

O

3 32

(1) (ii) Give the reagents and the observable result of a single reaction which is common to

both butan-2-ol and the organic product in (d)(i)


iodine + NaOH / KI + NaCIO (1)

full names or formulae required

not iodine and alkali / iodide and hypochlorite

cream / pale yellow / yellow ppt (1)


31. (a) The following data concerns the redox chemistry of halogen elements and sodium thiosulphate.

E /V

Cl2 + 2e– 2Cl– +1.36

Br2 + 2e– 2Br– +1.07

2SO42– + 10H + 8E– S2O32– + 5H2O +0.57

I2 + 2e– 21– +0.54

S4O62– 2S2O32– +0.09

(i) What is the maximum change in oxidation number of sulphur which can be brought about by the action of sodium thiosulphate with iodine and chlorine:

Write full ionic equations for each reaction. reaction with iodine: +2 +2.5 or 0.5 (1)

I2 + 2S2O32– S4O62– + 2I– (1)

allow S4O62– + 2I– I2 + 2S2O32– but not S4O62– + 2I– I2 + 2S2O32–

reaction with chlorine: +2 +6 OR 4 (1)

4XCI2 + S2O32– + 5H2O 8CI– + 25042– + 10H

allow 8CI– + 2SO42– + 10H+ 4CI2 + S2O32– + 5H2O

but not 8CI– + 2SO42– + 10H+ 4CI2 + S2O32– + 5H20

species (1)

balanced (1) this mark is conditional on correct species

if the equations are given the wrong way round with an but the oxidation number changes are given correctly for these equations as –0.5 and –4 then the 2 oxidation state marks can be given (ie (2))

(5)

(ii) Suggest a reason why bromine cannot be estimated by direct titration with sodium thiosulphate in the same way that iodine can. the colour change is not sharp / not easy to see (1)

no suitable indicator (1)

OR

bromine is volatile / is lost from the system (1)

therefore results are not accurate / reliable (1)

(2)


(b) Bromine can be used to convert ethene to 1,2-dibromoethane which can then undergo

further reactions as shown below:

CH CH BrCH CH Br NCCH CH CN2 2 2 2 2 2

Step 1 Step 2

Step 3

Step 4

2 2 2 2HO CCH CH CO HC

2 2 2 2 2 2H NCH CH CH CH NHD

A B

(i) Give the reagents and conditions required for each of steps 2, 3 and 4, clearly identifying the step to which you are referring. STEP 2:

KCN/NaCN in ethanol / propanone / alcohol / aq ethanol (1)

not HCN

heat / boil (under reflux) (1)

not just ‘reflux’ / warm

STEP 3:

dil HCI / conc HCI (1) NaOH followed by acid is acceptable

not conc. H2SO4


if warm is given in both steps 2 and 3 or if reflux is given in both Steps 2 and

3 then penalise only once

STEP 4:

Na + EtOH / LiAIH4 / H2 (1)

not NaBH4 / Sn/HCI

room temperature (for Na + EtOH) / dry ether (for LiAIH4)

Pt or Ni or Pd (for H2) (1)

for all steps conditions are conditional on correct reagents

(6)


(ii) Write a mechanism for the reaction in step 2. Give the type of mechanism and

identify the rate of determining step. substitution of just one Br is enough

SN1:

arrow on C–Br (1) must start on bond and end on Br: C Br

correct carbonium ion (1)

production of carbonium ion marked as rate determining step (1)

arrow from CN– to carbonium (1) C :CN

+ –

must be from C and not from the charge or from N

lone pairs not essential

nucleophilic substitution (1)

‘SN1’ alone is not adequate

formation of Br– (1)

OR SN2:

arrow from CN– (1) (as above)

arrow on C–Br (1) (as above)

this can appear in the transition state

structure of transition state including charge (1)


nucleophillic substitution (1)


formation of transition state as r.d.s. or single step so this is r.d.s. (1)

(6)

(iii) C and D can react together to give a polymer. Give the structural formula of this polymer showing clearly the repeat unit.

CC C C N C C C C N

O OH H H H H H

H H H H H H H H peptide link (1)

correct total unit (1) must be multiple of the above unit

open ended (1)

if no H’s (max 2)

(3)


(iv) The polymer in (b)(iii) could be made more rapidly in the laboratory if C was first

converted into another more reactive compound. Give the formula of this compound and show how it could be formed from C formula of acyl chloride (1)

PCI5 / PCI3 / SOCI2 (1)

equation (for whichever of the 3 reagents given above) (1)

RCOOH + PCI5 etc is acceptable.


32. Bromine can be used to convert ethene to 1,2–dibromoethane which can then undergo further reactions as shown below:

CH CH BrCH CH Br NCCH CH CN2 2 2 2 2 2

Step 1 Step 2

Step 3

Step 4

2 2 2 2HO CCH CH CO HC

2 2 2 2 2 2H NCH CH CH CH NHD

A B

(i) Give the reagents and conditions required for each of steps 2, 3 and 4, clearly identifying the step to which you are referring.

STEP 2:

KCN/NaCN in ethanol / propanone / alcohol / aq ethanol (1)

not HCN


not just ‘reflux’ / warm

STEP 3:

dil HCI / conc HCI (1) NaOH followed by acid is acceptable

not conc. H2SO4


if warm is given in both steps 2 and 3 or if reflux is given in both Steps 2 and

3 then penalise only once

STEP 4:

Na + EtOH / LiAIH4 / H2 (1)

not NaBH4 / Sn/HCI

room temperature (for Na + EtOH) / dry ether (for LiAIH4)

Pt or Ni or Pd (for H2) (1)

for all steps conditions are conditional on correct reagents

(6)


(ii) Write a mechanism for the reaction in step 2. Give the type of mechanism and identify

the rate of determining step. substitution of just one Br is enough

SN1:

arrow on C–Br (1) must start on bond and end on Br: C Br

correct carbonium ion (1)

production of carbonium ion marked as rate determining step (1)

arrow from CN– to carbonium (1) C :CN

+ –

must be from C and not from the charge or from N

lone pairs not essential

nucleophilic substitution (1)



OR SN2:

arrow from CN– (1) (as above)

arrow on C–Br (1) (as above)

this can appear in the transition state

structure of transition state including charge (1)


nucleophillic substitution (1)


formation of transition state as r.d.s. or single step so this is r.d.s. (1)

(6)

(iii) C and D can react together to give a polymer. Give the structural formula of this polymer showing clearly the repeat unit.

CC C C N C C C C N

O OH H H H H H

H H H H H H H H peptide link (1)

correct total unit (1) must be multiple of the above unit

open ended (1)

if no H’s (max 2)

(3)


(iv) The polymer in (b)(iii) could be made more rapidly in the laboratory if C was first

converted into another more reactive compound. Give the formula of this compound and show how it could be formed from C

formula of acyl chloride (1)

PCI5 / PCI3 / SOCI2 (1)

equation (for whichever of the 3 reagents given above) (1)

RCOOH + PCI5 etc is acceptable.


33. (a) A buffer solution of pH =3.87 contains 7.40 g dm–3 of propanoic acid together with a quantity of sodium propanoate. Ka for propanoic acid = 1.35 x 10–5 mol dm–3 at 298 K.

(i) Explain what a buffer solution is and how this particular solution achieves its buffer function. resists changes in pH (1) not pH is constant when (contaminated by) small amounts of acid or base (1)

weak (ionisation of) acid (1) must be stated

equation for ionisation (1)

HA allowed but is required

(suppressed by) strong ionisation of salt / completely ionised (1)

hence giving solution with relatively high concentrations of anion and acid molecules (1)

equation for removal of H+ (1)

OR DESCRIBED IN WORDS

equation for removal of OH– (1)

allow correct interpretation of the Henderson equation for the last 2 marks

(8)

(ii) Calculate the concentration in g dm–3 of sodium propanoate, C2H5CO2Na, in the solution, stating any assumptions made. pH = 3.87 [H+] = 1.35 10–4 (1)

1.35 10–5 = 1.35 10–4 [

.SALT]01 (1)

[SALT] = 0.01 mol dm–3 (1)

[SALT] = 0.01 96 = 0.96 g dm–3 (1)

(if correct answer and some working (4))

assumptions:

all anion comes from salt (1)

ionisation of acid negligible (1)

(6)


(iii) If the sodium propanoate were to be replaced by anhydrous magnesium propanoate

calculate the concentration of magnesium propanoate in g dm3, required to give a buffer of the same pH. magnesium propanoate has two propanoate ions per mole (1)

half the molar concentration required = 0.005

0.005 170 = 0.85g dm–3 (1)

(2)

(b) Suggest a sequence of reactions by which propanoic acid, C2H5CO2H, could be converted into 2-animopropanoic acid, stating the important conditions

chlorine / bromine (1)

u.v. with sunlight / phosphorus (red) (1)

intermediate ie 2–chloropropanoic acid or the bromo if bromine used (1)

(conc excess) ammonia (1)

2 × equations (2) OR free radical mechanism (1) and reference to necessity of separation of mixtures at stage 1 (1)

(5)

(c) Pure 2-aminopropanoic acid has a relatively high melting point and is insoluble in hydrocarbon solvents.

(i) Sketch a structure for this compound which would account for these observations.

2

3

|

CHCOCH

NH

3 (1) must show charges

(1)

(ii) Write ionic equations to show the action of an aqueous solution of 2-aminopropanoic acid with:

A. an excess of acid;

B. an excess of alkali.

CH3CH( N

H3)CO2– + H+ CH3CH( N

H3)CO2H

CH3CH( N

H3)CO2– + OH+ CH3CH( N

H2)CO2– + H2

mark consequentially on the answer to (i) but can allow both equations to start with the molecule

(2)

(iii) Hence explain how 2-aminopropanoic acid might behave as a buffer in solution. thus product of amino acid and OH– or H+ will be mixture of salt + acid/base (1)

removes excess acid or alkali so that pH changes little / able

to react with H+ and OH– and control pH (1) (max 4)

(consequential on first statement)



34. Suggest a sequence of reactions by which propanoic acid, C2H5CO2H, could be converted into

2-animopropanoic acid, stating the important conditions chlorine / bromine (1)

u.v. with sunlight / phosphorus (red) (1)

intermediate ie 2–chloropropanoic acid or the bromo if bromine used (1)

(conc excess) ammonia (1)

2 × equations (2) OR free radical mechanism (1) and reference to necessity of separation of mixtures at stage 1 (1)

(Total 5 marks)

35. A red solid A (Pb3O4) on treatment with dilute nitric acid gave a brown solid B and a colourless solution. Solid B when treated with concentrated hydrochloric acid gave a greenish-yellow gas C, which bleached damp red litmus paper. The resulting solution, on cooling, gave a white precipitate D.

Identify B, C and D and write equations for the conversion of A to B and B to C. What can you deduce about the nature of A from its reaction with dilute nitric acid?

B = PbO2 (1) allow correct name

C = CI2 (1) allow correct name

D = PbCI2 (1) allow correct name

equation A to B:

species (1) balancing (1) conditional on correct species

equation B to C:

PbO2 + 4HCI PbCI2 + CI2 + 2H2O

OR with 6HCI PbCI42– + 2H+

species (1) balancing (1) – conditional on correct species

mixed oxide / compound oxide / two oxidation states for lead (1)

(8)


(b) A purple solid salt F dissolved in water to form a purple solution G. When excess

acidified iron(II) sulphate solution was added to G and the precipitate allowed to settle the purple solution was seen to have changed to a pale yellow solution II. F imparted an apple-green colour to a bunsen flame.

Identify the species responsible for the colours in F, G and H and explain, with the aid of equations, any reactions occurring. What is the formula of F?

F = MnO4 (1) G = MnO4– (1) allow names

H = [Fe(H2O)6]3+ (1) (must be hexaaquo)

apple green flame shows barium (1)

F is Ba(MnO4)2 (1) formula only

manganate(VII) oxidises iron(II) to iron (III) (1)

Ba2+ + SO2– + BaSo4

MnO4– + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O (1)

can get oxidation of iron(II) to iron(III) mark from equation

OR

Ba(MnO4)2 + 10FeSO4 + 8H2SO4 5Fe2(SO4)3 + 2MnSO4 + 8H2O + BaSO4 (2)

(8)

(c) Compound L (molecular formula C5H1002) on boiling with dilute sodium hydroxide gave two substances M and N which both contain carbon. After separation from the mixture, N was found to produce steamy acidic fumes when treated with PCI5. N also gave a pale yellow precipitate when treated with iodine and dilute sodium hydroxide. N contains an asymmetric carbon atom.

Identify L, M and N, giving your reasoning. PCI5 shows OH / –OH group / hydroxyl group (1)

not OH–

undergoes hydrolysis (1)

L is an ester (1)

I2 / NaOH shows CH3CH(OH)–R (1)

chiral R must be CH3CH2–/ can not be CH3CH2CH2– / can not be CH3– (1)

+ reason eg ‘can not be latter since M must contain carbon’ (1)

N is CH3CH(OH)C2H5 (1) allow correct name

M is HCO2Na (1) allow correct names

L is HCO2CH(CH3)C2H5 (1) allow correct names


36. A red solid A (Pb3O4) on treatment with dilute nitric acid gave a brown solid B and a colourless solution. Solid B when treated with concentrated hydrochloric acid gave a greenish-yellow gas C, which bleached damp red litmus paper. The resulting solution, on cooling, gave a white precipitate D.


Identify B, C and D and write equations for the conversion of A to B and B to C. What can you

deduce about the nature of A from its reaction with dilute nitric acid? B = PbO2 (1) allow correct name

C = CI2 (1) allow correct name

D = PbCI2 (1) allow correct name

equation A to B:

species (1) balancing (1) conditional on correct species

equation B to C:

PbO2 + 4HCI PbCI2 + CI2 + 2H2O

OR with 6HCI PbCI42– + 2H+

species (1) balancing (1) – conditional on correct species

mixed oxide / compound oxide / two oxidation states for lead (1)

(8)

37. A purple solid salt F dissolved in water to form a purple solution G. When excess acidified iron(II) sulphate solution was added to G and the precipitate allowed to settle the purple solution was seen to have changed to a pale yellow solution II. F imparted an apple-green colour to a bunsen flame.

Identify the species responsible for the colours in F, G and H and explain, with the aid of equations, any reactions occurring. What is the formula of F?

F = MnO4 (1) G = MnO4– (1) allow names

H = [Fe(H2O)6]3+ (1) (must be hexaaquo)

apple green flame shows barium (1)

F is Ba(MnO4)2 (1) formula only

manganate(VII) oxidises iron(II) to iron (III) (1)

Ba2+ + SO2– + BaSo4

MnO4– + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O (1)

can get oxidation of iron(II) to iron(III) mark from equation

OR

Ba(MnO4)2 + 10FeSO4 + 8H2SO4 5Fe2(SO4)3 + 2MnSO4 + 8H2O + BaSO4 (2)

(8)


38. Compound L (molecular formula C5H1002) on boiling with dilute sodium hydroxide gave two

substances M and N which both contain carbon. After separation from the mixture, N was found to produce steamy acidic fumes when treated with PCI5. N also gave a pale yellow precipitate when treated with iodine and dilute sodium hydroxide. N contains an asymmetric carbon atom.

Identify L, M and N, giving your reasoning. PCI5 shows OH / –OH group / hydroxyl group (1)

not OH–

undergoes hydrolysis (1)

L is an ester (1)

I2 / NaOH shows CH3CH(OH)–R (1)

chiral R must be CH3CH2–/ can not be CH3CH2CH2– / can not be CH3– (1)

+ reason eg ‘can not be latter since M must contain carbon’ (1)

N is CH3CH(OH)C2H5 (1) allow correct name

M is HCO2Na (1) allow correct names

L is HCO2CH(CH3)C2H5 (1) allow correct names

(9)

39.

H C C C C Br

H C C C C OH

H C C C C

H C C C C H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

Br

H

H

H

A

B

C

E

OH

O

(1) allow CH CH CH CH Br

(1)

(1)

(1)

3 2 2 2

Note if A is 2 bromobutane B and E also consequential to maximum of 2 marks if tert–but then B consequentially to maximum of 1 mark if A is 1–bromo–2–methylpropane B and C consequential to maximum of 2 marks if no A is given then B, C, E can score a maximum of 2 marks 4

(b) (solution) of bromine (1) decolourised / loses its colour (1) not goes clear not discoloured OR alkaline manganate (VII) (1) green (1) 2


(c) CH3CH2CH2COOH + NaHCO3 CH3CH2CH2COONa + N2O + CO2

formula of sodium salt (1) not C4H7O2 Na but C3H7CO2Na is permissible equation (1) equation mark may be awarded if use C4H7O2Na if correct acid structure for C is given in (a) R can be used for CH3CH2CH2 2

(d) CH CH CH COCH CH CH CH

O

3 22

2 2 2 3

(1)

allow C3H7 for CH3CH2CH2

allow C4H9 for CH3CH2CH2CH2

the use of RCOOR' is permissible here if C is given correctly in (a) 1

(e) potassium (di)chromate (½) + dilute named strong acid (½) OR potassium manganate (VII) (½) + dilute H2SO4 (½) not HCl allow correct formulae but not just ions or ‘acidified' heat under reflux (1) condition mark conditional on an oxidising reagent 2

(f) (i) [H+] = cKa or Ka = d][

]][[ –

Aci

AH

(1)

= 0.1 10 51.1 5– (1) = 1.23 10–3 mol dm–3 pH = 2.9 or 2.91 (1) 3

(ii) buffer (1) or an explanation of pH protection not pH constant 1

(iii) Number of moles of salt = 1105.5

(1)

[salt] = 110

5.5 0.5 = 0.1 (1)

either pH = pKa + lg ]acid[]salt[

or [H+] = ][

][

Salt

AcidaK (1)

pH = 4.8(2) (1) 4 [19]

40. (a) (i) use of mixed acid increases concentration of NO2+ / creates NO2+ (1) electrophile or some comment on the way H2SO4 does this (1)

OR full equation (2) 2

(ii) (reaction exothermic and) temperature must not rise (1) some indication that the temperature must be kept down not too exothermic 1

(iii) too much conversion to dinitrobenzene otherwise / to prevent further nitration (1) if this is not mentioned in (iii) but then given in (iv) the mark can be awarded 1


(iv) optimum rate / good yield (1) 1

(v) acid and benzene immiscible / there are 2 layers / maintain mixing of reactants (1) so rate slow if not mixed or ensure reactants are in contact (1) 2

(vi) removes/dilutes most of the acid remaining (1) if this is not mentioned in (vi) but then given in (vii) the mark can be awarded 1

(vii) removes residual acid dissolved in nitrobenzene (1) 1

(viii) separates nitrobenzene from other volatile constituents / this is the temperature over which pure nitrobenzene will distill (1) 1

(b) tin / tin(II) chloride / Fe (1)

(concentrated) HCl (1) ignore any reference to concentration of HCl this mark is condit–ional on tin / tin(II) chloride / Fe or Sn2+ being given heat (1) this mark is conditional on both reagents being correct not LiAlH4 which gives other products 3

(c) (i)

NH CONHCO

enough to make structure clear amide links (1) correct 1,3 position of ring (1) some idea of extension required (ie the end groups must not be terminated) brackets not necessary 2

(ii) polyamide or condensation (1) 1

(iii) acid chlorides (much) more reactive / more easily reacted (1) give faster polymerisation / better yield (1) 2

(iv) add SOCl2 / PCl5 / PCl3 (1)

COCl

COCl

(1)

2 [20]

41. (a) (i) >C=C< / carbon–carbon double bond / alkene (1) 1

(ii)

CH CH CHH

CHH H HC C C C

3 3 3

3

(1) (1)

if the structure of but–1–ene is given as well as one (or both) stereoisomer(s) of but–2–ene then (max 1) 2


(iii) no/restricted rotation about C=C bond (1)

some reference to the relevance of bonds or overlapping p–orbitals/ substit groups on each carbon atom not the same (1) 2

(iv) CH3CH2CH=CH2 (1) accept C2H5CH=CH2 1

(b) (i)

CH CH

CH

C O

23

3

(1) accept C2H5 for CH3CH2

1

(ii)

CH C O3 and CH CH3

OH

(1) (1) 2

(iii) (D:) CH3CH2COO–(Na+) (1) (E:) CH3CH2COOH (1) D and E labels not required 2

(c) (i)

3 32

OH

CH CH CHCH

1

(ii) CH3CH2+ / C2H5+ 29 (1) CH3CHOH+ 45 (1) not C2H4OH+ or C2H5O+ if charge(s) is (are) absent or incorrect then penalise (–1) since these are the only products it must be assumed that these are more stable than the molecular ion / if both 29 and 45 are present this indicates total fracture of the molecule (1) 3

(d) H–bonding responsible for variable frequency (1) H–bonding broken as A is mixed with benzene (1) a recognition that i.r. measures vibration/stretching of the OH group (1) (some recognition of a modification of OH vibrations due to H–bonding which is then disrupted can score (2)) max 2

[17]

42. (a) (i) Step 1: Reagents - potassium cyanide / sodium cyanide (1) not HCN but HCN is near enough to allow award of correct conditions Conditions - boil/warm/heat under reflux (½) in ethanol / aq ethanol / alcohol / propanone (½)

Step 2: Reagents - identified strong acid (1) allow conc. HCl Conditions - heat under reflux not warm (½) dil or aq (½) if use conc H2SO4 or HNO3 (0) NaOH allowed as long as it is followed by acidification

Step 3: Reagents - bromine / Br2 (1) not Br2(aq) KOH (1) Conditions - heat / warm (½) conc (KOH) (½) 7

(ii) C2H5COCl or name (1) 1


(b) (i)

2 2 224 6Cl C (CH ) C Cl

O O

(1) (1)NH (CH ) NH

or diacid / dimethyl ester / diethyl ester 2

(ii) condensation or polypeptide (1) 1

(c) (i)

C C

F F

F F n(2)

(fluorine atoms must be shown) 2

(ii) non-stick coatings or plumbers tape or coating on clothing/ upholstery / carpets or lab taps (1) 1

(d) comment about polymer in (b) hydrolysis /amide or peptide breaking (1) and about polymer in (c) strength of C-F bond (1) not just comment on relative biodegradability 2

[16]

43. (a) (i)

A

OH

NHCOCH 3

(1) -OH on 2nd, 3rd, or 4th C acceptable

B

OH

NH Cl3–+

(1) ignore charges

C CH3CO2H (1)

D

OH

N Cl2

(1)

E

OH

OH

N N


(1) for diazo coupling process to a phenolic ring, but it does not need

to be a 1,4 linkage if no OH’s on benzene ring penalise (-1) once only 5

(ii) Any of the following ten points can score to maximum of seven: linking fizz to -CO2H (1) linking m/e 76 to -C6H4 (1) linking reaction with HNO2 to presence of -NH2 (1) linking lack of reaction with HNO2 in A to acylation of -NH2 (1) linking weak acidity to -OH (1) linking of coloured ppt to diazo compound (1) C6H4(OH)NHCOCH3 + H2O + HCl C6H4(OH)NH3+Cl– + CH3CO2H (1) 2CH3CO2H + CO32– 2CH3CO2– + H2O + CO2 (1) C6H4(OH)NH3+Cl– + NaNO2 + HCl C6H4(OH)N2+Cl– + 2H2O + NaCl (1) or using HNO2 itself C6H4(OH)N2+ Cl– + 2H2O C6H4(OH)N2+Cl– + C6H5OH (OH)C6H4-N=N-

6H4(OH)+HCl (1) or C6H4(OH)N2+Cl– + C6H5O– (OH)C6H4-N=N-

6H4(OH)+Cl (1) max (7) equations must be balanced and can be ionic quality of language (1) 8

(b)

(2) (2) (1)

C2H5Br aqNaOH/KOH (1)

heat under reflux(1)

C2H5OH K Cr O

dil H SO2

2 2 7

4

distil immediately (1)

CH3CHO and dry the

product

C2H5Br + Mg 2

ether

dry or I catalyst(1) C2H5MgBr (2)

C2H5MgBr + CH3CHO

H

OMgBr

CC H CHH O (1)

2 32

+or HCl or H (aq)

(1)

C2H5CH(OH)CH3(+

Mg(OH)Br)

final production of pure product by distillation (1) max 12 [25]

44. (a) (i) Tin or iron (1) HCl (1) Heat (1) consequential on correct reactants 3

(ii)

NH3+

1

(iii) Liberates free phenylamine, or removes proton from – +NH3 (1) 1

(b) (i) Sodium nitrite and HCl (1) Temp in range 0 – 5C (1) conditional on correct reactants or HNO2 2

(ii) Not so low that reaction is too slow (1) Not so high that product is decomposed (1) 2


(iii) C6H5NH2 C6H5N2+Cl–

NH

+

+

+

N N

HNO +HCl H O22

2Cl–

2

(2) diazonium cation (1) remainder (1) Need not have chloride ions present but the equation must balance 2

(iv)

N N

HO

If the coupling and OH are on the same ring but in the wrong positions then (1) 2

(c) Structure explained, or drawn (1) Can show cis/trans isomerism (1) If phenylazo–2–naphthol is used allow (2) but use of 1,2– diphenylethene scores (0) 2

[15]

45. (a) C68.2/12 H13.6/1 O18.2/16 = C5.68 H13.6 O1.14 (1) (1) = C5H12O (1) If calculation is based on C and H only max (1) 3

(b) (i)

CH CH CH CH

CH

3

3

3

OH 1

(ii) e.g .CH3CH2CH2CH2CH2OH (1) 1

(iii)

CCH OH

CH

CH CH

3

2 3

3

1

(c) (i) CH3CH2CH2CH2CHO or CH3CH2CH2CH2COOH (1) or other structures consequential on (b) (i) 1

(ii) CH3CH(CH3)CH2COCH3 (1) (only) 1

(iii) unchanged (1) or no product 1

Answers in c(i),(ii),and (iii) are consequential on answers in question part b.

If pentan – 3 – ol is given in (b) (ii) allow CH3CH2CH=CHCH3 cis – and trans– for (2), but no credit for (d) (ii).


(d) (i) (CH3)2CH=CHCH3 (1)

(CH3)2CHCH=CH2 (1)

Answers in d are consequential on answer in b(ii) (1) 2

(ii) Identification of CH3CH+CH3 must have a sign but its position is not important (1) Relation to correct structure in d(i) (1) Consequential on d(i) 2

(e) (i) 2–bromo–2–methylbutane (1) or other halogen aqueous KOH(1) and heat, or some reference to temperature (1) ‘Heat’ is conditional on correct solvent Max 2

(ii) 2–methylbut–2–ene (1) or – 1– ene conc H2SO4 (1) followed by H2O (1) conc H2SO4 and H2O (1) or phosphoric acid(1) steam or water as gas (1) or HBr (1) then aqueous NaOH (1) Max 2

(iii) methylmagnesium iodide (1) and butanone (1) or ethylmagnesium iodide (1) propanone(1) If candidate produces a long list of reagents mark first on list 2

[19]

46. (a) (i)

CH CH OH

KMnO / H SO orK Cr O /H SO (1)

heat (1)CH COOH (1)

CH COCl

2

2 2 2 4

4 4

7

3

3

3 2

5PCl r.t (1) (1)

5

Other possible reagents for step 2 include PCl3 and heat or SOCl2 at room temperature; not P + Cl2

Candidate can answer question in words, formulae not needed.

(ii) C=O electron withdrawing or reference to electronegativity of oxygen(1) C–Cl more polar in acyl halide than in chloroethane or C is more ä+ (1) Correct argument based on steric hindrance can score full marks 2

CH3COOH + CH3CH2OH CH3COOCH2CH3 + H2O

(b) (i)

CH3COCl + CH3CH2OH CH3COOCH2CH3 + HCl

(1)

not H+ + Cl_ in second equation 2

(ii) not an equilibrium or not reversible (1) or HCl is a gas and its consequence 1

[10]

47. (a) (i) linear axes correctly labelled (1) points and line plotted correctly (1)


Deduction of first order kinetics (2) e.g. titre graph shows 2 t½ (1) leading to first order (1) lg graph straight line leading to first order (2) 4

(ii) experiment done with double conc (say) of halogenoalkane (1) observe change in rate ; if rate doubles then order with respect to halogenoalkane is first (1) or Experiment done with double OH– concentration and rate does not change (1) then first order with respect to halogenoalkane (1) Quality of language (1) 3

(iii)

CH C Cl CH

CH CH CH

CH CH CH

C + Cl CH C OHOH

3 3

3 33

3 3 3

3–

–

··

+

(1)

rds

(1) (1) (1)

identifies rds (1)

first arrow from C–Cl bond to Cl (1)

correct intermediate (1) charge may be outside []

arrow from O NOT negative charge (1)

If SN2 max (2) marks for first two points 4

(iv)

CH

CH

CH

C 3

3

2 (1)

(1)

(2) methylpropene or (2) methylprop–1–ene or isobutene (1) ethanolic solution or dissolved in ethanol or alcohol (1) 3

(b) (i) R O H

H····

××+

(2) (need to show that one hydrogen is atively bonded for full marks) (any mistake or omission –1 mark) water is lost (1) water much weaker nucleophile than Cl– or OH– (1) 4


NaHCO3 removes residual acid (1)

H2O removes residual salts or other water–soluble components or inorganic materials (1) anhydrous CaSO4 removes water (1)

moles of alcohol = 1–gmol74

g25 = 0.34 mol

74gmol–1

1 mol halogenoalkane from 1 mol alcohol (1) theoretical yield = 0.34 mol × 92.5gmol–1

= 31.3g (31.25g) (1)

( % yield = 3.31

28 × 100 = 89.5% (89.6%) (1)

competing reactions or handling losses (1) NOT experimental error

[nb 25/28 = 89.3% is wrong ] 7 [25]

48. (a) Correct arrows involving C=C and H–Br(1) Correct intermediate with positive charge(1) Correct arrow from Br– towards carbon atom (1)

C CCH CH CH CHBrCHH H

H HH H

H

C C3 3 3 3

:Br–H Br 3

Note: if product is 1-bromo product max 2 for full correct mechanism.

(b) (i) Reagent: magnesium(1) dry ethoxyethane / ether(1)

iodine catalyst / warm under reflux (or at room temperature)(1) 3

(ii) Reagent: aqueous or dilute sodium hydroxide(1) Conditions: boil / heat under reflux(1) 2

(c) Reagent: carbon dioxide (solid)(1) Reagent second stage: water or dilute named acid (1) Note: a route using methanal followed by oxidation can score (2) marks 2

(d)

CCH

CH

CHCH

H

O

C C HO3

3

33 (2)

2

Note: if one branched and one straight chain on correct ester linkage (1) [12]

49. (a) (i)

CH CHCl CH OH

CH CH(OH) COO Na

CH CHCl COOH

COOH

COOH

3

3

32

+–

A

C

B

D

4 × 1 = 4


B consequential on A C consequential on B

(ii) 2-chloropropan-1-ol (mark consequentially) 1

(b) C H COCl C H CONH C H CN C H CH NH5 5 5 522 2 2 2 2 2

NH P O Na/C H OH3 5 5

4

2 2

or LiAlH

(1) (1) (1) (1) (1)

or

C H COCl C H COOH C H CH OH C H CH Br

C H CH NH

5 5 5 5

5

2 2

2 2

3

2 2 2 2

2

H O LiAlH KBr/concH SO2 4 42

(1)

(1)

(1) (1)

(1)NH

5

(c) (i)

NH + NaNO + 2HCl N NCl + 2H O + NaCl22 2+ –

1 for diazo cation + (1) for rest of equation 2

(ii) if greater than 5C X decomposes (1)

if less than 0C rate (of formation of X) is too slow (1) 2

(iii) N N

OH (1)(1)

for N=N link for link to correct ring

next to –OH 2

[16]


50. (a) (i) (moist red) phosphorus + iodine / sodium iodide + phosphoric acid / PI3 /

phosphorus iodide (1) 1

(ii) (concentrated )sulphuric acid / aluminium oxide / (concentrated) phosphoric acid (1) 1

(b) (i) General scheme of marking for a valid sequence Correct intermediates score 2 marks Correct reagents score 3 marks Any correct condition scores 1 mark

C H OH CH CHO CH

CH

CH(OH)CN

CH(OH)COOH

eg H SO / K Cr O Step 22 2 24

52

7

3 3

3

Step 2Accept for reagentHCN/KCN and acid/KCN at pH 5 to 9

Specified acidheat under refluxorsodium hydroxideheat under refluxthen add acid

Correct sequence up to Propanoic acid can score maximum of 4 marks as follows: 2 marks for all intermediates 1 mark for all correct reagents 1 mark for two correct conditions

e.g.

C H OH C H Br C H CN C H COOH5 5 5 52 2

2 4

2 2

KBr /(conc)H SO KCN dil(named)acid

aq ethanol heat

Candidate could get to propanoic acid via a Grignard. This would also score 4 marks max.

BUT following score full marks if all correct

C H OH C H Br C H CN C H COOH5 5 5 52 2 2

2 4

2

2

33

KBr /(conc)H SO KCN dil(named)acid

aq ethanol heat

CH CH(OH)COOH CH CHBrCOOH1. dil NaOH2. dil named acid

Br and UV

If scheme does not work can score 1 mark for correct conversion of nitrile to acid as final step 6

(c) No, equal amounts (1) of both isomers /some reference to the fact that one of the intermediates can be attacked from either side in the reaction sequence. (1)

Reference to racemic mixture scores first mark. 2


(d) dry ether / ethoxyethane (1)

CH3–CH(OH)–CH2OH (1) 2 [12]

51. (a) (i) Marking Scheme Diagram flask (1) reflux condenser (1) apparatus safe and works (1)

Conditions ether (solvent) (1) dry reagents (could be implied by marking calcium chloride in drying tube) (1) plus 1 mark for either iodine as catalyst or drying tube (of calcium chloride)

CaCl

(1)

(1)

2

Safety precaution must relate to how the candidate deals with the flammablity of the ether eg. no naked flames (1) The diagram drawn must not contradict the precaution given. 7

(b) (i)

CH CH CH CH(OH)CH

or

CH CH CH CH

OH

CH

3

3 32 2

2 2 3

( )1

pentan–2–ol (1) can score mark for correct name of an incorrect secondary alcohol if drawn 2

(ii) CH3CH2CH2COOH (1)

butanoic acid (1) 2 [11]

52. (a) (i) Formation of electrophile eg C2H5Br + AlBr3 C2H5

+ +[ AlBr4]– or from below (1)


if candidate uses aluminium chloride ignore error

C H Br AlBr2 2

22

5 5

55

3

+

+

C H

C H C H

H

H

( )1

( )1

( )1

+ H +

Kekuleintermediate

4

(b)

correct shape and labelled ( )bpts correct ( )zigzag lines drawnstarting at 40% benzene ( )

11

1

136Vapour

80

0 100% ethylbenzene

60%

vapour from 60% liquid richer in benzene / more volatile/ lower boiling point (1) vapour condensed and then reboiled (1) pure benzene distilled first from top then ethyl benzene distilled off or left as residue (1) If candidate draws a diagram that shows an azeotrope max 4 6

(c) Reagents: potassium manganate(VII) / potassium permanganate (1) any named alkali e.g. potassium hydroxide /carbonate sodium hydroxide / carbonate (1) or formula in each case Conditions: heat (under reflux) (1) 3

[13]

53. (a) (i) potassium / sodium dichromate(VI) and (dilute) sulphuric acid/ hydrochloric acid (1)

Use of correct formula acceptable 1


(ii) Drop potassium dichromate(VI) into acidic solution of alcohol /

drop potassium dichromate alcohol mixture onto acid / controlled addition of potassium dichromate solution to mixture (1)

This part is consequential on the oxidising agent used in (i) distil aldehyde out of reaction mixture(1)

The second mark can de awarded as a stand alone mark provided an oxidising agent is given in the first part

Reference to reflux scores (0) marks since this would produce the acid 2

(iii) CH3MgI / CH3MgCl / CH3MgBr (1) magnesium plus iodomethane / bromomethane / chloromethane (1) in dry ether (1) 3

(b) (i) Variable oxidation state / variable valency /interstitial hydride / surface hydride / empty ‘d’ orbitals / partly filled ‘d’ orbitals (1) 1

(ii) The answer needs three essential points. Please read the whole answer first.

1. (Electrons) delocalised (1)

2. Comparison with alkene or triene e.g. 1.3.5 cyclohexatriene or makes bonds much less electron dense than in alkenes (1)

3. Comment on energy difference / resonance stabilisation energy / activation energy (1)

Unqualified references to ‘stable’ do not score individual points 3

(iii) Lithium aluminium hydride / sodium borohydride (1) in dry ether / in water or ethanol (1)

HO

HO

OCH

OCH

CH CHCHCH

CH CH CHCH

3

3

3

32 2

3OH

OH

OH

R CH CHCHCH

( )1

If candidates interpret questions as reduction of product of step 4 then answer will be as above

If candidates give Pt/Hydrogen at high temperature as conditions for reduction they will need to show in the diagram of the product that they have also reduced the benzene ring to score the reducing agent marks 3

(c)

OCH3

2 2

OH

C C C C

H CH OH H CH OH

R H H

or


Correct repeating unit (1) Extension (1) 2

[15]

54. (a) (i) Names or full formulae required Step 1: Reagent: either NaBH4 or LiAlH4 or Na or H2 (1) Conditions: aqueous dry ether ethanol nickel heat (1) ethanol methanol Step 2: Reagent: KBr(s) or PBr3 / P+Br2 or HBr (1) Conditions: conc/50% H2SO4 room temp hot/ conc (1) Step 3: Reagent: KCN (1) Conditions: ethanolic/alcoholic soln or aqueous ethanolic (1) Heat (under reflux) allow warm (1) Note:*Conditions must match reagent 7

(ii)

3 3

3

CH CO

O

OH

CN

CN

H

H

–

–

CN

CH C H

CH C H

+

(1)

(1)

(1) intermediate

(arrows must start from Cof CN and from bond)

3

H+ could be H – C = N / H – O – H

(iii) Nucleophilic addition 1

(iv) Slightly alkaline / slightly acidic / buffered at stated pH 5 to 9 /mixture of HCl and KCN (1) So that there are (significant concentrations of) both CN– and HCN (or H+) / correct reference to equil. position (1) 2


(b) CH CHO CH CHOOH CH COCl CH CONH CH CN3 3 3 3 32

1 2 3 4

(1) (1) (1)

Step 1: any named oxidising agent (1) Step 2: PCl5 / PCl3 / SOCl2 (1) Step 3: NH3 (1) Step 4: P4O10 / P2O5 (1)

Allow STEP 2/3 ammonium salt amide

via add NH3 and then heat with (NH4)2 CO3 or CH3COOH(l) 7 [20]

55. (a) (i) Benzocaine is

H N CO C H2 2 2 5

or H2N – C6H4 – CO2C2H5

(1) (1)

(NH2 group in any position on ring) (1 for each group);

Either: RNH2 + H+ RNH3

+ (1) RNH2 + CH3COCl RNHCOCH3 + HCl (1) RCO2C2H5 + NaOH RCO2Na + C2H5OH (1) RCOONa + HCl RCO2H + NaCl (1) Or: amines are bases and react with acids (1) Amines react with acid chlorides to form amides (1) Esters are hydrolysed by alkali to salt and alcohol (1) The carboxylic acid is formed on acidification. (1) 6

(ii) RCOOH + NaHCO3 RCOONa + H2O + CO2 (1)

NH CO H + 2H OCO H + HNO + H N 222 2 22++

(1)

2 22+HO C HO C N N OHN + OH

(1)

Coupling in any position on phenol ring 3

(iii) Substance X forms a zwitterion / dipolar molecule/ compound of formula +H3N– –COO– (1) Ionic substances tend to be soluble in water and have high melting temperatures (1) Quality of language (1) 3


(b) (i) [HA] = 21.37g / 137g mol–1 (1) = 0.156 mol dm–3

Ka = 1.20 × 10–5 mol dm–3 (1)

[H+] = [A–] = (Ka × [HA]) (1) = 1.37 × 10–3 mol dm–3

pH = 2.86 (1) 4

(ii) [A–] = 0.0500mol dm–3 (1) [acid] = (moles of original acid – moles NaOH)/0.100 (1) = 0.0280 mol dm–3

[H+] = Ka [HA]

[A–] = 1.20 × 10–5 × (0.0280/0.0500) (1) = 6.72 × 10–6 pH = –lg[6.72 × 10–6 ] (1) = 5.17 Alternative answer route via Henderson Equation acceptable 4

(c) (i) Strong acid/strong base: H+ + OH– H2O or H3O+ + OH– 2H2O (1)

HA + H2O H3O+ + A– H = + 8.3 kJ mol–1

H3O+ + OH– 2H2O H = – 57.2 kJ mol–1

HA + OH– A– + H2O H = + 8.3 – 57.2 = – 48.9 kJ mol–1 (2)

Answer alone scores 1 mark only 3

(ii) Ionisation is endothermic (1) Ka increases and pH falls (1) 2 [25]

56. CH3CHO C2H5OH C2H5Br C2H5MgBr 1 1 2 1 3 1

step 1: correct reducing agent + conditions (e.g. LiAlH4/dry ether, NaBH4/aq, H2/ Ni at 100°C (2)

step 2: add KBr(s) (1) and conc or 50% H2SO4 (1) (or PBr3 (1) at R.T. (1) / Conc HBr (1) warm (1) / P + Br2 (1) R T (1))

step 3: add magnesium (1) in dry ether (1)

C2H5MgBr + CH3CHO CH3CH2CH(OH)CH3 CH3COCH2CH3

1 1

step 4: (to Grignard add ethanal), then hydrolyse with dil acid (1)

step 5: oxidise e.g. heat under reflux (1) with dilute sulphuric acid and potassium dichromate(VI) (1) Alternative correct routes can score marks up to 5 for intermediates and up to 8 for steps

[max 13]

57. (a) (i) C2H5C(CH3)2OH (1) 1

C C CH CH

H

H H OH

H CH3

3

(ii) CH3COCH3 (1) and CH3CH2MgX (X= Cl, Br, I) (1) or CH3MgX (1) and CH3CH2COCH3 (1) Dry ether (1) 3


(b) (i) C2H5CH(C2H5)OH (1)

H

H H H

H

OH H

H H H

C C C CCH H

1

(ii) (C2H5)2CO (1)

H

H H H HO

H H H

H HC C C C C

1

This is consequential on answer to b(i) Primary will give aldehyde or carboxylic acid

(iii) ketone (1) consequential on (ii) 1

(iv) 2,4 –dinitrophenylhydrazine / Brady’s reagent (1) result (1)

consequential on (iii) 2

(v) no CH3CO / CH3CH(OH) / methyl ketone / methyl secondary alcohol present 1

[10]

58. (a) (i) reducing agent or hydrogenating agent (1) 1

(ii) provide activation energy (1) reaction exothermic / prevent reaction getting out of control (1) 2

(iii) condenses vapours and returns reagents to flask / prevents loss of solvent / prevents loss of reactants or product / enables prolonged heating because reaction is slow (1) 1

(iv) precipitates tin oxides / forms ammonium salt of acid which is soluble / liberates free amine (1) 1

(v) increases yield (1) 1

(vi) product is salt (1) adding ethanoic acid gives free acid (1) 2

(b) Theoretical yield = 15/167 = 0.0898 (1) Actual yield = 12/137 = 0.0876 (1)

percentage yield = 0898.0

1000876.0 = 97.6% (1) 3

(c) oxidation of methyl group reagents (1) conditions (1) e.g. Oxidation of side chain with potassium manganate(VII) and sulphuric acid / sodium carbonate / sodium hydroxide (1) heat under reflux / heat (1)

introduction of nitro group e.g. React product with mixture of conc nitric and sulphuric acid (1) Warm / if specified, temperature must be between 50 and 60oC (1) 4


(d) (i) ethanol (1)

conc sulphuric acid / H2SO4 (1) 2

(ii) esterification / condensation / addition –elimination (1) 1

(e) ionic (1) ionic compounds tend to be soluble in aqueous media/ water soluble or ions hydrated by water (1) 2

[20]

59. (a) (i) conc sulphuric acid / H2SO4 / H3PO4 / P2O5 / P4O10 (1)

and heat / temperature over 100oC if specified (1) or vapour over alumina (1) heat (1) 2

(ii) HBr (1)

Any suggestion of aq loses mark

C C

C

C C

C

(1)

(1)

(1)

H H H

H

H

H

H

H

H CH

CH

H

H

CH

H Br

Br

Br

33

3

(1) for carbocation (1) for attack by bromide ion 4

(iii) Secondary carbocation more stable than primary (1) or activation energy argument (1) 1

(b) (i) CH3CH2CH2CN or CH3CH2CH2MgBr (1)

(Accept C3H7– ) 1

(ii) Step 1 HBr + ZnBr2 / sulphuric acid (not dilute) + KBr / KBr + phosphoric acid (1) heat (1) or P + Br2 or PBr3 (1) room temperature (1) (2)

Step 2 KCN (1) alcohol or aqueous alcohol (1) heat or warm (1) or if step 3 is addition to carbon dioxide Mg (1) dry ether (1) trace of iodine / hydrolysis with HCl (in correct order) (1) (3)

5


(iii) Step 1 Substitution (1)

But not electrophilic Step 3 Hydrolysis if cyanide (1) or addition if carbon dioxide to Grignard (1) 2

[15]

60. (a) CH3CO+ (1)

Candidate may not identify electophile but may score this mark if they use the correct electrophile in the mechanism CH3COCl + AlCl3 CH3CO+ + AlCl4

– (1) 2

H

COCH3

COCH3

3

4

+ HCl

(1)(1)

(1)+OCCH

-AlCl

In the mechanism the electrophile can be shown as + – CH3CO – Cl– AlCl3

(b) Notes: The arrow for the first mark should start inside the ring and go to the carbon of the CO group. The arrow for the last mark should start on the bond and finish inside the ring.

H

COCH3

COCH3

3

4

+ HCl

(1)(1)

(1)+OCCH

-AlCl

3


(c) There are several routes through this; all can score full marks

3

3

2

3

((

11

))

C C

C

OCH

CH

CH

OH

OH

+ HCN

CN

CO H(1)

Slightly acidic / slightly alkaline /buffered at name pH between 5and 9 / mixture of KCN and acid. /HCN plus KCN

Dilute HCl Heat 9underreflux)

Dilute NaOH followedby acid

(1)(1)

or(1)

(1)

5 [10]

61. (a) (i) Bond breaking (1) with each atom taking one electron (1) 2

(ii) Reaction 3 or reaction 1 (1) 1

(iii) A species / a known species is produced at the same rate that it is removed / rate of breakdown of ozone equals rate of formation / no overall change of concentration with time (1)

Choice of equations to show this 1,2 and 4 or 1, 2 and 3 or 2 and 3 (1)

Kinetic argument about rate of formation equals rate of depletion in stated equations (1) 3

(b) O + O2 O3 H = –100 Reaction 2 O3 + O 2O2 H = –390 Reaction 4

2O + O2 2O2 H = –490 kJ mol–1 (2)

or 2O O2 H = –490 kJ mol–1

2O O2 exact reverse of reaction 1 (1) This is exothermic and the light taken in during the initial reaction will have been changed to heat (and the temperature will rise) (1) 4

(c) (i) ClO + O O2 + Cl (1) 1

(ii) Cl is regenerated (1) this is a chain reaction so it can react again with more ozone (1) 2


(iii) Rate = k[Cl][O3] (1)

since Cl atoms /radical (not molecules) have been constantly increasing due to CFC’s (1)

since Cl is regenerated in the cycle and is not used up concentration of Cl rises (1) rate has been (constantly) increasing (1) Quality of Language (1) 5

(iv) upper atmosphere is very cold (1) low pressure / concentration (1) 2

(d) (i) (3 methoxyphenol +) benzoyl chloride (1) + AlCl3 (1) dry (1) 3

(ii) More CFCs lead to less ozone (1) less ozone leads to more UV and rise in sunburn (1) 2

[25]

62. (a) The idea of a unique spectra for each molecule / ‘fingerprint’ / implication that spectra of two compounds will be different (1) Citral will show carbonyl peak (1) Geraniol will show hydroxyl peak (1) 3

(b) (i) Potassium dichromate (1) sulphuric acid (1) Some recognition of the need to stop the reaction at the aldehyde such as distil out the product or do not use excess oxidising agent (1) 3

(ii) Test for carbonyl - reagents (1) Result (1) 2

(c) (i) Ester (1) 1

(ii) 3-methylbutan-1-ol (1) 1

(iii) Rate depends on the concentration of the reactant to the power 1 (1) To show [ester]a

1. Mix known amount of ester with OH– (1) (*)

2. Remove sample(s) at known time(s) (1) (*)

3. Quench (1) (*)

4. Titrate with acid (1) (*)

(*) Follow change of pH or conductivity (1) with time (1)

Then either 5. Use data to find initial rate (1) and then double conc of ester and repeat (1) or 5. plot graph of conc against time (1) and show half-life constant (1)

6. Double [OH–] and repeat 8


(iv) (CH3)2CHCH2CH2OH + PCl5 (CH3)2CHCH2CH2Cl + POCl3 + HCl (1)

halogenoalkane / replacement of OH by Cl (1)

This is consequential on the answer to part (ii) 2

(v) Acid has hydrogen bonding (1)

Ester has van de Waal forces (1)

Strength of (intermolecular) forces compared (1)

Relationship to boiling temperature (1)

Quality of language (1) 5 [25]

63. (a) Substance A is CH3CH2COCH3 (1) Substance B and C are CH3CH2CH2CHO (1) and CH3CH(CH3)CHO (1) 3

(b) (i) C2H5Br + Mg C2H5MgBr (1)

(ii) dry ether solvent (2)

(iii) CH3CH2C(OH)(CH3) C2H5 (1) 4

(c) step 1 PCl5 /PCl3 SOCl2 step 2 NH3 2

(d) (i) two isomers drawn with an attempt at a 3-d diagram (1) each = 2

C H C H

C C

H HCH CHOH HO

2 25 5

3 3

They have an asymmetric carbon atom (4 different groups on a carbon) (1) mirror image non-superimposable (1) (4)

(ii) rotate the plane of plane polarised light in opposite directions (1) 5

(e) (i) butanone CH3CH2COCH3 (2)

(ii) green (1) 3 [17]

64. (a) Mr 2-bromobutane = 137 (1) moles = 13.7/137 = 0.10 (1) allow 0.1 moles KOH = 9.0156 = 0.16 (0.1607 or 0.161) (1) KOH present in excess consequential (1) 4

(b) Mr butan-2-ol = 74 (1) max moles butan-2-ol obtainable = 0.10 (1) consequential on (a) max mass = 0. 10 × 74 = 7.4 g (1) answer with units If candidate has calculated that the 2-bromobutane is in excess the calculation is based on 9 g of KOH, this gives 0. 16 mol of KOH and 11.9 g of product 3


(c) lone pair donor / electron pair donor / lone or electron pair can

form co-ordinate / dative bond (1)

hydroxide ion / OH (1) 2

(d) rate increased (1) C-I bond weaker (than C-Br bond) / lower bond energy (1) 2

[11]

65. (a) diagram 1 (heating under) reflux (1) diagram 2 distillation (1) 2

(b) (i) reaction is slow / time needed for reaction to reach completion (1) 1

(ii) condenses vapours and returns liquid to flask / vapour turns to 1iquid and returns to flask (1) (it allows reaction at boiling point of reactants) without loss / escape of material/reactants prevents loss/escape of materials/reactants/products (1) 2

(c) heat the mixture (slowly) (1) collect only fraction/distillate (1) produced at 102 C / around 102 C / between 100-104 C / at the boiling temperature of the 1-bromobutane (1) Need to make clear that only distillate at this temperature is collected for second mark 3

(d) (i) 2.7

1.3 (1) × 100 = 43.1 % (1) Allow 2-4 significant figures 2

(ii) two reasons from: side reactions (1) reaction incomplete (1) Max product lost in purification / transfers (1) 2

[12]

66. (a) A – F – G – C – D – E – B 2 2 1 mark if / letter out of sequence but rest correct. 0 mark if 2 or more letters out of sequence. nb if adjacent pair inverted this is one error.

(b) Little – to produce a saturated solution / to prevent loss of solid / because all solid will not crystallise / to prevent loss of yield. (1) Small – (if large volume used) solid would be lost / dissolved (1) 2

[4]

67. (a) Spectra shows three types / environment of hydrogen (1) Each peak related to type of hydrogen in ethanol (1) References to mass spec score zero 2

(b) (i) Conc. sulphuric acid would produce (some) bromine (1) 50% acid produces HBr only (1) 2

(ii) Ammonia (1) heat in sealed tube / under pressure (1) or Concentrated ammonia (1) at room temperature (1) 2


(iii) Acid protonates -NH2 (to -NH3

+) / -NH2 reacts with H+ ( not HCl) (1) 2 ionic more soluble (1)

(c) If 3 steps that works Correct intermediates score 2 marks Any 3 correct reagents score 3 × 1 marks Any one correct condition scores 1 mark

CH CH OH CH COOH

CH COONH

CH COCl

CH CONH

3 3

3 4

3

3 2

2

(1)

(1) (1)

(2)

(4) (3)

(5)

-

Step Reagent Conditions 1 Potassium dichromate +

named mineral acid or potassium manganate(VII) with sulphuric acid

Heat under reflux

2 PCl3 / PCl5 / SOCl2 Dry / room temperature

3 Ammonia Room temperature 4 Ammonia Room temperature 5 Heat

6

If scheme breaks down mark from start to breakdown and from end back to breakdown and score best of these two marks names or correct formulae allowed. Name scores even if formulae then given and wrong

(d) (i)

2

2

2

2

or

O

O

O

O

O

O

C

C

N

N

C

C

(CH )n

(CH )n

(CH )m

(CH )m

C

C

N

N

H

H

H

H(1) 1

(CH2)n or m represents a sensible unit which may be a benzene ring

(ii) H2N(CH2)n NH2 and ClOC–(CH2)m–COCl

(1) (1) 2

Consequential on (i) Allow caprolactan (2 marks) or cyclic compound H2N–(CH2)n– COCl scores 2 marks


(iii) Toxic if burnt/ non-biodegradable (1) 1

[18]

68. (a) Apparatus to show round or pear shaped flask (1) not conical Reflux condenser - must have inner tube and inlet and outlet for water (1) Controlled source of heating e.g. electric heater/ hot plate (1) Reasonable drawing (1) of an apparatus that would work. 4

Eg not sealed apparatus, water must flow correctly through condenser, joint shown between flask and condenser (no obvious gaps), no extras Show as Q on the script for this mark

(b) (i) (Fractional) distillation

(ii) The mixture may be separated because the boiling temperatures are different / 1-bromobutane has lower boiling temperature (than butan-1-ol) (1) The 1-bromobutane will distil over / vaporise first (and can be collected) (1) allow butan-1-ol is left in the flask 2

(c) (i) Mr = 74 (1) 11.1/Mr = correct answer (1) [0.150 mol] 2

(ii) Mr = 137 (1) Answer to (i) Mr = correct answer (1) [20.55 g] 2

(iii) (ii)toanswer

10012.4 = correct answer [60.3 or 60.2] (1) 1

(iv) Any one of: competing reactions side reactions incomplete reaction product lost in purification product lost in transfers. 1

[13]

69. (a) (i) Aluminium chloride or AICl3 or iron(III) chloride or FeCl3 (1) catalyst (1) 2


(ii)

+

CH CO(1)

(1)

(1)

(1)

COCH

COCH

(AlCl )

+ HCl (+ AlCl )

3

3 3 3+ –

4

3

3

3

4–

+

H

CH COCl AlCl CH CO + AlCl

Marks: formation of electrophile attack on electrophile from ring intermediate removal of proton 4

(b) Esterification / condensation (1) O

CCH O3

(1) 2

(c) (i)

OH OH

OH

OHOH

OH

Three correct 2 marks. 2 correct 1 mark 2

(ii) O O

C CCHO O 2

Ester link (1) polymer / with correct benzene ring links (1) 2

(d) Benzene diazonium chloride (solution) / ion shown or C6H5N+N (1) Sodium nitrite and hydrochloric acid (1) 010C (1) alkaline solution (of phenol) (1)

[16]

70. (a) (i) Homolytic - even separation of electrons when a bond breaks / one electron from bonding pair goes to each atom when (a covalent) bond breaks (1) could be shown as diagrams Free radical - species with one unpaired electron (1) 2


(ii)

C C

H H

H

(2)

1 mark for correct repeating unit 1 mark for evidence of 2 continuing structure

(iii) Combustion - produces large volume of carbon dioxide contributing to global warming / may produce toxic products (1) Landfill - may result in

very slow decomposition producing flammable gases which must be managed / unsightly / uses up large areas of land (1) 2

(iv) Low softening point (1) makes it easy to mould into new shapes / can be turned into expanded polystyrene (1) Or Glassy like appearance (1) Makes it suitable for attractive contexts (1) 2

(b) (i)

H H

H

H

H

H H

H

OSO H

H H

H

H

C C

O

C

C C

SO H

C

O

H H

H

SO H

+

–

3

3 3

(1)

(1)for intermediate

(1)

3

(ii) Add bromine water / bromine in organic solvent (1) No change of specified colour (1) Allow valid alternative such as KMnO4 with named acid or alkali 2

(c) (i) 3 marks for developing evidence: Formula suggests that benzene has three double C-C bonds

and three single C-C bonds (1) Summation of bond enthalpy of three double and three single

bonds plus six carbon hydrogen (1) gives value for enthalpy of formation less than measured value (1)

2 marks for actual structure Actual structure has electrons delocalised in a orbital (1) Each carbon bonded to two other carbons by bond (1) 5

(ii) Substitution enables delocalisation to be preserved (1) energetically more favourable than addition (1) 2


(d) Reagent mixture of concentrated nitric and sulphuric acid (1)

Conditions Warm (under reflux) on water bath (1) at specified temperature at or below 30 - 55C. Purification Separate organic layer using separating funnel (1) Distil mixture to remove benzene / leave nitrobenzene behind/ distil at 211C (1) NB if no reference to separation of two layers max 3 5

[25]

71. (a) (Wear)gloves / add slowly / cool while adding ONLY (1)

IF multiple answers correct and incorrect (0) 1 NOT lab coats / be careful not to spill on hands / do in a fume cupboard

(b) (i)

HEAT NOT arrow on its own

Flask (round or pear shaped ONLY) + heat (1)

Vertical condenser (double surface and water inlet and outlet) (1) 3 NOT sealed , no gaps, joint shown between flask and condenser, safe,

NOT distillation, if water in out labelled they must be correct (1)

3rd mark can be awarded even if minor slip eg no heat

(ii) (Fractionally) distil (1)

Collect fraction that distils over at about 56 C / stated range 55-57 or 54

–58 °C (1) – a temperature value / range MUST be quoted 2 Mark independently

NOT “collect fraction above 56C


(c) Reagent Result

PCl5 / SOCl2 (1) No steamy fumes / no gas which turns blue litmus red (1)

OR

K2Cr2O7 + H2SO4 (1) No colour change / stays orange (1)

OR

KMnO4 + H2SO4 (1) No colour change / stays purple (1)

OR

Na (1) No bubbles / no effervescence (1) Name or formula can be given.

IGNORE oxidation numbers UNLESS they are incorrect

Results may be given as negative tests Or as “if propan-2-ol was present there would be steamy fumes”

Result mark is dependent on correct reagent but may have result mark if minor error in reagent.

MUST be a chemical test eg NOT “check that boiling point is 56 C exactly”

2 [8]

72. (a) (i) A molecule that is non-superimposable on its mirror image (1) or has no plane of symmetry (1) or has a single asymmetric carbon atom 1

(ii) Rotation (of the plane of polarisation) of plane-polarised (monochromatic) light 1

(b) bromine (1) allow Br2, not Br, not bromine water and sodium hydroxide (1) heat or warm (1) conditional on the correct reagents 3

(c) (i) CH3CH2CH(MgBr)CH3 (1) CH3CH2CH(COOH)CH3 (1) CH3CH2CH(COCl)CH3 (1)

If substituents are on the end max. (2) 3

(ii) Grignard reagents react with water (1) 1

(iii) ammonia (1) room temperature (1) 2

(d) (i)

CH CH CHCH

OH

3 2 3

1

(ii) contains C=O so reacts with 2,4 dnp (1) but cannot be oxidised so no reaction with Fehlings’ solution (1) 2


(iii)

(1)

C CHCH CH CH

O OH

3 3if included then

zero

1

(iv) CHI3 (1)

CH3CH2COO– (1) Or CH3CH2COONa 2 [17]

73. (a) (i) Conc. sulphuric acid(1) Conc. nitric acid (1) 2

[Conc. must be stated, or implied, for both acids]

(ii) HNO3 + H2SO4 H2O + HSO4– + NO2

+ (1) Can be shown in two stages

Or HNO3 + 2H2SO4 H3O+ + 2HSO4

– + NO2+ (1)

NO NO

NO

H

2 2

2

+

+

+

(1) (1)

(+ H )

(1) for intermediate

I.e. curved arrow from benzene ring of electrons towards N in NO2+

ion (1) Intermediate correctly drawn, including positive charge (1) Curved arrow from C-H bond back into benzene ring (1) 4

(iii) Electrophilic substitution 1

(b) 3

NO

NO NO NO

NO

NO NO NOO N

2

2 2 2

2

2 2 22

(1)(1) (1)

Vertical/right hand substituents must be shown with C to N bond [Mark consequentially on structural formula given for “nitrobenzene” in (a)(ii)]

(c) Tin / iron and concentrated hydrochloric acid/conc. HCl (1) Heat (under reflux) (1) 2

Second mark consequential on correct / “near miss” reagents

(d) (i) H H

N C C H

O H

1


Allow:

NH C CH

O

3

(ii) Dissolve in minimum volume (1) Of boiling/hot solvent (or any specified solvent other than water)

(1) Filter through a heated funnel (1) Cool or leave to crystallise (1) Filter under suction/filter using Buchner funnel (1) Wash crystals with cold solvent (1) 6

NB If no solvent used, no marks available at all in part (d)(ii) [19]

74. (a) (i) Reagent: ethanoyl chloride / CH3COCL (1) Conditions step 1 anhydrous aluminium chloride/AlCl3 (as catalyst) or anhydrous FeCl3(1) step 3 no AICl3 / catalyst or step 1 needs heating, step 3 does not (1) 3

(ii) Reagents: potassium cyanide + named acid / potassium cyanide buffered between pH 6 to 8 / hydrogen cyanide + named base / hydrogen cyanide buffered between pH 6 to 8 / HCN + KCN (1) 1 Names or full formulae acceptable Mechanism:

(1)

(1)

(1)

CH

CN

CCH

:CN

OC HorO CNH

33

–

– +

:

Marks awarded for: Curly arrow from C (not the minus) of -CN ion towards C of C=O (1) Curly arrow from C=O bond to O atom (1) Intermediate with its charge and a curly arrow from O (not the minus) to H+ or to H of HCN (1) 3

Type: nucleophilic addition (1) 1


(iii) Calculation:

Amount of benzene = 10.0 g / 78 g mol-1 = 0. 1282 mol (1) Maximum amount of product = 0.1282 mol (1) Maximum mass of product = 189 g mol–1 (1) × 0.1282 mol = 24.2(3) g Yield = (3.00 / 24.23) × 100 = 12.4 % (1) 4 Or by mole ratio Moles benzene = 10.0/78 = 0.1282 (1) Moles product = 3.00/189 = 0.01587 (1) They react in 1: 1 ratio (1) Yield = 0.01 587 × 100/0. 1282 = 12.4% (1) If the erratum had not been read out, the Mr of the product would

be 177 g mol-1, giving a yield of 13.2% Explanation: Yield low because so many / 3 steps / side reactions at each step (1) 1

(b) (i) Urea has + H and / or - O and / or - N atoms (1) so it can form hydrogen bonds with water (1) 3 to - O and / or + H in water/diagram showing charges and dotted lines(1)

Do not give all 3 marks unless the candidate has expressed their ideas clearly

(ii) Any two of: Lowers pH (1) Low % nitrogen (1) (high osmotic pressure) can cause scorching of foliage (1) leached out / very soluble (1) 2

[18]

75. (a) (Heating under) reflux (1) Distillation/simple distillation (1) NOT fractional distillation 2

(b) (i) 74

137 × 3.70 (1) = 6.9/6.85(g) (1) 2

(ii) (i) answer to

60.4 × 100 = 67 / 66.67 / 66.7 % 1

(iii) Slow/reaction takes a long time / high activation energy. 1

(iv) Measure boiling temperature/point (1) Compare with data book/literature/known value (1) 2

(c) (i) Orange to green 1

(ii) Oxidation continues (1) carboxylic acid formed (1) 2

(iii) Aldehyde/first product distilled off as it forms/removed from reaction mixture 1

[12]


76. (a) Potassium / sodium dichromate(VI)/K2Cr2O7/Na2Cr2O7

Allow potassium manganate(VII)/permanganate/KMnO4/potassium chromate /K2CrO4 1

(b) (i) Exothermic 1

(ii) (Cold) water moving through the condenser/water cools the vapour 1

(iii) To prevent ethanal vaporising / ethanal is volatile 1

(iv) (Remove ethanal) because fumes/vapour/gas flammable/irritant/harmful, not toxic 1

(c) 46

44 × 5.0 (1) = 4.8 g

4.8 × 100

40 = 1.9 g (1)

The second mark can be scored if candidates make use of 2 Mr values, 5.0g and 40% and the answer is less than 5.0g. 2

[7]

77. (a)

Cl2(1)

Refluxcondenser

SUITABLE HEAT SOURCElamp

Marking points: Reflux condenser – but must not be part of the flask (1) Cl2 passed into liquid in a non–closed system, without leaks (1) Source of U.V. light (e.g. lamp etc.) (1) Specified source of heat (e.g. electrical heating mantle, hot plate, liquid oil bath, but NOT a water bath) (1) 4


(b) Flammability of methylbenzene (1)

Avoid naked flames / use electrical heating (1) Or Poisonous nature / toxicity of chlorine (1) Use a fume cupboard (1) Or Hydrogen chloride is corrosive / irritant (1) Use a fume cupboard (1), Or U.V. light harmful to skin / eyes (1) Suitable protection (1) Measure to minimise risk dependent on correct hazard (or near miss) 2

(c) (i)

179 °C

VAP

LIQ

110 °C

(Boilingpoint)

100 %(chloromethyl)benzene

100 %methylbenzene

(COMPOSITION) 4

Both boiling points shown and at least one component’s 100% composition shown (1)

Boiling points in correct order related to composition axis (1) Vapour and liquid lines reasonably drawn (no max or min) (1) Liquid and vapour lines / areas labelled (1)

(ii) Draws tie lines (1) States that vapour is richer in more volatile component (methylbenzene) (1) Uses tie lines to describe repeated distillations (1) until pure methylbenzene produced / (chloromethyl)benzene is left in the flask (1) Third mark dependent on use of a correct series of tie lines Fourth mark dependent on a reasonable attempt at explanation of the process QWC* 4


(d) (i) chlorine atom / radical or Cl / Cl 1

(ii)

C

C C

C

H

H

H

H

H

ClCl Cl

Cl ClOR

OR

1

Allow all or part substitution of four central H atoms by Cl atoms

(e) both arrows (1)

HO(:) C Cl C Cl

H

HH

HH H

HO

HO C (+Cl )

(1)

(1) 3

SN1 mechanism can score only product formula mark [19]

78. (a) (i) Ag+(aq) + Cl–(aq) AgCl(s) No spectator ions, state symbols required ACCEPT multiples 1

(ii) So that all the chloride (ions are) precipitated / reacted NOT chlorine ions NOT “all the carnallite has reacted” 1


(iii) Molar mass AgCl = 143.5 g mol–1

Amount AgCl = 8.61 g + 143.5 g mol–1 = 0.06(00) mol (1) Amount of Cl– ions = 0.0600 mol Amount of carnallite = 1/3 × 0.0600 = 0.02(00) mol If multiplied by 3, mark consequentially as far as possible. (1) Molar mass carnallite = 5.55 g ÷ 0.0200 mol = 277.5 g mol–1 (1) if not divided by 3, molar mass would be 92.5 g mol–1

Mass of KCIMgCl2 in 1 mol of carnallite = 74.5 + 95 = 169.5 g (1) Mass of water in 1 mol of carnallite = 277.5 – 169.5 = 108 g (1) Number of molecules of water of crystallisation = 108 ÷ 18 = 6 (1) 6

(b) Dip a (platinum/nichrome) wire/test rod/spatula [of suitable material NOT glass] into the solid/solution and place in flame. NOT above flame IGNORE any mention of acid 1

Carnallite would give a lilac/purple/mauve flame, rock salt a yellow/orange flame. Both colours needed NOT red(pink for potassium, BUT if blue glass used potassium is red (1) 2

If “white” included then 2nd mark lost. [10]

79. (a) (i) 74

137 × 4.0 = 7.4 / 7.41 g 7.40 g is an s.f. error

2 or 3 SF 1

(ii) 5.9 × 100 = 80 % ALLOW 79.7 / 79.6 conseq on (a) (i) 2 or 3 SF 1

(b) (i) (Turns hot) vapour into liquid / condenses (1) NOT just ‘cooled’ NOT just ‘product vapour’ Which returns to reaction mixture / allows reaction to go to completion / minimises loss of reactants or products (1) 2

(ii) Two layers shown and upper layer is water 1

(iii) To dry (1-bromobutane) / as a drying agent NOT ‘to prevent reaction with 1-bromobutane’ 1

(iv) Heated flask (round or pear shaped ONLY) (1) Condenser (1) Thermometer in correct position (1) Quality – workable and safe (1)Q (1)

(NOT scored if: wrong direction of water flow in condenser gaps in apparatus sealed apparatus delivery tube in product no joints whatsoever condenser not sloping downwards water bath used for heating 4

(c) Wear gloves, 1-bromobutane harmful (by skin absorption) NOT ‘corrosive’ NOT ‘irritant’


H2SO4 is corrosive NOT ‘irritant’ Electrical heater / heating mantle, 1-bromobutane flammable Fume cupboard, 1-bromobutane harmful (vapour) NOT ‘irritant’ 1

[11]

80. (a) (i) CH3CH(OH)CH3 OR

H C C CH

OH

H

3 3

1

(ii) CHI3 (1) CH3COONa (1) OR

H

C

l

ll

CH CO

O Na3

– +

Penalise a covalent bond between O and Na Either no charges or both charges needed 2

(b) (i) HCN (+ base) OR KCN + acid OR HCN + KCN OR either HCN or KCN pH 5–8 – or any no. or range within 5–8 ACCEPT names 1


(ii) (hydrolysis with) dilute acid /H+(aq) (1)

ALLOW NaOH(aq) / OH–(aq) / dilute alkali followed by acid [if acid not added still carry on marking forwards]

to (CH3)2C(OH)COOH/ (CH3)2C(OH)CO2H (1)

PCl5 / PCl3 / SOCl2 or names (1) OR

any alcohol

to (CH ) C(Cl)COCl (1) (CH ) C(OH)COCl ALLOW correct formula of ester from that alcohol

3

3

32 2

(Aqueous) NH / ammonia (1)IGNORE attempt to hydrolise Cl

OR

(hydrolysis with) dilute acid /H+(aq) (1)

to (CH3)2C(OH)COOH/ (CH3)2C(OH)CO2H (1)

ammonia / ammonium carbonate (1)

to (CH3)2C(OH)COONH4 (1)

heat (1)

OR State/imply partial hydrolysis (1) NaOH (1) aq (1) or names Warm (1) ACCEPT temperatures 40–60 °C NOT heat / boil H2O2 / stopped to prevent further/complete hydrolysis (to acid) (1)

OR State/imply partial hydrolysis (1) HCl (1) aq (1) or names Warm (1) ACCEPT temperatures 40–60 °C NOT heat / boil H2O2 / stopped to prevent further/complete hydrolysis (to acid) (1) OR Any other alternative correct route 5

(iii) Bromine/ Br2 (1) NOT Br (aqueous) sodium / potassium hydroxide / NaOH / KOH (1) 2


(c)

C (CH ) C + n N (CH ) N

H

H H

H

H H

Cl Cl

O O

O O

2

2

2

2

x

x

y

y

(1) (1)

C (CH ) C N (CH ) N + 2nHCl

ne.g.

n

Ignore n, x, y values, 2nHCl and ( ) in polymer

ALLOW Any amino acid OR Any amino acyl chloride for first 2 marks 3

[14]

81. (a) Upper layer - must attempt a reason to get this mark (1) Can be shown on a diagram

The organic product has a lower density than water (or than 1g cm–3) / has (1) density of 0.84 (g cm–3) 2 If three layers correctly argues from data then (1 out of 2)

(b) (i) Carbon dioxide / CO2

If name and formula given, both must be correct 1

(ii) H+ / H3O+ 1

(c) (i) Bunsen burner should not be used (to heat flammable substances) OR (1) electrical heater / water bath / oil bath should be used Use anti-bumping granules (1) System is sealed OR no outlet for gases OR no vent (1) Water flow in and out of jacket is wrong way round (1)

NOT No thermometer adaptor NOT use cylinder / beaker NOT do it in a fume cupboard Max 3

(ii) 51 (°C) 1

(d) (i) 0.74

5.92 × 8.00 (1)

= 10.0 (g) – ALLOW for rounding errors (1) 2 IGNORE SF

(ii) )(

99.6

ians × 100 = 69.9 / 70% 1


(e) Reagent: PCl5/ SOCl2 (1)

Result: No steamy / misty / cloudy / fumes / no gas which turns blue litmus (1) paper red

ALLOW steamy white NOT “white” on its own If PCl3, or PCl5 solution is used then (0)

OR

Reagent: Na (1) Result: no bubbles / no effervescence (1)

Names or formulae can be given for reagents Results may be given either as negative tests or in the form “if 2- methylpropan-2-ol were present, there would be steamy fumes” MUST be a chemical test (NOT boiling point check) 2

[13]

82. (a) Ethylmagnesium bromide of formula, or any other halide NOT C2H5BrMg, (1) Dry ether / ethoxyethane Followed by hydrolysis / acid / water (1)

Grignard reagent / named reagent with incorrect alkyl group scores (0) for (1) reagent but can score both condition marks. If halogenoalkane given as reagent, can score 1st mark if Mg included under 3 conditions.

(b) (i) Observation effervescence/ bubbles/ fizzing (1) NOT gas evolved

2C2H5COOH + Na2CO3 2C2H5COONa + CO2 + H2O (1) 2

(ii) Observation

steamy/ misty/ white fumes (1) NOT smoke C2H5COOH + PCl5 C2H5COCl + POCl3 + HCl (1) 2

(c) Reagents potassium dichromate ((VI)) / K2Cr2O7, (1) sulphuric acid / H2SO4 / hydrochloric acid / HCl but conseq. on an oxidising (1) agent ALLOW acidified potassium dichromate / H+ and Cr2O7

2– (2) ALLOW acidified dichromate ions (2) Acidified dichromate (without ion) scores just (1) ACCEPT Potassium manganate(VII) / potassium permanganate / KMnO4 / Tollens’* / Fehling’s* (1) Acidified / alkaline* / neutral (1) 2 (*) need to acidify to liberate free acid for 2nd mark


(d) (i) Reagent Condition

(any one of) (to match) HCN and KCN HCN or KCN (buffered between) pH between 6 and 9 KCN + acid / H+ NOT excess HCN + base / OH– NOT excess (2)

Type of reaction

Nucleophilic addition - both words needed (1) 3

(ii) Reagent Condition

(any one of) (to match) Hydrogen Pt / Ni / Pd (catalyst) – IGNORE ref to temp. Sodium (in) ethanol Lithium aluminium hydride dry ether / ethoxyethane Sodium borohydride (in) aqueous / water / ethanol / methanol (2)

Type of reaction

Reduction ACCEPT redox / hydrogenation (not addition) ACCEPT nucleophilic addition if metal hydrides used (1) 3

(e) (i) 2


(ii)

OR

OR

NOT

1

(f) Optical NOT stereo (1)

ALLOW – C2H5 for – CH2CH3


MUST show the two as object and mirror image (2)

but NOT

C must not be bonded to H in OH group Near-miss molecule plus mirror image (1) 3 The two solid lines in 3D structure must not be at 180 °

[21]

83. (1)1 Distillation of crude ester NOT fractional (1) OR add water

(1)2 Separating funnel (1)

(1)3 Wash with Na2CO3 / NaHCO3 – need not specify solution (1)

(1)4 Wash with CaCl2 solution (1)

(1)3 and (1)4 can be in reverse order

(1)5 Any mention of ester as upper layer [not enough to just say ester is less (1) dense] OR add water and note which layer increases.

(1)6 Anhydrous / solid CaCl2 / Na2SO4 / MgSO4 (1)

(1)7 Clear (when dry) (1)

(1)8 Final distillation ALLOW fractional – can be a diagram (need not be perfect) (1)

(1)9 Collect in specified boiling range – range 2° - 4°C to include boiling (1) temperature IGNORE re-crystallisation

If solvent extraction method used, give it credit where appropriate for (1)1 to (1)7 and (1)9

(1)8 distil off ether Max 8 [8]


84. (1)1 Distillation of crude 1-bromopentane NOT fractional (1)

OR add water

(1)2 Separating funnel (1)

(1)3 Wash with concentrated HCl(1)

(1)4 Wash with Na2CO3 / NaHCO3 – need not specify solution (1) Do NOT award if this precedes HCl wash

(1)5 Any mention of 1-bromopentane as lower layer [not enough to say (1) 1-bromopentane is more dense] OR add water and note which layer increases.

(1)6 Anhydrous / solid CaCl2 / Na2SO4 / MgSO4 (1)

(1)7 Clear (when dry) (1)

(1)8 Final distillation ALLOW fractional – can be a diagram (need not be perfect) (1)

(1)9 Collect in specified boiling range – range 2° - 4°C to include boiling (1) temperature IGNORE re-crystallisation

If solvent extraction method used, give it credit where appropriate for (1)1 to (1)7 and (1)9

(1)8 distil off ether Max 8 [8]

85. (a) KC =

2322

22

COOHCHOHCHHOCH

OHformulaitsor ester

(1)

both molar masses (1) HOCH2CH2OH + 2CH3COOH Ester +2H2O Moles at start 24.8 / 62 = 0.400 66.0 / 60 = 1.10 (1) Moles at equilibrium 0.400 – 0.320 = 0.080 1.10 – 0.640 = 0.460 0.320 0.640 (1)

Concentration at equilibrium divide above by 0.0900 dm3 (1)

0.080 / 0.0900 = 0.889 5.11 3.56 7.11 OR explain why volume cancels in this case

KC = 2

2

11.5889.0

11.756.3

=

2.23

180

= 7.76 / 7.8 / 7.74 etc. (1) There are no units for K (1) 7

(b) Amount of ethan-1, 2-diol = 1054 / 62 = 17 mol

Amount of ethene = 28

560 = 20 mol or

1240

1054 × 100 (1)

Yield = 17 × 100 20 = 85% (1) 2


(c) (i) Any acid with two COOH groups or its acid dichloride or its dimethyl

ester (1) Accept HOOCCOOH

(Where R = the hydrocarbon part of their diacid). for correct ester linkage drawn out (1) for remainder with continuation (1) 3

(ii) No, because the acid would hydrolyse / is a catalyst for the hydrolysis of the ester. OR Yes, not hydrolysed at low temperature / only hydrolysed at high temperature 1

(d) The ester cannot form (intermolecular) hydrogen bonds but the acid can (1) The ester does not have a +hydrogen atom OR the acid has +hydrogen / polar OH / O and H have a large difference in electronegativities (1) thus less energy / heat is required to separate molecules of the ester (1) but as ethanoic acid has fewer electrons than propanoic acid, (1) it has weaker intermolecular instantaneous induced dipole / induced dipole forces / van der Waals / dispersion / London forces (1) Allow vdW 5

[18]


docshare02.docshare.tipsdocshare02.docshare.tips/files/23041/230415604.pdf1. (i) when calcium oxide...

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