CAPE Chemistry Study Paper 001α

Paper Reference(s)

Edexcel GCEChemistry

Advanced Subsidiary Organic Chemistry QuestionsUnit Test 1-5Wednesday 6 June 2007 and Monday 18 June – Morning

Time: 1 hour

Materials required for examination Items included with question papersNil Nil

Candidates may use a calculator.

Instructions to Candidates

Answer ALL the questions. Write your answers in the spaces provided in this question paper.

).Show all the steps in any calculations and state the units.

Information for Candidates

Marks are shown in round brackets: e.g. (2)

A Periodic Table is printed on the back cover of this question paper.

Advice to CandidatesYou are reminded of the importance of clear English and careful presentation in your answers.

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Team Leader’s use only

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Paper Reference

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2. (a) Chlorofluorocarbons, CFCs, are useful compounds as they are non-flammable and unreactive. In the stratosphere, the C—Cl bonds are broken producing free radicals.

(i) Draw the structure of 1,1,2-trichloro-1,2,2-trifluoroethane.

(1)

(ii) Suggest why C—F bonds are not broken in the stratosphere, whereas C—Cl bonds are.

................................................................................................................................(1)

(iii) What are free radicals and what is needed to produce them from CFCs?

................................................................................................................................

................................................................................................................................(2)

(b) (i) Draw the repeating units in

poly(chloroethene)

poly(tetrafluoroethene).

(2)

(ii) Give ONE important use of each of these polymers.

poly(chloroethene)

................................................................................................................................

poly(tetrafluoroethene)

................................................................................................................................(2) Q2

(Total 8 marks)




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3. (a) Name each of the following organic compounds and the homologous series to which it belongs.

Name .................................................... Name ....................................................

Homologous Series .............................. Homologous Series ..............................(4)

(b) (i) Draw the TWO structural isomers of C2H2Cl2.

(2)

(ii) Identify which of the two structural isomers of C2H2Cl2 can exist as geometric isomers. Explain why geometric isomerism can occur.

Identity

Explanation

................................................................................................................................

................................................................................................................................

................................................................................................................................(2)

H

HO

H H

H

H—C—C—C—H

H

H

H

H—C—C—C—C—H

H H H

HHH—C—H

H—C—H

H




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(c) Classify the following organic reactions.

(i) C2H4 + HBr → C2H5Br

................................................................................................................................(1)

(ii) C2H5Br + CN– → C2H5CN + Br–

................................................................................................................................(1)

(d) A hydrocarbon has the empirical formula CH2 and a relative molecular mass of 70.

Write the molecular formula of the hydrocarbon.

.......................................................................................................................................(1) Q3

(Total 11 marks)




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4. (a) When excess chlorine and methane are mixed at room temperature and pressure no reaction takes place but when ultraviolet light is shone into the mixture an explosion occurs, producing carbon and hydrogen chloride.

uvCH4(g) + 2Cl2(g) C(s) + 4HCl(g) ∆H = –219 kJ mol–1

Calculate the mass of methane needed to produce 1000 kJ of energy.

(2)

(b) Draw a labelled reaction profile for the reaction between methane and chlorine and use it to explain why the reaction does not take place unless ultraviolet light is present.

Explanation

.......................................................................................................................................

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.......................................................................................................................................

.......................................................................................................................................

.......................................................................................................................................(5)

Progress of reaction

Enthalpy




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5. (a) Define the term standard enthalpy of combustion.

.......................................................................................................................................

.......................................................................................................................................

.......................................................................................................................................

.......................................................................................................................................

.......................................................................................................................................(3)

(b) The following standard enthalpies of combustion are needed to calculate the standard enthalpy of formation of ethanol, C2H5OH.

Substance Standard enthalpy of combustion /kJ mol–1

carbon, C (s, graphite) –394

hydrogen, H2 (g) –286

ethanol, C2H5OH (l) –1371

(i) Complete the Hess’s Law cycle by filling in the box and labelling the arrows with the enthalpy changes.

∆Hf○

2C(s) + 3H2(g) + 3½O2(g) C2H5OH(l) + 3O2(g)

......................................................

........................... ...........................

............................... + ...............................

(3)



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(ii) Use your Hess’s Law cycle to calculate the standard enthalpy of formation of ethanol.

(2)

(c) Ethanol, C2H5OH, can be converted into propanenitrile, CH3CH2C≡N in two steps.

Step 1 Step 2C2H5OH C2H5I CH3CH2C≡N

Give the reagents for the steps.

Step 1

Reagents

.......................................................................................................................................

.......................................................................................................................................

Step 2

Reagent

.......................................................................................................................................(3) Q5

(Total 11 marks)




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3. A halogenoalkane, Y, has the molecular formula C4H9X, where X represents a halogen atom.

When Y is heated with excess aqueous sodium hydroxide, it is converted into Z,C4H10O.

Complete the tables below.

(a) Test Observation Inference

To the solution remaining after

heating Y with excess aqueous

sodium hydroxide,

add ...........................................

..................................................

followed by aqueous silver

nitrate.

White precipitateThe atom X

is ..............................

(2)

(b) Test Observation Inference

Add phosphorus

pentachloride to pure Z.

Test the gas evolved with

damp blue litmus paper.

................................. fumes

were seen at the mouth of

the test tube.

The litmus paper turned

red.

The gas evolved is

......................................

Z is an alcohol.

(2)

(c) Test Observation Inferences

Warm Z with acidified

aqueous potassium

dichromate(VI).

............................................

............................................

Z is not oxidised.

Z is a ............................

alcohol.

(2)




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(d) Based on the observations and inferences in (a) to (c), draw the structural formula of Y.

(1) Q3

(Total 7 marks)




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2. Ethanol can be converted into ethylamine by two different routes.

(a) Identify organic compounds V and W by writing their full structural formulae showing all bonds.

V

W

(2)

(b) Identify the reagents used in Steps A to E.

Step A ............................................................................................................................

Step B ...........................................................................................................................

Step C ...........................................................................................................................

Step D ...........................................................................................................................

Step E ............................................................................................................................(6)

Compound V → CH3CH2NH2

CH3CH2OH

CH3COOH → CH3COCl → Compound W → CH3CN

Step A

Step E

Step DNH3Step C

Step B

KBr and H2SO4




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(c) (i) What type of organic compound would be formed when ethylamine, CH3CH2NH2reacts with ethanoyl chloride, CH3COCl?

................................................................................................................................(1)

(ii) A polymer is formed when the two monomers X and Y shown below react together under suitable conditions.

H2N(CH2)6NH2 ClOC(CH2)4COClX Y

Draw sufficient of the polymer chain to make its structure clear.

(2)

(d) The compound CH3CH2NH2 has a distinctive smell. When dilute hydrochloric acid is added to an aqueous solution of CH3CH2NH2, the distinctive smell disappears. On the addition of excess aqueous sodium hydroxide, the smell returns.

Give an equation to explain each of these observations.

Loss of smell:

.......................................................................................................................................

Return of smell:

.......................................................................................................................................(2) Q2

(Total 13 marks)




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6. (a) Draw the structural formula of the organic product of each of the following reactions of propanal. Classify the type of reaction in each case.

(i) Propanal with sodium tetrahydridoborate(III) (sodium borohydride) in water.

Type of reaction .....................................................................................................(2)

(ii) Propanal with Fehling’s solution, followed by acidification of the product.

Type of reaction with Fehling’s solution ................................................................(2)

(iii) Propanal with hydrogen cyanide.

Type of reaction .....................................................................................................(2)

(b) Magnesium reacts with bromoethane to form a Grignard reagent.

(i) Write the equation for this reaction.

................................................................................................................................(1)

(ii) State the necessary conditions for the reaction.

................................................................................................................................

................................................................................................................................(1)




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(c) (i) Give the structural formula and name of the final organic product formed when the Grignard reagent produced in (b)(i) is reacted with the following reagents and the intermediate is hydrolysed.

Reagent Structural formula of finalorganic product

Name of final organicproduct

propanone

butanal

(4)

(ii) Explain why a racemic mixture and not a single optical isomer is obtained when butanal reacts.

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................................................................................................................................(2)

TOTAL FOR PAPER: 75 MARKS

END

Q6

(Total 14 marks)




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Answer ALL the questions. Write your answers in the spaces provided.

1. A chemist has synthesised a compound W believed to be

(a) State and explain what you would see if W is reacted with:

(i) sodium carbonate solution

................................................................................................................................

................................................................................................................................

................................................................................................................................(2)

(ii) bromine water.

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................(3)

HO CH CH CH C

Cl O

OH




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(b) W shows both types of stereoisomerism.

(i) How many stereoisomers of W are there? Briefly explain your answer.

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................(2)

(ii) Explain why W shows optical isomerism.

................................................................................................................................

................................................................................................................................

................................................................................................................................(2)

(c) Describe how you would show that W contains chlorine.

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.......................................................................................................................................(5) Q1

(Total 14 marks)




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4. Phenylethanoic acid occurs naturally in honey as its ethyl ester: it is the main cause of the honey’s smell.

The acid has the structure

Phenylethanoic acid can be synthesised from benzene as follows:

(a) State the reagent and catalyst needed for step 1.

.......................................................................................................................................

.......................................................................................................................................(2)

CH2COOH

step 1 step 2

step 3KCN

step 4CH2COOH Compound A

CH3

Cl2, uv

CH2Cl




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(b) (i) What type of reaction is step 2?

................................................................................................................................(1)

(ii) Suggest a mechanism for step 2. You should include the initiation step, the two propagation steps and a termination step.

You may use Ph to represent the phenyl group, C6H5.

(4)




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(iii) Draw an apparatus which would enable you to carry out step 2, in which chlorine is bubbled through boiling methylbenzene, safely.

Do not show the uv light source.

(3)




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(c) (i) Give the structural formula of compound A.

(1)

(ii) Give the reagent and the conditions needed to convert compound A into phenylethanoic acid in step 4.

................................................................................................................................

................................................................................................................................(2)

(iii) Suggest how you would convert phenylethanoic acid into its ethyl ester.

................................................................................................................................

................................................................................................................................(2)




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(d) (i) An isomer, X, of phenylethanoic acid has the molecular formula C8H8O2.

This isomer has a mass spectrum with a large peak at m/e 105 and a molecular ion peak at m/e 136.

The ring in X is monosubstituted.

Suggest the formula of the ion at m/e 105 and hence the formula of X.

(2)

(ii) Another isomer, Y, of phenylethanoic acid is boiled with alkaline potassium manganate(VII) solution and the mixture is then acidified. The substance produced is benzene-1,4-dicarboxylic acid:

Suggest with a reason the structure of Y.

Reason ....................................................................................................................

................................................................................................................................(2)

COOH

COOH




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(e) Benzene-1,4-dicarboxylic acid can be converted into its acid chloride, the structural formula of which is

This will react with ethane-1,2-diol to give the polyester known as PET.

(i) What reagent could be used to convert benzene-1,4-dicarboxylic acid into its acid chloride?

................................................................................................................................(1)

(ii) Give the structure of the repeating unit of PET.

(2)

(iii) Suggest, with a reason, a type of chemical substance which should not be stored in a bottle made of PET.

................................................................................................................................

................................................................................................................................

................................................................................................................................

................................................................................................................................(2)

TOTAL FOR PAPER: 75 MARKS more or less ; )

END

Q4

(Total 24 marks)

COCl

COCl




b 50 AS-Level Revision Notes

AS Chemistry – Revision Notes Unit 3 – Introduction To Organic Chemistry

Nomenclature And Isomerism 1. There are two types of organic compounds:

a. Aliphatic – has a functional group attached to an alkyl group. b. Aromatic – has a functional group attached to an aryl group (containing benzene).

2. A functional group is an atom or group of atoms which, when present in different molecules, causes them to have similar chemical properties.

3. Each functional group will have a homologous series, with the main carbon chain increasing by one carbon atom in each successive member (they differ by a CH2 unit):

Homologous Series Prefix / Suffix Functional Group Alkane -ane C – C Alkene -ene C = C

Haloalkane halo- – Cl, – Br, – I Alcohol -ol, hydroxy- – OH

Ether alkoxy- – O – Aldehyde -al

Ketone -one, oxo- C = O

Carboxylic acid -oic acid

Amine amino-, -amine – NH2 Amide -amide

Nitrile -nitrile – C ≡ N

4. There are three types of formula: a. Molecular – shows the actual number of atoms in a single molecule. b. Structural – shows how the atoms are grouped in the molecule. c. Displayed – shows all the atoms and all the bonds (drawn out).

5. To name an hydrocarbon from the displayed formula: a. Look for the longest carbon chain containing the functional group – this gives the stem. b. Add the position of the functional group (lowest number) to the stem (e.g. hex-2-ene). c. For any side-chains, write down the chain position (using the same numbering as for the

functional group) followed by a dash, followed by the name (e.g. 2-ethyl…). d. For more than one of the same alkyl, use the di- and tri- prefixes (e.g. 2,3-dimethyl…).

6. To draw a hydrocarbon from the name: a. Draw the longest carbon chain as shown in the stem – number it from the right. b. Add the functional group (if any) in the correct position. c. Add the side-chains in the correct positions. d. Fill in the rest of the bonds with hydrogens.

7. Isomers are compounds that have the same molecular formula but different structures: a. Structural isomers – have the same molecular formula but different structural formulae. b. Sterioisomers – have the same structural formula, but different arrangements of the bonds

in space. 8. There are three types of structural isomer:

a. Chain isomers – the carbon chain can be arranged in different formations of chains and side chains, i.e. the carbon skeleton is different (e.g. butane and 2-methylpropane).

b. Positional isomers – the functional groups are in different positions (e.g. but-1-ene and but-2-ene).

c. Functional group isomers – these have different functional groups, and belong to different homologous series’ (e.g. ethanol and methoxymethane – CH3OCH3).

9. There are two types of stereoisomer: a. Geometric isomers – there is no rotation about the double bond in an alkene, so two

different groups at each end of the double bond can either be on the same side, i.e. cis, or on opposite sides, i.e. trans (e.g. cis-but-2-ene and trans-but-2-ene).


base.5 AS-Level Revision Notes

b. Optical isomers – have the same geometric and structural formula, but cannot be superimposed onto one another, i.e. they are mirror images (e.g. 2-bromobutane).

Petroleum And Alkanes 1. The general formula of an alkane is CnH2n+2. 2. Alkanes are unreactive with acids, alkalis, nucleophiles and electrophiles and are non-polar. 3. The boiling points of alkanes increase as the chain length increases, because there are more Van

der Waals forces holding the longer molecules together, which require more energy to break. 4. Alkanes can be extracted from crude oil (a mixture of hydrocarbons) using fractional distillation:

a. The crude oil is heated until vaporised, and passed into the fractionating column. b. The temperature decreases going up the fractionating column. c. Hydrocarbons will pass up the column until they reach their boiling point, at which they

condense and can be tapped off. The bubble caps aid this by causing the vapour to bubble through the liquid at each fraction, and for the larger hydrocarbons to condense.

5. The following fractions are obtained from fractional distillation (increasing boiling points): Fraction Carbon Atoms Uses

LPG (liquefied petroleum gas) 1-4 calor gas, camping gas Petrol (gasoline) 4-12 petrol

Naphtha 7-14 petrochemicals Kerosine (paraffin) 11-15 jet fuel, petrochemicals

Gas oil (diesel) 15-19 central heating fuel Mineral oil (lubricating oil) 20-30 lubricating oil, petrochemicals

Fuel oil 30-40 fuel for ships / power stations Wax / grease 40-50 candles, grease, polish

Bitumen >50 roofing, road surfacing 6. After primary distillation, vacuum distillation can be used to separate the fractions further,

particularly the residue (the last four in the above table are from the residue). 7. Larger fractions that are not in high demand can be cracked into smaller fractions that are in much

higher demand by cracking: a. Cracking will result in a long alkane being broken into a shorter alkane, and an alkene

(e.g. ethene) being produced. Hydrogen may also be produced. b. Thermal cracking – temperatures from 400°C to 900°C and pressures of up to 7000kPa

are used to cause the cracking to take place as a free radical mechanism (the conditions will initiate the homolytic fission).

c. Catalytic cracking – a zeolite catalyst is used (acts as a Lewis acid and involves carbocation formation), at a temperature of about 450°C and a slight excess pressure. This produces more branched hydrocarbons, and also some cyclic and aromatic hydrocarbons.

8. Alkanes undergo complete combustion to form CO2 and H2O (g), which are both greenhouse gases. The enthalpy of combustion increases as the carbon chain length increases.

9. In a limited supply of oxygen, incomplete combustion of alkanes will take place, producing carbon as soot, or carbon monoxide gas, which is toxic and causes carbon monoxide poisoning.

10. Impurities in petrol can result in the formation of SO2 gases in car engines. Also the high temperature causes the oxidation of nitrogen gas to form NOx gases. These gases cause acid rain.

11. Catalytic converters are used to reduce the emissions of these gases, using a platinum, rhodium and palladium mixture spread in a thin layer – )g(N)g(CO2)g(NO2)g(CO2 22 +→+

12. There are two ways in which a covalent bond can break: a. Homolytic fission – one electron goes to each atom to form free radicals. b. Heterolytic fission – one atom takes both electrons, forming ions.

13. A free radical is a highly reactive species with an unpaired electron, signified with a • next to the atom that contains the unpaired electron.

14. The chlorination of methane is a substitution reaction and a free radical mechanism: a. Initiation (requires UV light): • → Cl2Cl lightUV

2 b. Propagation (chain reaction): HClCHClCH 34 +•→•+

•+→+• ClClCHClCH 323 c. Termination (free radicals react): 2ClCl2 →•

ClCHClCH 33 →•+•

623 HCCH2 →•


b 50 AS-Level Revision Notes

15. In an excess of chlorine, further substitution can occur to form CH2Cl2, CHCl3 and CCl4. 16. The reaction will occur between all alkanes and all halogens, but the rate will decrease going down

the group. Fluorine doesn’t require UV light, as it is so reactive. Alkenes And Epoxyethane 1. The general formula of an alkene is CnH2n. 2. In naming alkenes, the number signifies the first carbon atom in the chain that forms the double

bond. If there is more than one double bond, then add “a” to the end of the stem, and use the prefixes di-, tri- etc. in front of the -ene ending (e.g. hexa-1,2-diene).

3. An alkene has strong single covalent bonds between all the carbon atoms (σ bonds), with an extra weaker covalent bond between the two carbons held by the double bond (π bond). A π bond is formed due to the overlapping of two p orbitals parallel to one another, so that the delocalised π bond electrons lie above and below the σ bond and are therefore held less strongly.

4. Tests for alkenes: a. Shake with bromine water; alkenes decolourise the mixture. b. Shake with acidified potassium manganate (VII) (aq); alkenes decolourise the mixture.

5. Alkenes combust in the same way as alkanes, but are more likely to undergo incomplete combustion. They are not used as fuels because of this, and because they are so important in the manufacture of other chemicals.

6. An electrophile is an electron-deficient compound that can form a new covalent bond, using an electron pair provided by the carbon compound. This can be either a positive ion (e.g. H+) or a polar molecule with a δ+ charge (e.g. BF3).

7. Ethene reacts with bromine in the dark (not free radical), and is an electrophilic addition reaction. This must take place dissolved in an organic solvent:

a. Initiation (dipole induced by π electrons, and electrophile approaches double bond):

b. Formation of intermediate (the electrophile attacks the bond, forming a covalent bond and

resulting in the formation of a carbocation or carbonium ion and a Br– ion):

c. Termination (the carbocation and Br– ion react to form 1,2-dibromoethane):

8. Ethene reacts with hydrogen halides in the same way (electrophilic addition), except that the

dipole is permanent, and therefore doesn’t need to be induced. They react as aqueous solutions. 9. For unsymmetrical hydrocarbons, Markownikoff’s rule states which of the two possible products

will be formed: a. An alkyl group tends to donate electrons slightly to any carbon atom it is attached to –

this is the inductive effect. b. A primary carbocation will have one alkyl group doing this, whereas a secondary

carbocation will have two, and a tertiary carbocation will have three. c. As the inductive effect increases for increasing order of carbocation, so does stability. d. The highest order carbocation will always be formed, as this is the most stable. e. When a molecule adds across a double bond, the more electropositive atom or group (the

electrophile) adds to the carbon atom with more hydrogen atoms already attached. 10. Catalytic hydration of alkenes:

a. Generally – CnH2n + H2O (g) ℑ CnH2n+1OH – i.e. an alcohol is formed. b. A phosphoric (V) acid (H3PO4) is used, along with a temperature of 300°C and a pressure

of 60 atmospheres.


questionb 50m AS-Level Revision Notes

c. This reaction is reversible, and so can also be used to produce alkenes from alcohols. 11. An alternative method for the production of alcohols uses concentrated sulphuric acid:

a. Sulphuric acid should be thought of as:

b. The reaction will be electrophilic addition, and will form an alkyl hydrogen sulphate. It

takes place in the cold. c. The alkyl hydrogen sulphate is then added to water and warmed, causing it to be

hydrolysed to an alcohol and sulphuric acid. d. The sulphuric acid is regenerated, and has therefore acted as a catalyst.

12. Catalytic hydrogenation of alkenes: a. Generally – CnH2n + H2O → CnH2n+2 – i.e. an alkane is formed. b. A nickel (Ni) catalyst is used, along with a temperature of 150°C to 300°C. c. The mechanism for this reaction is not electrophilic addition; it is a catalytic reduction. d. This reaction is used to convert polyunsaturated fatty acids in vegetable oils into more

saturated fatty acids. This makes them less rigid, so they can pack closer together and the intermolecular forces increase, increasing the melting point. They will therefore be solid, and are used as margarine.

13. Polymerisation involves a large number of alkenes bonding together into addition polymers that are saturated, in the presence of a catalyst. This involves free radicals.

14. Epoxyethane (a cyclic ether) is produced by the direct partial oxidation of ethene:

15. Epoxyethane has the following properties:

a. It is a colourless gas at room temperature, with a boiling point of 10°C. b. It is flammable and explosive, due to the instability of its structure so it will readily react. c. It is highly reactive towards nucleophiles, due to the polar C–O bonds. d. It has a strong tendency to polymerise, particularly in contact with an alkali. e. It is highly toxic.

16. Epoxyethane is hydrolysed with a ten molar excess of water to produce ethane-1,2-diol: a. Steam is used at 200°C and a pressure of 14 atmospheres. b. Alternatively concentrated sulphuric acid catalyst at 60°C is used. c. Ethane-1,2-diol is used in antifreeze. It is 30% sweeter than sugar, but is very toxic.

17. Epoxyethane can be hydrated with less water to produce polyethylene glycols, that can be used as solvents and lubricants, and in the manufacture of plasticisers, polyurethanes and polyester resins.

18. Epoxyethane will react with alcohols to form monoalkyl ethers of ethane-1,2-diol. These can be used in paints, printing inks, plasticisers and surfactants.

19. An organic acid with two COOH groups will react with a diol to form a monomer with one ester linkage. These can then be polymerised to form a polyester.

20. Ethane-1,2-diol is used in the production of poly(ethyleneterephthalate), i.e. PET. Condensation reactions with benzene-1,4-dicarboxylic acid and two ethane-1,2-diol molecules form the monomer (230–250°C, 3–4 atm) and this is then polymerised to form PET (270–300°C, zero pressure, antimony catalyst).

Halogenoalkanes 1. In naming halogenoalkanes:

a. The name of the halogen (fluoro, chloro, bromo, iodo) precedes any alkyl groups. b. The prefixes di, tri, tetra etc. are used for more than one of the same halogen.

2. The boiling point of a halogenoalkane is affected by: a. The chain length – longer chains have greater Van der Waals forces and will therefore

have a higher boiling point. b. The size of the halogen atom – the smaller the halogen atom, the lower the boiling point. c. The number of halogen atoms – the more halogen atoms, the higher the boiling point.

3. Chlorofluorocarbons are haloalkanes whereby every hydrogen atom has been replaced by either a chlorine or a fluorine atom. They are very stable, so persist in the atmosphere for many years.

4. Halogenoalkanes have polar bonds, because the halogens are very electronegative, and so create a dipole. This leaves the δ+ carbon atom open to nucleophilic attack.


question 0m AS-Level Revision Notes

5. Increasing the size of the halogen atom: a. Decreases the bond strength, as the reactivity decreases down the group. b. Increases the reactivity of the haloalkane, as the weaker bond is easier to break. c. Decreases the polarity of the bond, as the electronegativity decreases down the group.

6. A nucleophile is an electron-rich species with a lone pair of electrons. This attacks a carbon atom that is electron deficient by donating a pair of electrons.

7. Primary and secondary haloalkanes will undergo nucleophilic substitution in the following mechanism (SN2):

8. Warming a halogenoalkane with an aqueous alkali will result in the formation of alcohols (SN2),

where the OH– ion is the nucleophile. 9. Warming a halogenoalkane with an aqueous solution of potassium cyanide will result in the

formation of nitriles (SN2), where the CN– ion is the nucleophile. This increases the length of the carbon chain by one carbon atom.

10. Nitriles can be hydrolysed to carboxylic acids by heating under reflux (a condenser is used to stop gases from escaping) with an aqueous alkali or with mineral acid:

a. Nitrile to amide – 223223 CONHCHCHOHCNCHCH →+ . b. Amide to carboxylic acid – 3232223 NHCOOHCHCHOHCONHCHCH +→+ .

11. Warming a halogenoalkane in a sealed container with an excess of ammonia will result in the formation of primary amines (SN2), where the NH3 molecule is the nucleophile:

a. Generally – R–Hal + NH3 → R–NH2 + H–Hal. b. The H–Hal will immediately react with NH3 to form NH4Hal. c. The mechanism occurs in two steps:

12. When a halogenoalkane reacts with OH– ions, the hydroxide ions may function as a base rather

than a nucleophile, causing an elimination reaction to take place. This results in the formation of an alkene, with one hydrogen atom and one halogen atom eliminated:

13. Primary haloalkanes tend to favour substitution reactions, whereas tertiary haloalkanes tend to

favour elimination reactions. With secondary haloalkanes, both reactions occur concurrently. 14. In order to favour one reaction, the conditions need to be altered:

a. Elimination is favoured by hot ethanolic conditions, and a stronger base. b. Substitution is favoured by warm aqueous conditions, and a weaker base.

Alcohols 1. When naming alcohols, use the prefixes di- and tri- to signify dihydric and trihydric alcohols

respectively. 2. Alcohols are industrially produced for alcoholic drinks by fermentation:

a. Anaerobic respiration in yeast is used to ferment sugar into alcohol and CO2. b. A series of enzymes called zymase catalyses the reaction. c. It is a batch process – if the ethanol gets too concentrated it denatures the enzymes.


base.50m AS-Level Revision Notes

d. Stronger alcohols are produced by distilling or by adding more ethanol (fortified wines). 3. Alcohols can also be industrially produced by the direct hydration of ethene, and this tends to be

preferred as gives a higher yield and is continuous, although it is more expensive to run. 4. The boiling points of alcohols are affected by:

a. Chain length – the longer the chain length, the higher the boiling point as the intermolecular forces are greater.

b. Order of alcohol – primary alcohols have higher boiling points than tertiary alcohols as the molecules can get closer together.

c. Hydrogen bonding – this is experienced by hydroxyl groups, and increases the boiling point. The more hydroxyl groups, the higher the boiling point will be.

5. The solubility of alcohols is affected by increasing the carbon chain length – this lowers the solubility, as there will be more interactions between the alcohol molecules than between alcohol and water molecules.

6. Alcohols can be oxidised in the presence of an oxidising agent (shown as [O]). This can be either acidified potassium dichromate (VI) solution (potassium dichromate (VI) with sulphuric acid) or acidified potassium manganate (VII) solution:

a. Primary alcohols – these will oxidise to form an aldehyde. In this reaction one hydrogen is removed from the hydroxyl group, and one from the carbon to which the hydroxyl group is attached. The aldehyde can be further oxidised to a carboxylic acid by refluxing the mixture at a higher temperature.

b. Secondary alcohols – these will oxidise to form a ketone. These cannot be oxidised any further.

c. Tertiary alcohols – these will not oxidise, as there is no other hydrogen attached to the carbon to which the hydroxyl group is attached.

7. Alcohols can be tested for with an oxidising agent, as the oxidising agent will be reduced during oxidation of the alcohol. If a colour change occurs then either a primary or secondary alcohol is present:

a. Acidified potassium dichromate changes from orange (Cr2O72– ions) to green (Cr3+ ions)

when reduced. b. Acidified potassium manganate changes from purple (MnO4

– ions) to colourless (Mn2+ ions) when reduced.

8. To test for a primary or secondary alcohol, either Tollen’s reagent or Fehling’s solution can be used. Both react to oxidise aldehydes to carboxylic acids but do not react with ketones, so 10 and 20 alcohols can be distinguished.

9. Tollen’s reagent will cause silver atoms to coat the test tube in the presence of an aldehyde: a. This is a mixture of ammonia and silver nitrate, containing [Ag(NH3)2]+ complex ions. b. Silver (I) ions are reduced are reduced to Ag when the aldehyde is oxidised.

10. Fehling’s solution will produce a brick red precipitate in the presence of an aldehyde: a. This contains copper (II) ions in a complex. b. Blue copper (II) ions are reduced to brick red copper (I) ions, producing Cu2O precipitate.

11. Aldehydes and ketones can be reduced to form 10 and 20 alcohols respectively, in the presence of a reducing agent (shown as 2[H]). This can be either sodium tetrahydridoborate (NaBH4) dissolved in aqueous alcohol, or lithium tetrahydridoaluminate (LiAlH4) dissolved in epoxyethane.

12. The dehydration of an alcohol to form an alkene works at a temperature of 180°C and a catalyst of either concentrated sulphuric acid or concentrated phosphoric acid:

a. An H+ ion from the acid is attracted to the δ– charge on the oxygen of the hydroxyl group

and the OH group is protonated. b. A molecule of H2O is lost from the protonated alcohol to form a carbocation. c. The carbocation loses a proton (H+ ion) to form an alkene.


questionbase.50m A2-Level Revision Notes

Nomenclature And Isomerism 1. Functional groups in addition to those in CHM3:

Homologous Series Prefix / Suffix Functional Group Acid chloride -oyl chloride

Acid anhydride -oic anhydride

Nitro nitro- – NO2

Phenyl phenyl-

Secondary amine N-alkylamino- – NH – R Tertiary amine N,N-dialkylamino-



2. When naming cyclic compounds, add the prefix cyclo- to the name of the carbon chain that forms the ring. Cycloalkanes can exhibit geometric isomerism, if two adjacent carbon atoms each have two different groups attached. As with alkenes, this leads to cis and trans isomers.

3. When functional groups are placed around a cycloalkane or an aromatic, the lowest number carbon atom should be used for the principle functional group. For a benzene ring, the positions can be indicated relative to the principle functional group:

a. Ortho (o-) – on the carbon next to the principle functional group. b. Meta (m-) – on the second carbon from the principle functional group. c. Para (p-) – on the carbon opposite the principle functional group.

4. Some aromatic compounds can be given alternative names for simplification: a. Methylbenzene à toluene. b. Hydroxybenzene à phenol.

5. If a beam of light is passed through a polaroid filter, the light emerging will have vibrations only in a single plane. Some compounds have the ability to rotate the plane of plane-polarised light, and are optically active. This can be measured using a polarimeter.

6. An optically active compound must be chiral (having no centre of symmetry), due to having a chiral carbon atom that has four different functional groups attached. The two optical isomers (enantiomers) are non-superimposable mirror images of one another, and have the same chemical properties, but often react differently in biological systems:

a. Dextrorotary (D-, +) – rotation to the right (usually used in biological systems). b. Laevorotary (L-, –) – rotation to the left.

7. A racemate (racemic mixture) is mixture containing equal concentration of both optical isomers. There is no net rotation of plane-polarised light.

8. Diastereomers are stereoisomeric structures that are not enantiomers (usually having substantially different chemical and physical properties).

9. Tautomerism is where two isomeric forms of a compound form a dynamic equilibrium, due to the changing position of an atom or feature. For a keto-enol taut, the equilibrium occurs between a ketone, and an alkene with an OH group.

The Carbonyl Group 1. Compounds containing the carbonyl group (C=O) can be formed from the oxidation of alcohols.

This requires the presence of either acidified KMnO4 or acidified K2Cr2O7 oxidising agent, and is can be carried out in a pear-shaped flask using a condenser:

a. Primary alcohols: i. Heat gently, with the condenser slanting into a collecting flask, to distil off the

product. This results in the formation of an aldehyde. ii. Boil gently (with an excess of oxidising agent and acid) under reflux, with the

condenser upright, to completely oxidise the alcohol into a carboxylic acid. The product can then distilled off.

b. Secondary alcohols – can be oxidised to a ketone. c. Tertiary alcohols – cannot be oxidised.

2. Alcohols have higher boiling points than aldehydes or ketones, due to hydrogen bonding caused by the OH group. Carboxylic acids have an even higher boiling point, as the C=O group withdraws more electrons from the OH oxygen, increasing the polarity of the bond and the strength of the hydrogen bonding.

3. To test between aldehydes and ketones, use Tollen’s reagent or Fehling’s solution – an aldehyde will be oxidised to a carboxylic acid, and reduce the species in the reagent, whereas a ketone cannot be oxidised.

4. To reduce an aldehyde or a ketone, use a reducing agent of either NaBH4 (aq) or LiAlH4 (in dry ether, then add H2O to hydrolyse the product). Alternatively a reducing agent of H2 / Ni catalyst can be used (although this will also reduce alkenes). NaBH4 and LiAlH4 provide a source of hydride (H–) ions, which are able to carry out a nucleophilic attack of the carbonyl carbon atom:



5. Hydrogen cyanide can react with an aldehyde or a ketone with a nucleophilic addition mechanism. This results in the formation of a racemic mixture of a hydroxyalkannitrile. KCN is normally used in preference to HCN, due to HCN being extremely toxic:

6. A nitrile can undergo two main reactions:

a. It can be hydrolysed to a carboxylic acid in the presence of water and an acid catalyst (usually HCl). This results in the production of an ammonium salt.

b. It can be reduced to a primary amine by catalytic hydrogenation with H2 and Ni. 7. Carboxylic acids are weak acids, so they are only partially dissociated in solution. This produces a

low concentration of H3O+ ions, but the concentration is sufficient to displace CO2 from aqueous carbonate ions.

8. Esters can be formed by the reaction of a carboxylic acid and an alcohol, in the presence of a strong acid catalyst (concentrated H2SO4). This is a condensation, or esterification reaction. The structure is that of the carboxylic acid, but with the hydrogen replaced by the alkyl group from the alcohol. Esters have no free OH groups, so cannot form hydrogen bonds – they are immiscible with water, volatile, and have lower melting and boiling points than carboxylic acids.

9. Uses of esters: a. Solvents – for polar organic compounds. They can be easily separated from solutes, due

to their high volatility (e.g. in nail varnish, the ester evaporates quickly). b. Plasticisers – they are incorporated into thermoplastic polymers, allowing movement of

the polymer chains, and greater flexibility of the material. c. Food flavourings – some esters smell nice (there are also a large number that do not!), and

can be used to provide artificial fruit flavourings. 10. Esters can be hydrolysed to an alcohol and a carboxylate salt using excess aqueous NaOH, and

heating. Naturally occurring esters (oils and fats) can be hydrolysed to produce soaps and glycerol. 11. Soaps are salts of long-chain fatty acids (e.g. stearic acid with 18C). The hydrolysis of fats and oils

to produce soaps is saponification. Soaps act as anionic detergents, as the negative carboxyl group is hydrophilic and the carbon chain is hydrophobic, so forms a layer around fats and oils to emulsify them. A strong acid (e.g. HCl) will convert soaps to fatty acids.

12. Acid chlorides can be synthesised from carboxylic acids, using PCl5: R–COOH (l) + PCl5 (s) à R–COCl (s) + PCl3O (l) + HCl (g)

13. Acid chlorides and acid anhydrides will undergo acylation reactions, with a nucleophilic addition-elimination mechanism. For the reaction of an acid chloride with ammonia (in excess):

14. The major acylation reactions of acid chlorides are as follows (these must all be carried out in

anhydrous conditions, except for the reaction with water!): a. Water à carboxylic acid RCOCl + H2O à RCOOH + HCl b. Alcohol à Ester RCOCl + R2OH à RCOOR2 + HCl c. Ammonia à Amide RCOCl + 2NH3 à RCONH2 + NH4Cl d. Primary amine à N-substituted amide RCOCl + 2R2NH2 à RCONHR2 + CH3NH3

+Cl– 15. For acid anhydrides, a carboxylic acid is produced rather than HCl, and an alkanoate salt rather

than a chloride salt. Note that the ammonium salts are only produced if an excess of ammonia or the primary amine is used – otherwise HCl (or a carboxylic acid) will be produced.

16. In the preparation of aspirin, acylation is used in the conversion of salicylic acid (2-hydroxy benzoic acid) into aspirin. The hydroxyl group is acylated with ethanoic anhydride. Salicylic acid can be produced by the reaction of phenol with CO2 (the Kolbe process).

17. In industrial processes, acid anhydrides are used rather than acid chlorides, because: a. They are cheaper than the acid chlorides. b. Carboxylic acid is produced rather than HCl (difficult and expensive to deal with). c. Acid anhydrides react more slowly, so the reaction is easier to control (not violent).


questionb 50m A2-Level Revision Notes

Aromatic Chemistry 1. Benzene is a colourless, volatile liquid, and is immiscible with water (although it can be used as an

organic solvent). Kekulé proposed a structure for benzene with alternating single and double bonds, that occurred as a resonance structure – this was shown to be wrong:

a. The Kekulé structure predicts four isomers of dibromobenzene, however there are only actually three, as 1,6-dibromobenzene and 1,2-dibromobenzene are not separate isomers.

b. Benzene is not as reactive as would be expected if it contained three double bonds. It does not decolourise bromine water, as alkenes do.

c. The enthalpy of hydrogenation of benzene would be expected to be about three times the value for cyclohexene, due to containing three double bonds. In actuality, it is considerably lower, and hence benzene is more stable than expected.

d. Using X-Ray diffraction to calculate the bond lengths in benzene, it was found that they are all the same length (half way between double and single bonds) – thus benzene cannot contain a mixture of single and double bonds.

2. Linus Pauling developed the delocalised structure of benzene – all the carbons are bonded together with single σ bonds, with a cloud of delocalised π electrons above and below the ring.

3. The high electron density around the benzene ring leaves it open to electrophilic attack. It does not undergo electrophilic addition, as this would disrupt the ring structure and is not energetically favoured. Instead, it will undergo electrophilic substitution (E+ represents the electrophile):

4. The nitration of benzene requires concentrated nitric acid and a concentrated sulphuric acid

catalyst, carried out at 50°C: a. HNO3 + H2SO4 H2NO3

+ + HSO4–

b. H2NO3+ NO2

+ + H2O c. E+ = NO2

+; Nu– = HSO4–

5. Nitrobenzene can be converted to aminobenzene (phenylamine) by reducing the NO2 group to an NH2 group, using a reducing agent of Fe or Sn with moderately concentrated HCl. Phenylamine is used in the production of azo dyes (with a functional group of –N=N–), by converting it into benzenediazonium chloride, and then reacting these together in a coupling reaction. The aromatic groups and the azo group together are called a chromophore (chromophoric group).

6. The nitration of benzene is used in the explosives industry, for the production of TNT (2,4,6-trinitrotoluene) from toluene (methylbenzene). The positive inductive effect of the methyl group favours substitution of the ortho and para positions.

7. The Friedel-Crafts alkylation of benzene requires a haloalkane, and an AlCl3 or FeCl3 catalyst, carried out at about 40°C in anhydrous conditions:

a. RCl + AlCl3 à R+ + AlCl4–

b. E+ = R+; Nu– = AlCl4–

8. Uses of Friedel-Crafts alkylation: a. Methylbenzene is used in the manufacture of explosives (TNT). b. Ethylbenzene is used in the production of polystyrene:

i. Benzene does not react directly with chloroethane, so a mixture of ethene and HCl is used.

ii. Dehydrogenation of ethylbenzene (with Fe2O3 at 600°C) produces phenylethene. iii. Polymerisation of phenylethene produces polystyrene.

c. 2-phenylpropane (cumene) is oxidised to form phenol and propanone: i. Propanone is used as a solvent in medicine and cosmetic applications.

ii. Phenol is used as a coupling agent in the dye-making industry. 9. The Friedel-Crafts acylation of benzene requires an acid chloride, and an AlCl3 or FeCl3 catalyst,

carried out at about 40°C in anhydrous conditions: a. RCOCl + AlCl3 à RCO+ + AlCl4

– b. E+ = RCO+; Nu– = AlCl4

– 10. Ethylbenzene can be prepared from the acylation of benzene with ethanoyl chloride, which first

produces phenylethanone. This is then reduced using H2/Ni to form ethylbenzene. Amines 1. An amine is defined as being primary, secondary or tertiary according to the number of alkyl

groups bonded to the nitrogen atom. Side chains are given the prefix N- (rather than a number).


base.50m A2-Level Revision Notes

2. Primary and secondary amines (with short alkyl groups) are soluble in water as they are able to form hydrogen bonds with water molecules, due to the presence of the N–H bond. As the order of the amine increases, the boiling point decreases (as the molecules cannot pack as closely together). The boiling points of amines are lower than those of the respective alcohols, due to weaker hydrogen bonding.

3. Ammonia and amines act as Brønsted-Lowry bases, as the nitrogen atom is able to accept a proton. Amines are weak bases, as they only partially react with water in solution:

a. RNH2(aq) + H3O+(aq) RNH3+(aq) + H2O(l)

b. Primary aliphatic amines are stronger bases than ammonia, due to the positive inductive effect of the alkyl group. This provides a greater electron density around the nitrogen atom, increasing the availability of the lone pair, so that it can react more readily.

c. Aromatic aryl amines are weaker bases than ammonia, as the lone pair of electrons on the nitrogen atom interacts with the delocalised π electrons in the benzene ring. This decreases the availability of the lone pair for accepting a proton.

d. Aromatic compounds where the amine group is not directly bonded to benzene have a similar base strength to the aliphatic amines, as the nitrogen electrons are not dissociated.

4. Ammonia and amines both have a lone pair of electrons on the nitrogen atom, so they are able to act as nucleophiles and ligands (electron pair donors). Amines can react in two main ways:

a. Nucleophilic substitution with a haloalkane – further substitution can take place to form a secondary amine, tertiary amine, and finally a quaternary ammonium salt (where four groups are bonded to the nitrogen to form a cation).

b. Acylation with an acid chloride or anhydride – this produces an N-substituted amide. 5. An excess of haloalkane will result in a high yield of quaternary ammonium salts. These can be

manufactured using long chain haloalkanes to produce cationic surfactants (the positive charge on the nitrogen is attracted to negatively charged surfaces such as glass, hair, fibres, metals and plastics). These can be used in fabric and hair conditioners, leather softeners, sewage flocculants (bringing particles to the surface), corrosion inhibitors, emulsifiers and disinfectants.

6. Primary aliphatic amines can be prepared in two main ways: a. Nucleophilic substitution of ammonia with a haloalkane – an excess of ammonia is used

to give a high yield of the primary amine, although further substitution can take place. b. Nucleophilic substitution of cyanide ions with a haloalkane – this produces a nitrile,

which can be reduced to an amine using either catalytic hydrogenation (H2/Ni) or LiAlH4 in dry ether (NaBH4 is not a strong enough reducing agent). This gives a higher yield (no further substitution), and increases the length of the carbon chain by one.

7. Aromatic amines can be produced by the nitration of benzene to form nitrobenzene. This is then reduced to aminobenzene by catalytic hydrogenation (H2/Ni) or Sn/Fe with moderately concentrated HCl (in this case, NaOH must be added afterwards, to deprotonate the phenylamine).

Amino Acids 1. Amino acids are bifunctional organic compounds, containing both the amino group and the

carboxylic acid group. There are 20 naturally occurring amino acids, which have the amino group on the carbon next to the carboxyl group (α-amino acids). These differ only in their R group:

2. All amino acids have optical isomers (except for glycine, where R = H), but only the D-isomers

occur naturally in body proteins. 3. Amino acids act as both acids and bases, and so in alkaline solution the carboxylic acid group is

deprotonated, and in acidic solution the amino group is protonated. Amino acids can never exist as uncharged compounds, so at its isoelectric pH (unique for each amino acid), where it has no net charge, it forms a zwitterion (or amphion), whereby the amino group is protonated and the carboxylic acid group is deprotonated:

4. Two amino acids can react in a condensation reaction to form a peptide (or amide) group – the

CONH group – containing a peptide bond (C–N). Many condensation reactions can take place to form a polypeptide or polyamide, which is the primary structure of a protein. Hydrogen bonding


questionb 50m A2-Level Revision Notes

can occur between N–H and C=O groups to form a helical shape (the secondary structure). The tertiary structure is the characteristic three-dimensional shape of the protein (fibrous or globular).

5. Polypeptides can be hydrolysed into their constituent amino acids by hydrolysing the peptide bonds between them. This requires heating the polypeptide with 5M HCl for 24 hours.

Polymers 1. There are two types of polymer (synthetic polymers are plastics):

a. Addition (chain-growth) – a double bond in the monomer is broken to bond with other monomers, forming a long chain polymer.

b. Condensation (step-growth) – the monomers have two different functional groups that react together, resulting in the loss of small molecules (e.g. H2O).

2. Thermosetting polymers char or burn on heating, whereas thermoplastic polymers melt or soften on heating.

3. Poly(ethene) is an addition polymer, and has two types: a. LDPE (low density) – there are many branches from the polymer chains, resulting in a

softer material, with a low melting point. It is used for packaging and electrical insulation. b. HDPE (high density) – there is very little branching, so the straight chains can get closer

together, resulting in a stronger and more rigid material, with a higher melting point. It is produced using a Ziegler catalyst (organometallic).

4. PVC (polyvinyl chloride) is stronger than poly(ethene), as the chlorine provides permanent dipoles in the molecules, resulting in stronger intermolecular forces. PVC is brittle though, as the big chlorine atoms mean that the chains cannot move past one another.

5. There are three different types of addition polymer: a. Isotactic – the functional groups are evenly spaced on the same side of the chain. b. Syndiotactic – the functional groups are evenly spaced and alternate between sides. c. Atactic – the functional groups are randomly positioned.

6. A polymer can be made stronger by adding an electronegative atom (stronger intermolecular forces) or by increasing the number of cross-linkages (e.g. using sulphur – vulcanised rubber).

7. Addition polymerisation usually involves a free radical mechanism, and is carried out at a high temperature and pressure, and with a catalyst. The repeating unit is enclosed in square brackets, with the letter n to denote a large number of repeats. Common addition polymers are:

a. Poly(ethene) or polythene – bags, packaging, insulating electric wires. b. Poly(phenylethene) or polystyrene – packaging, insulation. c. Poly(chloroethene) or PVC – window frames, guttering. d. Poly(tetrafluoroethene) or Teflon – waterproofing, non-stick surfaces. e. Poly(ethenyl ethanoate) or PVA – paints, adhesives.

8. There are two main types of condensation polymer: a. Polyesters – A dicarboxylic acid and a diol react to form the monomer. b. Polyamides (polypeptides) – the monomer is either an amino acid, or a mixture of a

dicarboxylic acid and a diamine. 9. Nylon-6,6 is a polyamide formed from the reaction of hexanedioic acid and hexane-1,6-diamine at

250°C. Nylon-6 is formed from the amino acid 6-aminohexanoic acid. In the condensation reaction, water is formed from the OH of the carboxylic acid and an H from the amino group.

10. Terylene, PET, or poly(ethyleneterephthalate) is a polyester formed from benzene-1,4-dicarboxylic acid, and ethane-1,2-diol. The water is produced from the acid OH and an H from the alcohol. If the dimethyl ester of the acid is used, methanol is evolved as a gas.

11. Addition polymers are non-biodegradable, as they are chemically inert, but they are highly flammable, so can sometimes be recycled (if the different types are separated first). Condensation polymers can undergo hydrolysis to break them down into their component monomers, so they are biodegradable.

Organic Analysis 1. Saturated hydrocarbons (alkanes) tend to burn with a clean flame, whereas unsaturated

hydrocarbons (alkenes and aromatics) tend to burn with a sooty flame. 2. Alkenes decolourise bromine water (orange to colourless), whereas alkanes and aromatics do not. 3. Carboxylic acids dissolve in water as a weakly acidic solution, whereas amines are weak alkalis. 4. Alcohols and carboxylic acids react with Na (s) to produce hydrogen gas. 5. Carboxylic acids will react with Na2CO3 to produce CO2 gas. 6. Aldehydes and ketones react with Brady’s reagent (2,4-dinitrophenylhydrazine in sulphuric acid)

to produce an orange-red precipitate (carboxylic acids and alcohols do not react).


base.50m A2-Level Revision Notes

7. Aldehydes will form a brick red precipitate or a silver precipitate when warmed with Fehling’s solution or Tollen’s reagent respectively. Ketones do not react.

8. Primary and secondary alcohols and aldehydes can be oxidised by acidified K2Cr2O7 (orange to green) or KMnO4 (purple to colourless) respectively. Tertiary alcohols and ketones do not react.

9. Carboxylic acids react with ethanol and a concentrated H2SO4 catalyst to form an ester. 10. Alcohols react with ethanoic acid and a concentrated H2SO4 catalyst to form an ester. 11. Haloalkanes can be hydrolysed with dilute NaOH (aq), then after adding nitric acid and silver

nitrate solution, a precipitate will appear (fluoroalkanes are not detected, as AgF is water soluble): a. Chloroalkanes – white precipitate (AgCl), dissolves in dilute ammonia. b. Bromoalkanes – cream precipitate (AgBr), dissolves in concentrated ammonia. c. Iodoalkanes – yellow precipitate (AgI), insoluble in dilute and concentrated ammonia.

12. Acyl chlorides react vigorously with AgNO3 (aq) to form a white AgCl precipitate. 13. Amines and amino acids produce a dark-blue soluble complex with copper (II) ions and excess

ammonia. Aromatic amines tend to form green, insoluble complexes. 14. Adding hydroxide ions to an amine causes the indicator to change colour more rapidly (with less

hydroxide) than with an amino acid, as there is no carboxylic acid group to react with the alkali. Spectroscopy 1. Mass spectrometry uses the varying mass-to-charge (m/z) ratios of ions, and their varying degrees

of deflection in a magnetic field, in order to determine their relative abundances: a. The main peaks on the mass spectrum are caused by the parent molecule and fragments of

it, although some minor peaks can be caused by rearrangements. b. The peak at the maximum m/z value is the parent molecular ion, M+ • (a radical cation),

which has an m/z value equal to the Mr of the molecule. c. The parent ion can fragment into a cation and a radical – M+ • à X+ + Y • – only the

cation is detected, as it can be accelerated by the electric field and deflected by the magnetic field. Further fragmentation of X+ (or Y+ if it is formed) can take place.

d. The notation (M – R)+ refers to the parent ion losing the group R (as R • ). e. The heights of peaks on the mass spectrum are shown as percentages of the height of the

largest peak (the base peak). The more abundant the fragment is, the more stable it will be, and the easier it will be to form it (the weaker the broken bond is).

f. Molecular ions that are two m/z units apart in a 3:1 intensity ratio will be due to the presence of chlorine, occurring as roughly 75% 35Cl and 25% 37Cl.

g. Molecular ions that are two m/z units apart in a 1:1 intensity ratio will be due to the presence of bromine, occurring as roughly 50% 79Br and 50% 81Br.

h. Fragmentation usually occurs as the weaker bonds break (e.g. C–C, C–O and O–H), whereas the stronger bonds remain unbroken (e.g. C–H and C=O). Some common stable ions formed as a result of fragmentation are as follows:

i. CH3 (m/z =15) ii. NH2 (m/z = 16)

iii. OH (m/z = 17) iv. C2H5 (m/z = 29) v. C6H5 (m/z 77)

i. Alcohols will readily eliminate water from the parent ion, to form a peak at (M – H2O)+ •. j. Alkanes tend to give peaks at (M – R)+, where R is an alkyl group. k. Ketones tend to form stable acylium cations (RCO+).

2. Infra-red spectroscopy uses the characteristic absorption wavelengths of photons for different bonds in order to identify a compound:

a. Covalent bonds are able to vibrate, by either stretching or bending, and they do this at a resonant frequency for each pattern of vibration. This corresponds to a specific wavelength in the infra-red spectrum, which can be absorbed by the bond electrons. Thus the intensity of each wavelength of infra-red light passing through a sample is recorded and compared to that through a blank (the background), to give an absorption spectrum.

b. The scale for an infra-red spectrum is given in terms of the wavenumber ν, which is defined as 1/λ. The units are given in cm–1.

c. The region of the spectrum below 1500 cm–1 (to the right) is the fingerprint region, which can be very complex. This is unique to each compound, and so comparison with a database can be used to determine the identity of the compound.

d. The higher wavenumber region of the spectrum is very useful in determining particular functional groups, which can aid identification:


base.50 A2-Level Revision Notes

i. O–H (alcohol) gives a broad absorption around 3300 cm–1, due to H-bonding. ii. O–H (acid) gives a very broad absorption around 2800 cm–1.

iii. C=O gives a sharp absorption at about 1700 cm–1. iv. C≡N gives a small, sharp absorption at about 2250 cm–1.

e. The most useful information can often be the absence of particular functional groups, as this eliminates possibilities as to the identity of the compound.

3. Nuclear magnetic resonance (NMR) spectroscopy gives the most information of the spectroscopic analyses, with regard to the positions and numbers of protons (hydrogen atoms) in the molecule:

a. Nuclei spin, and thus create a magnetic field around them, as they are charged. If they are placed within a magnetic field, they all line up either in the same direction (lower energy) or in the opposite direction (higher energy) to the magnetic field. The very small energy gap between these states corresponds to radio frequency waves. Normally a fixed magnetic field is used, and the frequency of applied radio waves (causing resonance) is modified, in order to scan the sample and detect the effects.

b. The chemical shift, δ, is measured in ppm (parts per million), and depends upon the amount of shielding of the protons by electrons – the more shielding, the higher the field strength needed for resonance, and the lower the δ value.

c. The amount of shielding depends upon the environment of the protons – if they are next to an electronegative atom (e.g. in COOH) then there will be deshielding and a high δ value (downfield), whereas if they are next to an atom with low electronegativity (e.g. in CH3) there will be more shielding and a lower δ value (upfield).

d. The δ value is determined as relative to a standard of TMS (tetramethylsilane – Si(CH3)4): i. This gives a strong signal due to 12 protons in the same environment.

ii. Si is less electronegative than carbon, so the signal is upfield of other organic molecules (δ is defined as 0.0ppm).

iii. It is inert and non-toxic, cheap, and is volatile (can be evaporated afterwards). e. The area under each peak on the NMR spectrum is proportional to the number of protons

it represents (in the same environment), and this ratio is given by the integration curve. f. A high resolution NMR spectrum is able to show splitting of individual peaks, due to

being adjacent to other protons in a different environment: i. The number of splits is n + 1 where the number of adjacent protons is n (n + 1

rule). The splitting pattern is given by the (n + 1)th line of Pascal’s triangle. ii. A triplet (1:2:1) indicates that the protons are adjacent to a CH2 group.

iii. A quartet (1:3:3:1) indicated that the protons are adjacent to a CH3 group. iv. An ethyl group (CH2CH3) is indicated by a triplet and a quartet, with integration

curves in a 3:2 ratio. v. If there is no splitting, then the protons are not adjacent to any different protons

(a singlet) – e.g. the H in an OH group. vi. If the protons are next to two different types of proton, then there will be

splitting of the splitting pattern (e.g. a quartet of triplets). vii. The splitting of peaks is called the coupling effect.

4. It is often most useful to compare the results of all three spectroscopic analyses, as they give different types of information, and to also use chemical analytical tests, in order to determine the identity of a compound.

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CAPE Chemistry Study Paper 001α

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