Question Bank.pdf

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Multiple-choice Questions

1. Which of the following statements about the Period 3 oxides (from sodium to chloride) is true?A The bonding between the molecules becomes covalent across the period.B Silicon dioxide reacts with hydrochloric acid and sodium hydroxide.C Magnesium oxide readily dissolves in water.D Sodium oxide dissolves forming a solution with the highest pH.

2. An oxide of X dissolves in water forming a solution Y which readily absorbs carbon dioxide gas. X can form Y because X A is very reactive with water.B has very negative standard reduction potential.C forms a cation with a high charge density.D has many valence electrons.

3. Which of the following about Group 2 elements is true?A Beryllium chloride is ionic.B Barium sulphate is insoluble in water.C The nitrates when heated form nitrogen dioxide gas and metal oxides.D Barium has high charge density.

4. Which of the following oxides does not dissolve in water?A SO

2 C Al

2O

3

B SiO2 D P

4O

10

5. Which of the following electronic configuration of elements is most likely to react with steam than water?A 1s22s22p63s1 C 1s22s22p63s23p1 B 1s22s22p63s2 D 1s22s22p63s23p2

Structured Question

1. (a) Strontium is below calcium in the Periodic Table. (i) Write the valence electronic configuration of strontium atom. (ii) State how strontium reacts with water and write the equation for the reaction. (iii) Write the equation for the heating of strontium nitrate. State whether strontium nitrate

has a higher thermal stability compared to calcium nitrate.(b) The oxides of aluminium and sulphur have different physical and chemical properties. (i) State one difference in the physical properties of both of them. (ii) Write equations for the reactions if any, of the oxides of these elements with aqueous

acid and aqueous alkali.

Essay Question

1. (a) Describe the combustion in oxygen of sodium and phosphorus, writing equations for both reactions.

(b) The oxides of sodium, silicon and phosphorus differ in their reactions with water.Write equations for the reactions and explain the difference in the reactions by referring to their structure and bonding.

(c) Explain how you would differentiate two separate test tubes, one containing magnesium chloride and the other, barium chloride. State the reagent, observation and write equations for both.

1 Period 3 & Group 2 – Selected Properties

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2 © Oxford Fajar Sdn. Bhd. (008974-T) 2008


1. D Sodium oxide, Na2O is basic. It dissolves

readily in water forming an alkaline solution, NaOH, which has a very high pH.

2. B X is a basic oxide as it dissolves in water to form an alkaline solution which absorbs the acidic gas, carbon dioxide. X must have a very negative standard reduction potential as it is a very strong reducing agent.

3. B The sulphates of Group 2 elements are insoluble going down the group.

4. C Aluminium oxide is amphoteric. Therefore it does not dissolve in water due to its covalent nature.

5. B Group 2 element, magnesium has its oxide reacting with steam rather than water.

Structured Question

1. (a) (i) 5s2

(ii) Strontium reacts vigorously with water liberating hydrogen gas.Sr + 2H

2O → Sr(OH)

2 + H

2

(iii) 2Sr(NO3)

2 → 2SrO + 4NO

2 + O

2

Strontium nitrate has a higher thermal stability than calcium nitrate.

(b) (i) Aluminium oxide has a high melting point but sulphur dioxide has a low melting point.

(ii) Aluminium oxide is amphoteric. It reacts with both acid and alkali.

Al2O

3 + 6HCl → 2AlCl

3 + 3H

2O

Al2O

3 + 2NaOH + 3H

2O → 2NaAl(OH)

4

Sulphur dioxide is an acidic gas. It reacts only with alkali.

SO2 + 2NaOH → Na

2SO

3 + H

2O

Essay Question

1. (a) When burning sodium metal is lowered in a jar with excess oxygen, sodium glows more brightly with a yellow flame, a yellow substance, sodium peroxide, an ionic compound

is formed and in limited oxygen supply, sodium oxide is formed. Both the oxides are basic.

2Na(s) + O2(g) → Na

2O

2(s)

4Na(s) + O2(g) → 2Na

2O(s)

Phosphorus burns readily in air to form acidic oxides. White phosphorus burst into flames in the presence of oxygen. In the presence of excess oxygen, phosphorus pentoxide, P

4O

10

is formed. In a limited supply of oxygen, phosphorus trioxide, P

4O

6 is formed. Both the

oxides are acidic.

P4(s) + 5O

2(g) → P

4O

10(s)

P4(s) + 3O

2(g) → P

4O

6(s)

(b) Sodium peroxide and sodium oxide are basic oxide. They have giant ionic lattice structures. The bonds are ionic and they are easily broken when dissolved in water. The reaction is exothermic so the heat released is enough to break the bonds. Sodium peroxide readily dissolves in water forming sodium hydroxide (an alkaline solution) and hydrogen peroxide whereas sodium oxide dissolves in water forming an alkaline solution and water.

Na2O

2 + 2H

2O → 2NaOH + H

2O

2

Na2O + 2H

2O → 2NaOH + H

2O

P4O

10 is a simple covalent molecule with

weak van der Waals forces between them. It readily dissolves in water forming an acidic solution.

P4O

10 + 6H

2O → 4H

3PO

4

Silicon dioxide exists as giant covalent molecules with strong Si – O covalent bonds. These bonds do not break in water, hence they are insoluble in water.

(c) Add dilute sulphuric acid to both the test tubes and shake them. A white precipitate will form in the test tube that contains barium chloride as barium sulphate is insoluble in water but no precipitate will form with magnesium chloride as magnesium sulphate is soluble in water.

MgCl2 + H

2SO

4 → MgSO

4(aq) + 2HCl

BaCl2 + H

2SO

4 → BaSO

4(s) + 2HCl

ANSWER

CHAPTER 1 Period 3 & Group 2 – Selected Properties




1. The element that does not belong to Group 13 of the Periodic Table isA boron C gallium B aluminium D rubidium

2. Which metal is protected by a layer of its own oxide?A Al C Au B Ag D Fe

3. Which of the following does not affect aluminium?A Oxygen C Alkali B Pure water D Acid

4. Cryolite is used in the electrolytic extraction of aluminium A to obtain pure aluminium. C to lower the melting point of bauxite.B to protect the anode. D as a reducing agent.

5. Most of the chemical properties of aluminium is due to the strong polarising power of Al3+ ion. From the following characteristics seen in aluminium compounds, which one cannot be explained by the polarising power factor? A High lattice energy of aluminium fluorideB The covalent characteristics seen in aluminium chloride bondingC The ability of the Al3+ ion to form complex compoundD High hydration energy of Al3+ ion

6. In the manufacture of aluminium through the electrolysis process, the electrolyte is made up of a mixture containing cryolite, Na

3AlF

6, with aluminium oxide, Al

2O

3. One theory suggests

that the reaction between the two compounds takes place as shown below.

Na3AlF

6 + Al

2O

3 → 3Al3+ + 3Na+ + 3O2– + 6F–

The above reaction equation shows thatA aluminium oxide behaves as an acid.B aluminium oxide behaves as a base.C aluminium oxide and cryolite undergo redox reactions.D cryolite helps aluminium oxide to form ions.

7. Although aluminium is a very reactive metal, as shown by its standard redox potential value of – 1.66 V, it is able to withstand corrosion. Which of the following statements below explains the above observation? A Al3+ ion has a high charge density.B Aluminium atoms are closely-packed.C Aluminium forms an oxide layer on its surface.D Aluminium forms an amphoteric compound.

Structured Questions

1. (a) Aluminium chloride, Al2Cl

6, is produced when aluminium reacts with chlorine gas.

(i) Draw the structural formula for aluminium chloride. (ii) Write a balanced equation for the above reaction. (iii) State one use of aluminium chloride.(b) If 1.0 g of aluminium is used in the reaction in (a)(ii), 2.3 g of aluminium chloride will be

produced. Calculate the percentage of product formed.

2 Group 13 - Aluminium



(c) Explain the following statements: (i) Aluminium chloride sublimes at 180 °C. (ii) The aqueous solution of aluminium chloride shows acidic characteristics.

2. (a) Aluminium hydroxide is precipitated out when carbon dioxide gas is allowed to flow through an aluminate solution.

(i) Write an equation to represent the precipitation of aluminium hydroxide. (ii) Write an equation to show the heating of aluminium hydroxide.(b) Aluminium can be obtained from aluminium oxide by electrolysis method. (i) What is the electrolyte used? (ii) What is the function of the electrolyte. (iii) What are the ions which will carry electrical charges during the electrolysis of

aluminium oxide?

Essay Question

1. (a) With reference to the physical and chemical characteristics of the aluminium ion, describe the characteristics of the aluminium salt aqueous solution.

(b) Explain the following using correct equations. (i) Reaction between anhydrous aluminium chloride with water. (ii) Why aluminium carbonate cannot be prepared from a solution containing Al3+ ions.




1. D Rubidium is in Group 1. The others are in Group 13.

2. A Al is protected from corrosion by Al2O

3.

3. B Aluminium does not react with pure water at room temperature.

4. C Cryolite lowers the melting point of pure bauxite from 2000 °C to 900 °C.

5. A High lattice energy of aluminium fluoride is due to high ionic charge and also small size of fluoride ion.

6. D 7. C Aluminium reacts with air to form a protective

layer of aluminium oxide, Al2O

3 on its surface.


1. (a) (i) Cl Cl Cl

Al Al

Cl Cl Cl (ii) 2Al + 3Cl

2 → Al

2Cl

6

(iii) As catalyst(b) From the equation

2Al + 3Cl2 → Al

2Cl

6

2 mol Al produces1 mol Al

2Cl

6

∴ 54 g Al produces 267 g Al2Cl

6

1.01.0 g Al produces —— � 267 54

= 4.94 g Al2Cl

6

2.3% of product = ———— � 100% = 46.6% 4.94 (c) (i) Aluminium chloride gas has strong

covalent bonding within the molecule but weak van der Waals forces of attraction between molecules.

(ii) The aqueous solution of aluminium chloride contains [Al(H

2O)

6]3+ complex

ions. The complex ions react with water and forms H

3O+ ion.

[Al(H2O)

6]3+ + H

2O →

[Al(H2O)

5OH]2+ + H

3O+

2. (a) (i) 2[Al(OH)4]– + CO

2 →

2Al(OH)3 + CO

32– + H

2O

(ii) 2Al(OH)3 → Al

2O

3 + 3H

2O

(b) (i) Cryolite / Sodium hexafluoroaluminate (ii) To decrease the melting point of

aluminium oxide so that electrolysis can be carried out at a lower temperature.

(iii) Al3+, O2–

Essay Question

1. (a) Al3+ ion is surrounded by six molecules of H

2O forming [Al(H

2O)

6]3+ complex ion.

3+H2O OH2OH2

OH2H2O OH2

Al

The Al3+ ion has a high charge and a small size, therefore the charge density is high. This causes Al3+ ion to have high polarising power, enabling it to pull the electrons from water molecules and weakening the O – H bond in water. The H+ ion can be easily released causing the aqueous solution containing Al3+ ion to be acidic.

(b) (i) Anhydrous Al2Cl

6 is hydrolysed in

water to form a cation [Al(H2O)

6]3+

Al2Cl

6 + 12H

2O → 2[Al(H

2O)

6]3+ + 6Cl–

(ii) CO32– is a strong base so the following

reaction occurs.

[Al(H2O)

4(OH)

2]+ + CO

32– →

Al(H2O)

3(OH)

3 + HCO

3–

ANSWER

CHAPTER 2 Aluminium

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1. When CO2 gas is passed through limewater, a white precipitate of CaCO

3 is formed. If an

excess of CO2 is passed through, the white precipitate dissolves because

A as more and more CO2 gas dissolves, the solution becomes increasingly acidic, hence

CaCO3 changes to CaHCO

3 which is soluble.

B as more and more CO2 gas dissolves, the solution becomes increasingly acidic and CaCO

3

will decompose.C as more and more CO

2 gas dissolves, a soluble complex is formed.

D the precipitate CaCO3 will react with an excess of CO

2 to form CaO.

2. Muscovite, KAl2(Si

3Al)O

10(OH)

2, is a type of mica which can be broken into thin layers. What

type of force holds the thin layers together in muscovite?A Electrostatic force of attraction C Hydrogen bonds between layersB Van der Waals forces of attraction D Metallic bonds between layers

3. The advantages of ceramic over iron and steel are as follows except thatA the density of ceramic is lower.B ceramic is an electrical conductor.C ceramic cannot be oxidised.D ceramic is less brittle.

4. The table below shows a few types of glass and their uses. Which of the following combination is incorrect?

A

B

C

D

5. Which industrial use reflect the hardness of diamond and the softness of graphite?

A

B

C

D

Structured Question

1. (a) Elements of Group 14 can react with chlorine to form tetrachlorides. (i) In the table below, name the process (if any) and write the equation for the reaction

that has taken place

Reaction with water Reaction equation

CCl4

SiCl4

PbCl4

3 Group 14 – C, Si, Ge, Sn, Pb

Glass Use

Soda glass

Lead glass

Borosilicate glass

Quartz glass

Making normal mirrors, windows, bottles and light bulbs

Making decorative glass, prisms and lenses

Making laboratory apparatus

Making cookware

Diamond

Drill head

Insulator

Glass cutter

Abrasive

Graphite

Lubricant

Pencil lead

Electrode

Moderator



(ii) Explain the differences in the reaction with water between the three tetrachlorides.(b) The composition of two types of glass X and Y is shown below.

Type of glass Composition

Glass X

Glass Y

14% Na2O, 12% CaO, 74% SiO

2

5% Na2O, 4% CaO, 17% B

2O

3, 74% SiO

2

(i) Glass X is called soda glass. State the name of glass Y. (ii) State two reasons for the addition of B

2O

3 to glass Y.

Essay Question

1. (a) What is the relative stability of the +2 oxidation state compared to the +4 oxidation state for the oxides of Group 14 elements when going down the group?

(b) (i) Draw the structure of carbon dioxide and silicon dioxide. (ii) Although the bonding in carbon dioxide and silicon dioxide is covalent, carbon

dioxide is a gas whereas silicon dioxide is a solid with high melting point. Explain this statement.

(c) Write all the formulae for the oxides of carbon and lead and discuss the respective oxides with respect to their

(i) acid-base characteristics (ii) redox characteristics

Write one reaction equation for each of the above characteristics.




1. A CO2 + Ca(OH)

2 → CaCO

3 + H

2O

CaCO3 + CO

2 + H

2O → Ca(HCO

3)

2

2. B Van der Waals forces exist between the layers of molecules.

3. D Ceramic is not less brittle but more brittle and breaks easily when dropped.

4. D Ouartz glass cannot withstand high temperature hence is not suitable for making cookware.

5. A Hardness of diamond enables the diamond to be used in drill heads. The ability of layers in the graphite to slide over one another gives the soft property of graphite and it is used as a lubricant.

Structured Question

1. (a) (i)

Reaction with water Reaction equation

CCl4

SiCl4

PbCl4

No reaction

Hydrolysis

Hydrolysis

–

SiCl4 + 2H

2O →

SiO2 + 4HCl

PbCl4 + 2H

2O →

PbO2 + 4HCl

(ii) For SiCl4 and PbCl

4, the central atom

has empty d orbitals which can accept and share the lone pair electrons from O atoms of water molecules.For CCl

4, the carbon atom does not

have an empty d orbital and the C – Cl covalent bond is very strong.

(b) (i) Borosilicate glass (ii) To increase the softening level of glass

To decrease the thermal expansion coefficient so that the glass does not break even if there is a sudden increase in temperature

Essay Question

1. (a) When going down the group, the stability of the +2 oxidation state increases whereas the stability of the +4 oxidation state decreases.

(b) (i) O = C = O

⏐ O ⏐ Si

O O O Si ⏐ Si Si ⏐ (ii) Carbon dioxide is a simple molecule and

has weak intermolecular van der Waals forces of attraction. Silicon dioxide has a giant molecular structure.

(c) Oxides of carbon: CO, CO2

Oxides of lead: PbO, PbO2

(i) CO is neutral and does not dissolve in water.CO

2 is a weak acid which partially

dissociates in water to form carbonic acid.

CO2 + H

2O H

2CO

3

PbO and PbO2 are amphoteric. Both

oxides can dissolve in dilute acid and dilute base.

PbO + 2H+ → Pb2+ + H2O

PbO + 2OH– + H2O → Pb(OH)

42–

(ii) CO is a strong reducing agent. The +2 oxidation state will easily be oxidised to +4.

CO + PbO → Pb + CO2

CO2 can behave as an oxidising agent.

CO2 + 2Mg → C + 2MgO

PbO is an oxidising agent.PbO + H

2 → Pb + H

2O

PbO2 is also an oxidising agent.

PbO2 + C → Pb + CO

2

ANSWER

CHAPTER 3 Group 14 – C, Si, Ge, Sn, Pb




1. Most metals burn readily in oxygen to form metal oxides. However metal nitrites are only formed when metal reacts with nitrogen at high temperature. Which of the following best explains this phenomenon?A Nitrogen atom has a stable half-filled p orbital in its electron arrangement.B Nitrogen has very high atomisation energy.C Oxygen is more electronegative than nitrogen.D Oxygen has more valence electrons than nitrogen.

2. Which of the following compounds of nitrogen has an odd electron?A NO

2 C HNO

3

B NH3 D N

2O

4

3. Which of the following shows the bonding in nitrogen dioxide?A N C N

O O O OB N D N

O O O O 4. Carbon and nitrogen both form dioxides. Which of the following is true of both carbon dioxide

and nitrogen dioxide?A Both are linear molecules.B Both are non-polar molecules.C Both dissolve in sodium hydroxide solution.D Both central atoms achieve the octet electronic configuration.

A B C D

Only 1 is correct Only 1 and 2are correct

Only 2 and 3are correct

1, 2 and 3are correct

5. Nitrogen dioxide exists in equilibrium with its dimer, nitrogen tetroxide in the following equation:

2NO2(g) N

2O

4(g)

The intensity of brown colour decreases when the temperature is lowered. Which of the following statements is/are true of the equilibrium?1 The dimerisation of nitrogen dioxide is exothermic.2 The equilibrium of the reaction shifts to the left with increased temperature.3 The equilibrium of the reaction shifts to the right with increased pressure.

6. When excess aqueous ammonia is added to copper(II) ions solution, a dark blue solution is formed. What is the role of ammonia in this reaction?1 As ligand 3 As a Bronsted-Lowry base2 As a Lewis base

Structured Question

1. Nitrogen forms three common oxides: nitrogen oxide, nitrogen dioxide and dinitrogen tetroxide. (a) Name the oxides of nitrogen that have an unpaired electron at the central atom.(b) Name the oxides that have delocalisation of electrons between the N atom and the O atom.

4 Group 15 – Nitrogen and its Compounds



(c) Name the oxide that is formed in the internal combustion of automobiles. Write a balanced equation showing its formation.

(d) Name the oxide that dissolves in rain water to produce acid rain. Write a balanced equation for the formation.

(e) Name the oxide that is formed when ammonia reacts with oxygen in the presence of a catalyst. Write an equation for the formation.

Essay Question

1. (a) Nitrogen gas is the major component in atmospheric air. However, relatively few compounds exist as nitrides compare to oxides. Explain this phenomenon.

(b) The oxides of nitrogen cause environmental pollution. (i) Identify one oxide of nitrogen that causes acid rain. (ii) Identify one oxide of nitrogen that causes photochemical smog.

Explain how these oxides of nitrogen cause pollution. Illustrate your answer with balanced equations.




1. B Nitrogen gas has triple covalent bond between two nitrogen atoms, N ≡ N. Hence it has a very high atomisation energy.

2. A Lewis structure of NO2

•N

O O 3. D Resonance occurs in NO

2

• • N ↔ N → NO O O O O O

4. C Structure of CO2 and NO

2 are as follows:

N O = C = O O O

CO2 can dissolve in water to form acidic

solution, H2CO

3. NO

2 can dissolve in water to

form acidic solution, HNO3. Hence both CO

2

and NO2 can react with NaOH to form salts.

5. D When the temperature is lowered, NO2 which

is a brown gas, dimerises to form N2O

4. When

N2O

4 absorbs heat, it dissociates to NO

2.

Dimerisation of NO2 to form N

2O

4 loses heat

(exothermic). Increased temperature shifts equilibrium to the left to absorb heat. Increased pressure shifts equilibrium to the right to decrease the pressure by decreasing the number of moles of gases.

6. B A complex ion [Cu(NH3)

4]2+ is formed. Lone

pair of electrons on N of NH3 is shared with

Cu2+ ion, hence NH3 acts as a ligand and Lewis

base.

H3N

Cu

H3N

NH3

NH3

2+

Structured Question

1. (a) NO, NO2

(b) NO2, N

2O

4

(c) NO; N2 + O

2 → 2NO

(d) NO2; 2NO

2 + H

2O → HNO

2 + HNO

3

(e) NO; 4NH3 + 5O

2 4NO + 6H

2O

Essay Question

1. (a) Nitrogen gas is relatively inert compared to oxygen gas.This is because the two N atoms in nitrogen gas are held by a very strong triple bond.The N ≡ N bond has very high bond energy / very short bond length.Furthermore, the nitrogen molecule is linear and non-polar.A lot of energy is required to dissociate N

2

molecule / very high atomisation energy to produce N atoms for reaction.

(b) (i) NO2, nitrogen dioxide causes acid rain.

NO2 dissolves in rain water to produce

nitrous acid and nitric acid.

2NO2 + H

2O → HNO

2 + HNO

3

4NO2 + 2H

2O + O

2 → 4HNO

3

(ii) NO2, nitrogen dioxide causes

photochemical smog.Ultraviolet light causes the formationof oxygen radical from nitrogen dioxide.

NO2(g) → NO(g) + O•(g)

The oxygen radical produced then reacts with oxygen molecule to form ozone molecule.

O•(g) + O2(g) → O

3(g)

The ozone combines with nitrogen dioxide and the unsaturated hydrocarbon from incomplete combustion of fuels to form smog.

ANSWER

CHAPTER 4 Group 15 – Nitrogen and its Compunds




1. Astatine is the last member of Group 17 elements. Which of the following statements is true?A Astatine has the lowest melting point compared to other halogens.B Astatine is the strongest oxidising agents compared to other halogens.C Hydrogen astatide, HAt, is a colourless gas.D Silver astatide is soluble in dilute ammonia solution.

2. NaClO3 is obtained by heating NaClO.

3NaClO ⎯→ 2NaCl + NaClO3 ∆

What are the oxidation states of these chlorine compounds?

A

B

C

D

3. Which of the following hydrides decomposes most easily when heated?A H – FB H – ClC H – BrD H – I

4. Which of the following is not produced when potassium iodide is heated with concentrated sulphuric acid?A SO

3 C I

2

B NaHSO4 D HI

5. Which of the following chlorine compounds is a covalent molecule?A BaCl

2 C AgCl

B AlCl3 D KClO

3

A B C D




6. Which of the following property / properties decrease(s) on descending Group 17 from chlorine to iodine?1 The boiling points of the halogens2 The electronegativity of the halogens3 The oxidising power of the halogens

Structured Question

1. (a) When chlorine is passed into an aqueous solution of sodium hydroxide, the smell of chlorine disappears.

(i) Explain why the smell of chlorine disappears. (ii) Write an equation for the reaction. (iii) How do you classify this type of reaction?

5 Group 17: Cl, Br, I

NaClO NaCl NaClO3

–1

–1

+1

+1

+1

+1

–1

–1

+5

+7

+5

+7



(b) When a red-hot platinum wire is placed inside a gas jar of hydrogen iodide, violet fumes are produced.

(i) Explain the observation. (ii) Write an equation for the reaction. (iii) How do you classify this type of reaction?

Essay Question

1. Explain the following:(a) The reactivity between hydrogen and halogens decreases in the order:

Cl2 > Br

2 > I

2

(b) When chlorine is passed into sodium iodide solution, the solution turns brown.(c) Concentrated sulphuric acid is not suitable for the preparation of hydrogen bromide from

sodium bromide.




1. C Hydrogen halides, HF, HCl, HBr and HI are all colourless gases.

2. C 3. D The longer the bond length, the weaker the

bond. 4. A H

2SO

4 is reduced to SO

2 and not SO

3.

5. B Metal chlorides are usually ionic compounds. AlCl

3 is covalent and exists as Al

2Cl

6 at room

conditions. 6. C

Structured Question

1. (a) (i) The chlorine gas is converted to the salts, sodium chloride and sodium chlorate(I), which are odourless.

(ii) Cl2(g) + 2NaOH(aq) →

NaCl(aq) + NaClO(aq) + H2O(l)

(iii) This is a disproportionation reaction. Chlorine is simultaneously oxidised to ClO– and reduced to Cl–.

(b) (i) At temperatures above 200 °C, H – I bonds break to form hydrogen and the violet fumes of iodine.

(ii) 2HI(g) H2(g) + I

2(g)

(iii) This is a thermal decomposition reaction.

Essay Question

1. (a) Under suitable conditions, hydrogen reacts with halogens to form hydrogen halides.

H2(g) + X

2(g) 2HX(g)

This is a redox reaction. The halogens oxidise hydrogen to hydrogen halides.The order of reactivity decreases on descending the group because the oxidising power of the halogens decreases as the sizes of the halogens increase.

(b) When chlorine is passed into sodium iodide solution, the solution turns brown because chlorine oxidises iodide ions to iodine. Aqueous sodium iodide is colourless but aqueous iodine is brown.

Cl2(g) + 2NaI(aq) → 2NaCl(aq) + I

2(aq)

(c) When concentrated sulphuric acid reacts with sodium bromide, hydrogen bromide is produced.

NaBr(s) + H2SO

4(l) → NaHSO

4(s) + HBr(g)

But concentrated sulphuric acid is also an oxidising agent. HBr is oxidised by concentrated sulphuric acid to form bromine.

2HBr(g) + H2SO

4(l) →Br

2(g) + 2H

2O(l) + SO

2(g)

So, instead of getting hydrogen bromide, a mixture of gases is produced.

ANSWER

CHAPTER 5 Group 17: Cl, Br, I




1. Which of the following does not indicate the properties of a transition element?A Iron reacts with dilute acid to liberate hydrogen gas.B Oxides of iron include FeO and Fe

2O

3.

C Iron is used in the manufacture of ammonia.D A blood red colouration is formed when potassium thiocyanate is added to ferric chloride

solution.

2. Which of the following statements is correct regarding transition elements?A All transition elements exhibit oxidation states of +1 to +7.B All transition elements have more than one valency.C All transition elements have completely filled 4s orbitals.D All transition elements react with dilute acid to produce hydrogen gas.

3. Which one of the following elements exhibits only one oxidation state in its compounds?A CopperB TitaniumC TinD Scandium

4. A bidentate ligand always A has bonds formed to two metal ions.B has a charge of +2 or –2.C forms complex ions with a charge of +2 or –2.D has two donor atoms.

5. Which one of the following best explains why water is a monodentate ligand? A The oxygen atom in a water molecule has only one lone pair of electrons that can be used

to form a coordinate bond to a metal ion.B Each hydrogen atom in a water molecule has only one electron to form a coordinate bond

to a metal ion.C The oxygen atom in a water molecule has two lone pairs of electrons but both pairs are

used to form a single coordinate covalent bond to a metal atom. D The oxygen atom in a water molecule has two lone pairs of electrons but the second lone

pair is not close enough to a second coordination site to form a coordinate bond.

6. What are the oxidation number and coordination number in the coordination compound, [Cu(en)(NH

3)

2]Cl

2?

A

B

C

D

7. When a concentrated solution of NH3 is added to a solution containing Zn(NO

3)

2, a colourless

solution containing the complex ion Zn(NH3)

42+ results. Which of the following best explains

why this solution is colourless?A The 3d orbitals are full, thus cannot be split. B The 3d orbitals on the Zn atom are completely filled, thus, an electron cannot be transferred

between the split 3d orbitals.

6 Introduction to the Chemistry of d-block Elements

Oxidation number Coordination number

+1

+2

+2

+2

2

2

4

6



C The Zn atom does not have any 4s electrons, thus an electron cannot be transferred between the split 4s orbitals.

D The Zn atom does not have any unpaired electrons which are necessary in order to transfer an electron between the split 3d orbitals.

8. When solid CoCl2 is dissolved in water, the following equilibrium is established:

Co(H2O)

62+(aq) + 4Cl– (aq) CoCl

42–(aq) + 6H

2O(l)

In aqueous solution, the Co(H2O)

62+ ion is pink and the CoCl

42– ion is blue.

Which of the following best describes what will happen when a concentrated solution of HCl is added to the CoCl

2 solution without changing the volume significantly?

A The colour of the CoCl2 solution will not change because the solution of HCl is

colourless. B The colour of the solution will become blue. C The shape of Co(H

2O)

62+ is octahedral whereas the shape of CoCl

42– is tetrahedral.

D The number of unpaired electrons in Co(H2O)

62+, but not CoCl

42–, will change because HCl

is a strong acid.

A B C D




9. Which of the following reactions show(s) the formation of complex ion by the replacement of ligand?1 [Pb(OH)

2]2– + 2H

2O

2 → PbO

2 + 2H

2O + 2OH–

2 [Cu(H2O)

4]2+ + 4NH

3 → [Cu(NH

3)

4]2+ + 4H

2O

3 [Fe(H2O)

6]3+ + SCN– → [Fe(SCN)(H

2O)

5]2+ + H

2O

10. The hexaaquairon(III) ion hydrolyses as shown below.

[Fe(H2O)

6]3+ + H

2O [Fe(H

2O)

5OH]2+ + H

3O+

Which of the following statements about the above equilibrium is/are correct?1 The corresponding iron(III) ion [Fe(H

2O)

5OH]2+ is less likely to undergo hydrolysis.

2 The iron ion undergoes a change in oxidation state.3 The above hydrolysis is favoured under low pH conditions.


1. Copper exhibits oxidation states of +1 and +2 in its compound.(a) Write the electronic configuration for Cu2+ ion and Cu+ ion.(b) Explain why hydrated copper(II) sulphate is coloured whereas copper(I) sulphate is white.(c) In aqueous solutions, Cu+ disproportionates to Cu2+ and Cu. (i) What do you understand by the term disproportionation? (ii) Write an equation to show the disproportionation of Cu+ in water.

2. (a) Complete the table.

Ion

Number of electrons in the 3d subshell

Number of unpaired electrons

Colour of the aqueous ion

Ti3+ Co2+ Cu2+

(b) The metals – platinum, rhodium and palladium are transition metals that are used in catalytic converters to reduce pollution from vehicle exhausts. State and explain the role of these transition elements in this process.



Essay Questions

1. (a) Explain what do you understand by the term stability constant?(b) When ammonia is added to a solution of nickel(II) sulphate, the solution turned from

green to pale blue. Discuss the stability constant of the complex [Ni(NH3)

6]2+.

(c) When aqueous ammonia is gradually added to aqueous copper(II) sulphate, a blue precipitate is formed initially but dissolves in excess aqueous ammonia to give a deep blue solution. This deep blue solution lightens considerably on the addition of an aqueous solution of EDTA. Explain the observations as fully as you can.

2. (a) Two common oxidation states of chromium are +3 and +6. (i) Explain why Cr 6+(aq) ion does not exist but Cr(VI) exists as CrO

42– and Cr

2O

72–

ions. (ii) Explain with an equation, the effect of acid on the aqueous chromate(VI) ion.(b) (i) Use the standard electrode potential values from the Data Booklet to discuss the

oxidation of Cr 3+ to dichromate(VI) ions by hydrogen peroxide. (ii) This oxidation of Cr 3+ by hydrogen peroxide is more satisfactorily achieved under

alkaline conditions. Use your equation in (b)(i) to suggest why alkaline conditions are used.




1. A Other metals besides transition metals react with dilute acids to liberate hydrogen gas.

2. B Transition elements have electrons which can be removed from s orbitals and also d orbitals hence they have variable oxidation states and more than one valency.

3. D Scandium is a d-block element but it is not a transition element. It shows only one oxidation state, +3 in the Sc3+ ion.

4. D A bidentate ligand has two lone pairs of electrons on two donor atoms for sharing with central metal ion.

5. D Water molecules has two lone pairs of electrons in a tetrahedron shape orbital, hence the second lone pair of electrons is too far away (big angle) to form another coordinate bond with the central metal ion.

6. C The compound has the following structure

Cu 2 Cl–enNH3

NH3

2+

CC

Oxidation number of Cu = +2Coordination number = 4

7. B Valence electronic configuration of Zn2+ ion is 3d10. In the complex ion, the split 3d orbitals are completely filled, hence there is no transfer of electrons between the split energy levels of 3d orbitals.

8. B When HCl is added to CoCl2 solution, the Cl–

from HCl reacts with Co(H2O)

62+(aq) to form

the blue complex CoCl42–.

9. C In 1, the oxidation state of Pb charges from 0 to +4, it is a redox reaction.In 2 and 3, the ligands H

2O in the complexes

are replaced by ligands NH3 and SCN–. Hence

2 and 3 are correct. 10. A Fe2+ has a lower charge density hence less

likely to undergo hydrolysis (1 is correct).Fe is still in +3 oxidation state in [Fe(H

2O)

6OH]2+ (2 is incorrect).

If the pH is low, equilibrium is shifted to the left and hydrolysis is depressed (3 is incorrect).


1. (a) Cu2+ : 1s22s22p63s23p63d 9

Cu+: 1s22s22p63s23p63d10 (b) The 3d orbitals of Cu+ are all fully filled.

Hence no d-d transition can take place.However, the 3d orbitals of Cu2+ are not fully filled and d-d transition can occur.

(c) (i) Disproportionation is a reaction in which a substance is both oxidised and reduced simultaneously.

(ii) 2Cu+ → Cu + Cu2+

2. (a)

Ion

Number of electrons inthe 3d subshell

Number of unpaired electrons

Colour of the aqueous ion

Ti3+

1

1

Violet

Co2+

7

3

Pink

Cu2+

9

1

Blue

(b) These transition metals act as heterogeneous catalysts.Dative or coordinate bonds can be formed as these metals have empty d orbitals to allow adsorption to take place.

Essay Questions

1. (a) Stability constant, K, is the equilibrium constant for the formation of the complex ion from its constituent ions/atoms/molecules. The higher the K value, the more stable the complex.

M(aq) + nL MLn(aq)

[MLn]

K = —————— [M][L]n

at constant temperature(b) When ammonia is added to nickel(II) sulphate,

equilibrium occurs.

[Ni(H2O)

6]2+(aq) + 6NH

3(g)

green[Ni(NH

3)

6]2+(aq) + 6H

2O(l)

pale blue [Ni(NH

3)

62+]

K = ——————————————— [Ni(H

2O)

62+][NH

3]6

K is the stability constant for [Ni(NH3)

6]2+

complex ion. The change of colour from green to pale blue shows that the [Ni(NH

3)

6]2+

complex ion has a high stability constant and that the equilibrium lies to the right. NH

3 is

a stronger ligand than H2O. [Ni(NH

3)

6]2+ is

more stable than [Ni(H2O)

6]2+.

(c) CuSO4 dissolves in water forming a blue

coloured solution of [Cu(H2O)

6]2+.

When aqueous ammonia is added to copper(II) sulphate solution, a light blue precipitate Cu(OH)

2 is formed.

Cu2+(aq) + 2OH– → Cu(OH)2(s)

The light blue precipitate Cu(OH)2 dissolves

in excess ammonia to form [Cu(NH3)

4]2+

complex ion which is soluble in water to give a deep blue solution.

ANSWER

CHAPTER 6 Introduction to the Chemistry of d-block Elements



Cu(OH)2(s) + 4NH

3(aq) →

[Cu(NH3)

4]2+(aq) + 2OH–(aq)

When an aqueous solution of EDTA is added to the copper complex, the deep blue solution of [Cu(NH

3)

4]2+ turns lighter due to formation

of the lighter coloured Cu(EDTA)2– complex ion.

[Cu(NH3)

4]2+(aq) + EDTA4–(aq) →

[Cu(EDTA)]2–(aq) + 4NH3(aq)

The reaction shows the displacement of weaker NH

3 ligand by the stronger EDTA4– ligand.

This implies that the [Cu(NH3)

4]2+ complex is

less stable than the [Cu(EDTA)]2– complex. 2. (a) (i) Cr 6+ ion has very high charge density.

Formation of Cr 6+ requires very high ionisation energy that results in high instability.

CrO42– and Cr

2O

72– have delocalised

bonding, thus are more stable. (ii) 2H+ + 2CrO

42– Cr

2O

72– + H

2O

In acidic condition, CrO42– converts to

Cr2O

72–.

(b) (i)H

2O

2(aq) + 2H+ + 2e– → 2H

2O(l); Eθ

red = +1.77 V

2Cr3+(aq) + 7H2O(l) → Cr

2O

72–(aq) + 14H+(aq) + 6e–;

Eθox

= – 1.33 V

2Cr3+(aq) + 3H2O

2(aq) + H

2O(l) → Cr

2O

72–(aq) + 8H+(aq);

Ecell

> 0

Cr3+ is oxidised to Cr2O

72– while H

2O

2 is

reduced to H2O.

(ii) In alkaline condition, the hydrogen ion produced will be removed by the alkali. The equilibrium will shift to the right. Oxidation of Cr3+ would be easier.




1. One mole of a hydrocarbon produces four moles of water when burnt in excess oxygen. The formula of the hydrocarbon isA CH

3CH

2CH

3 C CH

3CH = CH

2

B CH3CH

2CH

2CH

3 D CH

3C

6H

4CH

3

2 When 60 cm3 of a hydrocarbon were burnt completely in oxygen, 240 cm3 of carbon dioxide and 0.0125 mol of water were produced. What is the molecular formula of the hydrocarbon?(The volumes of all gases were measured at room conditions. The molar volume of any gas at room conditions is 24.0 dm3)A C

2H

6 C C

3H

6

B C3H

8 D C

4H

10

3. The compound R has the structural formula shown below.

C = O

OCH3

The functional group present in R isA carboxylic acid C esterB aldehyde D ketone

4. What is the bond angle of the C – C – C bonds in CH3C ≡ CH?

A 120 ° C 107 °B 180 ° D 109.5 °

5. Which type of isomerism is shown by each of the following organic compounds?

A

B

C

D

A B C D




6. The reaction between hydroxide ion and the organic compound, R – X can be represented by the equation

R – X + OH– → R – OH + X–

where R is the alkyl group and X is an electronegative element. In the above reaction, the hydroxide ion acts as a1 Bronsted-Lowry base2 Lewis base3 nucleophile

7 The Chemistry of Carbon

CH3CH = CHCH

3

Optical

Optical

Geometric

Geometric

CH3CH(NH

2)COOH

Geometric

Geometric

Optical

Geometric

C2H

5CH = CHBr

Geometric

Optical

Geometric

Optical



7. Which of the following hydrocarbons contain(s) delocalised electrons?1 CH3

2 CH3 – C ≡ C – CH

3

3 C(CH3)

4

8. Two compounds with the molecular formula, C2H

2Br

2, are expected to

1 exhibit structural isomerism2 exhibit cis-trans isomerism3 exist in d- and l- forms

Structured Question

1. Limonene has a lemon odour and is found in oils from citrus fruits. The structural formula of limonene is shown below.

CH3

CH2CH3

C

(a) Write the displayed formula of the limonene molecule.(b) Write (i) the empirical formula, (ii) the molecular formula of limonene.(c) What type of stereoisomerism is shown by limonene?(d) In which ways are the isomers (i) similar to each other, (ii) different from each other?

Essay Question

1. (a) The mass spectrum of a straight chain hydrocarbon shows the main peaks at m/e 56, 29, 27, 15.The percentage composition (by mass) of the hydrocarbon is as follows.

C = 85.7% H = 14.3 %

Write the structural formula of a hydrocarbon that can give the results described above. Explain your answer.

(b) The pKa values for three carboxylic acids are given below.

Compound pKa

Benzoic acid

4-methylbenzoic acid

4-nitrobenzoic acid

4.20

4.34

3.43

Explain the relative strengths of these acids.




1. A CxH

y → 4H

2O

Hence, y = 8 2. D Step 1 Calculate the number of carbon atoms

per molecule of hydrocarbon60 cm3 of C

xH

y produces 240 cm3 CO

2.

240That is, 1 mol of C

xH

y produces ———

60= 4 mol of CO

2.

∴ 1 molecule of CxH

y contains 4 carbon

atoms.Step 2 Calculate the number of hydrogen

atoms per molecule of hydrocarbon

6060 cm3 of C

xH

y contains —————————

24 � 1000= 0.0025 mol of C

xH

y.

0.0025 mol CxH

y produces 0.0125 mol

H2O.

∴ 1 mol of CxH

y produces 5 mol of

H2O.

∴ 1 molecule of CxH

y contains 10

hydrogen atoms.Step 3 Formula of hydrocarbon

Formula of hydrocarbon is C4H

10.

3. C The structural formula can be written as C

6H

5COOCH

3.

The ester group is – COO–. Thus R is methyl benzoate.

4. B The carbon atom in the C ≡ C bond uses sp hybrid orbitals for bond formation.Hence, the three carbon atoms are arranged linearly.

5. C 6. C δ+ δ–

R ⎯ X + OH– → R – OH + X–

OH– is a nucleophile and attacks the alkyl group which has low electron density.A nucleophile is also a Lewis base.OH– ion does not act as a Bronsted-Lowry base in this reaction because it does not accept a proton from RX.

7. A Only benzene ring contains delocalised electrons.

8. B H H H Br ⏐ ⏐ ⏐ ⏐

Structural isomerism C = C C = C ⏐ ⏐ ⏐ ⏐ Br Br H Br H H H Br ⏐ ⏐ ⏐ ⏐

Cis-trans isomerism C = C C = C ⏐ ⏐ ⏐ ⏐ Br Br Br H cis trans

Structured Question

1. (a) C

H

H

H

HHH

HH

H H

H

H

HH

HC

C

CCC

C

C

C

C

H

*

(b) (i) C5H

8

(ii) C10

H16

(c) Limonene has a chiral carbon, *C. Hence it exhibits optical isomerism.

(d) Limonene exists as d- and l-forms. (i) Similarity

Both the d- and l-forms have the same molecular and structural formulae.They have the same physical and chemical properties.

(ii) DifferenceThe d-form rotates the plane of plane-polarised light to the right.The l-form rotates the plane of plane-polarised light to the left.

Essay Question

1. (a) Empirical formula from composition by mass

Element

Number of moles

Simplest ratio

C

85.7————

12= 7.14

1

H

14.3————

1= 14.3

2

Empirical formula is CH2.

Molecular formula from mass spectrum:M

r as shown in the mass spectrum = 56

nCH2 = 56

14n = 56; ∴ n = 4∴ Molecular formula is C

4H

8.

Fragments from mass spectrumm/e 29 : C

2H

5+

m/e 27 : CH = CH2+

m/e 15 : CH3+

Structural formula: CH3CH

2CH = CH

2.

(b) The smaller the pKa value, the stronger the

acid.Thus, the acid strength increases in the order:4-methylbenzoic acid < benzoic acid <4-nitrobenzoic acid4-methylbenzoic acid is a weaker acid than benzoic acid because the methyl group is electron-donating (+I effect). The methyl

ANSWER

CHAPTER 7 The Chemistry of Carbon



group destabilises the carboxylate anion by increasing the charge density of oxygen atom in the C – O bond. This makes the loss of proton more difficult. Thus the acid strength decreases.

4 – nitrobenzoic acid is a stronger acid than benzoic acid because the nitro group is electron-withdrawing (–I effect). The nitro group stabilises the carboxylate ion and thus the acidity increases.




1. A mixture containing a little ethane and a large quantity of bromine gas is exposed to ultraviolet light. Which of the following bromoalkanes is the principle product of reaction?A CH

3CH

2Br C BrCH

2CH

2Br

B CH3CHBr

2 D CBr

3CBr

3

2. The following hydrocarbons are isomers of hexane. Which of these isomers has the lowest boiling point? A (CH

3)

2CHCH(CH

3)

2 C CH

3CH

2CH(CH

3)CH

2CH

3

B (CH3)

2CHCH

2CH

2CH

3 D C

2H

5C(CH

3)

3

3. 10 cm3 of methane and 10 cm3 of ethane were mixed with excess oxygen and the mixture was exploded. After cooling to room temperature, the residual gases were passed through aqueous sodium hydroxide. What was the volume of gas absorbed by aqueous sodium hydroxide?A 15 cm3 C 25 cm3

B 20 cm3 D 30 cm3

4. When X (a gas) reacts with Y, a mixture of CH

2BrCH

2Br and CH

2BrCH

2OH are produced.

X and Y are probably

A

B

C

D

5. The enthalpies of hydrogenation of cyclohexene and 1,3-cyclohexadiene are given below.

+ H2

+ 2H2

cyclohexane ; ∆H1 = –120 kJ mol–1

cyclohexane ; ∆H2 = –232 kJ mol–1

The enthalpy of hydrogenation when 1 mol of benzene reacts with three moles of hydrogen isA – (3 � 120) kJ mol–1 C –208 kJ mol

3B – (— � 232 ) kJ mol–1 D –152 kJ mol–1

2

A B C D




6. The relative molecular mass of a hydrocarbon, P, is 54. 0.25 g of Q required 207.4 cm3 of hydrogen at s.t.p. for complete hydrogenation. The structural formula of P could be1 CH

2 = CH – CH = CH

2

2 CH3C ≡ C – CH

3

3 CH2 – C = CH

2 CH

2

8 Hydrocarbons

X Y

Ethene

Ethene

HBr(g)

Br2 vapour

Bromine water

Sodium bromide solution

Ethene

Ethene



7. Which of the following reagents react(s) with benzene by addition reaction?1 Concentrated sulphuric acid2 Hydrogen gas in the presence of nickel at 150 °C3 Chlorine in the presence of ultraviolet light

8. When a monosubstitued benzene, – X, reacts with an electrophile, E+, the major organic products obtained are as follows:

XEX andE

The substituent, X, may be1 –NH

2

2 –COCH3

3 –NO2

Structured Question

1. Essential oil is a natural oil with a distinctive scent produced by certain plants. Geraniol and linalool are found in many essential oils. They are used as flavouring agents and in perfumes. The structural formulae of geraniol and linalool are shown below.

(CH3)

2C = CHCH

2CH

2C(CH

3) = CHCH

2OH

geraniol

(CH3)

2C = CHCH

2CH

2C(CH

3)(OH)CH = CH

2

linalool

(a) The IUPAC name for geraniol is 3,7-dimethyl-2,6-octadien-1-ol.Explain why it is named this way.

(b) Based on your explanation given above, suggest the IUPAC name of linalool. (c) Give two similarities and one difference between geraniol and linalool in terms of their

structures.(d) Geraniol reacts with bromine in CCl

4 in the absence of sunlight to form an organic

compound. What is the molecular formula of this organic compound?(e) Would you expect linalool to have any effect on plane-polarised light? Why?

Essay Question

1. (a) State three ways in which (i) cyclohexane is similar to hexane, (ii) cyclohexane is different from hexane.

(b) (i) 1,1,1-trichloroethane is used in ‘liquid paper’ as a solvent. Write the displayed formula of this compound.Name and write the structural formula of an isomer of 1,1,1-trichloroethane.

(ii) State two ways in which the isomers are similar to each other.(c) Outline the steps for the conversion of benzene to (i) (chloromethyl)benzene, (ii) benzoic

acid.




1. D With a little bromine, the main product is CH

3CH

2Br.

With a little ethane but a large amount of bromine, CH

3CH

2Br produced undergoes further

bromination. The main product is therefore CBr

3CBr

3.

2. D The greater the number of side chains, the smaller the surface area of the hydrocarbon molecule and the smaller the attractive forces between the molecules.

3. D CH4 + 2O

2 → CO

2 + 2H

2O

7C

2H

6 + — O

2 → 2CO

2 + 3H

2O

2Volume of CO

2 = 10 + (2 � 10) = 30 cm3

4. A Bromine water contains Br2 and HOBr.

CH2 = CH

2 + Br

2 → CH

2BrCH

2Br

CH2 = CH

2 + HOBr → CH

2OHCH

2Br

5. C The actual structure of benzene is considerably more stable than the Kekule structure of three carbon-carbon double bonds. Hence the enthalpy of hydrogenation of benzene is lower (less exothermic) than – ( 3 � 120) kJ mol–1. It is slightly lower than the enthalpy of hydrogenation of 1,3-cyclohexadiene but not as low as 152 kJ mol–1.

207.4 6. B 207.4 cm3 of H

2 = ————— = 9.26 � 10–3 mol

22 400 0.25

Number of moles of P = ———— = 4.6 � 10–3

54Thus, P contains two double bonds or one triple bond.

7. C The reaction between fuming sulphuric acid and benzene is electrophilic substitution.

8. A The – NH2 group is ortho- and para-directing.

Structured Question

CH3 CH

3

⏐ ⏐ 1.

(a) CH

3 – C = CHCH

2CH

2C = CHCH

2OH

8 7 6 5 4 3 2 1

3,7-dimethyl means two methyl groups at C3

and C7.

2,6 – octadien- means an alkene with two double bonds at C

2 and C

6 in the longest

continuous chain of eight carbon atoms.– 1-ol means an alcohol (–OH) group at C

1.

(b) CH3 CH

3

⏐ ⏐ CH

3 – C = CHCH

2CH

2C – CH= CH

2

8 7 6 5 4 3 ⏐ 2 1

OH3,7-dimethyl-1,6-octadien-3-ol

(c) Similarities Both geraniol and linalool have two carbon-carbon double bonds and one –OH group per molecule.Both have the same molecular formula, C

10H

18O.

DifferenceThey differ in the positions of double bonds and –OH group, that is, they have different structural formulae. Hence, they are structural isomers.

(d) Molecular formula of geraniol is C10

H18

O.C

10H

18O + 2Br

2 → C

10H

18OBr

4

CH3 CH

3

⏐ ⏐ (e) CH

3 – C = CHCH

2CH

2C* – CH = CH

2

⏐

OH

Linalool contains a chiral carbon atom. It is therefore optically active and will rotate the plane of polarised light.

Essay Question

1. (a) (i) Cyclohexane and hexane are saturated hydrocarbons.They have similar chemical reactions.They both have six carbon atoms per molecule.

(ii) Cyclohexane is a cyclic hydrocarbon but hexane is a straight chain hydrocarbon. The general formula of cyclohexane is C

nH

2n. The general formula of hexane

is CnH

2n+2.

Cyclohexane and hexane have different physical properties such as boiling point and density.

(b) (i) Cl H ⏐ ⏐

Cl – C – C – H ⏐ ⏐ Cl H

1,1,1-trichloroethane

Cl Cl ⏐ ⏐

H – C – C – H ⏐ ⏐ Cl H

1,1,2-trichloroethane

(ii) They have the same molecular formula, C

2H

3Cl

3. They have similar chemical

properties (c) (i) Step 1 Conversion of benzene to

methylbenzene by Friedel-Crafts reaction

ANSWER

CHAPTER 8 Hydrocarbons



+ CH3ClAlCl3 − CH3 + HCl

Step 2 Chlorination of side chain

− CH3 + Cl2uv − CH2Cl + HCl

(ii) Step 1 Conversion of benzene to methylbenzene by Friedel-Crafts reaction

+ CH3ClAlCl3 − CH3 + HCl

Step 2 Oxidation of side chain to benzoic acid

– CH3 + 3[O]acidified − COOH + H2OKMnO4




1. Three haloalkane compounds are shown below. CH

3

⏐ CH

3CH

2CH

2CH

2Cl CH

3CHCH

2CH

3 CH

3CCH

3

⏐ ⏐

Cl Cl I II III

Which sequence shows the haloalkanes arranged in ascending order of their boiling points?A I < II < III C III < II < IB II < I < III D II < III < I

2. Chloroethane, C2H

5Cl, and bromoethane, C

2H

5Br, are hydrolysed by aqueous potassium

hydroxide to form ethanol. Which combination shows correctly the more reactive compound and the reason for the reactivity?

A

B

C

D

3. When CH3CH(Br)CH

3 reacts with cyanide ions, compound X is formed. What is the compound

formed when dilute sulphuric acid is added to X?A (CH

3)

2CHCN C CH

3CH

2COOH

B (CH3)

2CHCOOH D CH

3CH

2CH

2COOH

4. An aromatic compound X has the molecular formula C8H

9Cl. When X is heated with acidified

potassium manganate(VII) solution, compound C8H

6O

4 is formed. X is

A C

CH2CH2Cl

ClCH3CH2

B D

CH2ClCH3

CH3

CH3

Cl

5. The structure of 3-methyl-3-pentanol is shown below.

OH ⏐

CH3CH

2 – C – CH

3

⏐ CH

2

⏐ CH

3

Which of the following compounds can be used to prepare 3-methyl-3-pentanol? O �

A CH3MgBr and CH

3CH

2 – C – CH

3

9 Haloalkanes

More reactive compound Reason

Bromoethane

Bromoethane

Chloroethane

Chloroethane

Higher relative molecular mass

C – Br bond is weaker

C – Cl bond is more polar

Chlorine is a stronger oxidising agent



O �

B CH3CH

2MgBr and CH

3CH

2 – C – CH

3

O �

C CH3CH

2CH

2MgBr and CH

3 – C – CH

3

O �

D CH3CH

2MgBr and CH

3CH

2 – C – CH

2CH

3

6. Which of the following reagents can be used to differentiate between chloropropane and iodopropane?A Aqueous potassium hydroxide C Ethanolic silver nitrateB Ethanolic sodium hydroxide D Chlorine gas

7. (Chloromethyl)benzene reacts with aqueous potassium hydroxide according to the following equation.

CH2Cl + OH– CH2OH + Cl–

What reaction mechanism is involved in the reaction?A Nucleophilic substitution reaction C Electrophilic substitution reactionB Nucleophilic addition reaction D Electrophilic addition reaction

8. The rate equation for the hydrolysis of a haloalkane P is as follows.

Rate = k [P] [OH–]Which haloalkane could be P?A CH

3CH

2CH

2CH

2Br

B CH3CHCH

3

⏐ Br

C CH2CH2CHBrCH3

D (CH3)

3CBr

9. Which haloalkane can be used as a refrigerant?A CCl

4 C CH

3CH

2CH

2Cl

B CF2Cl

2 D

CHCl2

Structured Question

1. (a) A compound Z has the following composition by mass.C, 29.76 %; H, 4.18 %; Br, 66.06 %

(i) Determine the empirical formula of Z. (ii) Given that the relative molecular mass of Z is 242, determine the molecular formula

of Z. (iii) Draw the structural formula of an alkene that can be used to prepare Z and name the

alkene.(b) Chlorofluorocarbons are organic compounds that act as pollutants in the environment. (i) Name a typical chlorofluorocarbon. (ii) State one property of the chlorofluorocarbon named in (b)(i) and one use related to

this property. (iii) Chlorofluorocarbons dissociate in the presence of light. Write an equation to show

the photochemical dissociation of the compound in (b)(i).



Essay Question

1. A haloalkane has the molecular formula of C4H

9Br.

(a) Write the structural formulae of four isomers with the molecular formula C4H

9Br.

(b) One of the isomers from (a) reacts with ethanolic sodium hydroxide to form three compounds with the molecular formula C

4H

8 according to the following equation.

C4H

9Br + NaOH → C

4H

8 + NaBr + H

2O

(i) Identify the isomer that reacts in this reaction. (ii) Give the IUPAC name of the three compounds with the molecular formula C

4H

8 that

can be obtained in this reaction.(c) Another isomer in (a) reacts with NaOH(aq) according to the following equation.

C4H

9Br + NaOH → C

4H

9OH + NaBr

(i) Explain why haloalkanes can be attacked by the OH– ion. (ii) Identify the isomer that would give the highest rate of reaction when reacted with

aqueous sodium hydroxide solution. Give a reason for your answer. Write the reaction mechanism for this isomer.

(d) Which isomer in (a) shows optical isomerism? Identify the chiral carbon with ‘*’.




1. C I is a straight chain molecule with the largest size and highest boiling point.III is a branched chain molecule with the smallest size and lowest boiling point.

2. C Cl is more electronegative than Br. Hence chloroethane is more reactive because of the C – Cl bond being more polar than C – Br bond.

H H ⏐ ⏐ 3. B CH

3 – C – CH

3 + CN– → CH

3 – C – CH

3 + Br–

⏐ ⏐ Br CN

H H ⏐

H+ ⏐

CH3 – C – CH

3 + H

2O ⎯→ CH

3 – C – CH

3

⏐ ⏐ CN COOH B

4. B

CH3 – – CH2Cl + [O] HOOC – – COOH

X C8H6O4

Oδ–

� δ– δ+

5. B CH3CH

2 – Cδ+ – CH

3 CH

3CH

2MgBr

OMgBr ⏐ H

2O, H+

→ CH3CH

2 – C – CH

3 ⎯⎯⎯⎯→

⏐ hydrolysis

CH2CH

3

OH ⏐

CH3CH

2 – C – CH

3

⏐ CH

2CH

3

6. C Chloropropane reacts with ethanolic silver nitrate to produce AgCl (white precipitate). Iodopropane reacts with ethanolic silver nitrate to produce AgI (yellow precipitate).

7. A OH– ion is a nucleophile. It replaces Cl in the molecule hence the reaction mechanism is nucleophilic substitution reaction.

8. A The rate equation shows the reaction is second order, depending on the concentration of P and OH– ion in the slow, rate-determining step. P is a primary haloalkane, CH

3CH

2CH

2CH

2Br.

9. B CFC (chlorofluorocarbon) compound, CF2Cl

2

can be used as a refrigerant.

Structured Question

1. (a) (i) C3H

5Br

(ii) C6H

10Br

2

(iii) Structural formula:

Name: Cyclohexene (b) (i) Dichlorodifluoromethane (ii) Property : Low boiling point

Use : As refrigerant (iii) CCl

2F

2 → CF

2Cl• + Cl•

Essay Question

1. (a) Isomers of C4H

9Br:

CH3CH

2CHCH

3 CH

3CH

2CH

2CH

2Br

⏐ Br

Br ⏐

CH3CHCH

2Br CH

3CCH

3 ⏐ ⏐ CH

3 CH

3

(b) (i) CH3CH

2CHCH

3 ⏐ Br (ii) 1-butene, cis-2-butene, trans-2-butene (c) (i) The C – Br is polar with the carbon

atom being partially positive, thus it can be attacked by OH– ion, which is a nucleophile.

Br ⏐ (ii) CH

3CCH

3 reacts with the highest rate of

⏐ CH

3

reaction. This is because the reaction occurs via a more stable carbonium ion.

Mechanism(CH

3)

3CBr is a tertiary haloalkane. It

reacts via nucleophilic substitution SN1

reaction.

CH3 CH

3 ⏐

slow ⏐

CH3 – C – Br ⎯⎯→ CH

3 – C+ + Br–

⏐ ⏐ CH

3 CH

3

tertiary carbonium ion

CH3 CH

3 ⏐

fast ⏐

CH3 – C+ + OH— ⎯→ CH

3 – C – OH

⏐ ⏐ CH

3 CH

3

*(d) CH

3CH

2CHCH

3

⏐ Br

ANSWER

CHAPTER 9 Haloalkanes




1. Which of the following compounds will produce an alkene in hydration?A (CH

3)

3COH C

CH2OH

B D

OH

OH

CH3

2. Which of the following is the major product formed when 2-butanol is refluxed with concentrated sulphuric acid at 140 °C?A CH

3CHCHCH

3

B CH2CHCH

2CH

3

C CH3CH

2CH

2CH

2 – O – CH

2CH

2CH

2CH

3

D CH3CH

2CH(CH

3) – O – (CH

3)CHCH

2CH

3

3. Which of the following compound has the highest solubility in water?A CH

3CH

2CH

2CH

2OH C (CH

3)

3COH

B CH3CH

2CH(OH)CH

3 D

OH

4. Which sets of reagents can be used to prepare phenyl benzoate in the laboratory?A Phenol and benzoic acidB Phenol and sodium benzoateC Phenol and benzoyl chlorideD Sodium phenoxide and benzoic acid

5. Which of the following compound is formed when ethanoyl chloride is reacted with

CH2OHH2N

OH

A H2N

OH

CH2OCOCH3

BCH2OHH2N

OCOCH3

CH2N CH2OCOCH3

OCOCH3

DCH2OCOCH2

OCOCH3

CH3CONH

A B C D




6. Which of the following compound(s) can react with ethanoic acid to produce an ester?1 C

6H

5COCl

2 C6H

5OH

3 C6H

5CH

2OH

10 Hydroxy Compounds



7. Vanillin is the active ingredient of vanilla and has the following formula:

CHOHO

CH3O

Which of the following is/are true of vanillin?1 Produce hydrogen gas with sodium metal2 Produce an orange precipitate with 2,4-dinitrophenyl hydrazine3 Produce a yellow precipitate with aqueous alkaline iodine

Structured Question

1. C4H

10O forms two alcoholic isomers A and B as follows:

A: CH3CH

2CH

2CH

2OH and B: (CH

3)

3COH

(a) Draw the structural formulae of two other alcoholic isomers of C4H

10O besides A and B.

(b) Give a chemical test that can be used to distinguish A from B.(c) Draw the structural formulae of the products formed from A and B respectively when they

are reacted with concentrated sulphuric acid at 170 °C.(d) Compound C is the alcoholic isomer of C

4H

10O that is optically active. Give a chemical

test that can be used to distinguish A from C.

Essay Question

1. (a) 3-methyl-1-butanol is a primary alcohol. Name and draw the structural formulae of three primary alcohols that are isomers of 3-methyl-1-butanol. Identify the isomer that is optically active and mark the chiral centre.

(b) The ester formed from the reaction of 3-methyl-1-butanol and ethanoic acid has a flavour of ripe pears.

(i) Draw the structural formula of the ester formed. (ii) State the condition used in the laboratory for the preparation of this ester. (iii) Name another chemical that can replace ethanoic acid to produce the same ester.(c) Predict the reaction that will take place when 3-methyl-1-butanol is reacted with (i) sodium metal (ii) phosphorous pentachloride (iii) concentrated sulphuric acid at 170 °C. (iv) acidified potassium dichromate(VI)




1. A Aliphatic alcohol will produce alkene during hydration.

CH3 CH

3 ⏐ ⏐

CH3 – C – CH

3 → CH

3 – C = CH

2 + H

2O

⏐ OH 2. D When 2-butanol is refluxed with excess

concentrated H2SO

4, the major product is alkene.

When excess 2-butanol is refluxed with concentrated H

2SO

4, the major product is ether.

conc. H2SO

42CH3CH

2CH – CH

3 ⎯⎯⎯⎯⎯→

⏐ ∆

OH H H ⏐ ⏐

CH3CH

2 – C – O – C – CH

2CH

3 + H

2O

⏐ ⏐ CH

3 CH

3

3. A The –OH group at the end of the straight chain gives the strongest hydrogen bond between the alcohol and the hydrogen in water hence the highest solubility in water.

4. C Benzoyl chloride, an acyl chloride is very reactive and reacts rapidly with phenol in NaOH solution.

5. D Ethanoyl chloride, an acyl chloride reacts with hydroxyl group, –OH to form ester,R – C – O – R’.

� O 6. C 2 and 3 are correct because ethanoic acid

reacts with –OH group in alcohol and phenol to form ester.

7. B 1 is correct because the –OH group reacts with Na to produce hydrogen gas. 2 is correct because the aldehyde group – C – H reacts with

� O

2,4-dinitrophenyl hydrazine to produce an orange precipitate.

Structured Question

1. (a) CH3CH(OH)CH

2CH

3, CH

2(OH)CH(CH

3)

2

(b) Heat with acidified potassium dichromate(VI) solution / alkaline potassium manganate(VII) solution.A will change potassium dichromate(VI) from orange to green / potassium manganate(VII) purple to colourless but B has no visible change.

[O]CH

3CH

2CH

2CH

2OH ⎯→ CH

3CH

2CH

2COOH

(c) A will produce CH3CH

2CH = CH

2.

B will produce CH2 = C(CH

3)

2.

(d) Heat with alkaline iodine solution.C will produce a yellow precipitate but A has no visible change.

I2 / OH–

CH3 – CH(OH) – CH

2CH

3 ⎯⎯⎯→

CH3CH

2COO– + CHI

3

Essay Question

1. (a) CH3CH

2CH

2CH

2CH

2OH (1-pentanol)

(CH3)

3C – CH

2OH (2,2-dimethyl-1-propanol)

CH3CH

2CH(CH

3)CH

2OH (2-methyl-1-butanol)

*CH

3CH

2CH(CH

3)CH

2OH is optically active.

(b) (i) CH3COOCH

2CH

2CH(CH

3)

2

(ii) Reflux with concentrated sulphuric acid (iii) Ethanoyl chloride, CH

3COCl

(c) (i) H2 gas is released.

(CH3)

2CHCH

2CH

2OH + Na →

(CH3)

2CHCH

2CH

2ONa + H

2

(ii) HCl gas is liberated.(CH

3)

2CHCH

2CH

2OH + PCl

5 →

(CH3)

2CHCH

2CH

2Cl + HCl + POCl

3

(iii) An alkene is produced. conc. H

2SO

4(CH3)

2CHCH

2CH

2OH ⎯⎯⎯⎯→

(CH3)

2CHCH = CH

2

(iv) A carboxylic acid is produced. K

2Cr

2O

7/H+

(CH3)

2CHCH

2CH

2OH + 2[O] ⎯⎯⎯→

(CH3)

2CHCH

2COOH + H

2O

ANSWER

CHAPTER 10 Hydroxy Compounds




1. Propanal, CH3CH

2CHO and propanone, CH

3COCH

3 are isomers. Which pair of characteristics

is shown by the mass spectra of the two compounds?

A

B

C

D

2. Which compound will give a yellow precipitate with alkaline iodine solution? O O � �

A CH3 – C – OCH

3 C CH

3 – C – NH

2

O O � �

B CH3 – C – OH D CH

3 – C – CH

2OH

3. The structure of cyclohexanone is shown below.

O

Which statement is true about cyclohexanone?A It is an aromatic ketone.B Its molecular formula is C

6H

12O.

C It can be oxidised to benzoic acid.D It will give a precipitate with 2,4-dinitrophenylhydrazine.

4. A compound Q has the following composition by mass.

Carbon, 64.3%; hydrogen, 7.1%; oxygen, 28.6%.

Q reduces Fehling’s solution to copper(I) oxide. 0.1 mol Q reacts with 4480 cm3 hydrogen gas at s.t.p. with nickel as catalyst. Q is[Relative atomic mass: H, 1; C, 12; O, 16. 1 mol gas occupies 22.4 dm3 at s.t.p.]A CH

2 = CHCHO C CH

3CH = CHCH = CHCHO

B CH ≡ CCH2OH D CH

3COCH = CHCH = CHOH

5. Citral is used by honeybees as pheromone that attracts other bees.

CH3

⏐CH

3C = CHCH

2CH

2C = CHC = O

⏐ ⏐ CH

3 H

Which reagent is decolourised by citral?A Fehling’s solutionB 2,4-dinitrophenylhydrazine solutionC Hot, dilute acidified potassium dichromate(VI) solutionD Cold, dilute acidified potassium manganate(VlI) solution

11 Carbonyl Compounds

m/e of molecular ion m/e of molecular fragments

Different

Different

Same

Same

Different

Same

Different

Same



6. The reaction between cyanide ion and phenylethanone is shown in the equation below.

COCH3 + CN– C

CN

OH

CH3

Which statement is true about the reaction?A The CN– ion acts as a nucleophile where the N atom on the CN– ion donates a lone pair of

electrons to a carbon atom in phenylethanone.B The reaction can be carried out by using an acidic solution of sodium cyanide.C The product formed exists as geometric isomers.D The reaction is a nucleophilic substitution reaction.

7. The structure of fructose is given below.

HO – CH2 – C – CH – CH – CH – CH

2OH

� ⏐ ⏐ ⏐ O OH OH OH

Which statement is not true about fructose?A It has six optical isomers.B It is a reducing sugar.C It exists in equilibrium with a pentagonal ring form.D It forms an orange precipitate with 2,4-dinitrophenylhydrazine solution.

A B C D




8. An organic compound gives an orange precipitate with 2,4-dinitrophenylhydrazine but does not form a silver mirror when heated with diamminesilver(I) ions (Tollen’s reagent). This compound could be1 C

6H

5COCH

3 3 C

6H

5CH

2CHO

2 C6H

5CHO

9. Phenylethanone can be converted to 1-phenylethanol as shown below.

C

O

CH3 C

H

OH

CH3

Which reagent(s) can be used to perform the above conversion?1 Sodium in ethanol 3 Hydrogen gas and hot nickel2 Zinc and dilute hydrochloric acid

Structured Question

1. Lactic acid is found in sour milk. It can be formed from ethanol as shown in the following reaction scheme.

OH Step I Step II Step III ⏐

CH3CH

2OH ⎯⎯⎯→ CH

3CHO ⎯⎯⎯→ Q ⎯⎯⎯→ CH

3 – CH – C = O

HCN ⏐ OH lactic acid

(a) In Step I, ethanol is converted to ethanal. (i) State the reagents used in Step I. (ii) Suggest a chemical test that can be used to differentiate between ethanol and

ethanal.



(b) In Step II, ethanal is allowed to react with HCN. (i) Draw the structure of compound Q. (ii) Write an equation for the reaction involved in Step II. (iii) Name the reaction mechanism involved in this reaction.(c) In Step III, compound Q is converted to lactic acid. (i) State the IUPAC name of lactic acid. (ii) State the reagent used in this conversion.(d) Lactic acid exhibits stereoisomerism. (i) Name the type of stereoisomerism shown by lactic acid. (ii) Draw the stereoisomers of lactic acid.

Essay Question

1. A compound P has the molecular formula C4H

6O

2. P reacts with hydrogen cyanide to give

compound Q, C6H

8O

2N

2. P reacts with warm acidified potassium dichromate(VI) to give an

acidic compound R, C4H

6O

3.

2.50 g of compound R was dissolved in 250 cm3 water. 25.0 cm3 of this solution required24.5 cm3 of 0.10 mol dm–3 sodium hydroxide solution for complete neutralisation.(a) (i) Use the molecular formula of R to calculate its relative molecular mass. (ii) Determine the basicity of acid R.(b) Write two structural formulae of compound P. (c) State a chemical test that can be used to distinguish between the two isomers of P in (b).(d) Using any one isomer of P, identify compounds Q and R.



Multiple-choice Objectives

1. C Propanol and propanone have the same molecular formulae, hence m/e of molecular ion will be the same. The structures of propanol and propanone are different, hence the fragments will have different m/e.

2. D Alkaline iodine will react with a ketone which has the CH

3 – C – group.

� O 3. D Cyclohexanone is a ketone hence it will react

with 2,4-dinitrophenylhydrazine to produce a yellow precipitate.

4. A Q reduces Fehling’s solution to Cu2O hence

Q is an aldehyde with the functional group, –CHO.

4480Number of moles of H

2 gas = ————— = 0.2

22 400

0.1 mol Q reacts with 0.2 mol H2 gas. Hence

Q contains one double bond, C = C besides the one aldehyde group.

5. D The presence of C = C double bonds in citral causes cold, dilute acidified KMnO

4 to be

decolourised when the KMnO4 reacts with

⏐ ⏐ ⏐ ⏐– C = C – to form diol, – C – C –.

⏐ ⏐ OH OH 6. B Acidic solution of sodium cyanide forms HCN

which will react with through

– COCH3

nucleophilic addition mechanism. 7. A Fructose does not have six optical isomers. B,

C and D are true. 8. A The compound does not react with Tollen’s

reagent, thus it is not aldehyde (2 and 3 are incorrect). 1 is correct because ketone can react with 2,4-dinitrophenylhydrazine to produce an orange precipitate.

9. D The reaction is a reduction reaction. 1, 2 and 3 are correct since all are reducing agents.

Structured Question

1. (a) (i) KMnO4/H

2SO

4

(ii) Ethanal gives an orange precipitate with 2,4-dinitrophenylhydrazine while ethanol does not.

(b) (i) CH3CH – OH

⏐ C ≡ N (ii) CH

3CHO + HCN → CH

3CH(OH)CN

(iii) Nucleophilic addition (c) (i) 2-hydroxypropanoic acid (ii) Dilute hydrochloric acid (d) (i) Optical isomerism (ii)

C

COOH

OH HH3C

C

COOH

OHHCH3

Essay Question

1. (a) (i) 102 (ii) [R] = 2.50/102 � 1000/250

= 0.0980 mol dm–3

aR + bOH → H2O

0.0980 � 25.0 0.10 � 24.5———————————— = ——————————

a b a 1

— = — b 1

Thus, basicity of R = 1

O O � �

(b) CH3 – C – CH

2 – C – H;

O O � �

CH3 – CH

2 – C – C – H

(c) Reagent: Iodine + NaOH(aq)Condition: HeatObservation: CH

3COCH

2CHO gives a yellow

precipitate while CH3CH

2COCHO does not.

OH OH ⏐ ⏐

(d) Q: CH3 – C – CH

2 – CH;

⏐ ⏐ CN CN O O � �

R: CH3 – C – CH

2 – C – OH

ANSWER

CHAPTER 11 Carbonyl Compounds




1. When chlorine gas is passed into pure ethanoic acid at 100 °C and in the presence of uv light, the organic compound obtained isA ClCH

2COOH C CH

3CH

2Cl

B CH3COCl D ClCH

2CHO

2. The reaction scheme for the synthesis of the organic compound, Y is shown below:

K2Cr

2O

7/H+ Cl

2/uv

CH3CH

2OH ⎯⎯⎯⎯→ X ⎯⎯→ Y

reflux

The pH values of ethanol, X and Y increases in the orderA pK

a of ethanol < pK

a of X < pK

a of Y

B pKa of ethanol < pK

a of Y < pK

a of X

C pKa of Y < pK

a of X < pK

a of ethanol

D pKa of X < pK

a of Y < pK

a of ethanol

3. Which of the following pairs of compounds cannot be used for the preparation of an ester?A Benzoic acid and phenol C Ethanoic acid and 2-propanolB Ethanoyl chloride and phenol D Benzoic acid and ethanol

4. The esterification of ethanoic acid with 2-propanol produces an ester with the formulaA CH

3COOCH(CH

3)

2 C CH

3CH

2COOCH(CH

3)

2

B CH3COOCH

2CH

3 D CH

3CH

2COOCH

2CH

3

5. The best method to prepare phenyl benzoate is byA refluxing benzoic acid with phenylmethanol.B refluxing benzoic acid with phenol.C mixing benzoyl chloride with phenylmethanol.D mixing benzoyl chloride with phenol dissolved in aqueous sodium hydroxide.

6. What is the molecular formula of benzenedicarboxylic acid and how many structural isomers can this molecular formula have?

A

B

C

D

A B C D




7. The structural formula of an organic compound is shown below.

OHCOOH

The organic compound is expected to react with1 bromine water2 aqueous sodium carbonate3 lithium tetrahydridoaluminate(III) in ether

12 Carboxylic Acids

Molecular formula Number of structural isomers

C7H

6O

4

C7H

8O

4

C8H

6O

4

C8H

8O

4

3

4

3

4



8. A compound with the formula CH3CH(OH)COOH will

1 act as an alcohol.2 act as a carboxylic acid.3 react with hot, concentrated sulphuric acid.

Structured Question

1. (a) The organic compounds, ethanol, phenol, ethanoic acid and benzoic acid, all contain the hydroxyl (–OH) group. From this list of compounds, select

(i) two compounds which will react with sodium hydrogen carbonate.Write equations to represent these reactions.

(ii) one compound which will react with aqueous sodium hydroxide but will not react with sodium hydrogen carbonate.

(iii) one compound which will react with sodium metal but will not react with sodium carbonate or sodium hydroxide.

(b) Aqueous sodium benzoate can be distinguished from aqueous sodium phenoxide by using hydrochloric acid.

(i) Name the organic product formed and describe what you would see in each case. (ii) Describe one chemical test which can be used to distinguish between these two

organic products.

Essay Question

1. Methacrylic acid, CH2 = C(CH

3)COOH, is used widely to form addition polymers.

(a) (i) What is the IUPAC name of methacrylic acid? (ii) How would you prepare methacrylic acid from HOCH

2CH(CH

3)COOH?

(b) Predict the reactions of methacrylic acid with (i) lithium tetrahydridoaluminate(III),(ii) potassium manganate(VII) solution and (iii) sodium.

(c) Write the formula of the polymer formed when methacrylic acid polymerises.




1. A Chlorine attacks the alkyl group in ethanoic acid.

2. C X is CH3COOH and Y is ClCH

2COOH or

Cl2CHCOOH or CCl

3COOH.

Ethanol is the weakest acid. Hence it has the highest pK

a value.

Chlorinated ethanoic acid is the strongest acid.Hence, it has the lowest pK

a value.

3. A Benzoic acid reacts with ethanol to form ethyl benzoate but benzoic acid does not react with phenol to form phenylbenzoate. Phenylbenzoate is prepared by Schotten-Baumann reaction.

O H � ⏐ 4. A CH

3 – C – O – H + H – O – C – CH

3

ethanoic acid ⏐ CH

3

2-propanol O H � ⏐

CH3 – C – O – C – CH

3 + H

2O

⏐ CH

3

5. D C6H

5COCl + Na+ –O – C

6H

5 →

C6H

5COOC

6H

5 + NaCl

6. C COOH

COOH

COOH

COOH COOH

COOH

7. D It contains a phenol group. Hence it will react with bromine water.It contains a – COOH group. Hence it liberates CO

2 from Na

2CO

3 and is reduced to – CH

2OH

by LiAlH4.

8. D –CH(OH) is a secondary alcohol group.–COOH is a carboxylic acid group.

conc. H2SO

4CH

3CH(OH)COOH ⎯⎯⎯⎯⎯→

CH2 = CHCOOH + H

2O

Structured Question

1. (a) (i) CH3COOH and C

6H

5COOH

2CH3COOH + Na

2CO

3 →

2CH3COONa + H

2O + CO

2

2C6H

5COOH + Na

2CO

3 →

2C6H

5 COONa + H

2O + CO

2

(ii) Phenol (C6H

5OH)

(iii) Ethanol (CH3CH

2OH)

(b) (i) For sodium benzoateOrganic product: Benzoic acidObservation: White precipitate of

benzoic acid formed.

For sodium phenoxideOrganic product: PhenolObservation: White emulsion of phenol

formed. (ii) Use neutral iron(III) chloride solution.

Benzoic acid: Buff-coloured precipitate formedPhenol: Purple solution formed

Alternative methodUse sodium carbonate solution.Benzoic acid: Effervescence and carbon

dioxide gas liberatedPhenol: No reaction

Essay Question

1. (a) (i) 2-methylpropenoic acid (ii) By dehydration, using concentrated

sulphuric acid at 180 °C.

H CH3

⏐ ⏐H

– C – C – COOH →

⏐ ⏐ OH H H CH

3

⏐ ⏐H

– C = C – COOH + H

2O

(b) (i) Reduction to CH2 = C(CH

3)CH

2OH

Comment Lithium aluminium hydride is the source of hydride ion (H–) which acts as the nucleophile. Hence, the C = C double bond is not affected because the C = C bond is susceptible to electrophilic attack and not nucleophilic attack.

(ii) KMnO4 decolourised.

In cold condition, mild oxidation CH

3

⏐produces a glycol, CH

2 – C – COOH

⏐ ⏐ OH OH

On heating, cleavage of the C = C double bond occurs and a mixture of aldehydes, ketones and carboxylic acids are formed.

(iii) Hydrogen gas is liberated andCH

2 = C(CH

3)COONa is formed.

(c) H CH3

⏐ ⏐–( C – C )–

n

⏐ ⏐ H COOH

ANSWER

CHAPTER 12 Carboxylic acids




1. When ethanoyl chloride is added to phenylamine, C6H

5NH

2, at room temperature, the organic

product obtained is A C

6H

5NHCH

3

B C6H

5N(CH

3)

2

C CH3COC

6H

4NH

2

D C6H

5NHCOCH

3

2. When acylation occurs, the hydrogen atom in the –OH group of alcohol is replaced withA R– C RCO

2–

B RCO– D RCH2–

3. An organic compound, W, is boiled with aqueous sodium hydroxide and the reaction mixture

is allowed to cool. Upon cooling, compound X and compound Y (an alcohol) are produced. On acidification, compound X is converted to compound Z (C

3H

6O

2). Compound Y gives a

positive triiodomethane (iodoform) test. Compound W is likely to have the structural formulaA CH

3CH

2COOCH

3

B CH3COOCH

3

C CH3CH

2COOCH

2CH

3

D CH3CH(OH)CH

2COCl

4. When ethyl ethanoate, CH3 – CO18O – C

2H

5, is hydrolysed with sulphuric acid, the organic

product obtained is A CH

3CH

2COOH

B CH3CO18OH

C CH3CH

2OH

D CH3CH

218OH

5. The mechanism for the reaction between ammonia and ethanoyl chloride is shown below.

δ–

O O– O � ⏐ –Cl– �

HNH2 + CH

3 – C – Cl → CH

3 – C – Cl ⎯→ CH

3 – C + Cl– + H+

δ+ ⏐ –H+ ⏐ H – N+H

2 NH

2

This shows that the mechanism A involves the formation of a planar intermediate.B involves elimination reaction followed by addition reaction.C is a nucleophilic addition reaction.D is a nucleophilic substitution reaction.

6. Calculate the number of isomers (structural and stereo) for the acyl chlorides with the molecular formulae, (i) C

4H

7OCl, and (ii) C

5H

9OCl.

A

B

C

D

13 Carboxylic Derivatives

C4H

7OCl C

5H

9OCl

2

2

3

3

3

5

4

5



A B C D




7. Ethanoyl chloride can be used to synthesise1 CH

3COOC

2H

5

2 CH3COOC

6H

4COOH

3 C6H

5NHCH

3

8. When a compound is boiled with aqueous sodium hydroxide, ammonia gas is liberated. The

compound(s) is/are likely to be1 CH

3CONH

2

2 CH3COONH

4

3 CH3CH

2CH

2NH

2

Structured Question

1. With reference to the reaction scheme given below, answer the following questions.

PCl5X (C

2H

4O

3) ⎯⎯⎯⎯→ 2 mol HCl

oxidation

PCl5Y (C

2H

2O

3) ⎯⎯⎯⎯→ 1 mol HCl

further

oxidation

PCl5Z (C

2H

2O

4) ⎯⎯⎯⎯→ 2 mol HCl

(a) State the reagents used for the conversion of X to Y. (b) Suggest the structural formulae for X, Y and Z.(c) State the reagents and conditions required to convert Z to a neutral compound with the

molecular formula, C4H

6O

4.

(d) How and under what conditions does Z react with (i) KMnO4, (ii) concentrated H

2SO

4.

Essay Question

1. An ester, X (C9H

10O

2), is boiled with excess aqueous sodium hydroxide. The solution is then

distilled. The distillate contains a neutral organic compound, Y (C8H

10O). When Y undergoes

mild oxidation, compound Z (C8H

8O) is obtained.

Compound Z gives a positive triiodomethane test but does not give a precipitate with aqueous solution containing the complex ion, [Ag(NH

3)

2]+.

(a) Name the class of organic compounds to which (i) Z and (ii) Y belong.(b) Hence, write the structural formula for (i) Y, (ii) Z.(c) From your answer in (b), deduce the structural formula of X.(d) How would you expect the residue from the distillation to react with (i) acidified potassium manganate(VII), (ii) concentrated sulphuric acid.




1. D CH3COCl + C

6H

5NH

2 →

C6H

5NHCOCH

3 + HCl

O � 2. B R – OH + R’COCl → R – O – C – R’ + HCl 3. C O �

CH3CH

2 – C – OCH

2CH

3 + NaOH →

WCH

3CH

2COONa + CH

3CH

2OH

H+ X Y

CH

3CH

2COOH

Z 4. D In hydrolysis of esters, the acyl-oxygen fission

occurs and the 18O isotope is present in the alcohol.

O �

CH3 – C – 18OC

2H

5 + H

2O →

CH3COOH + C

2H

518OH

5. D The intermediate is tetrahedral and not planar in shape. NH

3 is a nucleophile.

6. B Structural isomers of C4H

7OCl

CH3CH

2CH

2COCl ; CH

3CHCOCl

⏐ CH

3

Structural isomers of C5H

9OCl

CH3CH

2CH

2CH

2COCl;

CH3CH

2*CHCOCl;

⏐ CH

3

[This molecule exists as optical (d and l) isomers] CH

3 ⏐

CH3 – C – COCl ; CH

3CHCH

2COCl

⏐ ⏐ CH

3 CH

3

7. B CH3COCl + C

2H

5OH → CH

3COOC

2H

5 + HCl

CH3COCl + HOC

6H

4COOH →

CH3COOC

6H

4COOH + HCl

8. B An amide or an ammonium salt gives ammonia gas when heated with aqueous sodium hydroxide.An amine has no reaction with aqueous sodium hydroxide.CH

3CONH

2 + NaOH → CH

3COONa + NH

3

CH3COONH

4 + NaOH →

CH3COONa + H

2O + NH

3

Structured Question

1. (a) Potassium manganate(VII) solution, acidified with dilute sulphuric acid.

O � (b) X is HO – CH

2 – C – OH

Y is COOH Z is COOH ⏐ ⏐ CHO COOH

(c) Reagent: Methanol and a little concentrated sulphuric acid as catalyst.

Condition: Heat COOH COOCH

3 ⏐ + 2CH

3OH → ⏐ + 2H

2O

COOH COOCH3

(d) (i) When a mixture of Z (ethanedioic acid) and acidified KMnO

4 is heated,

decolourisation of KMnO4 occurs and

ethanedioic acid is oxidised to carbon dioxide and water. 5C

2O

42– + 2MnO

4– + 16H+ →

2Mn2+ + 8H2O + 10CO

2

(ii) When Z is warmed with concentrated sulphuric acid, dehydration occurs to produce carbon dioxide and carbon monoxide.COOH ⏐ → H

2O + CO

2 + CO

COOH

Essay Question

1. (a) (i) Z is a ketone (ii) Y is a secondary alcohol (b) (i) Y is

CH3 C

H

OH

(ii) Z is

CH3 C

O

(c) X is

O

CH3

C

H

C

O

H

(d) (i) Saponification of ester X gives sodium methanoate (residue) and Y.The residue decolourises KMnO

4 because

methanoate is a reducing agent. Methanoate ion is oxidised to carbon dioxide and MnO

4– is reduced to Mn2+ ions.

5HCOO– + 2MnO4– + 11H+ →

5CO2 + 8H

2O + 2Mn2+

(ii) Concentrated sulphuric acid reacts with methanoate to give methanoic acid.

HCOO– + H+ → HCOOH

Carbon monoxide is liberated when methanoic acid reacts with concentrated H

2SO

4 which acts as a dehydrating

agent.HCOOH → H

2O + CO

ANSWER

CHAPTER 13 Carboxylic Derivatives




1. Compound P is produced when bromoethane reacts with alcoholic potassium cyanide. Which of the following is formed when P is reacted with tetrahydridoaluminate(III)?A CH

3CH

2CN C CH

3CH

2CH

2OH

B CH3CH

2NH

2 D CH

3CH

2CH

2NH

2

2. Which of the following amines is most soluble in water?A CH

3CH

2NH

2 C (CH

3)

3N

B (CH3)

2NH D C

6H

5NH

2

3. Which of the following pairs of chemicals will react to produce N-phenylethanamide?A

NH2 and CH3CH2COCl

BNH2 and CH3COCl

CCOClCH3CH2NH2 and

DCOClCH3NH2 and

4. Which of the following product is formed when aniline is reacted with cold nitrous acid followed by p-nitrophenol?A N=N-O – NO2

B

N=N – OH

NO2

C HO

N=N

NO2

D

N=N

NO2

OH

5. When 1 mol of compound X is treated with nitrous acid, 1 mol of nitrogen gas is liberated. Which of the following compounds is likely to be compound X?A CH

3CH

2NH

2

B (CH3)

2NH

C (CH3)

3N

D H2NCH

2CH

2NH

2

A B C D




6. Which of the following chemicals can be used to distinguish from CH3CONH

2?

NH2

1 Nitrous acid 3 Bromine water2 Hydrochloric acid

14 Amines



7. Which of the following is true of aniline?1 Soluble in hydrochloric acid2 Soluble in sodium hydroxide solution3 Soluble in water

Structured Question

1. Benzocaine is an active ingredient found in many sunburn ointment. It can be prepared by the following reaction scheme.

CH3

I

CH3

NO2

II

CO2H

NO2 NH2

XIII IV

CO2CH2CH3

(a) Name the type of reaction that has occurred in step I. Hence state the reagent and the condition for step I.

(b) State the reagent and the condition for step II.(c) Ethanoyl chloride is added in step III. Draw the structural formula of the intermediate

compound X. Name the type of reaction that has taken place.(d) Suggest how step IV can be carried out. Name the type of reaction involved.

Essay Question

1. (a) 1-chloropropane is heated with alcoholic potassium cyanide to produce compound A. When compound A is reacted with tetrahydridoaluminate(III), compound B is formed. Compound B reacts with ethanoyl chloride to liberate white fumes and compound C is produced. Determine the structures and give the IUPAC names for compounds A, B and C. Write equations for all the reactions involved.Suggest how 1-butanamine can be prepared from compound C. Write an equation for the reaction involved.

(b) State a simple test to differentiate the following pairs of substances: (i) CH

3CH

2NH

2 and CH

3NHCH

3

(ii) NH2CH3CH2NH2 and




1. D CH3CH

2Br + CN– ⎯→ CH

3CH

2CN + Br–

∆ P

CH3CH

2CN + [H] → CH

3CH

2CH

2NH

2

2. A CH3CH

2NH

2 can form the most hydrogen

bonds with water. 3. B O

�– NH2 + CH3C – Cl

H O ⏐ �– N – C – CH3 + HCl

N-phenylethanamide

4. C – NH2 + HNO2 – N2+

– N2+ + NO2 – – OH

– N = N –

NO2

OH

5. A 1 mol of primary amine will react with HNO2

to produce 1 mol of N2 gas.

CH3CH

2NH

2 + HNO

2 →CH

3CH

2OH + N

2 + H

2O

6. C is basic and reacts with HCl to form – NH2

salt. CH3CONH

2 is neutral and does not react

with HCl (2 is correct).

Bromine will react with to form a

– NH2

white precipitate while CH3CONH

2 does not

react with bromine (3 is correct).

7. A Aniline, is basic since it can react

– NH2

with HCl to form soluble salt (1 is correct).

Structured Question

1. (a) NitrationReagent: Concentrated nitric acidCondition: In the presence of concentrated

sulphuric acid(b) Reagent: Potassium manganate(VII)

Condition: Acidified and reflux / heat

(c) CO2CH2CH3

NO2

Type of reaction: Esterification(d) React with tin / zinc with dilute hydrochloric

acidType of reaction: Reduction

Essay Question

1. (a) Compound A is butanenitrile,CH

3CH

2CH

2CN

Compound B is butamine,CH

3CH

2CH

2CH

2NH

2

Compound C is N-buthylethamide,CH

3CH

2CH

2CH

2NHCOCH

3

alcoholic KCNCH

3CH

2CH

2Cl ⎯⎯⎯⎯⎯⎯→

CH3CH

2CH

2CN

LiAlH4CH

3CH

2CH

2CN ⎯⎯⎯→

CH3CH

2CH

2CH

2NH

2

CH3COCl

CH3CH

2CH

2CH

2NH

2 ⎯⎯⎯⎯→

CH3CH

2CH

2CH

2NHCOCH

3 + HCl

Reflux with aqueous sodium hydroxide solution: ∆ with NaOH(aq)CH

3CH

2CH

2CH

2NHCOCH

3 ⎯⎯⎯⎯⎯⎯→

CH3CH

2CH

2CH

2NH

2 + CH

3COONa

(b) (i) React with nitrous acid and heat.CH

3CH

2NH

2 liberates N

2 gas, CH

3NHCH

3

has no visible change.

HNO2

CH3CH

2NH

2 ⎯⎯→ CH

3CH

2OH + N

2

(ii) Add bromine water.

produces a white precipitate,NH2

CH2CH

2NH

2 has no visible change.

NH2 Br

Br

BrNH2

Br2

ANSWER

CHAPTER 14 Amines




1. Which of the following is true about aminoethanoic acid?A It exists as a dimer in organic solvent.B It exists as a diazonium ion.C It is optically active.D It forms zwitterion.

2. The simplest amino acid, glycine, is soluble in both aqueous acid and aqueous alkali. When glycine is dissolved in aqueous sodium hydroxide solution at pH 12.5, it exists asA H

3N+CH

2COO– C H

3N+CH

2COOH

B H2NCH

2COO– D H

2NCH

2COOH

3. Tyrosine is a natural existing amino acid with the formula C9H

11NO

3. Which of the following

is the most likely structure of tyrosine?A

CH2CH(OH)CH3O2N

BCH2CH(NH2)COOHHO

CCH2CH(OH)COOHH2N

D

CH2 C COOH

OH

NH2

4. 1 mol serine, HOCH2CH(NH

2)COOH will

A react with 2 mol of hydrochloric acid.B react with sodium metal to liberate 2 mol of gas.C react with excess nitrous acid to liberate 1 mol of gas.D react with excess Na

2CO

3 solution to liberate 1 mol of gas.

5. Which of the following fibers is considered a protein fiber? A Silk C CottonB Hemp D Glycogen

A B C D




H3C

6. The structural formula for threonine is CH – CH – COOH ⏐ HO NH

2

Which of the following is/are true about threonine?1 Threonine has amphoteric properties.2 Threonine releases nitrogen gas when heated with nitric(V) acid 3 The repeating unit of threonine in a polypeptide chain is CH

3 O

⏐ � – O — CH – CH – C – . ⏐ NH

2

15 Amino Acids and Proteins



7. GABA with the chemical formula H2N(CH

2)

3COOH is the chief inhibitory neurotransmitter in

the central nervous system and also in the retina.Which of the following is/are true of GABA?1 It exists as zwitterion.2 It is found in protein.3 It is optically active.

8. Which of the following statements is/are true of proteins?1 Proteins are polyamides.2 Globular proteins are more soluble in water than fibrous proteins.3 When protein is denatured, the peptide sequence is destroyed.

Structured Question

1. Insulin is known to lower blood glucose level. Insulin is made up of a chain of 51 amino acids. Partial hydrolysis gives a peptide chain as shown.

O H O H O H � ⏐ � ⏐ � ⏐

Val-Glu-Ala-Leu H2N – CH – C – N – CH – C – N – CH – C — N – CH – COOH

⏐ ⏐ ⏐ ⏐ CH – CH

3 CH

2 CH

3 CH

2

⏐ ⏐ ⏐ CH

3 CH

2 CH – CH

3

⏐ ⏐ COOH CH

3

(a) With further hydrolysis, this peptide chain yields four amino acids. Draw the structures of the four amino acids obtained.

(b) Give the systematic names of these four amino acids(c) Classify these amino acids as either non-polar, polar, acidic or basic. Explain your

answers.

Essay Question

1. Amino acids has the general formula H2NCH(R)COOH.

(a) Draw the structural formulae of the first two amino acids, glycine and alanine.(b) Alanine is optically active. Explain what is meant by this term. (c) Describe what is observed and give balanced equations for the reactions of alanine with

solutions of (i) sodium bicarbonate (ii) nitrous acid(d) Using alanine as an example, explain how an amino acid can act as a buffer solution.(e) (i) How many different tripeptides are possible which contain two molecules of glycine

and one molecule of alanine? (ii) Draw the structures of the tripeptides formed. (iii) Discuss the reactions of the tripeptides with dilute sodium hydroxide solution under

different conditions.




1. D H2NCH

2COOH exists as zwitterion,

+

H3NCH

2COO–.

2. B H2NCH

2COOH + OH– → H

2NCH

2COO– + H

2O

+

or H3NCH

2COO– + OH– →

H2NCH

2COO– + H

2O

3. B General formula of naturally occurring amino acids: H

2NCH(R)COOH

4. C Only 1 –NH2 group to react with 1 HCl.

Na reacts with the 2 –OH groups to liberate 1 H

2.

With HNO2, 1 –NH

2 group liberates 1 N

2.

1 –COOH group reacts with Na2CO

3 to form

½ CO2.

5. A 6. A Threonine has –COOH group and NH

2 group

(1 is correct).The –NH

2 group will only release N

2 gas when

heated with nitric(III) acid, HNO2, not with

nitric(V) acid, HNO3 (2 is incorrect).

The structure is not a polypeptide chain (3 is incorrect).

7. A GABA is not α-amino acid, thus not found in protein.GABA does not have chiral carbon, hence not optically active.

8. B Proteins are polyamides (1 is correct).Globular proteins can form more hydrogen bonds with water hence more soluble in water than fibrous proteins (2 is correct).Denaturation causes the tertiary structure to open up and destroyed but the peptide sequence is not destroyed (3 is incorrect).

Structured Question

1. (a) H2N – CH – COOH H

2N – CH – COOH

⏐ ⏐ CH – CH

3 CH

2

⏐ ⏐ CH

3 CH

2

⏐ COOH (i) (ii)

H2N – CH – COOH H

2N – CH – COOH

⏐ ⏐ CH

3 CH

2

⏐ CH – CH

3

⏐ CH

3 (iii) (iv) (b) (i) 2-amino-3-methylbutanoic acid (ii) 2-aminopentanedioic acid

(iii) 2-aminopropanoic acid (iv) 2-amino-4-methylpentanoic acid

(c) 2-aminopentanedioic acid is acidic because it has two carboxyl groups.The other 3 amino acids are non-polar and neutral with hydrophobic aliphatic R groups.

Essay Question

O O � � 1. (a) H

2N – CH – C – OH H

2N – CH – C – OH

⏐ ⏐ H CH

3

glycine alanine(b) Alanine rotates the plane of polarised light.

Alanine has a chiral or asymmetrical C* that is bonded to four different groups.

(c) (i) Effervescence due to liberation of CO2

gas.

H2NCH(CH

3)COOH + NaHCO

3 →

H2N – CH(CH

3)COO–Na+ + CO

2 + H

2O

(ii) Effervescence due to liberation of nitrogen gas.

H2N – CH(CH

3)COOH + HNO

2 →

HO – CH(CH3)COOH + N

2 + H

2O

(d) Alanine can react with both acid and base.

H2NCH(CH

3)COOH + OH– →

H2NCH(CH

3)COO– + H

2O

H2NCH(CH

3)COOH + H+ →

H3N+CH(CH

3)COOH

(e) (i) 3 (ii)GAG O H O H � ⏐ � ⏐H

2N – CH

2 – C – N – CH – C – N – CH

2 – COOH

⏐ CH

3

GGA O H O H � ⏐ � ⏐H

2N – CH

2 – C – N – CH

2 – C – N – CH – COOH

⏐ CH

3

AGG O H O H � ⏐ � ⏐H

2N – CH – C – N – CH

2 – C – N – CH

2 – COOH

⏐ CH

3

(iii) With cold dilute alkali, a salt is formed.When boiled with dilute alkali, the tripeptide would be hydrolysed to amino acids.

ANSWER

CHAPTER 15 Amino Acids and Proteins




1. The polymer containing the following repeating unit,

O �

– C – C6H

4 – C

2H

4 – O –

is aA SBR rubber C detergentB protein D polyester

2. Concentrated nitric(V) acid oxidises cyclohexanol to an organic compound which can be used to makeA Terylene C polyetheneB nylon-6,6 D polystyrene

3. The structure of natural rubber can be represented by the formula

A –( CH2 – CH = C – CH

2 )–

n C –( CH

2 – CH = CH – CH

2 )–

n

⏐ CH

3

B –( CH2 = CH – CH

2 – CH

2 )–

n D –( CH

2 – CH = C – CH

2 )–

n

⏐ ⏐ CH

3 CH

3

4. O

Epoxyethane, CH2 – CH

2, can be polymerised under suitable conditions. The repeating unit for

this polymer is O

A – CH – CH – C – CH2 – CH

2 – O –

B – CH2 – O – CH

2 – O – D – CH – CH –

⏐ ⏐ OH OH 5. Which of the following statements about natural rubber is true?

A Its monomer is 2-methyl-1,3-butadiene. B It is the trans-isomer of polyisoprene. C It can be synthesised by the addition polymerisation of isoprene.D It is a saturated hydrocarbon.

6. Which of the following structures represents the repeating unit of nylon-6,6? O O � �

A – C – (CH2)

6 – C – O – (CH

2)

4 – O –

H O H O ⏐ � ⏐ �

B – N – (CH2)

6 – C – N – (CH

2)

4 – C –

H H O O ⏐ ⏐ � �

C – N – (CH2)

4 – N – C – (CH

2)

6 – C –

H H O O ⏐ ⏐ � �

D – N – (CH2)

6 – N – C – (CH

2)

4 – C –

16 Polymers



A B C D




7. nHO2CCH

2CH

2COOH + nHOCH

2CH

2OH → –( COCH

2CH

2CO

2CH

2CH

2O )–

n + 2nH

2O

PThe formation of polymer P in the above reaction can be considered as1 esterification2 condensation polymerisation3 hydrolysis

8. Which of the following can be used as the initiator for anionic polymerisation?1 Bu–Li+

2 Na+NH2–

3 Li+NH2–

Structured Question

1. (a) Write the structure of the copolymer formed from the following reactions. (i) CH

2 = CCl

2 + CH

2 = CHCN

(ii) CH2OH

⏐ + ClOC – (CH2)

6 – COCl

CH2OH

(b) The partial structure of polymer X is shown below.

O CH3 O CH

3

|| ⏐ � ⏐– C – (CH

2)

2 – C – O – C – (CH

2)

2 – C – O –

⏐ ⏐ H H

(i) Write the repeating unit for polymer X. (ii) Write the structural formula of the monomer for making polymer X.(c) The structural formulae of polymers Y and Z are shown below.

H H O H ⏐ ⏐ � ⏐

–( C – C )–n and –( C – (CH

2)

5 – N )–

n

⏐ ⏐ Z H CH

3

Y

(i) State one use of polymers Y and Z. (ii) Draw the structures of the monomers of Y and Z and name them. (iii) Which of these two polymers is not suitable for use as a container for concentrated

sodium hydroxide solution? Explain. (iv) Which polymer, Y or Z will create environmental problems? Why?

Essay Question

1. (a) Outline a reaction scheme to show how phenol can be used as starting material to make synthetic fibre.

(b) Explain the mechanism of free-radical polymerisation using chloroethene as an example.




O O � � 1. D – C – C

6H

4 – C

2H

4 – O – C – C

6H

4 – C

2H

4 – O –

ester linkage 2. B Cyclohexanol is oxidised to 1,6-hexanedioic

acid, one of the monomers making nylon-6,6. 3. A 4. C O

CH

2 – CH

2 → – CH

2 – CH

2 – O –

5. A Natural rubber is a cis-isomer. It is a polyunsaturated hydrocarbon.

6. D 7. B The reaction between a carboxylic acid and an

alcohol is called esterification.P is a condensation polymer.

8. D For anionic polymerisation, the initiator is a Lewis base (nucleophile).

Structured Question

1. (a) (i) H Cl H CN ⏐ ⏐ ⏐ ⏐

–( C – C – C – C )–n

⏐ ⏐ ⏐ ⏐ H Cl H H (ii) O O � �

–( O – CH2CH

2 – O – C – (CH

2)

6 – C )–

n

(b) (i) O CH3

� ⏐ – C – (CH

2)

2 – C – O –

⏐ H (ii) O CH

3

� ⏐HO – C – (CH

2)

2 – C – O – H

⏐ H (c) (i) Polymer Y: For making plastic bags

Polymer Z: As a synthetic fibre for making clothing

(ii) H H O H ⏐ ⏐ � ⏐

C = C and HO – C – (CH2)

5 – N – H

⏐ ⏐ monomer of Z H CH

3

monomer of Y (6-aminohexanoic acid)(propene)

(iii) Polymer Z because it contains the amide linkage. In the presence of concentrated NaOH, the amide linkage is broken up due to hydrolysis.

(iv) Polymer Y. It is non-biodegradable and causes land pollution.

Essay Question

1. (a) The synthetic fibre made from phenol is nylon-6,6.Step 1 Preparation of 1,6-hexanedioic acid

H2 HNO

3Phenol ⎯→ cyclohexanol ⎯⎯→ HOOC(CH

2)

4COOH

Step 2 Preparation of 1,6-hexanediamine NH

3HOOC(CH

2)

4COOH ⎯→

–H2O

ammonium 1,6-hexanedioate ⎯⎯→ H

2NC(CH

2)

4CN ⎯→ H

2N(CH

2)

6NH

2

Step 3 Reaction between HOOC(CH2)

4COOH

and H2N(CH

2)

6NH

2

HOOC(CH2)

4COOH + H

2N(CH

2)

6NH

2

→ nylon-6,6 + nH2O

(b) • Initiation step

O O � �

C6H

5 – C – O – O – C – C

6H

5 →

benzoyl peroxide2C

6H

5• + 2CO

2

• Chain propagation step

H H H H ⏐ ⏐ ⏐ ⏐

C6H

5• + C = C → C

6H

5 – C – C•

⏐ ⏐ ⏐ ⏐ H Cl H Cl

The chain propagates as successive monomers are added to the chain.

H H H H ⏐ ⏐ ⏐ ⏐

C6H

5 – C – C• + C = C →

⏐ ⏐ ⏐ ⏐ H Cl H Cl H H H H H ⏐ ⏐ ⏐ ⏐ ⏐

C6H

5 – C – C – C – C• → –( CH

2 – C )–

n

⏐ ⏐ ⏐ ⏐ ⏐ H Cl H Cl Cl

• Termination stepThe addition polymerisation stops when two free radicals combine.

H H ⏐ ⏐

2C6H

5 – C – C• →

⏐ ⏐ H Cl H H H H ⏐ ⏐ ⏐ ⏐

C6H

5 – C – C – C – C – C

6H

5

⏐ ⏐ ⏐ ⏐ H Cl Cl H

ANSWER

CHAPTER 16 Polymers


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