F324 M8

Front 1

The chemist observed that bromine decolourises when it reacts withphenol.What other observation would she have made?

Back 1

White precipitate OR white solid OR white crystals

Front 2

Explain the relative resistance to bromination of benzene compared tophenol and compared to cyclohexene.

Back 2

benzene electrons or π-bonds are delocalised.

phenol a lone or non-bonded pair of electrons on the oxygen orthe OH group is (partially) delocalised into the ring.

cyclohexene electrons are localised OR delocalised between twocarbons.

benzene has a lower electron density OR phenol has a higherelectron density OR cyclohexene has a higher electron density.

benzene cannot polarise or induce a dipole in Br2 OR phenolcan polarise the Br2 OR cyclohexene can polarise Br2 or theBr–Br bond.

Front 3

How does the halogen carrier allow the reaction to take place?

Back 3

Introduces a permanent dipole on Cl2 / forms Cl+/AlCl3 + Cl2 → AlCl4– + Cl+/AlCl3 + Cl2 → Clδ+ – AlCl3δ–

Front 4

In contrast to benzene, the reaction of an alkene with bromine does not need ahalogen carrier. Compare the different reactivities of benzene and alkenes towards chlorine.

Back 4

In benzene, π electrons are delocalised/spread out (1)In alkenes, π electrons are concentrated between 2 carbons (1)Electrophiles attracted more to greater electron density in alkenes (1)

Front 5

Describe with the aid of suitable diagrams the bonding and structure of a benzenemolecule.

Back 5

p-orbitals overlap (1) above and below the ring (1) (to form) π-bonds / orbitals (1)

Front 6

Phenol reacts much more readily with bromine than benzene does.

• Describe, with the aid of a diagram, the bonding in benzene.

• Explain why electrophiles, such as bromine, react much more readily with phenolthan with benzene.

Back 6

bonding in benzeneoverlap of p-orbitals / π bonds/electrons (or labelled) (1) electrons are delocalised (or labelled) (1)C–C bonds are: same length/strength / in between single anddouble / σ-bonded AW (1)greater reactivity of phenol(the ring is activated because …)lone pair from O is delocalised into the ring (1)so electron density (of the ring) is increased (1)so electrophiles are more attracted (to the ring) / dipole inelectrophile more easily induced (1)

Front 7

Compare the reagents and conditions for the nitration of phenol with those usedfor the nitration of benzene. State and explain the effect of the –OH group on the reactivity of the benzenering in phenol.

Back 7

conditions for nitration of benzene:

HNO3 is concentrated (1) conc H2SO4 is present (1) heating or stated temp above 50°C (1)

explanation for greater reactivity of phenollone pair from O atom is delocalised into the ring (1) greater (π) electron density around the ring (1) (the benzene ring in phenol) is activated (1) attracts electrophiles/+NO2 more / makes it moresusceptible to electrophiles AW (1)

Front 8

Explain what is meant by the term delocalised π-bond electrons.

Back 8

delocalised electronselectrons are spread over more than two atoms AW (1)π-bondformed by overlap of p-orbitals/ diagram to show (1)

Front 9

Carbon dioxide is normally a very poor electrophile. However, this reactiondoes occur because the benzene ring in the phenoxide ion is activated. Explain how the benzene ring in the phenoxide ion is activated.

Back 9

lone/electron pair from oxygen is delocalised into thering /interacts with π-electrons (1) increases π-electron density / negative charge(around the ring) (1) attracts electrophiles more (1)

Front 10

Name two types of commercially important material whose manufacture involvesthe nitration of benzene.

Back 10

any two of …fibres / dyes / explosives / pharmaceuticals etc (1)(1)

Front 11

The benzene ring and the ring in the intermediate formed after step 2 havedifferent structures shown below. Both structures have π-bonds.

benzene ring

ring in the intermediate

Deduce how many electrons are involved in the π-bonding in each structure anddescribe how their arrangements are different.

Back 11

π-bonding electrons are delocalised (1)six π-electrons in benzene (1)four π-electrons in the intermediate (1) π-electrons are not over one carbon atom /over five carbon atoms / p-orbitals in the intermediate (1)this must be stated in words to compare benzene and theintermediate π-electrons are over the complete ring / all around the ringall six carbon atoms/ p-orbitals overlapping (1)

Front 12

What is meant by the term nucleophile?

Back 12

electron pair donor (1)

Front 13

Butanal and butanone both react with 2,4-dinitrophenylhydrazine to producemixtures containing orange precipitates. Outline how the mixtures containing these orange precipitates can be used todistinguish between butanal and butanone.

Back 13

recrystallise /purify (the precipitate) (1) measure melting point (1) compare with known values (1)

Front 14

Describe a simple chemical test that would show that but-2-enal is analdehyde.

Back 14

heat with:Tollens’ reagent / ammoniacal silver nitrate (1) to give: silver mirror / precipitate (1)

Front 15

Explain why this test gives a different result with aldehydes than it doeswith ketones.

Back 15

aldehydes can be oxidised to a carboxylic acid ora/ aldehydes can reduce Ag+ to Ag (1)

Front 16

Tollens’ reagent can be used to identify the aldehyde group in cinnamaldehyde.

• Describe how you would make Tollens’ reagent and carry out this test in thelaboratory.

• Explain what happens to both the Tollens’ reagent and the cinnamaldehyde inthis reaction. Identify the organic product.

Back 16

method:

silver nitrate (1)ammonia / ammoniacal (1)warm / heat (1)silver (mirror) / brown ppt forms (1)

explanation:

silver ions reduced / Ag+ + e– → Ag (1)aldehyde oxidised to a carboxylic acid (1)correct structure – eg C6H5CHCHCOO–/COOH (1)

Front 17

Suggest one other reason why detergents such as sorbitan monolaurate areregarded as ‘environmentally friendly’.

Back 17

sorbitan monolaurate is made from a renewable resource/ not based on crude oil (1) AW

Front 18

Explain why triglycerides are soluble in non-polar solvents and not in water.

Back 18

Three of following points: (1)(1)(1)

• 1. There is van der Waals (IDID) between triglycerides.

• 2. There is van der Waals between triglycerides and (non-polar) solvent.

• 3.Triglycerides cannot hydrogen bond (to water)(enough).

• Because there are not enough suitable sites/oxygen atomsOr long hydrocarbon chains do not hydrogenbond/would interfere with hydrogen bonding in water

Front 19

Explain what is meant by the term stereoisomerism.

Back 19

same structural formula/order of bonds,different spacial arrangement AW (1)

Front 20

Suggest why esters are used in the manufacture of foods.

Back 20

flavouring / fruity smell

Front 21

Benzoic acid and phenylmethanol will react with each other in the presence of asuitable catalyst. State a suitable catalyst for this reaction.

Back 21

H+ /acid / named strong acid eg H2SO4 / HCl

Front 22

Coloured organic compounds also include azo dyes. Describe how an azo dye can be made from phenylamine.

Back 22

odium nitrite + HCl / nitrous acid (1)

<10°C (1)

phenol/named example (added to the products from above) AW (1)

alkaline conditions / OH–(1)

Front 23

Describe how you would prepare a sample of an azo dye in the laboratory froman amine, a phenol and any other necessary reagents.

Include in your answer

• essential reagents and conditions for each stage

• names of any functional groups formed during the process.

Back 23

1st stage

aromatic amine / named aromatic amine / structure (1)sodium nitrite / nitrous acid (1)HCl/H2SO4 (but not conc) /H+(1)at <10°C (1) which forms a diazonium salt / ion (1)

2nd stage

the product from the first stage mixed with the phenol AW (1)(in excess) hydroxide / alkali (1)

Front 24

Explain what is meant by the term 1,4-diamino in the name of this compound.

Back 24

Diaminotwo/2 amine groups (1)

1,4their position on the ring / numbering of carbonsaround ring (or shown on a diagram) (1)

Front 25

Explain how the amino groups in a primary amine such as1,4-diaminobenzene allow the molecule to act as a base.

Back 25

accepts H+ using the lone pair (on N) (1)

which is donated/forms a (dative) covalent bond (1)

Front 26

State whether you would expect hexane-1,6-diamine to be a stronger orweaker base than 1,4-diaminobenzene. Explain your reasoning.

Back 26

hexane-1,6-diamine is a stronger base because: electrons move towards the N (due to the inductive effect)(in hexane-1,6-diamine) (1)

the lone pair from N is (partially) delocalised around the ring(in diaminobenzene) (1)

so the electron pair is more easily donated /H+ more easily accepted (in hexane-1,6 diamine) ora (1)

Front 27

The mixture of tripeptides can be analysed by using gas chromatography,coupled with mass spectrometry.

Summarise how each method contributes to the analysis.

Back 27

gas/liquid chromatograph separates the tripeptides (1)

mass spectrometer produces a distinctive fragmentation pattern (1)

identification by computer using a spectral database (1)

Front 28

State what is meant by the term zwitterion

Back 28

molecule/ion/’it’ has both + and – charges

Front 29

State suitable reagents and conditions to break down a protein into aminoacids.

and

State the type of reaction occurring.

Back 29

heat/warm/reflux (1)

named strong acid/basean enzyme (which need not be named) (1)

hydrolysis (1)

Front 30

What is meant the term stereoisomer?

Back 30

molecules with the same structure / order of bonds … butdifferent arrangements in space / 3-D arrangment (1)

Front 31

State a use for Kevlar.

Back 31

fire resistant / bullet proof clothing / cycle tyres / tennis rackets (1)

Front 32

State three ways in which the monomer units of a protein differ from thoseof nylon-6,6.

Back 32

any three different chemically or biologically correct differencesbetween amino acids and the nylon monomers (1)(1)(1) - eg

• protein monomers are amino acids / nylon monomers are a(di)amine/base and a (di)acid

• protein monomers have different types/R groups / nylonmonomers are two types/no variation• protein monomers have stereo/optical isomers/are chiral

• protein monomers have higher melting points/form zwitterions

Front 33

Name the process by which TLC separates α-amino acids

Back 33

adsorption

Front 34

The chromatogram was treated to show the positions of the separated α-aminoacids.

Explain how the student could analyse the chromatogram to identify the threeα-amino acids that were present.

Back 34

measure how far each spot travels relative to the solvent front or calculate the Rf value

compare Rf values to those for known amino acids

ALLOW compare Rf values to database

ALLOW compare to known amino acids

Front 35

The mixture of tripeptides can be analysed by using gas chromatography,coupled with mass spectrometry.Summarise how each method contributes to the analysis

Back 35

gas/liquid chromatograph separates the tripeptides (1)mass spectrometer produces a distinctive fragmentation pattern (1)identification by computer using a spectral database (1)

Front 36

Define the following terms used in chromatography.

Rf value

Retention time

Back 36

Rf value is distance moved by a component/spot/solute dividedby distance moved by solvent. (1)

Retention time is the time between injection and emergence(or detection) of a component. (1)

Front 37

What name is given to the process by which components in a mixture areseparated during gas/liquid chromatography?

Back 37

Partitioning or adsorption

Front 38

What are the roles of the gas and liquid in gas/liquid chromatography?

role of gas

role of liquid

Back 38

Role of gas: carrier gas / mobile phase / to carry to samplethrough the chromatography column (1)

Role of liquid: stationary phase (1)

Front 39

Explain how the gas/liquid chromatogram could be used to determine thepercentage composition of each component in the mixture.

Back 39

Measure area under each peak (1)

Find total area (1)

% = (area of one peak/total area) × 100% (1)

Front 40

Name a type of chromatography that is used to separate and identify dissolvedsubstances.

Back 40

Accept paper, column or thin-layer chromatography