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1

2015 TJC H2 Chemistry Preliminary Examinations [Turn over

CHEMISTRY 9647/01 Paper 1 Multiple Choice 18th September 2015 1 hour Additional materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid.

There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate.

This document consists of 21 printed pages.

Write your name

and Civics Group

Write and shade your index number

2


Section A

For each question, there are four possible answers, A, B, C and D. Choose the one you consider to be correct.

1 Incomplete combustion of y dm3 of butane, C4H10, yielded a mixture of carbon dioxide and carbon monoxide in the ratio of 3:1, together with water vapour.

What is the volume of oxygen consumed?

A 5y dm3 C 9y dm3

B 6y dm3 D 12y dm3

2 0.005 mol of a metal oxide, Y2Ox, reacted exactly with 0.006 mol of acidified potassium manganate(VII) solution. The half-equation for the reduction of MnO4

– is shown below.

MnO4– + 8H+ + 5e– Mn2+ + 4H2O

Given that the oxidation state of Y in the product is +6, what is the value of x?

A 1 C 3

B 2 D 4

3 Use of the Data Booklet is relevant to this question.

Significant contributions to carbon dioxide levels in the atmosphere comes from the thermal decomposition of limestone (calcium carbonate) and the manufacture of cement and lime (calcium oxide).

Cement works roast 1000 million tonnes of limestone per year and a further 200 million tonnes is roasted in kilns to make lime.

What is the total annual mass output of carbon dioxide (in million tonnes) from these two processes?

A 440

B 528

C 660

D 880

4 Which of the following species has three unpaired electrons?

A Al B Ti3+ C Zn2+ D Cr3+

3


5 Which of the following describes a phenomenon which cannot be explained by hydrogen bonding?

A Ethanoic acid molecules form a dimer when dissolved in benzene.

B Ice has a lower density than water at 0 oC.

C The boiling point of alcohol increases with increasing molecular mass.

D 2-nitrophenol has a lower boiling point than 4-nitrophenol.

6 Which plot clearly exhibits ideal gas behavior?

A

C

B

D

T/K

V/d

m3

P = 3 atm

P = 1 atm

T/K

V/d

m3

P = 3 atm

P = 1 atm

T/K

V/d

m3

P = 3 atm

P = 1 atm

T/K

V/d

m3

P = 3 atm

P = 1 atm

T/oC

V/d

m3

P = 3 atm

P = 1 atm

T/oC

V/d

m3

P = 3 atm

P = 1 atm

T/K

V/d

m3

P = 3 atm

P = 1 atm

T/K

V/d

m3

P = 3 atm

P = 1 atm

4


7 Use of Data Booklet is relevant to this question.

Propane burns exothermically in air as follows.

CH3CH2CH3(g) + 5O2(g) 3CO2(g) + 4H2O(l) H = -2200 kJ mol-1

0.05 moles of propane from the burner was used to heat 0.9 dm3 of water. It was found that the temperature of the water increased to 40 oC after three minutes at room conditions.

What is the rate of heat loss to the surroundings in kJ min-1?

A 17.9 kJ min-1

B 36.7 kJ min-1

C 53.6 kJ min-1

D 103 kJ min-1


A cell involving aqueous potassium iodide and acidified sodium manganate(VII) is shown.

Which of the following statements is true?

A The e.m.f of the cell is -0.98 V.

B The purple colour of iodine will become more intense over time.

C The light bulb will be brighter when the concentration of MnO4-(aq) increases.

D The direction of electron flow will reverse when the MnO4-/Mn2+ half-cell is

replaced with a Cl2/Cl- half cell.

Pt Pt

1 mol dm-3 MnO4-(aq)

1 mol dm-3 Mn2+(aq) 1 mol dm-3 I-(aq) and I2 solid

5



In the commercial electrolysis of concentrated sodium chloride, the products are chlorine, hydrogen and sodium hydroxide.

What is the maximum yield of each of these products when 58.5 kg of sodium chloride was electrolysed?

Yield of chlorine /kg

Yield of hydrogen /kg

Yield of sodium hydroxide /kg

A 35.5 1 40

B 35.5 2 80

C 71 1 40

D 71 2 80

10 The equilibrium, A2(g) + 4B(g) ⇌ 2AB2(g), has an equilibrium constant Kc of 3.60. What will be the numerical value of the equilibrium constant for the following reaction at the same temperature?

AB2(g) ⇌ ½ A2(g) + 2B(g)

A 0.14 C 0.28

B 0.53 D 3.60

6


11 Sulfur dioxide is converted to sulfur trioxide in the Contact process as shown by the equation below.

2SO2(g) + O2(g) ⇌ 2SO3 (g)

To investigate the reaction, two experiments were conducted.

Experiment 1: Quantities of SO2 and O2 were placed in a sealed vessel and the reaction was allowed to proceed at a constant temperature.

Experiment 2: The experiment was repeated at a different temperature using quantities of reactants same as experiment 1.

The graph below shows the amount of SO2 present in the mixture.

These results show that experiment 2 was conducted at a

A higher temperature and the forward reaction is endothermic.

B lower temperature and the forward reaction is endothermic.

C higher temperature and the forward reaction is exothermic.

D lower temperature and the forward reaction is exothermic.

12 Steam reforming process is the most common method for industrial production of hydrogen.

CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g)

The value of Kc is found to be 5.27 for this reaction when 0.65 mol of methane is mixed with 0.30 mol of steam in a 1 dm3 vessel.

How many moles of steam must be added to the mixture to raise the equilibrium amount of hydrogen gas to 0.84 mol at the same temperature?

A 0.030 C 0.085

B 0.065 D 0.090

Number of moles of SO2

time

Experiment 2

Experiment 1

7


13 The numerical value of solubility product of magnesium hydroxide is 1.8 x 10-12. What is the solubility of magnesium hydroxide in a solution of sodium hydroxide at pH 12.5 and 25 °C?

A 5.69 x 10-11 C 1.80 x 10-9

B 4.50 x 10-10 D 7.66 x 10-5

14 The graph below shows the variation in the boiling points for eight consecutive elements in the Periodic Table, all with atomic number between 10 and 20.

Which of the following statements is true?

A Element A and barium are in the same group.

B Element G has a smaller atomic radius than element H.

C Element D has a lower first ionisation energy than element C.

D The oxide of G dissolves in water to give an acidic solution.

Atomic number

A

B

C

D

E

F

G

H

Bo

ilin

g p

oin

t

B

A

D

C E

H

G

F

8


15 X, Y and Z are elements in Period 3 of the Periodic Table.

A mixture containing the oxides of X, Y and Z was dissolved in excess dilute sulfuric acid and filtered. The oxide of Z was collected as a residue. When excess dilute sodium hydroxide was added to the filtrate, only a white precipitate of the hydroxide of Y was formed.

What are the possible identities of X, Y and Z?

X Y Z

A Mg Al P

B Al Mg P

C Mg Al Si

D Al Mg Si

16 Which property of Group II elements (magnesium to barium) decreases with increasing atomic number?

A reactivity with water

B solubility of hydroxides in water

C second ionisation energy

D reducing power

17 Two experiments were conducted to prepare pure hydrogen iodide from solid potassium iodide.

Experiment Reagent used

1 concentrated H2SO4

2 concentrated H3PO4

Which of the following describes the outcome of each experiment correctly?

A Experiment 1 was successful, as hydrogen iodide can be quickly isolated due to its low boiling point.

B Experiment 1 was unsuccessful, as hydrogen iodide formed will further react to give iodine and hydrogen sulfide.

C Experiment 2 was unsuccessful, as H3PO4 is a weaker acid than H2SO4, thus hydrogen iodide cannot be formed.

D Experiment 2 was successful, as H3PO4 is a weaker reducing agent than H2SO4.

9


18 The graph below shows the visible spectra of two complexes of transition metal M, MX6n+

and MY6n+.

The colours corresponding to the wavelengths in the visible light region are shown below.

colour wavelength / nm

violet 400 – 430

blue 430 – 490

green 490 – 570

yellow 570 – 590

orange 590 – 620

red 620 – 750

What is the colour of each complex and which ligand causes a larger d-orbital splitting?

Colour of MX6n+

Colour of MY6n+

ligand which causes a larger d-orbital splitting

A Violet Orange-red X

B Red Green-blue X

C Violet Orange-red Y

D Red Green-blue Y

MX6n+

MY6n+

10


19 Methylcyclohexane and Cl2 are mixed together and irradiated with light.

Which of the following compound is not formed?

A

B CH2Cl

C

D

20 A student wishes to synthesise compound Q from benzene.

NH2

Cl

HO2C

Compound Q

Which of the following routes would give the maximum yield of Q?

A Nitration alkylation reduction chlorination oxidation

B Chlorination nitration reduction alkylation oxidation

C Alkylation nitration oxidation chlorination reduction

D Alkylation nitration chlorination reduction oxidation

11


21 Deuterium, D, is an isotope of hydrogen.

Which reaction will not yield organic product(s) containing deuterium?

A

CH3CH2CH2OH

B CH3CH=CHCH3

C CH3CH2CH2CN

D CH3CH2CONHCH3

22 Which of the following shows the correct order of increasing acidity?

A

OH

< CH3CH2OCOH < CH3CH2CO2H < CH3CHFCO2H

B CH3CH2OCOH <

OH

< CH3CH2CO2H < CH3CHFCO2H

C CH3CHFCO2H < CH3CH2CO2H < CH3CH2OCOH <

OH

D CH3CHFCO2H < CH3CH2OCOH < CH3CH2CO2H <

OH

CD3CO2D, concentrated D2SO4,

heat

NaOD, D2O, heat

DBr

NaOD, D2O, heat

12


23 Part of the structure of a fungicide, strobilurin, is shown.

If strobilurin is first warmed with aqueous sulfuric acid, and its product then treated with hydrogen in the presence of a palladium catalyst, what could be the structure of the final product?

A

B

C

D

24 An organic compound R has the following properties:

(i) It decolourises Br2(aq).

(ii) It reacts with PCl5(s).

(iii) Upon heating with acidified Na2Cr2O7, it forms a product that gives a brick-red precipitate with an alkaline solution of complexed Cu2+(aq).

Which of the following is likely to be compound R?

A

B

C CO2H

D

13


25 The reaction scheme below shows the synthesis of compound U.

Which of the following can be U?

A

B

C

D

14


26 Which of the following statements is true regarding compound Z?

A There is only one sp hybridised carbon atom in a molecule of Z.

B A molecule of Z contains six bonds.

C After Z reacts with LiAlH4, all the carbon atoms in the product formed are sp3 hybridised.

D Z is a planar molecule.

27 Compound P reacts with concentrated H2SO4 at 180⁰C to form Compound Q. Compound P has the following structure.

Compound P

What is the total number of stereoisomers of Compound Q?

A 2 B 4 C 8 D 16

15


28 Unsaturated carbonyl compounds can undergo a useful reaction known as the Diels-Alder reaction with a diene. An example is shown below.

C

C

C

C

H C

H H

O

H

C

C

CH2

CH2

CH

CH2H

H

C

O

HCH2H

CH2H

+

A student reacted the following diene and carbonyl compound together in a Diels-Alder reaction.

C

C

CH2H3C

CH2H

O

O

CH2CH3

and

Which of the following would not be formed as a product?

A

O

O

CH2CH3

H3C

C

O

O

H3C

CH3CH2

B

O

O

CH3CH2

CH3

D

O

O

CH3

CH2CH3

16


30 "Helix breakers" are amino acids that disrupt the regularity of the α-helical backbone

conformation in a protein. These amino acids also cause breakages or deformations in -pleated sheets.

Which of the following amino acids is likely to be a “helix breaker”?

A

B

C

D

29 Partial hydrolysis of a tetrapeptide (containing four amino acid residues) produces the following three dipeptides.

NH2CH(CH(CH3)2)CONHCH(CH3)CO2H

NH2CH(CH(CH3)2)CONHCH(CH(CH3)2)CO2H

NH2CH(CH3)CONHCH(CH3)CO2H

Which of the following is the structure of the tetrapeptide?

A NH2CH(CH(CH3)2)CONHCH(CH3)CONHCH(CH(CH3)2)CONHCH(CH3)CO2H

B NH2CH(CH(CH3)2)CONHCH(CH(CH3)2)CONHCH(CH3)CONHCH(CH3)CO2H

C NH2CH(CH3)CONHCH(CH(CH3)2)CONHCH(CH3)CONHCH(CH(CH3)2)COOH

D NH2CH(CH3)CONHCH(CH3)CONHCH(CH3)CONHCH(CH3)CO2H

17


Section B

For each of the questions in this section, one or more of the three numbered statements 1 to 3 may be correct.

Decide whether each of the statements is or is not correct (you may find it helpful to put a tick against the statements which you consider to be correct).

The responses A to D should be selected on the basis of

A B C D

1,2 and 3 are correct

1 and 2 only are correct

2 and 3 only are correct

1 only is correct

No other combination of statements is used as a correct response.

31 Which of following statements are incorrect?

1 A covalent compound that contains both hydrogen and oxygen can form hydrogen bonds between its molecules.

2 Ionic compounds can be distinguished from metals by their electrical

conductivity in the liquid states.

3 All substances with covalent bonding have poor electrical conductivity.

32 The graph below shows the behaviour of two gases X and Y.

Ideal gas

P

What could be the identities of X and Y?

X Y

1 NH3 N2

2 CO2 O2

3 N2 at 270K N2 at 370K

X

Y

18


33 The reduction of copper(I) oxide with carbon is shown below.

Cu2O(s) + C(s) 2Cu(s) + CO(g)

Enthalpy change of formation of Cu2O(s) and CO(g) are -167 kJ mol-1 and

- 111 kJ mol-1 respectively. The numerical value of the S is 165 J K-1 mol-1.

Which of the following statements are true?

1 The entropy change for the reaction is positive.

2 The enthalpy change for the reaction is + 56 kJ mol-1.

3 Reaction is spontaneous at 340 K.

34 The table below shows the experimental results for the reaction of bromate(V) and bromide in the presence of acid.

BrO3– + 5Br

– + 6H+ → 3Br2 + 3H2O

Expt [BrO3–] [Br

– ] [H+] relative rate

1 0.10 0.15 0.40 1

2 0.20 0.15 0.40 2

3 0.10 0.30 0.40 2

4 0.30 0.10 0.20 0.5

5 0.01 0.60 0.80 x

Which of the following statements are correct for this reaction?

1 The value of x is 1.6.

2 The overall order of reaction is 4.

3 A plot of rate against [BrO3

–] will be a straight line graph with positive gradient for experiment 5.

19


35 Which property about X, Y and Z will give the trend shown below?

property

ZX Y

Property X Y Z

1 pH of the resultant solution when added to water

MgCl2 AlCl3 SiCl4

2 atomic radius S Cl Ar

3 boiling point HCl HBr HI

36 Vanadium occurs naturally in minerals and fossil fuel deposits.

Which of the following statements are not correct for vanadium and its compounds?

1 The maximum oxidation state of vanadium is found in VO2+.

2 V2O5 is used as a catalyst in the manufacture of ammonia.

3 Vanadium is less dense than calcium.

20


37 Consider the following reaction scheme.

OCl Step IO

Step IIOH

Step III

Step IVCl

Cl

L M N

PQ

AlCl3

Which of the following statements about the above reaction scheme is correct?

1 Step I requires a catalyst.

2 Compound M is a planar molecule.

3 Step IV is a substitution reaction.

38 Perspex is a clear, colourless polymer used for optical applications. It is made of the monomer, methyl methacrylate.

Which of the following reagents and conditions can be used to distinguish methyl methacrylate and compound Z?

1 Aqueous alkaline iodine, heat

2 2,4-DNPH, room temperature

3 NaOH(aq), heat

Methyl methacrylate Compound Z

21


39 Reserpine is an antipsychotic and antihypertensive drug used for the control of high blood pressure and relief of psychotic behaviours.

[The ether –OCH3 group is inert]

Which statements are correct?

1 The reserpine molecule does not give an orange precipitate with 2,4-DPNH.

2 The reserpine molecule has 6 chiral centres

3 The N atom labeled N1 has a lower pKb than the N atom labeled N2.

40 Valine is a naturally occurring amino acid. Which of the following statements about valine and 3-aminopropanoic acid is correct?

3-aminopropanoic acid

O

OH

NH2

CH3

alanine

OH

O

NH2

1 Valine and 3-aminopropanoic acid are not structural isomers.

2 Both valine and 3-aminopropanoic acids are soluble in water due to formation of hydrogen bonding.

3 Both compounds are able to react with ethanoic acid to give amide.

valine

2015 H2 Chemistry Prelim Paper 1 MCQ Answers

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

B C B D C D A C A B D B C C D C B B C D

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

C B C C A B B B B A A C A A D A D D B D

2015 TJC H2 Chemistry Preliminary Exam

s

CANDIDATE NAME

CIVICS GROUP /

CENTER NUMBER S INDEX

NUMBER

CHEMISTRY 9647/02

Paper 2 Structured Questions 1 September 2015

2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet

READ THESE INSTRUCTIONS FIRST

Write your Civics Group, centre number, index number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.


For Examiner’s Use 1 / 12 2 / 10 3 / 14 4 / 10 5 / 10 6 / 16

Paper 1 / 40 Paper 3 / 80

Total

PRELIMINARY EXAMINATIONS

HIGHER 2

2

2015 TJC H2 Chemistry Preliminary Exam [Turn over

For examiner’s

use 1 Planning (P)

Potassium hydrogen tartrate, KHC4H4O6 (KHT), is sparingly soluble in water with a solubility of approximately 0.030 mol dm-3 at 25 °C. It dissolves to give potassium ion and hydrogen tartrate ion, HT–, as shown below.

KHT(s) K+(aq) + HT–(aq) The hydrogen tartrate ion is a weak acid that can react with a strong base, NaOH, to form sodium tartrate. The solubility product of KHT can be determined by preparing a saturated solution of KHT at a given temperature. The concentration of HT– can be determined by titrimetric analysis from which the Ksp can be calculated.

(a) Calculate the minimum mass of KHT that needs to be weighed in order to obtain 100 cm3 of a saturated solution of KHT at 25 °C. [Mr of KHT = 188.1]

[1]

(b) Assuming the titre volume to be 25.00 cm3 of sodium hydroxide solution, calculate the concentration of the sodium hydroxide solution needed to determine the concentration of HT–.

[1]

(c) You are provided with a standard solution of 0.300 mol dm-3 sodium hydroxide. In view of your answer to (a) and (b), write the procedure for the determination of the concentration of HT– in the saturated solution. Your procedure should also include the preparation of 100 cm3 of the saturated solution of HT–.

3


For examiner’s

use Procedure

[6]

4


For examiner’s

use (d) y cm3 of z mol dm-3 sodium hydroxide solution is needed to react with 25.0 cm3 of the

saturated solution of KHT at 25 °C. Calculate the Ksp of KHT.

[2]

(e) If the experiment is repeated with a saturated solution of KHT prepared by adding solid KHT to 100 cm3 of 0.010 mol dm-3 KNO3 solution, suggest how the values of y will change, if any. Explain your answer.

[2]

[Total: 12]

5


For examiner’s

use

2 A research student was given a few unknown samples. He used different techniques such as instrumental analysis and titrimetric methods to identify the compounds.

(a) An acidified solution of an unknown salt A, KClOx, was reacted with Fe2+(aq) to give Fe3+(aq) and Cl-(aq).

When 0.150 g of the salt A, KClOx, reacted with 0.500 mol dm-3 Fe2+(aq) in the presence of H+(aq), 11.30 cm3 of Fe2+(aq) was needed for complete reaction.

(i) State the full electronic configuration of Fe3+.

(ii) Calculate the value of x of salt A, KClOx .

[4]

6


For examiner’s

use

(b) The research student was given the second ionisation energies of seven consecutive elements in the Periodic Table as shown below:

(i) Define, with the aid of an equation, what is meant by second ionisation energy of element B.

(ii) With reference to the graph above, deduce the Group of the Periodic Table to which

B is likely to belong. Explain your answer.

[4]

Sec

ond

ioni

satio

n en

ergy

/k

J m

ol-1

B C D E F G H

7


For examiner’s

use

(c) When the student passed a beam of protons through an electric field, it deflected 12° towards the negative plate.

Under identical conditions, the student passed a beam of doubly charged particles J through the electric field. The angle of deflection was found to be 1.5° towards the positive plate.

Identify the ion J.

[2]

[Total: 10]

8


For examiner’s

use 3 (a) The Earth's atmosphere contains 21% of oxygen. Oxygen from the atmosphere

dissolves in water as shown in the following equilibrium.

O2 (g) O2(aq) H = negative

The solubility of gases in water, in mol dm-3, is directly proportional to the partial pressure of the gas above the solution.

Solubility of gas = kH x Partial Pressure of gas

(where kH = 1.3 x 10-2 atm-1 mol dm-3 for oxygen at 298 K)

(i) Calculate the solubility of oxygen in water at 298 K and 1 atm.

(ii) Explain how concentration of oxygen in water can be affected by increasing the temperature of the water.

(iii) Predict the sign of SƟ for the dissolution of oxygen in water. Give a reason for your answer.

9


For examiner’s

use (iv) Dissolved oxygen in water is important for aerobic respiration for marine life.

Oxygen and glucose, C6H12O6, are required for aerobic respiration as follows:

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

Compound Gf

o / kJ mol-1 C6H12O6 -917.2

CO2 -394.4

H2O -237.2

Using your knowledge of Hess’ Law, calculate GƟ for the aerobic respiration.

[7]

10


For examiner’s

use (b) Diving tanks store a mixture of oxygen, nitrogen and helium gas at high pressures to

sustain divers for long periods underwater.

(i) Assuming gases behave ideally, 900 dm3 of gas could be filled into a 3 dm3 gas tank at high pressure and room temperature. However, in reality only 810 dm3 of gas fills the tank under the same conditions.

Explain why this is so.

(ii) When air bubbles are released underwater, they expand in size.

A 10 cm3 bubble was released from the diver, 120 m below the water surface at a pressure of 1300 kPa and temperature of 10 °C. It was estimated that pressure increases by 101 kPa with each 10 m depth.

Calculate the volume of the air bubble when it ascends towards the water surface by 60 m where the water temperature is 20 °C.

[3]

11


For examiner’s

use (c) Oxygen is required for many industrial processes and reactions. One use is in anodising

where aluminium objects are coated with a layer of oxide to resist corrosion.

(i) Write equations for the reactions at the anode to explain how the layer of oxide can be formed on the aluminium objects.

(ii) An aluminium object with a surface area of 300 cm2 is to be anodised.

How many coulombs of electricity will be required to increase its oxide layer by 1x10-3 cm in thickness? [Density of Al2O3 = 4.0 g cm-3]

[4]

[Total: 14]

Aluminium object

Dilute sulfuric acid

Graphite electrode

Power Source

12


For examiner’s

use 4 Amines have many uses in daily life. For instance, phenylamine is commonly used to make

herbicides and rubber processing chemicals.

(a) State the reagents and conditions needed to produce phenylamine from benzene in steps I and II. Identify the intermediate D.

Step I

D

Step II NH2

Step I: _____________________________________________________

Step II: ____________________________________________________

[3]

(b) Explain whether 4-nitrophenylamine is a stronger or weaker base than phenylamine.

[2]

13


For examiner’s

use

Phenylamines react with nitrous acid, HNO2, to form stable diazonium salts in a process called diazotization. This reaction can be illustrated by the equation below. a diazonium salt Diazonium salts are useful because they can undergo coupling reactions with activated aromatic rings such as phenols to yield brightly coloured azo compounds, Ar-N=N-Ar’, as follows:

Where Y = -OH

(c) (i) Name the type of reaction taking place in diazonium coupling reactions.

(ii) Explain why phenol is preferred over benzene to react with diazonium salts in the coupling reactions.

(iii) Based on the information given, suggest a use of azo compounds in daily life.

[3]

Y

NN

Y

an azo compound

+ HNO2 + H2SO4 + HSO4– + 2 H2O

+

cold N N

N N

+ H+

14


For examiner’s

use

(d) Hence, devise a two-step synthesis to convert phenylamine to compound X.

NN

N

compound X Suggest reagents and conditions for each step and draw the structure of the intermediate formed.

[2] [Total: 10]

15


For examiner’s

use

5 (a) Valinomycin is a dodecadepsipeptide antibiotic in which one or more of its amide groups are replaced by the corresponding ester groups.

Valinomycin

Valinomycin is highly selective for K+ ions which fit snugly into the coordination site in the centre of a valinomycin molecule.

(i) Explain why valinomycin has a higher selectivity for K+ than for Na+ ions.

Charged K+ ions are unable to diffuse across lipid bilayers of membranes. Valinomycin facilitates the movement of K+ ions across membranes by binding to the ion and diffusing through the hydrophobic core of lipid bilayers.

(ii) Suggest how valinomycin transport K+ ions across lipid bilayers with reference to the types of bonds formed between - valinomycin and K+ ions, - valinomycin and lipid bilayers.

16


For examiner’s

use

(iii) When treated with a peptidase enzyme, the peptide bonds in the ring of valinomycin are hydrolysed. Draw the structural formulae of the two compounds produced.

Hydrolysis of valinomycin produces the amino acid valine. Information on valine and two other amino acids are given below.

Amino acid aspartic acid glutamic acid valine

Structure C

CH2CO2H

H2N CO2H

H

C

CH(CH3)2

H2N CO2H

H

Isoelectric point 3.0 3.2 6.0

A mixture of valine, glutamic acid and aspartic acid can be separated by electrophoresis at pH 6.0. The following gel strip is obtained.

(iv) Draw the structural formula of the species found at position A.

Species at position A:

H

H2N CO2H

(CH2)2CO2H

(–)

Cathode

+)

Anode

Original position of sample

Gel strip at pH = 6.00

(

A B C

17


For examiner’s

use

(v) Identify the amino acid found at position B.

[7]

(b) During strenuous exercises, glucose is broken down to pyruvic acid, CH3COCO2H. The

pyruvic acid produced is reduced to lactic acid, CH3CH(OH)CO2H, while the coenzyme NADH is oxidised to NAD+. This ensures that energy production is maintained.

(i) Construct a half-equation for conversion of pyruvic acid to lactic acid.

After exercising, we do not stop panting immediately as we need to breathe in additional oxygen so that lactic acid can be reconverted back to pyruvic acid.

(ii) With aid of the Data Booklet, write a balanced equation for the conversion of lactic acid back to pyruvic acid.

(iii) The standard electrode potential for pyruvic acid/ lactic acid system is -0.185 V. Calculate the value of EƟ

cell for the reaction in (b)(ii).

[3] [Total: 10]

18


For examiner’s

use

6 Iron is by mass the most common element on Earth, forming much of Earth's outer and inner core. Iron is commonly used to make alloys with metals such as magnesium which is a Group II metal. Uses of iron alloys include air-plane parts and teeth fillings.

(a) What do you understand by the term transition element?

[1]

(b) The following diagram shows the reactions of iron and its compounds.

(i) Name the types of reactions taking place in Step I and Step II.

Step I: _____________________________________

Step II: _____________________________________

(ii) State the reagent that could be used for Step III.

(iii) Identify gas A and precipitate B.

Gas A: _____________________________________

Precipitate B: ________________________________

Deep red solution

gas A

Step III

[Fe(CN)6]4-(aq)

Deep blue precipitate, B

Fe3+

Yellow solution Step I Cl2

Fe2+(aq) green

KF [FeF5(H2O)]2-(aq) colourless

Step II

Mg

19


For examiner’s

use

(iv) Fe2+(aq) appears green while [FeF5(H2O)]2-(aq) appears colourless. I Explain why Fe2+(aq) is green. II Suggest a reason why [FeF5(H2O)]2-(aq) is colourless.

[9]

20


For examiner’s

use

(c) The thermal decomposition of Group II nitrates is investigated in an experiment. Equal amounts of the nitrates of magnesium, calcium and barium were heated separately for one minute and the gases produced were passed through aqueous potassium hydroxide. The volume of the remaining gas was measured using a gas syringe. The results are tabulated as follows:

Compound Mg(NO3)2 Ca(NO3)2 Ba(NO3)2

Gas syringe reading / cm3 89 21 8

(i)

Explain the purpose of potassium hydroxide. Hence, identify the gas collected in the syringe.

(ii) Using suitable data from the Data Booklet, explain the results obtained.

(iii) Hence, estimate the volume of remaining gas produced after 1 min if zinc nitrate

is used instead.

[6]

[Total: 16]

2015 TJC H2 Chemistry Preliminary Exam

s

CANDIDATE NAME ANSWERS

CIVICS GROUP /

CENTER NUMBER S INDEX

NUMBER

CHEMISTRY 9647/02

Paper 2 Structured Questions 1 September 2015

2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet


Write your Civics Group, centre number, index number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.


For Examiner’s Use 1 / 12 2 / 10 3 / 14 4 / 10 5 / 10 6 / 16

Paper 1 / 40 Paper 3 / 80

Total

PRELIMINARY EXAMINATIONS

HIGHER 2

2


For examiner’s

use 1 Planning (P)

Potassium hydrogen tartrate, KHC4H4O6 (KHT), is sparingly soluble in water with a solubility of approximately 0.030 mol dm-3 at 25 °C. It dissolves to give potassium ion and hydrogen tartrate ion, HT–, as shown below.

KHT(s) K+(aq) + HT–(aq) The hydrogen tartrate ion is a weak acid that can react with a strong base, NaOH, to form sodium tartrate. The solubility product of KHT can be determined by preparing a saturated solution of KHT at a given temperature. The concentration of HT– can be determined by titrimetric analysis from which the Ksp can be calculated.

(a) Calculate the minimum mass of KHT that needs to be weighed in order to obtain 100 cm3 of a saturated solution of KHT at 25 °C. [Mr of KHT = 188.1] Amount of KHT in 100 cm3 saturated solution = 100/1000 x 0.030 mol

= 3.00 x 10-3 mol Mass of KHT = 3.00 x 10-3 x 188.1 g = 0.564 g

[1]

(b) Assuming the titre volume to be 25.00 cm3 of sodium hydroxide solution, calculate the concentration of the sodium hydroxide solution needed to determine the concentration of HT–. Assume 25.0 cm3 of HT- pipetted and titre volume to be 25.00 cm3. Amount of HT- in 25.0 cm3 = 25/1000 x 3.00 x 10-2 mol

= 7.50 x 10-4 mol [NaOH] = 1000/25 x 7.50 x 10-4 mol dm-3 = 0.0300 mol dm-3

[1]

(c) You are provided with a standard solution of 0.300 mol dm-3 sodium hydroxide. In view of your answer to (a) and (b), write the procedure for the determination of the concentration of HT– in the saturated solution. Your procedure should also include the preparation of 100 cm3 of the saturated solution of HT–.

3


For examiner’s

use Procedure

From (b), the [NaOH] needs to be diluted by about 10 times. Dilution of NaOH

1. Pipette 25.0 cm3 of 0.300 mol dm-3 NaOH into a 250 cm3 standard flask. 2. Make up to the mark with deionised water. Stopper and shake thoroughly to obtain

a homogeneous solution.

Preparation of saturated solution of KHT

1. Using the electronic balance, weigh approximately about 1 g of KHT into a weighing bottle.

2. Using a burette, place 100 cm3 of deionised water into a 250 cm3 conical flask. 3. Transfer the KHT solid into the conical flask and swirl or stir the contents for about

5 minutes. Allow to stand and occasionally swirl for another 5 minutes to allow the solution to be saturated.

4. Filter the solution using a dry filter paper and dry filter funnel into a dry 250 cm3 conical flask.

5. Using a thermometer, record the temperature of the saturated solution.

Titration

1. Pipette 25.0 cm3 of the saturated solution of KHT into a 250 cm3 conical flask and add 1 to 2 drops of phenolphthalein indicator.

2. Fill a burette with the diluted NaOH and titrate the KHT solution until end point is reached ie when solution changes colour from colourless to pink.

3. Repeat the titration until consistent readings ie readings differ by ± 0.10 cm3 are obtained.

Dilution of NaOH

Preparation of saturated solution of KHT: 1m (steps 1 to 2). Can also accept adding solid until no more can dissolve to prepare saturated solution.

Filtration, Choice of apparatus + dry filter paper and funnel (step 4)

Titration, Indicator and colour change

[6]

(d) y cm3 of z mol dm-3 sodium hydroxide solution is needed to react with 25.0 cm3 of the saturated solution of KHT at 25 °C. Calculate the Ksp of KHT.

Amount of NaOH = yz/1000 mol Amount of KHT in 25.0 cm3 = yz/1000 mol [HT-] = 1000/25 x yz/1000 mol = yz/25 mol Ksp = [K+][HT-] = (yz/25)2 mol2 dm-6

[2]

4


For examiner’s

use (e) If the experiment is repeated with a saturated solution of KHT prepared by adding solid

KHT to 100 cm3 of 0.010 mol dm-3 KNO3 solution, suggest how the values of y will change, if any. Explain your answer.

The titre value would be < y due to common ion effect. In the presence of K+ from KNO3, the position of equilibrium shifts left according to Le Chatelier’s principle.

KHT(s) K+(aq) + HT-(aq) The solubility of KHT decreases, resulting in a smaller amount of HT-. Thus a

smaller amount of NaOH is needed for the titration.

[2]

[Total: 12]

2 A research student was given a few unknown samples. He used different techniques such as instrumental analysis and titrimetric methods to identify the compounds.

(a) An acidified solution of an unknown salt A, KClOx, was reacted with Fe2+(aq) to give Fe3+(aq) and Cl-(aq).

When 0.150 g of the salt A, KClOx, reacted with 0.500 mol dm-3 Fe2+(aq) in the presence of H+(aq), 11.30 cm3 of Fe2+(aq) was needed for complete reaction.

(i) State the full electronic configuration of Fe3+.

1s22s22p63s23p63d5 (ii) Calculate the value of x of salt A, KClOx .

No of moles of Fe2+ = 11.3 × 10–3 × 0.500 = 0.00565 mol

ClOx– + 2xH+ + 2xe Cl– + xH2O

Fe2+ Fe3+ + e

Mole ratio of ClOx– : Fe2+ = 1 : 2x

0.150 / (39.1 + 35.5 + 16.0x) = 1

0.00565 2x x = 2

[4]

5


For examiner’s

use

(b) The research student was given the second ionisation energies of seven consecutive elements in the Periodic Table as shown below:

(i) Define, with the aid of an equation, what is meant by second ionisation energy of element B.

Second ionisation energy of element B is the minimum energy required to completely remove one mole of valence electrons from one mole of ground state ions in the gaseous state to form doubly charged gaseous ions.

B+(g) → B2+(g) + e

(ii) With reference to the graph above, deduce the Group of the Periodic Table to

which B is likely to belong. Explain your answer.

Element H has a lower second ionisation energy than element G so the second electron is removed from outer quantum shell and element H is a group II element.

Element B is in Group IV. OR Element G has the largest second ionisation energy so the second

electron is removed from inner quantum shell and element G is a group I element.

Element B is in Group IV.

[4]

Sec

ond

ioni

satio

n en

ergy

/k

J m

ol-1

B C D E F G H

6


For examiner’s

use

(c) When the student passed a beam of protons through an electric field, it deflected 12° towards the negative plate.

Under identical conditions, the student passed a beam of doubly charged particles J through the electric field. The angle of deflection was found to be 1.5° towards the positive plate.

Identify the ion J.

since J is doubly charged and the deflection is towards the positive plate : J2-

angle of deflection ∝chargemass

mass of J2- = (12 x 2) / 1.5 = 16 J is O2-

[2]

[Total: 10]

3 (a) The Earth's atmosphere contains 21% of oxygen. Oxygen from the atmosphere dissolves in water as shown in the following equilibrium.

O2 (g) O2(aq) Ho = negative

The solubility of gases in water, in mol dm-3, is directly proportional to the partial pressure of the gas above the solution.

Solubility of gas = kH x Partial Pressure of gas

(where kH = 1.3 x 10-2 atm-1 mol dm-3 for oxygen at 298 K)

(i) Calculate the solubility of oxygen in water at 298 K and 1 atm.

푷푶ퟐ =ퟐퟏퟏퟎퟎ

× ퟏ = ퟎ.ퟐퟏ풂풕풎

[푶ퟐ] = 풌푯 × 푷푶ퟐ

●[푶ퟐ] = ퟏ.ퟑ × ퟏퟎ ퟐ × ퟎ.ퟐퟏ = ퟐ.ퟕ × ퟏퟎ ퟑ풎풐풍풅풎 ퟑ

(ii) Explain how concentration of oxygen in water can be affected by increasing the temperature of the water.

●When temperature increases, by Le Chatelier’s Principle, the position of equilibrium will shift to the left to favour the endothermic reaction and ● less oxygen will dissolve in water hence concentration of oxygen in water decreases.

7


For examiner’s

use (iii) Predict the sign of SƟ for the dissolution of oxygen in water. Give a reason for

your answer.

●So has a negative value. ●There is less number of moles of free oxygen gas hence less number of ways to arrange the particles and distribute its energy. OR

Go = Ho- TSo

● At equilibrium Go= 0

Thus Ho- TSo = 0

Ho = TSo

● Since T is positive and Ho is negative, hence So is negative

(iv) Dissolved oxygen in water is important for aerobic respiration for marine life. Oxygen and glucose, C6H12O6, are required for aerobic respiration as follows:

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

Compound Gf

o / kJ mol-1 C6H12O6 (s) -917.2

CO2 (g) -394.4

H2O (l) -237.2

Using your knowledge of Hess’ Law, calculate GƟ for the aerobic respiration.

●Go = Gfo (products) - Gf

o (reactants) = 6(-394.4) + 6(-237.2) - (-917.2) ● = -2870 kJ mol-1

OR using cycle

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

6C(s) + 6H2(g) + 9O2(g)

[7]

8


For examiner’s

use (b) Diving tanks store a mixture of oxygen, nitrogen and helium gas at high pressures to

sustain divers for long periods underwater.

(i) Assuming gases behave ideally, 900 dm3 of gas could be filled into a 3 dm3 gas tank at high pressure and room temperature. However, in reality only 810 dm3 of gas fills the tank under the same conditions.

Explain why this is so.

● Gas is less compressible at high pressure as the volume of the gas particle becomes significant.

(ii) When air bubbles are released underwater, they expand in size.

A 10 cm3 bubble was released from the diver, 120 m below the water surface at a pressure of 1300 kPa and temperature of 10 °C. It was estimated that pressure increases by 101 kPa with each 10 m depth.

Calculate the volume of the air bubble when it ascends towards the water surface by 60 m where the water temperature is 20 °C.

●● [1 mark working, 1 mark answer] PV = nRT Since number of moles of gas in the bubble remain the same,

푷ퟏ푽ퟏ푻ퟏ

=푷ퟐ푽ퟐ푻ퟐ

(ퟏퟑퟎퟎퟎퟎퟎ − (ퟔ × ퟏퟎퟏퟎퟎퟎ))푽ퟏ(ퟐퟕퟑ+ ퟐퟎ)

=(ퟏퟑퟎퟎퟎퟎퟎ)(ퟏퟎ× ퟏퟎ ퟔ)

(ퟐퟕퟑ+ ퟏퟎ)

Volume of the bubble at 60 m below the surface = 1.94 x 10-5 m3

(19.4 cm3)

[3]

9


For examiner’s

use

(c) Oxygen is required for many industrial processes and reactions. One use is in anodising where aluminium objects are coated with a layer of oxide to resist corrosion.

(i) Write equations for the reactions at the anode to explain how the layer of oxide can be formed on the aluminium objects.

2H2O(l) O2(g) +4H+(aq) + 4e-

3O2(g) + 4Al(s) 2Al2O3(s) ● 1m for both equations

(ii) An aluminium object with a surface area of 300 cm2 is to be anodised.

How many coulombs of electricity will be required to increase its oxide layer by 1x10-3 cm in thickness? [Density of Al2O3 = 4.0 g cm-3]

Volume of Al2O3 to be formed = 1x10-3 x 300 = 0.3 cm3 ●Mass of Al2O3 to be formed = 0.3 x 4 = 1.2g Number of moles of Al2O3 = 1.2 / (2(27) + 3(16)) = 0.0118 mol 2Al2O3(s) ≡ 3O2(g) ≡ 12 e- ● Number of moles of electrons required = 6 x 0.0118 = 0.0706 mol ● Coulombs of electricity required = n F = 0.0706 x 96500 = 6810 C

[4]

[Total: 14]

Aluminium object

Dilute sulfuric acid

Graphite electrode

Power Source

10


For examiner’s

use

(a) State the reagents and conditions needed to produce phenylamine from benzene in steps I and II. Identify the intermediate D.

Step I

D

Step II NH2

4 Amines have many uses in daily life. For instance, phenylamine is commonly used to make herbicides and rubber processing chemicals.

● Step I: Conc HNO3, Conc H2SO4, reflux at 60 oC ● Step II: Sn with Conc HCl, reflux, followed by aq NaOH to liberate the free

amine ● D is

NO2

[3]

(b) Explain whether 4-nitrophenylamine is a stronger or weaker base than phenylamine.

[2]

● 4-nitrophenylamine is a weaker base. ● –NO2 group is electron-withdrawing, increases the extent to which the lone pair of electrons on the N atom is delocalised into the benzene ring, making it less available for dative bonding to a proton.

11


For examiner’s

use

Phenylamines react with nitrous acid, HNO2, to form stable diazonium salts in a process called diazotization. This reaction can be illustrated by the equation below. a diazonium salt Diazonium salts are useful because they can undergo coupling reactions with activated aromatic rings such as phenols to yield brightly coloured azo compounds, Ar-N=N-Ar’, as follows:

Where Y = -OH

(c) (i) Name the type of reaction taking place in diazonium coupling reactions.

● Electrophilic substituition.

(ii) Explain why phenol is preferred over benzene to react with diazonium salts in the coupling reactions.

● The delocalisation of the lone pair of electrons on O atom into the benzene ring increases the electron density in the ring, making phenol more reactive than benzene towards electrophilic substitution. Note: Be aware of the difference between electron density and charge density and use it in the correct context.

(iii) Based on the information given, suggest a use of azo compounds in daily life.

[3] ● As dyes (for textiles) or colouring agents

Y

NN

Y

an azo compound

+ HNO2 + H2SO4 + HSO4– + 2 H2O

+

cold N N

N N

+ H+

12


For examiner’s

use

(d) Hence, devise a two-step synthesis to convert phenylamine to compound X.

NN

N

compound X Suggest reagents and conditions for each step and draw the structure of the intermediate formed.

NH2

NN

N

N

● 1 mark for step 1 and correct intermediate ● 1 mark for correct reagent for step 2

[2] [Total: 10]

HNO2, H2SO4

cold N N

13


For examiner’s

use

5 (a) Valinomycin is a dodecadepsipeptide antibiotic in which one or more of its amide groups are replaced by the corresponding ester groups.

Valinomycin

Valinomycin is highly selective for K+ ions which fit snugly into the coordination site in the centre of a valinomycin molecule.

(i) Explain why valinomycin has a higher selectivity for K+ than for Na+ ions.

Since K+ ions fit snugly into the coordination site in the centre of a valinomycin molecule, the site is too large for Na+ ions which have a smaller ionic radius.

Charged K+ ions are unable to diffuse across lipid bilayers of membranes. Valinomycin facilitates the movement of K+ ions across membranes by binding to the ion and diffusing through the hydrophobic core of lipid bilayers.

(ii) Suggest how valinomycin transport K+ ions across lipid bilayers with reference to the types of bonds formed between - valinomycin and K+ ions, - valinomycin and lipid bilayers.

Dative bonds are formed between carbonyl groups (C=O) of valinomycin and K+ ions.

Valinomycin transport K+ ions across lipid bilayers by forming Van der Waals forces of attraction (using the methyl and isopropyl side chains) with hydrophobic cores of lipid bilayers. (As such the valinomycin can dissolve in the lipid bilayers allowing the K+ to be transported).

14


For examiner’s

use

(iii) When treated with a peptidase enzyme, the peptide bonds in the ring of valinomycin are hydrolysed. Draw the structural formulae of the two compounds produced.

● ●

Hydrolysis of valinomycin produces the amino acid valine. Information on valine and two other amino acids are given below.

Amino acid aspartic acid glutamic acid valine

Structure C

CH2CO2H

H2N CO2H

H

C

CH(CH3)2

H2N CO2H

H

Isoelectric point 3.0 3.2 6.0

A mixture of valine, glutamic acid and aspartic acid can be separated by electrophoresis at pH 6.0. The following gel strip is obtained.

(iv) Draw the structural formula of the species found at position A.

A:

(v) Identify the amino acid found at position B.

Glutamic acid

[7]

H

H2N CO2H

(CH2)2CO2H

-+H

H3N CO2

CH(CH3)2

(–)

Cathode

+)

Anode

Original position of sample

Gel strip at pH = 6.00

(

A B C

15


For examiner’s

use (b) During strenuous exercises, glucose is broken down to pyruvic acid, CH3COCO2H.

The pyruvic acid produced is reduced to lactic acid, CH3CH(OH)CO2H, while the coenzyme NADH is oxidised to NAD+. This ensures that energy production is maintained.

(i) Construct a half-equation for conversion of pyruvic acid to lactic acid. [3]

2e + 2H+ + CH3COCO2H CH3CH(OH)CO2H

After exercising, we do not stop panting immediately as we need to breathe in additional oxygen so that lactic acid can be reconverted back to pyruvic acid.

(ii) With aid of the Data Booklet, write a balanced equation for the conversion of lactic acid back to pyruvic acid.

CH3CH(OH)CO2H CH3COCO2H + 2H+ + 2e O2 + 4H+ + 4e 2H2O 2CH3CH(OH)CO2H + O2 2CH3COCO2H + 2H2O

(iii) The standard electrode potential for pyruvic acid/ lactic acid system is -0.185 V.

Calculate the value of EƟcell for the reaction in (b)(ii).

EƟ = +1.23 – (-0.185) = +1.42V

[3] [Total: 10]

16


For examiner’s

use 6 Iron is by mass the most common element on Earth, forming much of Earth's outer

and inner core. Iron is commonly used to make alloys with metals such as magnesium which is a Group II metal. Uses of iron alloys include air-plane parts and teeth fillings.

(a) What do you understand by the term transition element? [1]

● A transition element is a d–block element which forms at least one stable ion with a partially filled d orbital.

(b) The following diagram shows the reactions of iron and its compounds.

(i) Name the types of reactions taking place in Step I and Step II.

● Step I: Oxidation ● Step II: Ligand exchange (ii) State the reagent that could be used for Step III.

● KSCN(aq) or SCN-(aq) (iii) Identify gas A and precipitate B.

● A is hydrogen gas. ● B is Fe4[Fe(CN)6]3 (iv) Fe2+(aq) appears green while [FeF5(H2O)]2-(aq) appears colourless.

I Explain why Fe2+(aq) is green. II Suggest a reason why [FeF5(H2O)]2-(aq) is colourless.

I ● Fe2+ has an electronic configuration of 1s22s22p63s23p63d6 and has partially-filled d-orbitals. ●● When the ligands approach the central Fe2+ ion, splitting of the d-orbitals occur. The energy gap, E, between the non-degenerate orbitals corresponds to the wavelength of light in the visible region of the

Deep red solution

gas A

Step III

[Fe(CN)6]4-(aq)

Deep blue precipitate, B

Fe3+

Yellow solution Step I Cl2

Fe2+(aq) green

KF [FeF5(H2O)]2-(aq) colourless

Step II

Mg

17


For examiner’s

use

electromagnetic spectrum. When d-d transition of electrons takes place, radiation in the visible region of the electromagnetic spectrum corresponding to E is absorbed. The light energy not absorbed will be seen as the colour of the complex ion. II ● The energy absorbed for d-d transition is out of the range of the visible spectrum. Hence white light is not absorbed/is reflected and [FeF5(H2O)]2-

appears colourless.

[9] (c) The thermal decomposition of Group II nitrates is investigated in an experiment.

Equal amounts of the nitrates of magnesium, calcium and barium were heated separately for one minute and the gases produced were passed through aqueous potassium hydroxide. The volume of the remaining gas was measured using a gas syringe. The results are tabulated as follows:

Compound Mg(NO3)2 Ca(NO3)2 Ba(NO3)2 Gas syringe reading / cm3 89 21 8

(i)

Explain the purpose of potassium hydroxide. Hence, identify the gas collected in the syringe.

● KOH absorbs the NO2 gas. ● The remaining gas is O2.

(ii) Using suitable data from the Data Booklet, explain the results obtained.

● Ionic radii: Mg2+(0.065 nm), Ca2+(0.099 nm), Ba2+(0.135 nm)

● Down the group, as ionic radii increases, charge density of the cation decreases and polarizes the nitrate anion to a lesser extent. The N-O bond breaks less easily and hence, the nitrate becomes more thermally stable and produces lesser gas.

(iii) Hence, estimate the volume of remaining gas produced after 1 min if zinc nitrate is used instead.

● Ionic radii of Zn2+ is 0.074 nm, in between that of Mg2+ (0.065 nm) and Ca2+ (0.099 nm). ● Hence, volume of gas produced is approximately 70 cm3. (Accept 60 – 80 cm3)

[6]

[Total: 16]

2015 TJC H2 Chemistry Preliminary Exam [Turn Over

1

CHEMISTRY (H2) 9647/03 Paper 3 Free Response 16th September 2015 2 hours Candidates answer on separate paper. Additional materials: Answer paper Graph Paper

Data Booklet


Write your name, Civics Group, Centre number and Index number in the spaces provided on

the cover page and on all the work you hand in.

Write in dark blue or black pen on both sides of the paper.

You may use a HB pencil for any diagrams, graphs or rough working.

Do not use staples, paper clips, glue or correction fluid.

Answer any four questions.

A Data Booklet is provided.

The use of an approved scientific calculator is expected, where appropriate.

You are reminded of the need for good English and clear presentation in your answers.

At the end of the examination, fasten all your work securely together.

The number of marks is given in brackets [ ] at the end of each question or part question.



2

1 (a) The halogens and their compounds are useful laboratory reagents and have many applications.

(i) Describe the trend in the colour of the halogens. (ii) Explain, in terms of structure and bonding, the trend in the volatility of the halogens.

(iii) Halogens are oxidising agents.

By quoting appropriate values from the Data Booklet, explain the reactions of the halogens with sodium thiosulfate. Write balanced equations for the reactions.

[6]

(b) Hydrocarbons react with halogens under different conditions to give a wide range of organic

compounds. The Wohl-Ziegler bromination is shown below.

In this reaction, a bromine atom is incorporated into a molecule at the position next to a carbon-carbon double bond. The reagent used for the reaction is N-bromosuccinimide (NBS).

The catalytically active species is bromine, which is present in trace amount in NBS samples. Subsequent amount of bromine is generated in situ when NBS reacts with hydrogen bromide.

(i) NBS reacts with hydrogen bromide in a 1:1 molar ratio to produce bromine and succinimide as the only products. Write a balanced equation for the reaction.

In one such reaction, cyclohexene is converted to 3-bromocyclohexene using NBS.

(ii) Describe the mechanism for the formation of 3-bromocyclohexene from cyclohexene

and bromine. (iii) State the type(s) of stereoisomerism that could be exhibited by 3-bromocyclohexene

and draw the isomers.

N-bromosuccinimide (NBS)


3

In another reaction, compound A can be converted to compound B in two steps. One of the steps is the Wohl-Ziegler bromination.

(iv) Suggest suitable reagents and conditions and the structural formula for the intermediate formed.

(v) State the total number of stereoisomers of compound B and draw the isomers.

[10] (c) The mineral fluorspar, which is mainly calcium fluoride, is a major source of fluorine.

The first stage in liberating the fluorine from calcium fluoride is to grind this compound up and

react it with concentrated sulfuric acid. The products are hydrogen fluoride and calcium sulfate.

(i) Define the standard enthalpy change of formation, ΔHf

o, of a substance.

(ii) Using the following data of ΔHf and lattice energy, construct an energy cycle to

calculate the enthalpy change for the reaction between calcium fluoride and concentrated sulfuric acid.

Hf [HF(g)] = -271 kJ mol-1 Hf [CaSO4(s)] = -1434 kJ mol-1 Hf [Ca2+(g)] = +1918 kJ mol-1

Hf [F-(g)] = -249 kJ mol-1

L.E. [CaF2(s)] = -2640 kJ mol-1 Hf [H2SO4(l)] = -814 kJ mol-1

[4] [Total: 20]

2 (a) In folk medicine, willow bark teas were used as headache remedies and other tonics. Its

analgesic property is due to salicylic acid. Nowadays, salicylic acid is administered in the

form of aspirin which is less irritating to the stomach.

CO2H

CO2H

OCOCH3

(i) In the preparation of aspirin from salicylic acid, water is added to precipitate aspirin.

Explain why aspirin is not soluble in water.

compound A compound B

Salicylic acid Aspirin


4

(ii) Aspirin has a pKa of 3.48 while benzoic acid has a pKa of 4.2. Explain the difference in

their pKa values.

(iii) Draw the structure of the 2 organic compounds formed when aspirin is heated with

NaOH(aq).

(iv) Acid Ka1 / mol dm-3 Ka2 / mol dm-3

Salicylic acid 1.07 x 10-3 1.82 x 10-14

Ethanoic acid 1.80 x 10-5 -

Carbonic acid 4.50 x 10-7 -

Using the information above, suggest the organic products formed if CO2 is bubbled

through the mixture formed in (a)(iii).

[8]

(b) Tautomers are isomers with the same carbon skeleton and differ only in the positions of

hydrogen atoms and electrons. Carbonyl compounds such as ethanal exhibits keto-enol

tautomerisation as shown below.

keto enol

Although these 2 isomers exist in equilibrium, the enol form is less stable than the keto form.

(i) Using data from the Data Booklet, calculate the enthalpy change for the conversion of

the keto to enol.

β-diketones such as pentan-2,4-dione also exhibits tautomerisation.

keto enol

It is found that the enol form is more stable than the keto form.

(ii) Suggest a reason for this.

[3]

(c) 4-hydroxypentan-2-one gives off a fresh herb smell and has the following structure.

4-hydroxypentan-2-one


5

(i) 4-hydroxypentan-2-one can be synthesised from 4-chloropent-1-ene in 3 steps.

4-chloropent-1-ene

Suggest the reagents and conditions used in each step and the structural formula for

the intermediate formed.

(ii) Compound R, C6H12O2, is neutral and does not decolourise Br2(aq). When compound R

is heated with NaOH(aq), one of the products obtained is HCO2Na.

Upon refluxing compound R with acidified KMnO4, compound S and carbon dioxide gas

were formed. Compound S is an isomer of 4-hydroxypentan-2-one. It is chiral and gives

effervescence with Na2CO3(s).

Suggest the structures of compounds R and S, explaining your answer.

[9] [Total: 20]

3 (a) Acid rain has harmful effects on aquatic animals, plants and infrastructure. It contains

sulfuric acid formed from sulfur trioxide. Sulfur trioxide is produced upon atmospheric oxidation of sulfur dioxide from coal burning.

(i) Write an equation for the formation of sulfuric acid from sulfur trioxide.

Even before the existence of acid rain, unpolluted rain water was slightly acidic due to dissolved CO2. The solubility of pure carbon dioxide gas in water is 88 cm3 per 100 cm3 of water under room conditions.

(ii) Given that air contains 0.033% by volume of carbon dioxide, calculate the

concentration in mol dm3 of carbon dioxide dissolved in unpolluted rain water.

Dissolved carbon dioxide forms carbonic acid, H2CO3, in water. The acid dissociation constant of carbonic acid is given below.

H2CO3 HCO3 + H+ Ka1

= 4.5 107 mol dm3

(iii) Using your answers in (a)(ii), calculate the pH of unpolluted rain water (ignore the

effect of Ka2 on the pH).

Some fishes and shellfish die at pH values of 4.5 to 5.0. Lakes with limestone-rich soil can

maintain a relatively stable pH even when acid rain falls due to the HCO3/CO3

2 buffer system.

(iv) With aid of an equation, explain how lakes with limestone-rich soil maintain a relatively stable pH.


6

An environmental chemist needs a carbonate buffer of pH 10.0 to study the effects of acid

rain on limestone-rich soils.

(v) The acid dissociation constant, Ka, of HCO3 is 4.7 1011 mol dm3. Calculate the

mass of solid Na2CO3 the chemist needs to add to 500 dm3 of 0.2 moldm3 NaHCO3 to form a buffer of pH 10.0.

[9] Oxides and halides of Period 3 elements have many applications.

(b) A farmer intends to add a magnesium-containing compound to his farmland to correct

magnesium deficiency and raise soil pH.

(i) Suggest whether the farmer should add magnesium oxide or magnesium chloride to his farmland.

(ii) Explain your answer in (b)(i) by describing the reactions of the compound suggested with water and acid. Write equations where appropriate and state the pH of the solution formed when the compound is dissolved in water.

[4]

(c) Phosphorus tribromide can be used to synthesise bromoethane from ethanol as shown below.

CH3CH2OH + P

Br

Br Br CH3CH2 O PBr2

H

+ Br-

CH3CH2Br + HOPBr2

(i) Name the reaction for the conversion of ethanol to bromoethane.

(ii) Suggest how the reactivity of ethanol is increased in the presence of PBr3.

(iii) Suggest why water cannot be used as a solvent for this reaction.

(iv) Draw a structure to illustrate the shape about the P atom in a molecule of the product,

HOPBr2. Suggest the shape about the P atom.


7

Both bromoethane and 1-bromo-2-methylpropane are primary alkyl halides that undergo

SN2 reaction with CN nucleophile.

CH2 CHBr CH3

CH3

1-bromo-2-methylpropane

(v) Explain the following observations.

Bromoethane undergoes SN2 reaction with CN- nucleophile at a faster rate than 1-

bromo-2-methylpropane.

Both compounds undergo SN2 reactions with CN at a slower rate when the solvent

is ethanol, CH3CH2OH, compared to using DMF, as solvent.

H C N

O CH3

CH3 DMF

[7] [Total: 20]

4 (a) Hydrogen sulfide is a colourless gas with a characteristic ‘rotten egg’ smell. It occurs naturally in

crude petroleum and reacts with methane according to the following equilibrium.

CH4(g) + 2H2S(g) CS2(g) + 4H2 H = +232 kJ mol-1

(i) State and explain the effect on the number of moles of methane when the volume of the vessel is reduced at a constant temperature.

(ii) Write an expression for the equilibrium constant, Kc, stating its units.

(iii) In an experiment, 1.00 mol of CH4, 2.00 mol of H2S, 1.00 mol of CS2 and 2.00 mol of H2 are mixed in a 250 cm3 vessel at 960 °C. The concentration of methane is found to be 5.56 mol dm-3 when the system reaches equilibrium. Calculate the value of Kc at 960 °C.

(iv) The equilibrium constant varies with temperature according to the van’t Hoff equation:

ln

H

(

T

T )

where K1 is the equilibrium constant at temperature T1

K2 is the equilibrium constant at temperature T2

H is the enthalpy change in J mol-1 R is the ideal gas constant and T is temperature in Kelvin.

Using your answer in (a)(iii), calculate the value of equilibrium constant at 480 °C.

(v) Hydrogen sulfide can be removed by passing it through nitric acid. 0.15 mol of H2S was

found to react with 0.10 mol of HNO3 to give a yellow solid and an oxide of nitrogen. Determine the oxidation state of nitrogen in the oxide. Hence, write a balanced equation for the reaction of H2S and HNO3.

[10]


8

(b) Adipic acid is one of the most important dicarboxylic acids and has a wide range of industrial applications. It is used as a monomer in the synthesis of nylon, and in the production of plasticizer and coatings.

adipic acid

cyclohexanone

(i) Show a three-step synthetic route to obtain adipic acid from cyclohexanone. Suggest reagents and conditions you would use and show the structures of any intermediates formed.

(ii) Suggest a simple chemical test to distinguish between adipic acid and cyclohexanone. State the observations and write a balanced equation for any reaction that occurs.

In the 1880s, German chemist Rainer Ludwig Claisen discovered that esters were able to react in the presence of a strong base to give a keto ester. In the first step of the mechanism, a proton is removed from the α carbon of the ester molecule.

(iii) Suggest a reason why the hydrogen atom on the carbon is more acidic than the other hydrogen atoms in the same molecule.

(iv) Name the type of reaction that has taken place between the ester molecules.

(v) Two molecules of an ester undergoes the above reaction to give methanol and the following organic product.

Suggest the structure of the ester.

(vi) Adipic acid undergoes esterification to form diethyl adipate.

diethyl adipate

The two ester functional groups of diethyl adipate can react intramolecularly to give a cyclic product and ethanol. Draw the structure of the cyclic product.

[10] [Total: 20]

OO

R

OO

O O

carbon


9

5 Aluminium chloride, AlCl3, exhibits properties which differ from chlorides of other Period 3

elements. It sublimes at a relatively low temperature of 180⁰C at atmospheric pressure due to the

original lattice structure of AlCl3 being converted into Al2Cl6 molecules. In the presence of excess

water, aluminium chloride forms an acidic solution of pH 3 to 4.

(a) Describe and explain the reactions of aluminium chloride with excess water, writing equations

where appropriate.

[2]

(b) Aluminium chloride reacts with sodium chloride to form sodium chloroaluminate, NaAlCl4.

Sodium chloroaluminate is one of the simplest compounds containing chloroaluminate anion

and has a melting point of 185 ⁰C.

(i) Explain why sodium chloroaluminate would be formed from the above reaction, stating

the type of bond that is formed during this reaction.

(ii) Draw the dot-and-cross diagram of chloroaluminate anion. Using the valence shell

electron pair repulsion theory, state its shape and bond angle.

(iii) A class of compounds called room temperature ionic liquids (RTILs) can be formed

from chloroaluminate anion and organic cations such as 1-ethyl-3-methylimidazolium

and pyridinium ions.

1-ethyl-3-methylimidazolium pyridinium ion

Explain the differences in physical states between RTILs and sodium chloroaluminate

in terms of bonding.

[6]

(c) To determine the rate equation of the following chlorate-chloride reaction, an experiment was

conducted using 0.000480 mol dm-3 of ClO3–, 0.1 mol dm-3 of Cl

– and 0.4 mol dm-3 of H+.

2ClO3– + 2Cl

– + 4H+ → Cl2 + 2ClO2• + H2O

At regular 5-minutes intervals, small samples of the reaction mixture were withdrawn,

quenched and placed into the UV-vis spectrometer to record its absorbance value. The

absorbance value corresponds to the concentration of the product ClO2•.

(i) The results of the above experiment were shown below.

Time/min 0 5 10 15 20 25

Absorbance/A 0.000 0.211 0.348 0.436 0.494 0.531

Plot the graph of absorbance/A against time/min.


10

(ii) Beer-Lambert’s Law states that the absorbance values, A, is directly proportional to

the concentration of absorbing species, c, as shown below.

A = cl

where is the molar extinction coefficient and l is the path length, which is usually

1.0 cm.

This equation can be used to calculate the absorbance value when maximum amount

of ClO2• was formed from the above chlorate-chloride reaction.

Prove that the maximum absorbance value in the above experiment is 0.600, given

that of ClO2• is 50 mol–1dm3cm–1. Show your working clearly.

(iii) Using your graph plotted in (c)(i) and the information given in (c)(ii), determine the

half-life of the experiment and hence the order of reaction with respect to [ClO3–].

A research student carried out a series of experiments to further investigate the order of reaction with respect to [Cl

–] and [H+] in the chlorate-chloride reaction.

The following graphs were plotted.

(iv) Using the above data, deduce the order of the reaction with respect to [Cl– ] and [H+].

(v) Hence, write a rate equation for the chlorate-chloride reaction.

(vi) Using the half-life obtained from Experiment 1 in (c)(iv), calculate a value for the rate

constant of the rate equation suggested in (c)(v).

(vii) The following mechanism was proposed by a research student.

Step 1: ClO3– + Cl

– + 2H+ HClO2 + HOCl (slow)

Step 2: HClO2 + ClO3– + H+

2ClO2• + H2O (fast)

Step 3: HOCl + Cl– + H+ Cl2 + H2O (fast)

From the rate equation in (c)(v), deduce if the proposed mechanism is to be

accepted or rejected.

[12]

[Total: 20]

Experiment 1

[ClO3–] = 0.12 mol dm

−3

[H+] = 0.5 mol dm

−3

0.012

0.008

0.004

[Cl–]

0.0000

0.002

0.006

0.010

5 10 15 20

Time / min 0

Experiment 2

Assuming that [ClO3–] and

[Cl–] are constant,

Initial Rate

[H+]2

0


1

CHEMISTRY (H2) 9647/03 Paper 3 Free Response 16th September 2015 2 hours Candidates answer on separate paper. Additional materials: Answer paper Graph Paper

Data Booklet


Write your name, Civics Group, Centre number and Index number in the spaces provided on

the cover page and on all the work you hand in.

Write in dark blue or black pen on both sides of the paper.

You may use a HB pencil for any diagrams, graphs or rough working.

Do not use staples, paper clips, glue or correction fluid.

Answer any four questions.

A Data Booklet is provided.

The use of an approved scientific calculator is expected, where appropriate.

You are reminded of the need for good English and clear presentation in your answers.

At the end of the examination, fasten all your work securely together.

The number of marks is given in brackets [ ] at the end of each question or part question.



2

1 (a) The halogens and their compounds are useful laboratory reagents and have many

applications.

(i) Describe the trend in the colour of the halogens.

• Increasing intensity or darker down the group. • Cl2 (g) yellowy-green; Br2 (l) reddish-brown; I2 (s) black

(ii) Explain, in terms of structure and bonding, the trend in the volatility of the halogens.

The halogens have simple molecular structure and weak van der Waals forces

of attraction between molecules.

Down the group, Mr of X2 increases, giving a larger electron-cloud, which can be more easily distorted, resulting in stronger van der Waals forces hence

requiring more energy to overcome. The volatility decreases from Cl2 to I2.

(iii) Halogens are oxidising agents.

By quoting appropriate values from the Data Booklet, explain the reactions of the halogens with sodium thiosulfate. Write balanced equations for the reactions.

[6]

½Cl2 + e Cl- E = +1.36 V

½Br2 + e Br- E = +1.07 V ½I2 + e I

- E = +0.54 V

(With quoting of Data) Oxidising power of the halogens decreases down the

group as can be seen from the less positive E value, hence iodine can only

oxidise thiosulfate to tetrathionate.

[•Bonus mark]

4Cl2 + S2O32- + 5H2O 8Cl- + 2SO4

2- + 10H+

4Br2 + S2O32- + 5H2O 8Br- + 2SO4

2- + 10H+

I2 + 2S2O32- S4O6

2- + 2I-

(b) Hydrocarbons react with halogens under different conditions to give a wide range of organic

compounds. The Wohl-Ziegler bromination is shown below.

In this reaction, a bromine atom is incorporated into a molecule at the position next to a carbon-carbon double bond. The reagent used for the reaction is N-bromosuccinimide (NBS).

N-bromosuccinimide (NBS)

2015 TJ Prelim H2 Paper 3 Answers


3

The catalytically active species is bromine, which is present in trace amount in NBS samples. Subsequent amount of bromine is generated in situ when NBS reacts with hydrogen bromide.

(i) NBS reacts with hydrogen bromide in a 1:1 molar ratio to produce bromine and succinimide as the only products. Write a balanced equation for the reaction.

In one such reaction, cyclohexene is converted to 3-bromocyclohexene using NBS.

(ii) Describe the mechanism for the formation of 3-bromocyclohexene from cyclohexene

and bromine.

Free radical substitution

•• Mechanism Initiation

Propagation

Termination

Br2 2Br

2Br Br2


4

(iii) State the type(s) of stereoisomerism that could be exhibited by 3-bromocyclohexene and draw the isomers.

Optical isomerism

In another reaction, compound A can be converted to compound B in two steps. One of the

steps is the Wohl-Ziegler bromination.

(iv) Suggest suitable reagents and conditions and the structural formula for the intermediate formed.

••

compound A compound B

* *


5

(v) State the total number of stereoisomers of compound B and draw the isomers. [10]

•• 3

(c) The mineral fluorspar, which is mainly calcium fluoride, is a major source of fluorine.

The first stage in liberating the fluorine from calcium fluoride is to grind this compound up and

react it with concentrated sulfuric acid. The products are hydrogen fluoride and calcium sulfate.

(i) Define the standard enthalpy change of formation, ΔHf

o, of a substance.

It is the enthalpy change when one mole of the substance is formed from its constituent elements in their standard states at 298 K and 1 atm.

(ii) Using the following data of ΔHf and lattice energy, construct an energy cycle to

calculate the enthalpy change for the reaction between calcium fluoride and concentrated sulfuric acid.

Hf [HF(g)] = -271 kJ mol-1 Hf [CaSO4(s)] = -1434 kJ mol-1 Hf [Ca2+(g)] = +1918 kJ mol-1

Hf [F-(g)] = -249 kJ mol-1

L.E. [CaF2(s)] = -2640 kJ mol-1 Hf [H2SO4(l)] = -814 kJ mol-1

[4]


6

CaF2(s) + H2SO4(l) CaSO4(s) + 2HF(g) Ca2+(g) + 2F-(g) + H2(g) + S(s) + 2O2(g) Ca(s) + F2(g) + H2(g) + S(s) + 2O2(g) By Hess’ Law,

Hr – 2640 + 1918 + 2(–249) + (–814) = –1434 + 2(–271)

Hr = +58 kJ mol-1

•Balanced equation for the reaction •working •Answer

[Total: 20]

2 (a) In folk medicine, willow bark teas were used as headache remedies and other tonics. Its

analgesic property is due to salicylic acid. Nowadays, salicylic acid is administered in the

form of aspirin which is less irritating to the stomach.

CO2H

CO2H

OCOCH3

(i) In the preparation of aspirin from salicylic acid, water is added to precipitate aspirin.

Explain why aspirin is not soluble in water.

Both aspirin and water have simple molecular structures.

Aspirin has both van der Waals interaction and hydrogen bonding between its

molecules. Water has hydrogen bonding between its molecules. The energy

released from the hydrogen bonding formed between aspirin and water

molecules release insufficient energy to overcome the van der Waals interaction

and hydrogen bonding between aspirin molecules and the hydrogen bonding

between water molecules. Therefore, aspirin is insoluble in water.

• 1m – explanation

• 1m – bonding and structure

Salicylic acid Aspirin

Hr

–1434 + 2(–271)

+1918 + 2(–249)

– 2640 – 814


7

(ii) Aspirin has a pKa of 3.48 while benzoic acid has a pKa of 4.2. Explain the difference in

their pKa values.

• As Aspirin has a lower pKa than benzoic acid, aspirin is a stronger acid.

• Aspirin has an electron-withdrawing –OCOCH3 group which further disperses

the negative charge on

CO2-

OCOCH3, making it more stable than

benzoate.

(iii) Draw the structure of the 2 organic compounds formed when aspirin is heated with

NaOH(aq).

•

CO2-Na+

O-Na+

• CH3CO2-Na+

(iv) Acid Ka1 / mol dm-3 Ka2 / mol dm-3

Salicylic acid 1.07 x 10-3 1.82 x 10-14

Ethanoic acid 1.80 x 10-5 -

Carbonic acid 4.50 x 10-7 -

Using the information above, suggest the organic products formed if CO2 is bubbled

through the mixture formed in (a)(iii).

[8]

Carbon dioxide dissolves in water to form carbonic acid. As carbonic acid is

stronger than the phenolic group in salicylic acid but weaker than carboxylic acid

group in salicylic acid and ethanoic acid, the products formed are

•

CO2-Na+

OHand • CH3CO2

-Na+.


8

(b) Tautomers are isomers with the same carbon skeleton and differ only in the positions of

hydrogen atoms and electrons. Carbonyl compounds such as ethanal exhibits keto-enol

tautomerisation as shown below.

keto enol

Although these 2 isomers exist in equilibrium, the enol form is less stable than the keto form.

(i) Using data from the Data Booklet, calculate the enthalpy change for the conversion of

the keto to enol.

Bonds broken: C=O, C-H, C-C

Bonds formed: C=C, C-O, O-H

• ∆H = +740 + 410 + 350 – 610 – 360 – 460

= • + 70 kJ mol-1

(ii) β-diketones such as pentan-2,4-dione also exhibits tautomerisation.

keto enol

It is found that the enol form is more stable than the keto form. Suggest a reason for

this.

[3]

• Intramolecular hydrogen bonding stabilises the enol form.

Alternative answer:

The enol is stabilised by the resonance of the conjugated double bonds (C=C and

C=O)

Hydrogen bond


9

(c) 4-hydroxypentan-2-one gives off a fresh herb smell and has the following structure.


(i) 4-hydroxypentan-2-one can be synthesised from 4-chloropent-1-ene in 3 steps.

4-chloropent-1-ene

Suggest the reagents and conditions used in each step and the structural formula for

the intermediate formed.


5 correct: 3m, 3-4 correct: 2m, 1-2 correct: 1m

Note: Oxidation cannot be the last step and cannot use KMnO4 for oxidation

which would otherwise oxidise the alkene. For hydration, also accept cold conc.

H2SO4, followed by heating with water at 80oC.

Alternative route: hydrate first, oxidise then hydrolysis of RX. In this route, can

use KMnO4.

(ii) Compound R, C6H12O2, is neutral and does not decolourise Br2(aq). When compound R

is heated with NaOH(aq), one of the products obtained is HCO2Na.

Upon refluxing compound R with acidified KMnO4, compound S and carbon dioxide gas

were formed. Compound S is an isomer of 4-hydroxypentan-2-one. It is chiral and gives

effervescence with Na2CO3(s).

Suggest the structures of compounds R and S, explaining your answer.

[9]

NaOH(aq) heat

K2Cr2O7/H+

heat

Steam, H3PO4, 300oC, 65 atm


10

● R is neutral and undergoes alkaline hydrolysis with NaOH(aq), it contains an

ester group.

● As R does not undergo electrophilic addition with Br2(aq), C=C is absent.

● After refluxing R with acidified KMnO4, R undergoes hydrolysis and oxidation

to form compound S and CO2 gas, R is an ester.

● S has a chiral carbon with 4 different groups, has no plane of symmetry and

non-superimposable mirror images.

● S has a –COOH (carboxylic acid group) as it gives effervescence with

Na2CO3(s).

● S is

● R is

[1 bonus mark; 7 marking points]

[Total: 20]

3 (a) Acid rain has harmful effects on aquatic animals, plants and infrastructure. It contains

sulfuric acid formed from sulfur trioxide. Sulfur trioxide is produced upon atmospheric oxidation of sulfur dioxide from coal burning.

(i) Write an equation for the formation of sulfuric acid from sulfur trioxide.

SO3 + H2O H2SO4

Even before the existence of acid rain, unpolluted rain water was slightly acidic due to dissolved CO2. The solubility of pure carbon dioxide gas in water is 88 cm3 per 100 cm3 of water under room conditions.

(ii) Given that air contains 0.033% by volume of carbon dioxide, calculate the

concentration in mol dm3 of carbon dioxide dissolved in unpolluted rain water.

If air contains 0.033% by volume of carbon dioxide,

Volume of CO2 dissolved per 100 cm3 water =

= 0.0290 cm3

No. of moles of CO2 dissolved per 100 cm3 water

=

= 1.21 106 mol

Concentration of dissolved CO2 in water =

= 1.21 105 mol dm3


11

Dissolved carbon dioxide forms carbonic acid, H2CO3, in water. The acid dissociation constant of carbonic acid is given below.

H2CO3 HCO3 + H+ Ka1

= 4.5 107 mol dm3

(iii) Using your answers in (a)(ii), calculate the pH of unpolluted rain water (ignore the

effect of Ka2 on the pH).

Dissolved CO2 forms carbonic acid, H2CO3.

CO2 + H2O H2CO3

H2CO3 HCO3 + H+

Ka = [ ]

[H+] = √ = 2.33 106 mol dm3

pH = -lg (2.33 106) = 5.63

Some fishes and shellfish die at pH values of 4.5 to 5.0. Lakes with limestone-rich soil can

maintain a relatively stable pH even when acid rain falls due to the HCO3/CO3

2 buffer system.

(iv) With aid of an equation, explain how lakes with limestone-rich soil maintain a relatively stable pH.

CO32 + H+ HCO3

The base, CO32 (from limestone), reacts with H+ to form HCO3

. Negligible changes in pH as H+ ions are removed.

An environmental chemist needs a carbonate buffer of pH 10.0 to study the effects of acid

rain on limestone-rich soils.

(v) The acid dissociation constant, Ka, of HCO3 is 4.7 1011 mol dm3. Calculate the

mass of solid Na2CO3 the chemist needs to add to 500 dm3 of 0.2 moldm3 NaHCO3 to form a buffer of pH 10.0.

[9]

pH = pKa + lg [

]

[ ]

10 = -lg (4.7 1011) + lg [

]

[CO32] = 0.0940 mol dm3

Mass of solid Na2CO3 = 500 x 0.094 106 = 4.98 kg

Oxides and halides of Period 3 elements have many applications.

(b) A farmer intends to add a magnesium-containing compound to his farmland to correct

magnesium deficiency and raise soil pH.

(i) Suggest whether the farmer should add magnesium oxide or magnesium chloride to his farmland.

Magnesium oxide

(ii) Explain your answer in (b)(i) by describing the reactions of the compound suggested with water and acid. Write equations where appropriate and state the pH of the solution formed when the compound is dissolved in water.

[4]


12

MgO reacts with water to a very small extent (or almost insoluble in water) due

to its high lattice energy, forming magnesium hydroxide, Mg(OH)2, which is

sparingly soluble. pH of solution formed is 9.

MgO(s) + H2O(l) Mg(OH)2(aq)

MgO also raises pH by removing/reacting with acid to form salt and water.

MgO(s) + 2H+(aq) Mg2+(aq) + H2O(l)

(c) Phosphorus tribromide can be used to synthesise bromoethane from ethanol as shown below.

CH3CH2OH + P

Br

Br Br CH3CH2 O PBr2

H

+ Br-

CH3CH2Br + HOPBr2

(i) Name the reaction for the conversion of ethanol to bromoethane.

Nucleophilic substitution

(ii) Suggest how the reactivity of ethanol is increased in the presence of PBr3.

The intermediate

CH3CH2 O PBr2

H formed from PBr3 is more electron-deficient and susceptible to nucleophilic attack by Br-.

OR HOPBr2 is a better leaving group than OH-.

(iii) Suggest why water cannot be used as a solvent for this reaction.

PBr3 undergoes hydrolysis in water to give H3PO3 and HBr.

PBr3(s) + 3H2O(l) H3PO3(aq) + 3HBr(aq)

(iv) Draw a structure to illustrate the shape about the P atom in a molecule of the product,

HOPBr2. Suggest the shape about the P atom.

HO

P

BrBr

Trigonal pyramidal

Both bromoethane and 1-bromo-2-methylpropane are primary alkyl halides that undergo

SN2 reaction with CN nucleophile.

CH2 CHBr CH3

CH3

1-bromo-2-methylpropane


13

(v) Explain the following observations. Bromoethane undergoes SN2 reaction with CN- nucleophile at a faster rate than 1-

bromo-2-methylpropane.

Both compounds undergo SN2 reactions with CN at a slower rate when the solvent

is ethanol, CH3CH2OH, compared to using DMF, as solvent.

H C N

O CH3

CH3 DMF

[7] [Total: 20] The nucleophile experiences less steric hindrance during the SN2 reaction

with bromoethane.

H on –OH group in ethanol forms hydrogen bond with the nucleophile CN-, making CN- less reactive.

4 (a) Hydrogen sulfide is a colourless gas with a characteristic ‘rotten egg’ smell. It occurs naturally in

crude petroleum and reacts with methane according to the following equilibrium.

CH4(g) + 2H2S(g) CS2(g) + 4H2(g) = +232 kJ mol-1

(i) State and explain the effect on the number of moles of methane when the volume of the vessel is reduced at a constant temperature.

When the volume of vessel is reduced, the pressure of the system increases. By

Le Chatelier’s Principle, the system counteracts the change by reducing the total

pressure by favouring the backward reaction which generates less number of

moles of gases.

Hence, the number of moles of methane increases (answer must come with

correct explanation to gain the credit).

(ii) Write an expression for the equilibrium constant, Kc, stating its units.

Kc =

[CS ][H ]4

[CH4][H S]

mol2dm-6

(iii) In an experiment, 1.00 mol of CH4, 2.00 mol of H2S, 1.00 mol of CS2 and 2.00 mol of H2

are mixed in a 250 cm3 vessel at 960 °C. The concentration of methane is found to be 5.56 mol dm-3 when the system reaches equilibrium. Calculate the value of Kc at 960 °C.

working

CH4(g) +

2H2S(g) CS2(g)+ 4H2(g)

Initial conc/mol dm-3 4 8 4 8 Change/ mol dm-3 +1.56 +3.12 -1.56 -6.24 Equilibrium/ mol dm-3 5.56 11.12 2.44 1.76

Kc = 2.44 x 1.764 / (5.56 x 11.122) = 0.0341 mol2dm-6


14

(iv) The equilibrium constant varies with temperature according to the van’t Hoff equation:

ln

H

(

T

T )

where K1 is the equilibrium constant at temperature T1

K2 is the equilibrium constant at temperature T2 ΔH is the enthalpy change in J mol-1 R is the ideal gas constant and T is temperature in Kelvin.

Using your answer in (iii), calculate the value of equilibrium constant at 480 °C.

working

lnKc

0 0341=

3 103

31(1

753

1

1 33)

Kc = 1.84 x 10-8 mol2dm-6

(v) Hydrogen sulfide can be removed by passing it through nitric acid. 0.15 mol of H2S was found to react with 0.10 mol of HNO3 to give a yellow solid and an oxide of nitrogen. Determine the oxidation state of nitrogen in the oxide. Hence, write a balanced equation for the reaction of H2S and HNO3.

[10]

3 H2S ≡ 3S ≡ HNO3 ≡ 6e-

HNO3 ≡ 3e

-

Oxidation state of nitrogen decreases from +5 in HNO3 to +2 in the product

3H2S + 2HNO3 → 3S + NO + 4H2O

(b) Adipic acid is one of the most important dicarboxylic acids and has a wide range of industrial applications. It is used as a monomer in the synthesis of nylon, and in the production of plasticizer and coatings.

adipic acid

cyclohexanone

(i) Show a three-step synthetic route to obtain adipic acid from cyclohexanone. Suggest reagents and conditions you would use and show the structures of any intermediates formed.


15

LiAlH4 in dry ether, rtp

Accept other reducing agents e.g. H2/Ni, high temp or NaBH4, alcohol, rtp

excess concentrated H2SO4, 180 °C

acidic KMnO4, reflux

[reagent & condition + intermediate = 1 mark]

(ii) Suggest a simple chemical test to distinguish between adipic acid and cyclohexanone. State the observations and write a balanced equation for any reaction that occurs.

2,4-dinitrophenylhydrazine, rtp

Orange ppt with cyclohexanone, no observable reaction with adipic acid

OR

Na2CO3 (aq), rtp

Effervescence of CO2 with adipic acid, no observable change with

cyclohexanone

HO2C-(CH2)4-CO2H + Na2CO3 → Na+ –O2C-(CH2)4-CO2–Na+ + CO2 + H2O

In the 1880s, German chemist Rainer Ludwig Claisen discovered that esters were able to react in the presence of a strong base to give a keto ester. In the first step of the mechanism, a proton is removed from the α carbon of the ester molecule.

(iii) Suggest a reason why the hydrogen atom on the carbon is more acidic than the other hydrogen atoms in the same molecule.

OO

R

OO

carbon


16

The anion formed by removal of the -proton (hydrogen atom on carbon next to the carbonyl carbon) is stabilised by delocalisation of electrons, hence the H+ is more likely to dissociate from the ester molecule.

(iv) Name the type of reaction that has taken place between the ester molecules.

condensation/ nucleophilic (acyl) substitution/ nucleophilic addition followed by

elimination

(v) Two molecules of an ester undergoes the above reaction to give methanol and the following organic product.

Suggest the structure of the ester.

CH3CH2CO2CH3

(vi) Adipic acid undergoes esterification to form diethyl adipate.

diethyl adipate

The two ester functional groups of diethyl adipate can react intramolecularly to give a cyclic product and ethanol. Draw the structure of the cyclic product.

[10]

[Total: 20] 5 Aluminium chloride, AlCl3, exhibits properties which differ from chlorides of other Period 3

elements. It sublimes at a relatively low temperature of 180⁰C at atmospheric pressure due to the

original lattice structure of AlCl3 being converted into Al2Cl6 molecules. In the presence of excess

water, aluminium chloride forms an acidic solution of pH 3 to 4.

(a) Describe and explain the reactions of aluminium chloride with excess water, writing equations

where appropriate.

[2]

AlCl3 undergoes hydration in water and dissolves to give an aqueous solution.

AlCl3 + 6H2O [Al(H2O)6]3+ + 3Cl

–

Hydrolysis also occurs due to the high charge density of Al3+ ions. It polarizes the

water molecule and weakens the O-H bond of water making it easier for a H+ ion to

leave the water molecule. An acidic solution of pH = 3 – 4 is obtained.

O O


17

[Al(H2O)6]3+ [Al(H2O)5OH]2+ + H+

(b) Aluminium chloride reacts with sodium chloride to form sodium chloroaluminate, NaAlCl4.

Sodium chloroaluminate is one of the simplest compounds containing chloroaluminate anion

and has a melting point of 185 ⁰C.

(i) Explain why sodium chloroaluminate would be formed from the above reaction, stating

the type of bond that is formed during this reaction.

Aluminium in AlCl3 is electron deficient (or has vacant orbitals) as it only has 6

electrons around it and Cl– anion has lone pairs of electrons and thus,

dative bond is formed in the chloroaluminate anion.

(ii) Draw the dot-and-cross diagram of chloroaluminate anion. Using the valence shell

electron pair repulsion theory, state its shape and bond angle.

1m – dot-and-cross

Shape: Tetrahedral

Bond angle: 109.5⁰

(iii) A class of compounds called room temperature ionic liquids (RTILs) can be formed

from chloroaluminate anion and organic cations such as 1-ethyl-3-methylimidazolium

and pyridinium ions.

1-ethyl-3-methylimidazolium pyridinium ion

Explain the differences in physical states between RTILs and sodium chloroaluminate

in terms of bonding.

[6]

RTIL and sodium chloroaluminate are held by ionic bonds. RTIL have a lower

melting point and exists as a liquid and sodium chloroaluminate exists as a

solid because the organic cation in RTIL is larger than sodium ion and thus,

weaker ionic bonds exist between the oppositely-charged ions in RTIL as

| | |

|


18

(c) To determine the rate equation of the following chlorate-chloride reaction, an experiment was

conducted using 0.000480 mol dm-3 of ClO3–, 0.1 mol dm-3 of Cl

– and 0.4 mol dm-3 of H+.

2ClO3– + 2Cl

– + 4H+ → Cl2 + 2ClO2• + H2O

At regular 5-minutes intervals, small samples of the reaction mixture were withdrawn,

quenched and placed into the UV-vis spectrometer to record its absorbance value. The

absorbance value corresponds to the concentration of the product ClO2•.

(i) The results of the above experiment were shown below.

Time/min 0 5 10 15 20 25

Absorbance/A 0.000 0.211 0.348 0.436 0.494 0.531

Plot the graph of absorbance/A against time/min.

(ii) Beer-Lambert’s Law states that the absorbance values, A, is directly proportional to

the concentration of absorbing species, c, as shown below. The general Beer-

Lambert’s Law is usually written as,

A = cl


19

where is the molar extinction coefficient and l is the path length, which is usually

1.0cm.

This equation can be used to calculate the absorbance value when maximum amount

of ClO2• was formed from the above chlorate-chloride reaction.

Prove that the maximum absorbance value in the above experiment is 0.600, given

that of ClO2• is 50 mol–1dm3cm–1. Show your working clearly.

Limiting reagent is ClO3–

ClO3– ≡ ClO2•

Maximum [ClO2•] = 0 0004 0 moldm-3

Applying Beer-Lambert’s Law, A = cl

Maximum absorbance = (1250)(0.000480)(1) = 0.600 (proven)

(iii) Using your graph plotted in (c)(i) and the information given in (c)(ii), determine the

half-life of the experiment and hence the order of reaction with respect to [ClO3–].

1m – dotted lines to read off half-life (t1/2) on the graph

Since half-life (t1/2) is constant at 8 min, the reaction is 1st order with respect to

[ClO3–].

A research student carried out a series of experiments to further investigate the order of reaction with respect to [Cl

–] and [H+] in the chlorate-chloride reaction. The following graphs were plotted. (iv)

Using the above data, deduce the order of the reaction with respect to [Cl

– ] and [H+].

From experiment 1, the graph of [Cl

–] shows a constant half-life of 5 min.

Rate [Cl–]1

1st order with respect to [Cl–]

From experiment 2, the graph of initial rate vs. [H+]2 is a straight line, hence

rate of reaction is directly proportional to [H+]2. Rate [H+]2

2nd order with respect to [H+].

Experiment 1

[ClO3–] = 0.12 mol dm

−3

[H+] = 0.5 mol dm

−3

0.012

0.008

0.004

[Cl–]

0.0000

0.002

0.006

0.010

5 10 15 20

Time / min 0

Experiment 2

Assuming that [ClO3–] and

[Cl–] are constant,

Initial Rate

[H+]2

0


20

(v) Hence, write a rate equation for the chlorate-chloride reaction.

Rate = k[ClO3–][Cl

–][H+]2

(vi) Using the half-life obtained from Experiment 1 in (c)(iv), calculate a value for the rate

constant of the rate equation suggested in (c)(v).

Rate = k[ClO3–][Cl

–][H+]2

Since [ClO3–] and [H+] are in large excess,

Rate = k’[Cl–] where •k’ = k[ClO3

–][H+]2

[ ][ ]

[ ][ ]

•k = 4.62 mol–3 dm9 min–1

(vii) The following mechanism was proposed by a research student.

Step 1: ClO3– + Cl

– + 2H+ HClO2 + HOCl (slow)

Step 2: HClO2 + ClO3– + H+

2ClO2• + H2O (fast)

Step 3: HOCl + Cl– + H+ Cl2 + H2O (fast)

From the rate equation in (c)(v), deduce if the proposed mechanism is to be

accepted or rejected.

Yes, the proposed mechanism could be accepted. Based on the rate

equation, it is 1 ion of ClO3–, 1 ion of Cl

– and 2 ions of H+ reacting in the slow

step which agrees with the proposed mechanism and the overall balanced

equation from all 3 steps in the mechanism matches the corresponding

equation (2ClO3– + 2 Cl

– + 4H+ Cl2 + 2ClO2• + H2O)

[12]

[Total: 20]

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