1 On heating an aqueous solution of 2,4,6-trinitrobenzoic acid, the following reaction occurs.

(a) The rate of reaction can be studied by measuring the change in concentration of acid over time. Describe briefly how this can be done.

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(b) Using the following data, obtained at 363 K, plot a suitable graph

(i) To establish the order of the reaction

(ii) To determine the value of the rate constant at 363 K.

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Class Reg Number

Candidate Name .......................................................................

Chemistry H2 9746Tutor TuteeRevision Exercise 18: Integrated Questions

(c) Suggest a method of following the progress of the reaction which does not involve measuring the acid concentration and state one advantage of the method you suggest.

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Revision Questions in ‘A’ Level Chemistry by EW Jenkins 2 In the Wurtz reaction, iodoalkanes react with sodium to form alkanes. The reaction is used

to make alkanes which have an even number of carbon atoms. For example, iodomethane and sodium react to produce ethane as shown below.

CH3–I + 2Na + I–CH3 CH3–CH3 + 2NaI

(a) Name the iodoalkane required to make octane.

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(b) Draw the structural formula of 2-iodopropane and the product when it undergoes the Wurtz reaction.

(c) Alkanes such as pentane, which have odd numbers of carbon atoms per molecule, can be produced by the Wurtz reaction.

C5H7I + 2Na + C2H5I C5H12 + 2NaI

However, two other alkanes are formed in this reaction. Name these alkanes.

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(d) It is known that the Wurtz reaction is a 2-step process, where the first step involves the generation of R– when R–I reacts with Na, and the second step involves nucleophilic substitution. Hence, propose a mechanism for the Wurtz reaction.

Revision Notes and Questions for New Higher Chemistry

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3 For people who suffer from bronchitis, even low concentrations of ozone, O3, irritate the lining of the throat and can cause headaches. NO2 gas from car exhausts reacts with hydrogen to form ozone as follows.

O2(g) + NO2(g) NO(g) + O3(g)

Car exhaust fumes also contain volatile organic compounds (VOCs), which can combine with NO2 gas.

(a) Explain how a rise in VOC concentration will change ozone concentration.

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(b) In an experiment to measure the ozone concentration of air in a Scottish city, 105 litres of air were bubbled through a solution of potassium iodide. Ozone reacts with potassium iodide, releasing iodine.

2KI(aq) + O3(g) + H2O(l) I2(aq) + O2(g) + 2KOH(aq)

The iodine formed was titrated with 0.01 mol dm-3 sodium thiosulfate solution, Na2S2O3(aq), using starch indicator.

The results of three titrations are shown in the table below.

Experiment Volume of thiosulfate / cm3

1 22.90

2 22.40

3 22.50

(i) What colour change would show that the titration was complete?

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(ii) Explain why the volume of sodium thiosulfate to be used in the calculation was taken to be 22.45 cm3 although this is not the average of the three titres in the table?

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(iii) Taking the volume of sodium thiosulfate solution to be 22.45 cm3, calculate the volume of ozone in one litre of air at room temperature and pressure.

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Revision Notes and Questions for New Higher Chemistry4 Perfumes normally contain three groups of components called the top notes, the middle

note and the end note.

(a) The top note components of a perfume form vapours most easily. Two compounds found in top note components are:

Describe a chemical test to distinguish between the two compounds, stating the observations.

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(b) The middle note compounds form vapours less readily than the top note compounds. A typical compound of the middle note is:

2-phenylethanol

Compare the relative melting points of 2-phenylethanol and p-cresyl acetate. Explain your answer.

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(c) The end note of a perfume has a long lasting odour which stays with the user. An example of an end note compound is:

Civetone can undergo the following reaction.

(i) State the reagents and conditions for Step I and Step III.

Step I ............................................................................................

Step III ............................................................................................

(ii) Hence, deduce the structures of X and Y.

New Higher Chemistry (modified)5(a) An optically active compound F, C9H12O, reacts with acidified K2Cr2O7 to give G,

C9H10O. F reacts with iodine in aqueous sodium hydroxide to give a yellow precipitate and the sodium salt of the acid H, C8H8O2. H is oxidised by acidified KMnO4 to form I, C8H6O4.

Suggest the structures of F, G, H and I, giving relevant equations to explain your reasoning.

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(b) J, C9H11I, on boiling with aqueous potassium hydroxide gave a compound K, C9H12O. With acidified sodium dichromate(VI), compound K yielded L, C9H10O. With hot acidified potassium manganate(VII), compound K yielded M, C7H6O2. K when warmed with potassium hydroxide and iodine is able to produce yellow precipitate.

Suggest the structures of J, K, L and M, giving relevant equations to explain your reasoning.

Excel in Chemistry blog

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6 Soda-lime glass is made by heating a mixture of calcium carbonate, sodium carbonate and sand in a furnace to a high temperature. Other glasses contain compounds called silicates. The simplified structures of a silicate and sodalime glass are shown.

(a) Describe two differences between the silicate and the soda-lime glass.

any 2 of: silicate has regular arrangement of atoms and soda-lime glass has irregular

arrangement; ALLOW: e.g. soda lime glass has a less regular arrangement of atoms silicate has no ions/named ion(s)/all atoms (covalently) bonded and soda lime

glass has calcium/sodium ions; [ALLOW: has oxygen ions] all the oxygen atoms are (covalently) bonded to two silicon atoms in silicate

but in soda lime some are only bonded by one (covalent) bond; silicate has larger spaces/an

open structure and soda-lime glass has a more compact structure/collapsed structure

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(b) When soda-lime glass is melted, it conducts electricity. Use the information in the diagram to explain this fact.

In melted soda-lime glass, the giant ionic lattice structure is broken down and Ca2+/Na+ ions can move about to conduct electricity under an applied potential difference.ALLOW: ions can move/ions are free to moveNOT: ions are delocalised/ions are free.................................................................................................................................................

(c) Calcium carbonate decomposes in the furnace. Compare the decomposition temperature of calcium carbonate and magnesium carbonate.

Decomposition temperature of CaCO3 is greater than that of MgCO3. Since Ca2+ has a lower charge density than Mg2+ due to higher ionic radius, its polarising power is

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lower and hence is less able to distort the large anionic charge cloud of CO32- to

decompose into CaO and CO2. Hence, a greater amount of energy is needed to decompose CaCO3 to CaO and CO2 than that of MgCO3 to MgO and CO2..................................................................................................................................................

(d) Some types of glass contain lead ions, Pb2+. Dishwasher powders are highly alkaline.

(i) When glasses containing lead ions are washed repeatedly in a dishwasher they go slightly white in colour. With the aid of an equation, explain this observation.

Pb2+ + 2OH- Pb(OH)2

Pb2+ ions react with the OH- ions present in the dishwasher powders to form the precipitate Pb(OH)2 which is white in colour..................................................................................................................................................

(ii) When crystals of PbCl2 are shaken with water at 25°C, it is found that 1.62 × 10–2 mol PbCl2 dissolves per cubic decimeter of solution. Find the value of Ksp at this temperature.

PbCl2 (s) ⇌ Pb2+ (aq) + 2Cl- (aq)

Ksp = [Pb2+][Cl-] = (1.62 x 10-2)(2 x 1.62 x 10-2)2 = 1.70 x 10-5 mol3 dm-9

5070_w07_qp2 Q B10 (modified)7 Tartaric acid is an organic acid. Volumetric analysis can be used to find out how many acid,

–COOH, groups each molecule of the acid contains. The formula of tartaric acid can be represented in this acid-base reaction as HxTa.

Ta represents the rest of the tartaric acid molecule, and x is the number of hydrogen atoms in the molecule which are part of the acid, –COOH, groups.

The equation for its reaction with sodium hydroxide can be written as follows:

HxTa + xNaOH NaxTa + xH2O

The value of x can be found by experiment. 25.0 cm3 of 0.110 mol dm–3 tartaric acid solution was titrated with 0.235 mol dm–3 sodium hydroxide solution, using phenolphthalein as the indicator. The following results were obtained.

(a) Given that there is an anomaly in one of the readings obtained, calculate the value of x.

Average vol of NaOH = ½ (23.45 + 23.35) = 23.40 cm3

No of moles of NaOH = (23.40/1000) x 0.235 = 5.499 x 10-3 mol

No of moles of HxTa = (25.0/1000) x 0.110 = 2.75 x 10-3 mol2.75 x 10-3 mol HxTa 5.499 x 10-3 mol NaOH

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1 mol HxTa 2 mol NaOH

x = 2

(b) Tartaric acid has a molar mass of 150 g mol–1. The percentage composition by mass is C 32%, H 4%, O 64%. Use these data to show that the molecular formula of tartaric acid is C4H6O6.

Molar formula C4H6O6 (shown)

(c) Tartaric acid is a straight chain compound with molecular formula C4H6O6. It contains two alcohol groups. Suggest a structural formula for the molecule.

Edexcel Jan 6243(02)8 The graph shows the boiling temperatures of the hydrogen halides, hydrogen fluoride, HF,

to hydrogen iodide, HI.

(a) Explain why HF has a higher boiling temperature than the other hydrogen halides and why there is an increase from HCl to HI.

(The boiling temperature of HF is higher) because the hydrogen bonding between HF molecules is stronger than the intermolecular forces in HCl

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The rise from HCl to HI is because the strength of the van der Waals’ forces (etc) increases with increase in number of electrons .................................................................................................................................................

(b)(i) Boron trichloride, BCl3, reacts with water to form HCl as one of its products.Calculate the mass of water that would be needed to react with 12.3 g of BCl3.

BCl3 + 3H2O 3HCl + H3BO3

(ii) Predict the pH of the solution obtained in (b)(i), explaining your answer.

pH will be about 3. Hydrogen ions/H+/H3O+/oxonium ions formed (from HCl and H3BO3). .................................................................................................................................................

(c) Hydrogen fluoride forms the weak acid hydrofluoric acid, HF(aq), when dissolved in water.

(i) Write an equation to show the partial dissociation of hydrofluoric acid in water and write the expression for the acid dissociation constant, Ka, for this reaction.

(ii) When sodium hydroxide solution is added to hydrofluoric acid, the following reaction occurs:

HF + OH– F– + H2O

A 10.0 cm3 portion of 0.120 mol dm–3 sodium hydroxide solution was added to 25.0 cm3 of 0.100 mol dm–3 hydrofluoric acid.

Calculate the concentrations of HF and of F– in the resulting mixture, and hence the pH of this mixture.

[The value of Ka for HF is 5.62 × 10–4 mol dm–3]

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(iii) With the aid of equations, explain how the solution in (c)(ii) is able to maintain a fairly constant pH when a few drops of HCl(aq) is added to it.

Large reservoir of both HF and F- is needed (to absorb both acid and base) however [F-] is small (so cannot absorb acid) .................................................................................................................................................

Edexcel 6246(02), 6241 (01)9 The reaction scheme below shows a two-stage synthesis of ethane-1,2-diamine,

H2NCH2CH2NH2.

(a) Suggest a reagent for Reaction 1. Name and outline a mechanism for this reaction.

(b) Suggest a reagent for Reaction 2. Name the type of mechanism involved and write an equation for the overall reaction.

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(c) Draw the structure of the complex ion formed when aqueous cobalt(II) ions react with an excess of ethane-1,2-diamine.

(d) Ethane-1,2-diamine can be coverted into the EDTA4- ion shown below.

Explain why the EDTA4- ion readily displaces monodendate ligands like water.

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Edexcel 10 Compound A, HCOOCH2CH2CH3, is an ester.

(a) Write down the equation for its reaction with aqueous sodium hydroxide.

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(b) The initial rate of reaction between ester A and aqueous sodium hydroxide was measured in a series of experiments at a constant temperature. The data obtained are shown below.

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Use the data in the table to deduce the order of reaction with respect to A and the order of reaction with respect to NaOH. Hence calculate the initial rate of reaction in Experiment 4.

(c) In a further experiment at a different temperature, the initial rate of reaction was found to be 9.0 × 10–3 mol dm–3 s–1 when the initial concentration of A was 0.020 mol dm–3 and the initial concentration of NaOH was 2.00 mol dm–3.

Under these new conditions with the much higher concentration of sodium hydroxide, the reaction is first order with respect to A and appears to be zero order with respect to sodium hydroxide.

(i) Write a rate equation for the reaction under these new conditions.

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(ii) Calculate a value for the rate constant under these new conditions and state its units.

(iii) Suggest why the order of reaction with respect to sodium hydroxide appears to be zero under these new conditions.

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AQA Jun 05 CHM-4-W-QP11 Lactic acid, 2-hydroxypropanoic acid, CH3CH(OH)CO2H, occurs in sour milk.

Glycollic acid, 2-hydroxyethanoic acid, HOCH2CO2H, occurs in sugar cane.

(a) Lactic acid may be synthesised from propene by the following sequence.

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(i) What reagent(s) and condition(s) are used for step I?

reagent(s) manganate(VII) ions

condition(s) cold, dilute

(ii) What type of reaction is step II?

oxidation.................................................................................................................................................

(b) Glycollic acid may be synthesised from ethanoic acid by the following sequence.

(i) Suggest the reagent(s) and condition(s) that are used for step III.

reagent(s) chlorine

condition(s) uvl or sunlight

(ii) What reagents and conditions are used in step IV?

reagent(s) NaOH(aq)/OH-(aq)

condition(s) heat

(c) Lactic acid and glycollic acid react differently when heated under reflux with acidified dichromate(VI) ions.

Draw the structural formula of the organic product in each case.

product from lactic acid product from glycollic acid

CH3COCO2H HO2CCO2H

(d) Lactic acid exhibits isomerism. Draw the displayed formulae showing the isomers of lactic acid.

central C shown as chiral (C*)

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Glycollic acid and lactic acid each give the reactions of an alcohol group and of a carboxylic acid group. Each compound will react with the other to give an ester.

(e) When one molecule of glycollic acid reacts with one molecule of lactic acid, it is possible to form two different esters.

Draw the structure of each of these esters.

CH3CH(OH)CO2CH2CO2H

HOCH2CO2CH(CH3)CO2H

Extension: structure of cyclic ester in the presence of a little H2SO4 (observation: no reaction with Na2CO3)

(f) Glycollic acid and lactic acid are reacted together to make the material for ‘soluble stitches’ (also known as ‘soluble sutures’) which are used in surgery. In this material, many molecules of each acid have been reacted to form a long chain ‘polyester’ molecule which contains many ester groups.

This polyester is used in surgery to sew up wounds inside the body. Over a period of time, the polyester undergoes a chemical reaction and breaks up to re-form the two individual hydroxy-acids.

(i) This reaction occurs where the pH of the body is about pH 5 to pH 6. With the aid of an equation, suggest the type of chemical reaction which causes the polyester material to break up.

Acid hydrolysis.H2O + RCOOR’ RCOOH + R’OH.................................................................................................................................................

(ii) Suggest why the products of this reaction are soluble in water.

Products formed can form hydrogen bonds with water molecules and are comparable in strength to the hydrogen bonding present between water molecules, allowing the products to be soluble in water..................................................................................................................................................

9701_w06_qp212 The rod cells in the retina at the back of the eye contain an alcohol called retinol which is

responsible for their sensitivity to light. Retinol is oxidised by an enzyme-catalysed reaction to the aldehyde retinal.

(a)(i) Deduce the molecular formula of retinal from its skeletal formula above.

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(ii) Suggest the structure of the alcohol retinol.

(b)(i) What reagents and conditions could be used to convert an alcohol to an aldehyde in a laboratory?

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(ii) How many moles of hydrogen molecules would you expect to react with one mole of retinol?

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(iii) How many additional optical isomers will there be in the product formed in (b)(ii)? Show your calculation clearly.

(c) When light shines on the rod cells, an enzyme-catalysed reaction occurs. This changes the arrangement around the double bond from trans to cis, as indicated in the structure below.

(i) Based on the structure of trans-retinal shown below, suggest the structure of cis-retinal.

(ii) Describe how the cis and trans isomers differ in their physical properties.

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(d) The cis-retinal binds to the protein opsin to form rhodopsin. Part of the mechanism of this reaction is shown below.

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(i) Name the type of reaction mechanism which starts in step 1 and is completed in step 2.

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(ii) Draw a ‘curly arrow’ on the cis-retinal molecule to complete the electron movements that occur in step 1.

(iii) Deduce a structure for compound X and draw it in the box above.

AS_Level_Chemistry_B_SAM_Unit F335

End of Paper

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