(total 1 mark) - the maths and science tutor

Use your understanding of the bonding in benzene to identify the compound that has the mostexothermic enthalpy of hydrogenation.

A

B

C

D

(Total 1 mark)

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Equations for the hydrogenation of cyclohexene and of benzene, together with the enthalpies ofhydrogenation, are shown.

(a) (i) Use these data to show that benzene is 152 kJ mol−1 more stable than thehypothetical compound cyclohexa−1,3,5−triene.

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of 84Catalyst Tutors

(ii) State, in terms of its bonding, why benzene is more stable thancyclohexa−1,3,5−triene.

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(1)

(b) Three carbon−carbon bonds are labelled on the structures shown.These bonds are of different lengths.

Write the letters w, x and y in order of increasing bond length.

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(1)

(c) The structures of two cyclic dienes are shown.

(i) Use the enthalpy of hydrogenation data given opposite to calculate a value for theenthalpy of hydrogenation of cyclohexa−1,4−diene.

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(1)

(ii) Predict a value for the enthalpy of hydrogenation of cyclohexa−1,3−diene.

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(1)


(iii) Explain your answers to part (i) and part (ii) in terms of the bonding in these twodienes.

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(3)

(Total 8 marks)


Each of the following conversions involves reduction of the starting material.

(a) Consider the following conversion.

Identify a reducing agent for this conversion.

Write a balanced equation for the reaction using molecular formulae for the nitrogen-containing compounds and [H] for the reducing agent.

Draw the repeating unit of the polymer formed by the product of this reaction with benzene-1,4-dicarboxylic acid.

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(5)

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(b) Consider the following conversion.

Identify a reducing agent for this conversion.

State the empirical formula of the product.

State the bond angle between the carbon atoms in the starting material and the bond anglebetween the carbon atoms in the product.

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(4)

(c) The reducing agent in the following conversion is NaBH4

(i) Name and outline a mechanism for the reaction.

Name of mechanism _____________________________________________

Mechanism

(5)


(ii) By considering the mechanism of this reaction, explain why the product formed isoptically inactive.

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(3)

(Total 17 marks)

This question is about nitrobenzenes.

(a) Nitrobenzene reacts when heated with a mixture of concentrated nitric acid andconcentrated sulfuric acid to form a mixture of three isomeric dinitrobenzenes.

Write an equation for the reaction of concentrated nitric acid with concentrated sulfuric acidto form the species that reacts with nitrobenzene.

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(1)

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(b) Name and outline a mechanism for the reaction of this species with nitrobenzene to form1,3-dinitrobenzene.

Name of mechanism____________________________________________________________

Mechanism

(4)


(c) The dinitrobenzenes shown were investigated by thin layer chromatography (TLC).

In an experiment, carried out in a fume cupboard, a concentrated solution of pure1,4-dinitrobenzene was spotted on a TLC plate coated with a solid that contains polarbonds. Hexane was used as the solvent in a beaker with a lid.

The start line, drawn in pencil, the final position of the spot and the final solvent frontare shown on the chromatogram in the diagram below

Use the chromatogram in the diagram above to deduce the Rf value of 1,4-dinitrobenzenein this experiment.

Tick (✔) one box.

A 0.41

B 0.46

C 0.52

D 0.62

(1)


(d) State in general terms what determines the distance travelled by a spot in TLC.

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(e) To obtain the chromatogram, the TLC plate was held by the edges and placed in thesolvent in the beaker in the fume cupboard. The lid was then replaced on the beaker.

Give one other practical requirement when placing the plate in the beaker.

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(1)

(f) A second TLC experiment was carried out using 1,2‑dinitrobenzene and1,4‑dinitrobenzene. An identical plate to that in part (c) was used under the sameconditions with the same solvent. In this experiment, the Rf value of 1,4‑dinitrobenzenewas found to be greater than that of 1,2‑dinitrobenzene.

Deduce the relative polarities of the 1,2‑dinitrobenzene and 1,4‑dinitrobenzene and explainwhy 1,4‑dinitrobenzene has the greater Rf value.

Relative polarities

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Explanation

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(2)


(g) A third TLC experiment was carried out using 1,2‑dinitrobenzene. An identical plate to thatin part (c) was used under the same conditions, but the solvent used contained a mixture ofhexane and ethyl ethanoate.

A student stated that the Rf value of 1,2‑dinitrobenzene in this third experiment would be

greater than that of 1,2‑dinitrobenzene in the experiment in part (f)

Is the student correct? Justify your answer.

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(2)

(Total 12 marks)

When methylbenzene reacts with ethanoyl chloride in the presence of aluminium chloride, theproduct, H, is formed.

(a) Deduce the molecular formula of H.

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(1)

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(b) Two other isomers are also produced in the reaction.

Draw the structure of one of the other isomers.

Name the type of structural isomerism shown by these three products.

Structure

Type of isomerism __________________________________________________

(2)

(c) Name and outline the mechanism for the reaction of ethanoyl chloride with methylbenzeneto produce H.

Include an equation for the formation of the reactive intermediate that is involved in thereaction.

Name _____________________________________________________________

Equation ___________________________________________________________

Mechanism

(5)

(Total 8 marks)


1,4-diaminobenzene is an important intermediate in the production of polymers such as Kevlarand also of polyurethanes, used in making foam seating.

A possible synthesis of 1,4-diaminobenzene from phenylamine is shown in the following figure.

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(a) A suitable reagent for step 1 is CH3COCl

Name and draw a mechanism for the reaction in step 1.

Name of mechanism __________________________________________________

Mechanism

(5)


(b) The product of step 1 was purified by recrystallisation as follows.

The crude product was dissolved in the minimum quantity of hot water and the hotsolution was filtered through a hot filter funnel into a conical flask. This filtration removedany insoluble impurities. The flask was left to cool to room temperature.The crystals formed were filtered off using a Buchner funnel and a clean cork was used tocompress the crystals in the funnel. A little cold water was then poured through thecrystals.After a few minutes, the crystals were removed from the funnel and weighed.A small sample was then used to find the melting point.

Give reasons for each of the following practical steps.

The minimum quantity of hot water was used

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The flask was cooled to room temperature before the crystals were filtered off

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The crystals were compressed in the funnel

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A little cold water was poured through the crystals

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(4)


(c) The melting point of the sample in part (b) was found to be slightly lower than a data-bookvalue.

Suggest the most likely impurity to have caused this low value and an improvement to themethod so that a more accurate value for the melting point would be obtained.

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(2)

The figure above is repeated here to help you answer the following questions.


(d) In an experiment starting with 5.05 g of phenylamine, 4.82 g of purified product wereobtained in step 1.

Calculate the percentage yield in this reaction.Give your answer to the appropriate number of significant figures.

Percentage yield = _______________%

(3)

(e) A reagent for step 2 is a mixture of concentrated nitric acid and concentrated sulfuric acid,which react together to form a reactive intermediate.

Write an equation for the reaction of this intermediate in step 2.

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(f) Name a mechanism for the reaction in step 2.

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(g) Suggest the type of reaction occurring in step 3.

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(h) Identify the reagents used in step 4.

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(Total 18 marks)


The hydrocarbons benzene and cyclohexene are both unsaturated compounds.Benzene normally undergoes substitution reactions, but cyclohexene normallyundergoes addition reactions.

(a) The molecule cyclohexatriene does not exist and is described as hypothetical.Use the following data to state and explain the stability of benzene compared with thehypothetical cyclohexatriene.

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(4)

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(b) Benzene can be converted into amine U by the two-step synthesis shown below.

The mechanism of Reaction 1 involves attack by an electrophile.

Give the reagents used to produce the electrophile needed in Reaction 1.

Write an equation showing the formation of this electrophile.

Outline a mechanism for the reaction of this electrophile with benzene.

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(6)


(c) Cyclohexene can be converted into amine W by the two-step synthesis shown below.

Suggest an identity for compound V.

For Reaction 3, give the reagent used and name the mechanism.

For Reaction 4, give the reagent and condition used and name the mechanism.

Equations and mechanisms with curly arrows are not required.

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(6)


(d) Explain why amine U is a weaker base than amine W.

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(3)

(Total 19 marks)

(a) Use the following data to show the stability of benzene relative to the hypotheticalcyclohexa-1,3,5-triene.

Give a reason for this difference in stability.

(4)

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(b) Consider the following reaction sequence which starts from phenylamine.

(i) State and explain the difference in base strength between phenylamine andammonia.


(ii) Name and outline a mechanism for the reaction in Step 1 and name the organicproduct of Step 1.

(iii) The mechanism of Step 2 involves attack by an electrophile. Give the reagents usedin this step and write an equation showing the formation of the electrophile.Outline a mechanism for the reaction of this electrophile with benzene.

(iv) Name the type of linkage which is broken in Step 3 and suggest a suitable reagent forthis reaction.

(17)

(Total 21 marks)

Which one of the following does not contain any delocalised electrons?

A poly(propene)

B benzene

C graphite

D sodium

(Total 1 mark)

9

Compound X (ClCH2COCl) is used as a reagent in organic synthesis.

(a) One important reaction of X is in the preparation of compound P as shown.

(i) Draw the structure of the electrophile formed by the reaction of X with AlCl3.

(1)

(ii) Outline the mechanism for the reaction of the electrophile from part (a)(i) withbenzene in the preparation of P.

(3)

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(b) Compound Q is an alternative product that could be formed when X reacts with benzene.

Describe how you could distinguish between P and Q by a test-tube reaction.Give the reagent used and the observation with each compound.

Reagent _________________________________________________________

Observation with P _________________________________________________

Observation with Q _________________________________________________

(3)

(c) X is also used to make the compound HOCH2COOH. This compound is polymerised toform the polymer known as PGA. PGA is used in surgical sutures (stitches).

(i) Draw the repeating unit of PGA.

(1)

(ii) Production of PGA occurs via a cyclic compound. Two HOCH2COOH molecules reacttogether to form the cyclic compound and two molecules of water.

Draw the structure of this cyclic compound.

(1)


(d) Poly(propene) is also used in surgical sutures.

(i) Draw the repeating unit of poly(propene).

(1)

(ii) Suggest an advantage of surgical sutures made from PGA rather than frompoly(propene).Explain your answer.

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(2)

(Total 12 marks)

Consider the following reaction sequence starting from methylbenzene.

(a) Name the type of mechanism for reaction 1.

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(b) Compound J is formed by reduction in reaction 2.

(i) Give a reducing agent for this reaction.

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(ii) Write an equation for this reaction. Use [H] to represent the reducing agent.

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(iii) Give a use for J.

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(c) Outline a mechanism for the reaction of bromomethane with an excess of compound J.You should represent J as RNH2 in the mechanism.

(4)


(d) Compound K (C6H5CH2NH2) is a structural isomer of J.

Explain why J is a weaker base than K.

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(3)

(Total 11 marks)

Kevlar is a polymer used in protective clothing.The repeating unit within the polymer chains of Kevlar is shown.

(a) Name the strongest type of interaction between polymer chains of Kevlar.

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(1)

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(b) One of the monomers used in the synthesis of Kevlar is

H2N NH2

An industrial synthesis of this monomer uses the following two-stage process starting fromcompound X.

Stage 1

Cl NO2 + 2NH3 H2N NO2 + NH4Cl

X

Stage 2

H2N NO2 H2N NH2

(i) Suggest why the reaction of ammonia with X in Stage 1 might be consideredunexpected.

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(2)

(ii) Suggest a combination of reagents for the reaction in Stage 2.

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(1)

(iii) Compound X can be produced by nitration of chlorobenzene.

Give the combination of reagents for this nitration of chlorobenzene.Write an equation or equations to show the formation of a reactive intermediate fromthese reagents.

Reagents _____________________________________________________

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Equation(s) ____________________________________________________

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(3)


(iv) Name and outline a mechanism for the formation of X from chlorobenzene and thereactive intermediate in part (iii).


Mechanism

(4)

(Total 11 marks)

This question is about acylium ions, [RCO] +

(a) The acylium ion

Write an equation for this fragmentation.

Include in your answer a displayed formula for the radical formed.

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(2)

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is formed in a mass spectrometer by fragmentation of themolecular ion of methyl ethanoate.

(b) The acylium ion

(i) Write an equation to show the formation of this acylium ion by the reaction of ethanoylchloride with one other substance.

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(2)

can also be formed from ethanoyl chloride. The ion reactswith benzene to form C6H5COCH3

(ii) Name and outline a mechanism for the reaction of benzene with this acylium ion.


Mechanism

(4)


(iii) Ethanoic anhydride also reacts with benzene to form C6H5COCH3

Write an equation for this reaction.

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(1)

(Total 9 marks)

Benzene reacts with ethanoyl chloride in a substitution reaction to form C6H5COCH3.This reaction is catalysed by aluminium chloride.

(a) Write equations to show the role of aluminium chloride as a catalyst in this reaction.

Outline a mechanism for the reaction of benzene.

Name the product, C6H5COCH3.

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(6)

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(b) The product of the substitution reaction (C6H5COCH3) was analysed by massspectrometry. The most abundant fragment ion gave a peak in the mass spectrum with m/z= 105.Draw the structure of this fragment ion.

(1)


(c) When methylbenzene reacts with ethanoyl chloride and aluminium chloride, a similarsubstitution reaction occurs but the reaction is faster than the reaction of benzene.Suggest why the reaction of methylbenzene is faster.

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(2)

(Total 9 marks)

Many aromatic nitro compounds are used as explosives. One of the most famous is 2-methyl-1,3,5-trinitrobenzene, originally called trinitrotoluene or TNT. This compound, shown below, canbe prepared from methylbenzene by a sequence of nitration reactions.

(a) The mechanism of the nitration of methylbenzene is an electrophilic substitution.

(i) Give the reagents used to produce the electrophile for this reaction.Write an equation or equations to show the formation of this electrophile.

Reagents ______________________________________________________

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Equation ______________________________________________________

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(3)

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(ii) Outline a mechanism for the reaction of this electrophile with methylbenzene toproduce 4-methylnitrobenzene.

(3)

(b) Deduce the number of peaks in the 13C n.m.r. spectrum of TNT.

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(C) Deduce the number of peaks in the 1H n.m.r. spectrum of TNT.

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(d) Using the molecular formula (C7H5N3O6), write an equation for the decomposition reactionthat occurs on the detonation of TNT. In this reaction equal numbers of moles of carbonand carbon monoxide are formed together with water and nitrogen.

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(1)

(Total 9 marks)


Many synthetic routes need chemists to increase the number of carbon atoms in a molecule byforming new carbon–carbon bonds. This can be achieved in several ways including

• reaction of an aromatic compound with an acyl chloride• reaction of an aldehyde with hydrogen cyanide.

(a) Consider the reaction of benzene with CH3CH2COCl

(i) Write an equation for this reaction and name the organic product.Identify the catalyst required in this reaction.Write equations to show how the catalyst is used to form a reactive intermediate andhow the catalyst is reformed at the end of the reaction.

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(ii) Name and outline a mechanism for the reaction of benzene with this reactiveintermediate.

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(4)


(b) Consider the reaction of propanal with HCN

(i) Write an equation for the reaction of propanal with HCN and name the product.

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(2)

(ii) Name and outline a mechanism for the reaction of propanal with HCN

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(5)

(iii) The rate-determining step in the mechanism in part (b) (ii) involves attack by thenucleophile.Suggest how the rate of reaction of propanone with HCN would compare with therate of reaction of propanal with HCNExplain your answer.

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(2)

(Total 18 marks)


Consider compound P shown below that is formed by the reaction of benzene with anelectrophile.

(a) Give the two substances that react together to form the electrophile and write an equationto show the formation of this electrophile.

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(3)

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(b) Outline a mechanism for the reaction of this electrophile with benzene to form P.

(3)


(c) Compound Q is an isomer of P that shows optical isomerism. Q forms a silver mirror whenadded to a suitable reagent.

Identify this reagent and suggest a structure for Q.

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(2)

(Total 8 marks)

Synthetic dyes can be manufactured starting from compounds such as 4-nitrophenylamine.

A synthesis of 4-nitrophenylamine starting from phenylamine is shown below.

(a) An equation for formation of N-phenylethanamide in Step 1 of the synthesis is shownbelow.

2C6H5NH2 + CH3COCl → C6H5NHCOCH3 + C6H5NH3ClN-phenylethanamide

(i) Calculate the % atom economy for the production of N-phenylethanamide(Mr = 135.0).

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(ii) In a process where 10.0 kg of phenylamine are used, the yield ofN-phenylethanamide obtained is 5.38 kg.

Calculate the percentage yield of N-phenylethanamide.

(iii) Comment on your answers to parts (i) and (ii) with reference to the commercialviability of the process.

(7)


(b) Name and outline a mechanism for the reaction in Step 1.

(5)

(c) The mechanism of Step 2 involves attack by an electrophile. Write an equation showing theformation of the electrophile. Outline a mechanism for the reaction of this electrophile withbenzene.

(4)

(Total 16 marks)

A possible synthesis of phenylethene (styrene) is outlined below.

(a) In Reaction 1, ethanoyl chloride and aluminium chloride are used to form a reactive specieswhich then reacts with benzene.Write an equation to show the formation of the reactive species.Name and outline the mechanism by which this reactive species reacts with benzene.

(6)

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(b) NaBH4 is a possible reagent for Reaction 2.Name and outline the mechanism for the reaction with NaBH4 in Reaction 2.Name the product of Reaction 2.

(6)

(c) Name the type of reaction involved in Reaction 3 and give a reagent for the reaction.

(2)

(Total 14 marks)


Ethanoyl chloride reacts with methylbenzene forming compound X according to the equationbelow.

If the experimental yield is 40.0%, the mass in grams of X (Mr = 134.0) formed from 18.4 g ofmethylbenzene (Mr = 92.0) is

A 26.8

B 16.1

C 10.7

D 7.4

(Total 1 mark)

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(a) Name and outline a mechanism for the reaction between propanoyl chloride,CH3CH2COCl, and methylamine, CH3NH2

Draw the structure of the organic product.

(6)

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(b) Benzene reacts with propanoyl chloride in the presence of aluminium chloride. Writeequations to show the role of aluminium chloride as a catalyst in this reaction. Outline amechanism for this reaction of benzene.

(5)

(c) Write an equation for the reaction of propanoyl chloride with water. An excess of water isadded to 1.48 g of propanoyl chloride. Aqueous sodium hydroxide is then added from aburette to the resulting solution.Calculate the volume of 0.42 mol dm–3 aqueous sodium hydroxide needed to react exactlywith the mixture formed.

(5)

(Total 16 marks)

In a reaction which gave a 27.0% yield, 5.00 g of methylbenzene were converted into theexplosive 2,4,6-trinitromethylbenzene (TNT) (Mr = 227.0). The mass of TNT formed was

A 1.35 g

B 3.33 g

C 3.65 g

D 12.34 g

(Total 1 mark)

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The following reaction scheme shows the formation of two amines, K and L, frommethylbenzene.

(a) (i) Give the reagents needed to carry out Step 1. Write an equation for the formationfrom these reagents of the inorganic species which reacts with methylbenzene.

Reagents ______________________________________________________

Equation ______________________________________________________

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(ii) Name and outline a mechanism for the reaction between this inorganic speciesand methylbenzene.


Mechanism

(7)

(b) Give a suitable reagent or combination of reagents for Step 2.

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(1)


(c) (i) Give the reagent for Step 4 and state a condition to ensure that the primary amine isthe major product.

Reagent _______________________________________________________

Condition ______________________________________________________

(ii) Name and outline a mechanism for Step 4.


Mechanism

(7)

(Total 15 marks)

In which one of the following reactions is the role of the reagent stated correctly?

Reaction Role of reagent

A TiO2 + 2C + 2Cl2 → TiCl4 + 2CO TiO2 is an oxidising agent

B HNO3 + H2SO4 → H2NO + HSO HNO3 is a Brønsted-Lowry acid

C CH3COCl + AlCl3 → CH3CO+ + AlCl AlCl3 is a Lewis base

D 2CO + 2NO → 2CO2 + N2 CO is a reducing agent

(Total 1 mark)

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(a) Outline a mechanism for the reaction of CH3CH2CH2CHO with HCN and name the product.

Mechanism

Name of product ____________________________________________________

(5)

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(b) Outline a mechanism for the reaction of CH3OH with CH3CH2COCl and name the organicproduct.

Mechanism

Name of organic product _______________________________________________

(5)

(c) An equation for the formation of phenylethanone is shown below. In this reaction a reactiveintermediate is formed from ethanoyl chloride. This intermediate then reacts with benzene.

(i) Give the formula of the reactive intermediate.

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(ii) Outline a mechanism for the reaction of this intermediate with benzene to formphenylethanone.

(4)

(Total 14 marks)

(a) Outline a mechanism for the formation of ethylamine from bromoethane. State why theethylamine formed is contaminated with other amines. Suggest how the reaction conditionscould be modified to minimise this contamination.

(6)

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(b) Suggest one reason why phenylamine cannot be prepared from bromobenzene in a similarway. Outline a synthesis of phenylamine from benzene. In your answer you should givereagents and conditions for each step, but equations and mechanisms are not required.

(5)

(Total 11 marks)

(a) The reaction between aqueous persulphate ions, , and iodide ions, I–(aq), iscatalysed by Fe2+(aq) ions. Suggest why this reaction has a high activation energy.Write equations to explain the catalytic action of Fe2+(aq) ions.Suggest why V3+(aq) ions will also act as a catalyst for this reaction but Mg2+(aq) ions willnot.

(6)

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(b) Outline a mechanism for the reaction between benzene and ethanoyl chloride and explainwhy AlCl3 acts as a Lewis acid catalyst for this reaction. Predict, with an explanation ineach case, the suitability of FeCl3 and of NH4Cl to act as a catalyst for this reaction.

(9)

(Total 15 marks)


Which one of the following can react both by nucleophilic addition and by nucleophilicsubstitution?

A

B

C

D

(Total 1 mark)

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Mark schemes

A

[1]1

(a) (i) 3(-120) − (-208) = -152OR3(120) − 208 = 152 (kJ mol−1)

Must show working and answer and maths must be correct, butignore sign

1

(ii) Electrons delocalised OR delocalisation (QOL)ORallow reference to resonance (QOL)

1

2

(b) x, y, w

Must be in this order1

(c) (i) -240 (kJ mol−1)

Must have minus sign1

(ii) between -239 and -121 (kJ mol−1)

Must have minus sign1

(iii) Must specify which diene:

Proximity − for 1,3 C=C bonds are close togetherallow converse for 1,4 diene

M11

Delocalisation − for 1,3 some delocalisationOR

some overlap of electrons, π clouds or p orbitals

allow converse for 1,4 diene

M21

some extra stability for the 1,3- isomer

M31

[8]


(a) Sn / HCl OR Fe / HCl not conc H2SO4 nor any HNO3

Ignore subsequent use of NaOH

Ignore reference to Sn as a catalyst with the acid

Allow H2 (Ni / Pt) but penalise wrong metal

But NOT NaBH4 LiAlH4 Na / C2H5OH1

3

Equation must use molecular formulae

C6H4N2O4 + 12 [H]

12[H] and 4H2O without correct molecular formula scores 1 out of 21

→C6H8N2 + 4H2O

Allow .... + 6H2 if H2 / Ni used

Allow −CONH− or −COHN− or −C6H4−1

Mark two halves separately: lose 1 each for

• error in diamine part

• error in diacid part

• error in peptide link

• missing trailing bonds at one or both ends

• either or both of H or OH on ends

Ignore n2

(b) H2 (Ni / Pt) but penalise wrong metal

NOT Sn / HCl, NaBH4 etc.1

CH21

In benzene 120°1

In cyclohexane 109° 28’ or 109½°

Allow 108° - 110°

If only one angle stated without correct qualification, no mark awarded1

(c) (i) Nucleophilic addition1


• M2 not allowed independent of M1, but allow M1 for correct attackon C+

• + rather than δ+ on C=O loses M2• M3 is for correct structure including minus sign but lone pair ispart of M4

• Allow C 2H5

• M1 and M4 include lp and curly arrow

• Allow M4 arrow to H in H2O (ignore further arrows)4

(ii) M1 Planar C=O (bond / group)

Not just planar molecule1

M2 Attack (equally likely) from either side

Not just planar bond without reference to carbonyl1

M3 (about product): Racemic mixture formed OR 50:50 mixture or each enantiomer equally likely

1

[17]

(a) HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4

−

Allow H2SO4 + HNO3 → NO2+ + HSO4

− + H2O

Allow a combination of equations which produce NO2+

Penalise equations which produce SO42−

1

4


(b) Electrophilic substitution.

Ignore nitration1

3

OR Kekule

M1 Arrow from inside hexagon to N or + on N (Allow NO2+)

M2 Structure of intermediate

• horseshoe centred on C1 and must not extend beyond C2 andC6, but can be smaller

• + in intermediate not too close to C1 (allow on or “below” aline from C2 to C6)

M3 Arrow from bond into hexagon (Unless Kekule)

• Allow M3 arrow independent of M2 structure• + on H in intermediate loses M2 not M3

(c) D1

(d) (Balance between) solubility in moving phase and retention by stationary phase

OR (relative) affinity for stationary / solid and mobile / liquid /solvent (phase)

(e) Solvent depth must be below start line

Ignore safety1


(f) 1,2- is more polar OR 1,4- is less polarOR 1,2 is polar, 1,4- is non-polar

1

1,4- ( or Less/non polar is) less attracted to (polar) plate / stationary phase / solidOR (Less/non polar is) more attracted to / more soluble in (non-polar) solvent / mobilephase / hexane

1

M2 dependent on correct M1

If M1 is blank then read explanation for possible M1 and M2

Allow converse argument for 1,2

(g) No CE = 0

Yes - mark on but there is NO MARK FOR YES

Mark independently following yes

Solvent (more) polar or ethyl ethanoate is polar1

Polar isomer more attracted to / more soluble in / stronger affinity to the solvent (thanbefore)

Penalise bonded to mobile phase in M21

[12]

(a) C9H10O15

(b)

1

Position (isomerism).

Allow Positional.1


(c) Electrophilic substitution1

CH3COCl + AlCl3 ⟶ CH3CO+ + AlCl4−

1

Mechanism 3 marks:

M1 arrow from circle or within it to C of CH3C+O (+ must be on C of

CH3C+O).1

M2 for Intermediate (must be 4-isomer)CH3CO must be correctly positioned and bonded to gain M2horseshoe must not extend beyond C2 to C6 but can be smaller+ not too close to C1.

1

M3 arrow into hexagon unless KekuleLoss of H+ (allow from incorrect isomer)Allow M3 arrow independent of M2 structureIgnore base removing H in M3.

Allow Kekule structures (which must be correct).1

[8]

(a) (nucleophilic) addition-elimination

Not electrophilic addition-elimination1

6


Allow C6H5 or benzene ring

Allow attack by :NH2C6H5

M2 not allowed independent of M1, but allow M1 for correct attackon C+

M3 for correct structure with charges but lone pair on O is part ofM4

M4 (for three arrows and lone pair) can be shown in more than onestructure

4

(b) The minimum quantity of hot water was used:

To ensure the hot solution would be saturated / crystals would form on cooling1

The flask was left to cool before crystals were filtered off:

Yield lower if warm / solubility higher if warm1

The crystals were compressed in the funnel:

Air passes through the sample not just round it

Allow better drying but not water squeezed out1

A little cold water was poured through the crystals:

To wash away soluble impurities1

(c) Water

Do not allow unreacted reagents1

Press the sample of crystals between filter papers

Allow give the sample time to dry in air1

(d) Mr product = 135.01


Expected mass = 5.05 × = 7.33 g1

Percentage yield = × 100 = 65.75 = 65.8(%)

Answer must be given to this precision

(e)

OR

C6H5NHCOCH3 + NO2+ C6H4(NHCOCH3)NO2 + H+

1

(f) Electrophilic substitution1

(g) Hydrolysis1

(h) Sn / HCl

Ignore acid concentration; allow Fe / HCl1

[18]

(a) M1 Benzene is more stable than cyclohexatriene

more stable than cyclohexatriene must be stated or implied

If benzene more stable than cyclohexene, then penalise M1 butmark on

If benzene less stable: can score M2 only1

M2 Expected ΔHο hydrogenation of C6H6 is 3(–120)

= –360 kJ mol-1

Allow in words e.g. expected ΔHο hydrog is three times the ΔHο

hydrog of cyclohexene1

7


M3 Actual ΔHο hydrogenation of benzene is

152 kJ mol-1 (less exothermic)

or 152 kJ mol-1 different from expected

Ignore energy needed1

M4 Because of delocalisation or electrons spread out or resonance1

(b) No mark for name of mechanism

Conc HNO3

If either or both conc missing, allow one;1

Conc H2SO4

this one mark can be gained in equation1

2 H2SO4 + HNO3 → 2 HSO4– + NO2

+ + H3O+

OR

H2SO4 + HNO3 → HSO4– + NO2

+ + H2O

OR via two equations

H2SO4 + HNO3 → HSO4– + H2NO3

+

H2NO3+ → NO2+ + H2O

Allow + anywhere on NO2+

1


M1 arrow from within hexagon to N or + on N

Allow NO2+ in mechanism

horseshoe must not extend beyond C2 to C6 but can be smaller

+ not too close to C1

M3 arrow into hexagon unless Kekule

allow M3 arrow independent of M2 structure

ignore base removing H in M3

+ on H in intermediate loses M2 not M33

(c) If intermediate compound V is wrong or not shown, max 4 for 8(c)

or chlorocyclohexane or bromocyclohexane1

Reaction 3

M2 HBr1

M3 Electrophilic addition

Allow M2 and M3 independent of each other1


Reaction 4

M4 Ammonia if wrong do not gain M51

Allow M4 and M6 independent of each other

M5 Excess ammonia or sealed in a tube or under pressure1

If CE e.g. acid conditions, lose M4 and M5

M6 Nucleophilic substitution1

(d) Lone or electron pair on N

No marks if reference to “lone pair on N” missing1

Delocalised or spread into ring in U1

Less available (to accept protons) or less able to donate (to H+)1

[19]

(a) Cyclohexane evolves 120 kJ mol–1

–1 (1) or 3 × 120

360 – 208 = 152 kJ (1) NOT 150

152 can score first 2

QofL: benzene lower in energy / more (stated) stable (1)Not award if mentions energy required for bond breaking

due to delocalisation (1) or explained4

8

(expect triene to evole) 360 kJ mol

(b) (i) phenylamine weaker (1)

if wrong no marks

lone pair on N (less available) (1)delocalised into ring (1) or “explained”

3


(ii) addition – elimination (1)

structure (1) M33 arrows (1) M4

N-phenyl ethanamide (1)6

(iii) conc HNO3 (1)conc H2SO4 (1)

HNO3 + 2H2SO4 → O2 + H3O+ + 2HSO4– (1)

6

(iv) peptide / amide (1)

NaOH (aq) (1)

HCl conc or dil or neither

H2SO4 dil NOT conc

NOT just H2O2

Notes

(a) • 360 or 3 × 120 or in words (1);• 152 NOT 150 (1); (152 can get first two marks)• Q of L benzene more stable but not award if ΔH values used to say that more energy is required by benzene for hydrogenation compared with the triene or if benzene is only compared with cyclohexene (1);• delocalisation or explained (1)

(b) (ii) or N-phenylacetamide or acetanilidemechanism: if shown as substitution can only gain M1if CH3CO+ formed can only gain M1

lose M4 if Cl– removes H+

be lenient with structures for M1 and M2 but must be correct for M3

alone loses M2


(iii) No marks for name of mechanism in this partif conc missing can score one for both acids (or in equation)allow two equations

allow HNO3 + H2SO4 → NO2+ + HSO4– + H2O

ignore side chain in mechanism even if wrongarrow for M1 must come from niside hexagonarrow to NO2

+ must go to N but be lenient over position of ++ must not be too near “tetrahedral” Carbonhorseshoe from carbons 2-6 but don’t be too harsh

(iv) reagent allow NaOHHCl conc or dil or neitherH2SO4 dil or neither but not concnot just H2O

[21]

A

[1]9

(a) (i)

Allow [ClCH2CO]+

1

(ii)

M1 for arrow from inside hexagon to C or + on C on correctelectrophile

M2 for structure of intermediate

• Horseshoe centred on C1;

• + in intermediate not too close to C1 (allow on or “below” a linefrom C2 to C6)

M3 for Arrow from bond to H into ring

• Allow M3 arrow independent of M2 structure

• + on H in intermediate loses M2 not M3

• Ignore Cl- removing H +

111

10


(b) Reagent

Water

(Aqueous) silver nitrate

NaOH followed by acidified silver nitrate

(Water +) named indicator

Named alcohol

Na2CO3 or NaHCO3

Ammonia1

P

No reaction

No reaction (or slow formation of ppt)

No reaction (or slow formation of ppt)

No colour change

NVC

NVC

No reaction

Do NOT award

No observation1

Q

Steamy /misty/ white fumes

White precipitate (immediately formed)

White precipitate (immediately formed)

Indicator turns to correct acid colour

Fruity or sweet smell or misty fumes

Fizzing or effervescence (not just gas produced)

White smoke1


(c) (i)

One unit only

Must have trailing bonds

Ignore n and brackets

allow

1

(i)

Allow CO for C=O1

(d) (i)

One unit only

Must have trailing bonds

Ignore n and brackets1

(ii) PGA sutures react/dissolve/break down/are biodegradable/are hydrolysed / attacked by water or nucleophiles /no need toremove

OR Polypropene not biodegradeable/ not hydrolysed / not attackedby water/nucleophiles

1

(Ester links have) polar bonds

polypropene contains non-polar bonds

ignore intermolecular forces1

[12]

(a) Electrophilic substitution

Both words needed

Ignore minor misspellings1

11


(b) (i) Sn / HClOR H2 / Ni OR H2 / Pt OR Fe / HCl OR Zn / HCl OR SnCl2 / HCl

Ignore conc or dil with HCl,

Allow (dil) H2SO4 but not conc H2SO4

Not allow HNO3 or H+

Ignore NaOH after Sn / HCl

Ignore catalyst1

(ii) CH3C6H4NO2 + 6[H] → CH3C6H4NH2 + 2H2O

OR

Allow molecular formulae as structures given

C7H7NO2 + 6[H] → C7H9N + 2H2O

Qu states use [H], so penalised 3H21

(iii) making dyes

OR making quaternary ammonium salts

OR making (cationic) surfactants

OR making hair conditioner

OR making fabric softener

OR making detergents1


(c)

M3

NO Mark for name of mechanism

Allow SN1

M1 for lone pair on N and arrow to C or mid point of space betweenN and C

M2 for arrow from bond to Br

M3 for structure of protonated secondary amine

M4 for arrow from bond to N or + on N

For M4: ignore RNH2 or NH3 removing H+ but penalise Br−

4

(d) lone or electron pair on N

If no mention of lone pair CE = 0

If lone pair mentioned but not on N then lose M1 and mark on

M11

in J spread / delocalised into ring (or not delocalised in K)

Ignore negative inductive effect of benzene

Allow interacts with Π cloud for M2

M21

less available (for protonation or donation in J)

M3

OR

in K there is a positive inductive effect / electron releasing)

M2

more available (for protonation or donation in K)

M31

[11]


(a) Hydrogen bond(ing)

Allow H bonding.

Penalise mention of any other type of bond.1

12

(b) (i) Ammonia is a nucleophile

Allow ammonia has a lone pair.1

Benzene repels nucleophiles

Allow (benzene) attracts / reacts with electrophiles.

OR benzene repels electron rich species or lone pairs.

OR C–Cl bond is short / strong / weakly polar.1

(ii) H2 / Ni OR H2 / Pt OR Sn / HCl OR Fe / HCl

Ignore dil / conc of HCl.

Ignore the term ‘catalyst’.

Allow H2SO4 with Sn and Fe but not conc.

Ignore NaOH following correct answer.

Not NaBH4 nor LiAlH4.1

(iii) conc HNO3

conc H2SO41

If either or both conc missed can score 1 for both acids.1

HNO3 + 2H2SO4 NO2+ + H3O+ + 2HSO4

−

OR using two equations

HNO3 + H2SO4 H2NO3+ + HSO4

−

H2NO3+ H2O + NO2

+

Allow 1:1 equation.

HNO3 + H2SO4 NO2+ + H2O + HSO4

−.1

(iv) Electrophilic substitution1


OR

• Ignore position or absence of Cl in M1 but must be in correct position for M2.

• M1 arrow from within hexagon to N or + on N.

• Allow NO 2+ in mechanism.

• Bond to NO 2 must be to N for structure mark M2.

• Gap in horseshoe must be centered around correct carbon (C1).

• + in intermediate not too close to C1 (allow on or “below” a line from C2 to C6).

• M3 arrow into hexagon unless Kekule.

• Allow M3 arrow independent of M2 structure.

• Ignore base removing H in M3.

• + on H in intermediate loses M2 not M3.3

[11]

(a)

OR

1

13

NOT penalise missing brackets.

If wrong ester, no further mark.


Must be displayed formula

Radical dot must be on OIgnore lone pair(s) on O in addition to single electron

Allow radical with brackets as

Ignore errors in acylium ion.1

(b) (i) AlCl3 or FeCl3If wrong no further marks.

1

Correct equation scores 2 - contrast with (b)(iii)Allow + on C or O in equation.

1

(ii) Electrophilic substitution

Ignore Friedel crafts.1


OR

• + must be on C of RCO here

• M1 arrow from within hexagon to C or to + on C

• Gap in horseshoe must approximately be centred around C1 and not extend towards C1 beyond C2 and C6

• + not too close to C1

• M3 arrow into hexagon unless Kekule

• allow M3 arrow independent of M2 structure, i.e. + on H in intermediate loses M2 not M3

• ignore base removing H for M33

(iii) (CH3CO)2O + C6H6 C6H5COCH3 + CH3COOH

OR

Correct equation scores 1 – contrast with (b)(i)

Not allow molecular formula for ethanoic anhydride or ethanoicacid.

1

[9]


(a) CH3COCl + AlCl3 CH3CO+ + AlCl4–

Allow RHS as

Allow + on C or O in equation but + must be on C in mechanismbelow

Ignore curly arrows in equation even if wrong.1

14

AlCl4– + H+ AlCl3 + HCl1

• M1 arrow from within hexagon

to C or to + on C

• + must be on C of RCO in mechanism

• + in intermediate not too close to C1

• gap in horseshoe must be centred approximately around C1


• allow M3 arrow independent of M2 structure

• ignore base removing H for M3

• NO mark for name of mechanism3

Phenylethanone ignore 1 in name, penalise other numbers

Note: this is the sixth marking point in (a)1


(b)

+ must be on C

But allow [C6H5CO]+

1

(c) M1 about electrons

methyl group has (positive) inductive effect OR increases electron density onbenzene ring OR pushes electrons OR is electron releasing

Ignore reference to delocalisation1

M2 about attraction

electrophile attracted more

or benzene ring better nucleophile

Allow intermediate ion stabilised

M2 only awarded after correct or close M11

[9]

(a) (i) Conc HNO3

If either or both conc missing, allow one;1

15

Conc H2SO4

this one mark can be gained in equation`1

2 H2SO4 + HNO3 2 HSO4– + NO2

+ + H3O+

1


OR H2SO4 + HNO3 HSO4– + NO2

+ + H2O

Allow + anywhere on NO2+

OR via two equations

H2SO4 + HNO3 HSO4– + H2NO3

+

H2NO3+ NO2

+ + H2O

(ii)

• ignore position or absence of methyl group in M1 but must be in correct position for M2

• M1 arrow from within hexagon to N or + on N

• Allow NO 2+ in mechanism

• Bond to NO 2 must be to N

• horseshoe must not extend beyond C2 to C6 but can be smaller

• + not too close to C1


• allow M3 arrow independent of M2 structure

• ignore base removing H in M3

• + on H in intermediate loses M2 not M33

(b) 51

(c) 21

(d) 2C7H5N3O6 5H2O + 3N2 + 7C + 7CO

Or halved1

[9]


(a) (i) C6H6 + CH3CH2COCl → C6H5COCH2CH3 + HClORC6H6 + CH3CH2CO+ → C6H5COCH2CH3 + H+

allow C2H5

penalise C6H5–CH3CH2COallow + on C or O in equation

1

Phenylpropanone

OR ethylphenylketone OR phenylethylketone

Ignore 1 in formula, but penalise other numbers1

16

AlCl3can score in equation

1

CH3CH2COCl + AlCl3 → CH3CH2CO+ + AlCl4–

allow C2H5

allow + on C or O in equation1

AlCl4– + H+ → AlCl3 + HCl1


(ii) electrophilic substitution

can allow in (a)(i) if no contradiction1

M1 arrow from circle or within it to C or to + on Chorseshoe must not extend beyond C2 to C6 but can be smaller+ not too close to C1

M2 penalise C6H5–CH3CH2CO (even if already penalized in (a)(i))



ignore base removing H in M33

(b) (i) CH3CH2CHO + HCN → CH3CH2CH(OH)CN OR C2H5CH(OH)CN

aldehyde must be –CHO brackets optional1

2-hydroxybutanenitrile OR 2-hydroxybutanonitrile

no others1


(ii) nucleophilic addition1

M1 includes lp and arrow to Carbonyl C and minus charge (oneither C or N)Not allow M2 before M1, but allow M1 to C+ after non-scoringcarbonyl arrowIgnore δ+, δ– on carbonyl group, but if wrong way round or full +charge on C lose M2

M3 for correct structure including minus sign. Allow C2H5

M4 for lp and curly arrow to H+

4

(iii) (propanone) slower OR propanal faster1

inductive effects of alkyl groupsORC of C=O less δ+ in propanoneORalkyl groups in ketone hinder attackOReasier to attack at end of chain

if wrong, no further marks1

[18]

(a) CH3CH2COCl OR CH3CH2CClO OR propanoyl chlorideOR (CH3CH2CO)2O OR propanoic anhydridepenalize contradiction in formula and name e.g. propyl chloride

could score in equation1

AlCl3 or FeCl3 or names

could score in equation1

CH3CH2COCl + AlCl3 → CH3CH2CO+ + AlCl4–

Allow RCOCl in equation but penalise above

allow + on C or O in equation1

17


(b)

M1 arrow from circle or within it to C or to + on C

Horseshoe must not extend beyond C2 to C6 but can be smaller +not too close to C1



Ignore base removing H in M33

(c) Tollens or ammoniacal silver nitrate1

penalise wrong formula1

[8]


MarkRange

The marking scheme for this part of the question includes an overallassessment for the Quality of Written Communication (QWC). Thereare no discrete marks for the assessment of QWC but thecandidates’ QWC in this answer will be one of the criteria used toassign a level and award the marks for this part of the question

Descriptoran answer will be expected to meet most of the criteria in the level

descriptor

4-5 – claims supported by an appropriate range of evidence

– good use of information or ideas about chemistry, going beyondthose given in the question

– argument well structured with minimal repetition or irrelevantpoints

– accurate and clear expression of ideas with only minor errors ofgrammar, punctuation and spelling

2-3 – claims partially supported by evidence

– good use of information or ideas about chemistry given in thequestion but limited beyond this

– the argument shows some attempt at structure

– the ideas are expressed with reasonable clarity but with a fewerrors of grammar, punctuation and spelling

0-1 – valid points but not clearly linked to an argument structure

– limited use of information or ideas about chemistry

– unstructured

– errors in spelling, punctuation and grammar or lack of fluency

18

(a) (i) Mr of C6H5NH2 = 93 Mr of CH3COCl = 78.5total Mr of reagents = 264.5

1

% atom economy = 1

= 1

× 100 QWC

× 100 = 51.0 %


(ii) expected yield = 1

% yield = 1

× 0.5 × 135 = 7.26 kg

× 100 = 74.1 %

(iii) Although yield appears satisfactory (74%) % atom economyis only 51% QWC

1

nearly half of the material produced is waste and must bedisposed of QWC

1

(b) (nucleophilic) addition-elimination1

QWC (2)4

(c) HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4

–

1

3

[16]

(a) CH3COCl + AlCl3 → CH3 O + AlCl

2

penalise wrong alkyl group once at first errorposition of + on electrophile can be on O or C or outside [ ]penalise wrong curly arrow in the equation or lone pair on AlCl3 else ignore

19

(1) equation (1)


Electrophilic substitution

NOT F/C acylation1

horseshoe must not extend beyond C2 to C6 but can be smaller

+ not too close to C1



M1 arrow from within hexagon to C or to + on C

+ must be on C of 3

(b) Nucleophilic addition

NOT reduction1

M2 not allowed independent, but can allow M1 for attack of H– onC+ formed

4

1–phenylethan(–1–)ol or (1–hydroxyethyl)benzenel

(c) dehydration or elimination1

(conc) H2SO4 or (conc) H3PO4

allow dilute and Al2O3

Do not allow iron oxides1

[14]

C

[1]20


(a) (nucleophilic) addition-elimination;

(M3 for structure)(M4 for 3 arrows and lone pair)(M2 not allowed independent of M1, but allow M1 for correct attackon C+ if M2 show as independent first.)(+on C of C=O loses M2 but ignore δ+ if correct)

(Cl– removing Ft loses M4)1

21

(If MS lost above for wrong C chain, do not penalise same erroragain here)

5

(b) CH3CH2COCl + AlCl3 → [CH3CH2CO]+ + AlCl4–;

(penalise wrong alkyl group once at first error)

(position of + on electrophile can be on O or C or outside [ ])(penalise wrong curly arrow in the equation or lone pair on AlCl3)

1

(M1 arrow from within hexagon to C orto + on C)

(don’t penalise position of + on C ofRCO+)

(horseshoe must not extendbeyond C2 to C6 but can besmaller)

(+ not too close to C1)

(penalise M2 if CH3 chain wrongagain but allow M1 and M3)

(M3 arrow into hexagon unlessKekule)

(allow M3 arrow independent ofM2 structure)

3

(or can be gained in mechanism);1


(c) M1 CH3CH2COCl + H2O → CH3CH2COOH + HCl 1(penalise wrong alkyl group once at first error)

1

M2 Mr of CH3CH2COCl = 92.5 1(if Mr wrong, penalise M2 only)

1

M3 moles of CH3CH2COCl = 1.48/92.5 = 0.016 11

M4 moles NaOH = 2 × 0.016 = 0.032 1(allow for × 2 conseq to wrong no of moles)

1

M5 volume of NaOH = 0.032/0.42 = 0.0762 dm3 or 76.2 cm3 1(with correct units)(if ×2 missed in M4 lose M5 also)

1

[16]

B

[1]22

(a) (i) conc HNO3

1

conc H2SO4

allow 1 for both acids if either conc missing1

HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4

–

or HNO3 + H2SO4 → NO2+ + H2O + HSO4

–

1

23

(iii) electrophilic substitution CH31

horseshoe must not extend beyond C2 to C6 but can be smaller+ must not be too close to Cl

3

(b) Sn or Fe / HCl (conc or dil or neither)or Ni / H2 not NaBH4 LiAlH4

1


(c) (i) NH3

1

Use an excess of ammonia1

(ii) nucleophilic substitution1

4

[15]

D

[1]24

(a) Mechanism

Allow C3H7 if structure shown elsewhere

penalise HCN splitting if wrong

Name of product: 2-hydroxypenta(neo)nitrile (1)

or 1-cyanobutan-1-ol5

25


(b) Mechanism

Name of organic product: methylpropanoate (1)5

(c) (i) ([) CH3CO (])+ (1)

(ii)

4

Notes

(abc) extra curly arrows are penalised

(a) be lenient on position of negative sign on : CN– but arrow must come from lp

(a)/(b) alone loses M2 but can score M1 for attack on C+, similarly

(a) allow 2-hydroxypentanonitrile or 2-hydroxypenta(ne)nitrile ... pentylnitrile

(b) in M4, allow extra: Cl– attack on H, showing loss of H+

(c) (i) allow formula in an “equation”(balanced or not)be lenient on the position of the + on the formula

(ii) for M1 the arrow must go to the C or the + on the Cdon’t be too harsh about the horseshoe, but + must not be close to the saturated CM3 must be final step not earlier; allow M3 even if structure (M2) is wrong

[14]


Organic points

(1) Curly arrows: must show movement of a pair of electrons,i.e. from bond to atom or from lp to atom / spacee.g.

(2) Structures

penalise sticks (i.e.

Penalise once per paper

allow CH3– or –CH3 or 3

or H3C–

) once per paper

or CH

(a)

Further reaction / substitution / formation of 2° / 3° amines etc (1)use an excess of NH3 (1)

6

26


5

(b) 3) (1) repels nucleophiles (such as NH

Notes

(a) allow SN1

penalise: Br– intead of NH3 removing H+ for M4not contamination with other amines (this is in the question) not diamines

(b) allow because NH3 is a nuclephile or benzene is (only) attacked by electrophilesor C–Br bond (in bromobenzene) is stronger / less polar or Br lp delocalized

HNO3 / H2SO4 without either conc scores (1) allow 20 – 60° for (1) (any 2 ex 3)

allow name or structure of nitrobenzene

other reducing agents: Fe or Sn with HCl (conc or dil or neither) not conc H2SO4 or conc HNO3

allow Ni/H2

Not NaBH4 or LiAlH4

ignore wrong descriptions for reduction step e.g. hydrolysis or hydration

[11]

Organic points

(1) Curly arrows: must show movement of a pair of electrons,i.e. from bond to atom or from lp to atom / spacee.g.


(2) Structures

penalise sticks (i.e.

Penalise once per paper

allow CH3– or –CH3 or 3

or H3C–

) once per paper

or CH

(a) High Ea: S2O82– repels I– or both ions negative (1)

2Fe2+ + S2O82– → 2Fe3+ + 2SO4

2– (1)2Fe3+ + 2I– → 2Fe2+ + I2 (1)

N.B. Ignore additional incorrect equations

Vanadium is a transition element or Magnesium is not a transition element (1)

Vanadium has variable oxidation states (1)

Magnesium only forms Mg2+, or has only one oxidation state (1)

N.B. Score two marks for “Only vanadium has variable oxidationstates”

6

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(b) AlCl3 + Cl-COCH3 → AlCl4– + CH3CO+ (1)

H+ + AlCl4– → AlCl3 + HCl (1)Lewis acid: AlCl3 accepts electron pair

N.B. penalise incorrect acyl chloride by oneN.B. penalise chloroethane by two marks i.e. first equation mark,attack on benzene mark


NH4Cl: Not a catalyst (1)

FeCl3: A catalyst (1)has a low energy vacant shellor has spaces or vacancies in d shellor has a partially filled d shellor able to accept an electron pairor can form FeCl4– (1)

9

[15]

B

[1]28


Examiner reports

The straightforward question in part (a)(i) was well answered, although some students did notgain the mark as a result of failing to use all three pieces of data or of writing incorrect mathssuch as –360 – (–208) = +152. As expected part (a)(ii) was high scoring but part (b) wasanswered less well. In part (c) the data was often referred to as hydration and also as energyrequired to break the bonds rather than energy released as cyclohexane was formed.The value for 1,4-cyclohexadiene in part (c)(i) was usually correct, but the possibility of partialdelocalisation in the 1,3 diene in part (c)(ii) was not at all well understood. The predicted enthalpyof hydrogenation of the 1,3-diene was often given as the same as the 1,4-diene. In part (c)(iii) amark for noting that the C=C bonds were closer in the 1,3-diene was often gained. Howevermany students then went on to say in error that adjacent C=C bonds repel and so the 1,3-dieneis less stable. It was rare to see a student apply their knowledge of delocalisation in benzene tothe 1,3-diene, to deduce that partial delocalisation can occur here and, as with benzene, thiscauses some extra stability and so the enthalpy of hydrogenation is less exothermic than -240

kJmol₋₁.

2

In part (a), many students thought NaBH4 or LiAlH4 could be used as the reducing agent. Mostdid not give the molecular formulae in the equation but could balance it correctly; the repeatingunit of the polymer was very well done.

Similarly, in part (b), NaBH4 was often suggested as the reducing agent and the empiricalformula was frequently wrong. Students found the bond angles difficult, with many failing todeduce that there is a tetrahedral arrangement around a C atom in cyclohexane and appearingsimply to guess any angle. Only the most able scored both marks for the bond angles.

There was a good spread of marks in the mechanism with over 40% gaining full marks.In part (c)(ii), most realised that the product was a racemic mixture and that it was produced as aresult of attack on the ketone from both sides, but often this was thought to be by an electrophile

or an H + ion. Some thought that a carbocation intermediate was attacked.

3


(a) This was well known; most students gave an equation for a 1:1 mole ratio, although a 1:2ratio and a two-step process were also seen and credited.

(b) This was also well known and 35% of students scored full marks. In the mechanism, somestudents incorrectly showed a lone pair on the electrophile, or the horseshoe wassometimes drawn in an incorrect orientation relative to C1. Some students started frombenzene and formed nitrobenzene first.

(c) Only the very weakest 7% of students failed to calculate the correct Rf value.

(d) This question tested understanding of the balance between solubility in the moving phaseand retention by the stationary phase. Both parts were needed in the answer.

(e) This question was found very easy by most, although weaker students discussed safetyissues.

(f) A third of students scored both marks, but sadly 7% of students made no attempt at thequestion. They were asked to deduce the answer to an unfamiliar practical situation, butone which is similar to those they will have met in Required Practical Activity 12. Manydeduced that the 1,4 isomer was less polar, but then attempted to explain this relativepolarity rather than answering the question asked.

(g) This final part to the question was not attempted by 10.2% of students. The best answersdemonstrated the ability of candidates to deduce that the student was correct, and toexplain their deduction.

4

In part (a), although most candidates scored at least half marks, some candidates did not use thedata as requested. Some answers also referred to cyclohexatriene but went on to talk about onedouble bond.

Part (b) was well done, especially the mechanism. The most common errors were to omitconcentrated for either or both acids or to omit the positive charge on the H2NO3 + when writingtwo equations.

The synthesis in part (c) was a stretch and challenge question and was, as expected, verydiscriminating. There were some excellent answers but also many which failed to score.

7

Many candidates were unable to identify V and those who did often named it as a halobenzeneeven if they drew a correct structure. Preparation of the amine by nucleophilic substitution usingan excess of ammonia was credited in Reaction 4 even if the answers to V and Reaction 3 werewrong.

A surprising number of candidates thought a base was a proton donor. Some otherwise goodanswers did not score as they failed to mention the lone pair on the nitrogen atom.

Many also simply stated that amine U contained delocalised electrons but failed to explain howinvolvement of the lone pair on nitrogen with these electrons lowered the base strength.


In part (a), not all candidates noted the requirement in the question to “Use the data”. Thenumerical difference of 152 kJ mol-1 between the expected enthalpy of hydrogenation of -360 kJmol-1 and the actual value of –208 kJ mol-1 was rarely noted. Weaker candidates also thoughtthat the data given was of the energy required to cause the reaction, rather than that evolved inthe reaction. In part (b), weaker candidates muddled the expected addition-elimination reaction ofan acid chloride and an amine with a Friedel Craft’s substitution in the benzene ring andfrequently gained no marks for the section. The correct name for the product,N-phenylethanamide was rarely seen. Part (c) was better answered; most candidates were ableto discuss nitration correctly and the mechanism was better answered than that in part (b)(ii). Theamide or peptide link was not always named correctly in part (iv) and many did not state thatacidic or alkaline hydrolysis is required to break the bond.

8

Apart from the occasional suggestion of AlCl4- as the electrophile, part (a)(i) was answered wellas was the mechanism in (a)(ii). Marks are still lost by some students who draw curly arrows thatlack the precise start and end positions required to gain the mark. About a third of studentsscored full marks in part (b). The vast majority of incorrect answers involved oxidation: usingTollens or Fehling’s or acidified potassium dichromate solution. Students should be reminded thatwhen asked to give an observation, those who write the word “nothing” or “no observation” willnot gain a mark.

Part (c)(i) was answered well although a few students drew more than one repeating unit. Part(c)(ii) was also answered well. About a third of students did not answer part (d)(i) correctly. Part(d)(ii) proved a good differentiator with many gaining a mark for noting that PGA isbiodegradable, but only the best students were able to link this to the presence of polar bonds inPGA. Some students included the term polar but applied it to the whole molecule and not aspecific bond.

10

In part (a) a mechanistic name was required not just nitration. In part (b)(i) many wrong reagentswere given such as NaBH₄, and incomplete reagents such as HCl. Many student gave correctanswers in part (b)(ii) but some equations were wrongly balanced and some included oxygen asa product.

Part (b)(iii) was answered correctly by few students. A common error was to miss out the‘making’ of the product, J is used to make dyes, it is not itself used as a dye. Wrong productswere also frequently given such as explosives, polymers or soap.

In part (c), the nucleophilic substitution mechanism, also in the AS specification, was answeredfairly well. Common errors were to put a negative charge on the amine (this lost M1), to usebromoethane instead of bromomethane, (this lost M3) and to show the arrow in M4 goingtowards H. Students who drew the complete structure for J rather than RNH₂ were not penalisedexcept in the extra time they spent in drawing it.

There were some good answers to part (d). The majority mentioned a lone pair, but did notalways state that it was on the nitrogen atom. The mark scheme allowed many to score M2 whenreferring to J and then M3 when relating to K. Some students used the word ‘dissociate’ insteadof ‘delocalise’.

11


In part (b)(i), only the highest-scoring students gave the simplest correct answer in terms of thebenzene repelling the nucleophile ammonia but alternative answers were allowed, including acomparison of the strength or polarity of the C–Cl bond in chlorobenzene with that in ahaloalkane.

There was also some confusion between identifying regions of high electron density such as lonepairs and delocalised electrons, and referring to benzene and ammonia as being negativelycharged. Some students also wrongly discussed further substitution. Part (b)(ii) was wellanswered. However, reagents were needed in full, so Sn with HCl was required not just H+. Parts(b)(iii) and (iv) were also well answered. In part (b)(iii), ‘concentrated’ was needed with both acidsto gain two marks, and M3 was often lost for the charge missing on H2NO3

+. The electrophilicsubstitution mechanism was generally well done although the positioning of the horseshoerepresenting delocalised electrons caused some problems in this example where the substitutiondid not occur at the carbon at the “top” of the benzene ring.

12

In part (a), the molecular ion was often shown incorrectly without the plus and dot. Many also didnot draw a displayed formula for the radical.

Part (b) was answered well, and although nearly half of the students gained full marks in themechanism in part (b)(ii), several lost a mark by drawing the + on the Wheland intermediate tooclose to carbon number 1. A significant number also gave Friedel-Crafts acylation as the name ofthe mechanism rather than electrophilic substitution. In part (b)(iii), the formula of the acidanhydride proved to be challenging to a considerable number.

13

In part (a), the mechanism was generally well answered, but the equation to show theproduction of aluminium chloride and hydrogen chloride was often missing. Very few studentswere able to name the product correctly in part (b).

Part (c) was well done but in part (d) some students appeared to think that in methylbenzene themethyl group would be substituted rather than one of the hydrogen atoms. However, the betterstudents were able to use their understanding of the inductive effect of alkyl groups to suggestthat the electron density on the ring in methyl benzene would be increased and go on to suggesthow this would increase the rate of reaction.

14

About half of the students scored all three marks in part (a)(i); the main errors were not statingthat both acids needed to be concentrated and also failing to balance the equation. Themechanism in part (a)(ii) was also very well done, although some students attempted to producethe trinitro– compound and often the intermediate shown had the nitro group attached to thewrong carbon atom. In part (d), many students assumed that this was a combustion reactionrather than the decomposition stated in the question, and, disappointingly, 6N often appearedrather than 3N2 in an otherwise correct equation.

15


As is usual, mechanism questions such as this discriminate well. In part (a)(i), 13% of thecandidates gained all five marks whereas a similar number scored only one mark. Manycandidates omitted the name of the product in an otherwise perfect answer. In part (a)(ii), twofifths of the candidates gained full marks.

In part (b)(i), a number of candidates lost a mark because they gave the formula of the aldehydepropanal as CH3CH2COH rather than as CH3CH2CHO. For clarification of acceptable formulaestyles, candidates and teachers are advised to consult the document “General principles appliedto marking CHEM4 papers by CMI+ January 2011” which is available with the Mark Scheme forthis paper.

The mechanism in part (b)(ii) was well known and two fifths of candidates gained full marks.

Part (b)(iii) was a How Science Works question where candidates were not required to know theanswer, but were expected to apply understanding gained from other areas of the specification.The inductive effect of alkyl groups is part of the explanation of the relative stability ofcarbocations. A similar effect using two alkyl groups will reduce the δ+ nature of the carbonylcarbon in propanone more than the single alkyl group does for the carbonyl carbon in propanal.Hence it can be predicted that the nucleophile will attack propanal more easily than propanone.About a third of the candidates gained both marks; most of these gave an answer based oninductive effects as above, but others correctly discussed steric effects and this answer was alsoaccepted.

16

This question was generally well done; part (a) and the structure of the aldehyde in part (c) werethe most challenging sections. In part (c), a number of candidates lost the mark for the reagentby giving an incorrect formula for the diamminesilver(I) ion even though the name Tollens¡¦reagent was also correctly given.

17

Part (a) was well answered and many candidates scored highly. The mechanism in part (b)

proved more difficult and some candidates tried to use BH4– instead of H– as expected. The

name of the product proved very difficult for most candidates and many made no attempt toanswer the question. In part (c), many confused Reaction 3, which is a dehydration or anelimination of water, with the removal of hydrogen from ethylbenzene using iron oxides.

19

The better candidates often gained full marks in part (a). The most common errors were similarto those described in Question 3(a) above. In part (b), despite the request to write equations toshow the role of aluminium chloride, many drew a mechanism often containing incorrect curlyarrows and were penalised. The electrophilic substitution mechanism was generally wellanswered although careless drawing of the electrophile with wrong position of the + charge oftenled to incorrectly linked COCH2CH3 side chains, with CH3 or O joined to benzene ring. Manyfailed to write an equation for the regeneration of the A1C13 catalyst. In part (c), few realised thathydrolysis of propanoyl chloride forms both propanoic and hydrochloric acids, both of which reactwith sodium hydroxide. Hence the most common answer was half the correct value. Adisappointing number also failed to calculate the correct Mr of propanoyl chloride, the formula forwhich was given at the start of the question.

21

Part (a) was answered well by most candidates, but many did not note that both acids used innitration are concentrated. Weaker candidates were unable to name the mechanism correctlyand several lost marks by drawing the curly arrow going to the oxygen rather than the nitrogen of

, by drawing the intermediate structure carelessly with the + charge shown too close to the

tetrahedral carbon or by drawing the curly arrow showing the loss of H+ the wrong way round.

23


Surprisingly, part (b) was badly answered: many incorrectly suggested NaBH4 and in part (c) theneed for an excess of ammonia was not well known. The first stage in the mechanism wasanswered well, but many were unable to show loss of H+ correctly from the protonated amine.

In organic mechanisms, a curly arrow must start at a lone pair or at a bond. This idea was stillforgotten by some candidates. In part(a), several used a lone pair on the nitrogen atom ratherthan one on the carbon of the cyanide ion. A few confused the nucleophilic addition in part (a)with the addition-elimination in part (b) and tried in part (a) to reform a carbon to oxygen doublebond. An appreciable number thought that the name of the product was a propanenitrile insteadof 2-hydroxypentanenitrile. In both parts (a) and (b) the C=O bond should not be shown breakingbefore the nucleophile attacks. The ester was often correctly named as methyl propanoate butmany other suggestions were also seen. Part (c) discriminated well. Only the better candidateswere able to identify the correct electrophile, join it correctly to the benzene ring, draw the correctintermediate and show the loss of H+ correctly.

25

There was evidence that some candidates had spent too long attempting to solve the problemsin Question 6 and had left themselves short of time for the descriptive aspects of this question.The standard of the mechanisms given in this question was similar to that in Question 4 andalthough many realised that further reaction could occur leading to ‘other amines’ many could notexplain that this process began with the reaction of the first product, ethylamine, withbromoethane to form a secondary amine. To reduce this problem an excess of ammonia is used;many suggested that an excess of bromoethane was needed. In part (b) the five marks could beearned without giving detail in the answers. Several candidates unnecessarily wrote mechanismsand equations. The idea that the electron rich benzene ring repels nucleophiles was notexpressed clearly by many. The two step synthesis of phenylamine via nitrobenzene wasgenerally well known.

26

The reaction between persulphate ions and iodide ions was well known and almost allcandidates stated correctly that the reaction has a high activation energy as these ions are bothnegative and repel each other. Good candidates gave correct equations explaining the catalyticaction of iron(II) ions but a significant number of weaker candidates lost marks when they failedto balance these equations.Many candidates completed their answer to part (a) by predicting correctly that the variableoxidation states of the transition element vanadium would enable V3+(aq) to function as a catalystfor the reaction whilst Mg2+(aq) would not. The role of AlCl3 in the reaction between benzene andethanoyl chloride was well known by many candidates. Some, however, lost marks when theyfailed to show the involvement of AlCl3 in the formation of the acylium intermediate and thereformation of AlCl3 at the end of the reaction. Marks were lost by candidates who drewinaccurate structures in the organic reaction mechanism. The link between the ability of AlCl3 tobehave as a Lewis acid and accept a lone electron pair was made by good students who thenpredicted correctly that FeCl3 would function as a catalyst but NH4Cl would not. Marks in thissection were not awarded to candidates who made vague predictions.

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(total 1 mark) - the maths and science tutor

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