AL-CHEM Chemistry of Carbon Compounds(03-06)

AS/AL-CHEM (2003-2006) Part XI Chemistry of carbon compounds

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NOTE: Questions with an * are NOT required by the HKDSE syllabus. They are included so as to enrich your knowledge and broaden your understanding in chemistry.

2003-AL-CHEM 2 5. (a) A student carried out an experiment to extract 4-nitrophenylamine and 2-hydroxybenzoic acid from a solution of

the two compounds in ethoxyethane. The solution contained 1.5 g each of the compounds in 20 cm3 of ethoxyethane.

The student shook the solution with 12 cm3 of 0.5 M NaHCO3(aq) in a separating funnel. The mixture in the separating funnel was allowed to settle, and the aqueous layer was then separated from the organic layer. Evaporation of the aqueous layer gave a solid residue, which is water-soluble and has a melting point higher than 200C. Evaporation of the organic layer gave another solid residue, which melts over a wide temperature range of 140C to 160C.

(i) Why was NaHCO3(aq) used in the extraction ?

(ii) While the student was shaking the separating funnel, the stopper of the funnel popped out, spilling part of the mixture on the bench. With the help of equation(s), explain why this happens.

(iii) Is the solid residue obtained from the organic layer a pure compound ? Support your answer by stoichiometric calculation.

(iv) State two reasons why ethoxyethane is commonly used to extract organic compounds from an aqueous solution.

(v) State one safety precaution necessary in the evaporation of ethoxyethane. (9 marks)

Mark Scheme .............................................................................................................................................. 5. (a) (i) To convert 2-hydroxybenzoic acid to the carboxylate salt which is soluble in water. 1 (ii) Carbon dioxide gas is formed.

1

Pressure therefore builds up inside the separating funnel. 1

or, Ethoxyethane (CH3CH2OCH2CH3) is highly volatile. Vaporisation of ethoxyethane leads to the building up of pressure in the funnel.

CH3CH2OCH2CH3(l) CH3CH2OCH2CH3(g)

(1)

(1) (iii) No. of moles of 2-hydroxybenzoic acid =

1385.1

= 0.011 mol 1

No. of moles of NaHCO3 = 0.5 12 10-3 = 6 10-3 mol 1

No. of moles of 2-hydroxybenzoic acid > No. of moles of NaHCO3 (0.011 0.006) = 0.005 mol of 2-hydroxybenzoic acid will remain in the organic layer. The residue is therefore not a pure compound.

1

(iv) Any TWO of the following: Ethoxyethane has a small dipole moment. It is a good solvent for most organic compounds. Ethoxyethane is immiscible with water. Ethoxyethane is volatile. It can easily be removed by distillation. Ethoxyethane is chemically unreactive.

2 (1) (1) (1) (1)

(v) Avoid naked flame / Extraction should be carried out in a fume cupboard (good ventilation) 1 [9]

EEEExam Practice Functional groups Do Brilliantly

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-------------------------------------------------- Take a break ----------------------------------------------------- 5. (b) Part of the structure of Super Glue is shown below:

(i) Draw the structure of the monomer of Super Glue.

*(ii) Give the conditions necessary for the formation of Super Glue from its monomer and outline the reaction mechanism.

(4 marks)

Mark Scheme .............................................................................................................................................. 5. (b) (i)

1

(ii) In the presence of a peroxide / free radical initiator / benzoyl peroxide + ultra-violet radiation / high temperature

1

Mechanism (free radical polymerisation):

2

[4]

-------------------------------------------------- Take a break ----------------------------------------------------- 6. (b) Compound A (C6H6O6) is an acyclic tribasic acid isolated from the leaves and tubers of Aconitum napellus.

Hydration of A gives two isomeric compounds, B and D. B is achiral, but D is chiral.

Deduce the structures of A, B and D.

(5 marks) Double Bond Equivalent (degree of unsaturation) of A (C6H6O6) = 2

614 = 4

Double Bond Equivalent (degree of unsaturation) of B (or D) = 2

814 = 3

NOTE: The degree of unsaturation is derived by comparing the given formula with CnH2n+2 C6H14 (when n = 6).


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Mark Scheme .............................................................................................................................................. 6. (b) The double bond equivalent of A and of B (or D) are 4 and 3 respectively.

Each molecule of A contains 3 CO2H groups and 1 C=C bond, while B (or D) contains 3 CO2H groups and no C=C bond. Three possible sets of answers:

1

4

[5]

-------------------------------------------------- Take a break ----------------------------------------------------- 7. *(b) Outline a synthetic route, in not more than four steps, to accomplish each of the following transformations. In

each step, give the reagent(s), conditions and the structure of the organic product.

(3 marks) Mark Scheme .............................................................................................................................................. 7. (b) (ii) Step 2: oxidation *Step 3: condensation 3

H3O+

heat

OH

heat

Cr2O72-/H3O+O H2NOH

NOH

Step 1: catalytic hydration (H2O/H+, heat) Alternatively, the addition of water to the carbon-carbon double bond can be accomplished by (1) H2SO4(l); then, (2) H2O, boil (see the sample paper Q.13 (b))

Understanding redox reactions in organic chemistry

Test Yourself

addition of [O] / removal of [H]

addition of [H] / removal of [O]


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-------------------------------------------------- Take a break ----------------------------------------------------- 7. *(c) Compound E (C15H14O) is a white solid with a melting point of 32C and displays the following infra-red

spectrum.

E gives a negative result when treated with Tollens reagent but reacts readily with LiAlH4 to give an achiral compound F. Drastic oxidation of F with acidified potassium permanganate solution gives an aromatic compound G (C7H6O2) with a pKa value of 4.2, as the major product.

Deduce the structures of E, F and G.

Hint: The aromatic compound G is benzoic acid, (6 marks)

Mark Scheme .............................................................................................................................................. 7.

NOTE: Oxidation of alkylbenzenes using acidified KMnO4(aq) DSE

Alkyl and unsaturated side chains of aromatic rings can be oxidized to benzoic acid using hot KMnO4, except those with a tertiary alkyl group because they do not have benzylic hydrogen atom.

CH(CH3)2 KMnO4/H+COOH

heat

COOH

COOHheat

KMnO4/H+

If the oxidation is carried out by the action of hot alkaline KMnO4 solution, benzoate ion is formed instead of benzoic acid. Adding dilute mineral acid such as dil. H2SO4 to benzoate can generate the benzoic acid (protonation).

COOH


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2004-AL-CHEM 1

5. (a) Identify K, L, M, N, P and Q in the following reactions:

(2 marks) Mark Scheme .............................................................................................................................................. 5. (a) (i) K : H2 + appropriate catalyst (Pt, Ni, Pd)

(Catalytic hydrogenation. Note that the carbonyl group (C=O) is NOT reduced.) 1

(ii) N : H2C=CH(CH2)3NH2 (Reduction of polar CN bond with LiAlH4. Note that the C=C is not affected.)

1

[2]

-------------------------------------------------- Take a break ----------------------------------------------------- 5. (c) What is the main chemical ingredient of rubbing alcohol for disinfection ? Write an equation to show the

industrial synthesis of this chemical.

(2 marks) Mark Scheme .............................................................................................................................................. 5. (c) propan-2-ol / 2-propanol / isopropanol 1

CH3CH CH2 + H2O CH3CHCH3OH

catalyst

( H+ )

1

(catalytic hydration of propene) [2]

-------------------------------------------------- Take a break ----------------------------------------------------- 6. (a) The equation below shows the reaction of benzoic acid with butan-1-amine at elevated temperatures to give

compound R:

*(i) Give the systematic name of R.

(ii) Name the type of reaction involved.

(iii) The reaction of benzoic acid with butan-1-amine at room temperature gives another product S. Give the structure of S.

(iv) Suggest why high temperature can facilitate the formation of R.

(v) Using benzoic acid as the starting material, devise another route for the synthesis of R, which does not involve reactions that take place at temperature much higher than room temperature.

(6 marks)


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Mark Scheme ..............................................................................................................................................

2004-AL-CHEM 2

5. (a) The active ingredient of household bleach is sodium chlorate(I), which is manufactured from chlorine.

(i) With the help of a chemical equation, suggest how sodium chlorate(I) can be obtained from chlorine.

(ii) Household bleach diluted by a volume ratio of 1:99 is widely used as an effective and inexpensive disinfectant during the recent SARS epidemic outbreak.

A certain brand of household bleach contains 6.0 g of sodium chlorate(I) per 100 cm3 of the bleach. Calculate the concentration of sodium chlorate(I), in mol dm-3, in the diluted bleach.

(iii) With the help of chemical equation(s), suggest why household should not be used together with

(I) toilet cleaner which contains sodium hydrogensulphate(VI),

*(II) nail vanish remover which contains propanone. (8 marks)

Mark Scheme ..............................................................................................................................................

Note: Acid-base neutralisation at room temperature RCO2H + NH3 RCO2-NH4+


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-------------------------------------------------- Take a break ----------------------------------------------------- 5. *(b) Phenolphthalein is an acid-base indicator with the following structure:

It can be synthesised from compound B (C8H10) via the following reaction pathway:

Under acidic conditions, D reacts with butan-1-ol to give compound F (C16H22O4).

Suggest structures for compounds B, D, E and F. (4 marks)

Mark Scheme .............................................................................................................................................. 5.

1

-------------------------------------------------- Take a break ----------------------------------------------------- 6. (b) Hydrocarbon G reacts with HBr to give J as the major product.

(i) Give the structure of J and its systematic name.

*(ii) Outline the mechanism of the above reaction and explain why J is the major product.

*(iii) A student commented that J obtained from the above reaction is optically active. Do you agree with the student ? Explain.

(7 marks)


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Mark Scheme .............................................................................................................................................. 6.

1

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (a) Ethene and 2-methylpropenoic acid react in the presence of benzoyl peroxide to give polymer N, which does not

have a definite repeating unit.

A portion of the structure of N is shown below:

(i) Name the type of polymerisation involved in the formation of N.

(ii) State, with explanation, whether or not N is a thermosetting polymer.

(iii) Explain why the strength of N obtained depends on the relative amounts of ethene and 2-methylpropenoic acid used in the polymerisation.

(iv) N reacts with excess sodium hydroxide to give another polymer P.

(I) Draw the structure of P which corresponds to the portion of the structure shown above.

(II) Compare the strengths of N and P. Explain your answer.

(7 marks)

CH2CH2


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Mark Scheme .............................................................................................................................................. 7. (a) (i) Addition polymerization / free radical polymerization 1 (ii) No. There is no cross-link between polymer chains. 1 (iii) The attraction between polymer chains is hydrogen bond. 1

Increase in the 2-methylpropenoic acid: ethene ratio gives rise to an increase in no. of H-bond per unit length. Therefore attraction between polymer chains increases.

1

(iv) (I)

1

(II) P is stronger than N. After treating with NaOH, the attraction between polymer chains would be changed from H-bond to ionic bond.

1 1

[7]

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (c) You are provided with four unlabelled bottles each containing one of the following white solids:

starch, sucrose, lactose and methyl -2,3,4,6-O-mthyl-D-glucoside

Suggest how you would carry out tests to distinguish the four substances from one another.

Hint: The structures of sucrose, lactose and methyl -2,3,4,6-O-mthyl-D-glucoside are shown below:

(6 marks)


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Mark Scheme .............................................................................................................................................. 7.

2005-AL-CHEM 1

5. (a) Give the structures of the major organic products A, B, C, D and E in the following reactions:

(2 marks) Mark Scheme .............................................................................................................................................. 5. (a)

1+1

NOTE: (i) condensation polymerisation nylon 6.6 (ii) formed by intramolecular condensation [2]

-------------------------------------------------- Take a break ----------------------------------------------------- 5. (b) Consider the following synthesis of compound F:

(i) Give the systematic name of F.

*(ii) The synthesis gives a side-product G with the following structure:

(I) Write the chemical equation for the formation of G.

(II) Suggest how the reaction conditions can be modified to minimise the formation of G.

*(iii) Suggest an alternative synthetic pathway for F, using any starting material with no more than 8 carbon atoms.

(5 marks)


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Mark Scheme .............................................................................................................................................. 5. (b) (i) 3-phenylpropan-1-amine

NOTE: The role of NH3 is like OH- (nucleophile). The reaction is called Nucleophilic Substitituion. 1

(ii) (I) Ph = Phenyl group (C6H5)

NOTE: The amine formed competes with NH3 for the substitution reaction. Multiple substitution occurs.

1

(II) use excess NH3 / add Ph(CH2)3I to NH3 dropwise. 1 (iii)

CH2CH2BrNaCN

CH2CH2CNLiAlH4 CH2CH2CH2NH2

Step 1: (nucleophilic) substitution (CN- acts like OH-) Step 2: reduction of CN group with LiAlH4

2

[5]

-------------------------------------------------- Take a break ----------------------------------------------------- 6. (b) Answer the following multiple-choice questions:

Mark Scheme .............................................................................................................................................. 6. (b) C 1

2005-AL-CHEM 2

5. (a) Most of the petroleum stock located on Earth is likely to be used up in 50 to 100 years if petroleum consumption is maintained at the current rate. With a view to cutting down petroleum consumption, some countries have adopted an alternative fuel for motor vehicles gasoline which contains ethanol.

(i) Based on the standard enthalpy changes of formation given below, calculate the standard enthalpy changes for the complete combustion of octane and ethanol respectively.

(ii) Assuming that gasoline contains only octane, compare the enthalpy change of combustion values, in kJ g-1, of gasoline and an alternative fuel containing gasoline and 10% ethanol by mass.

(iii) Besides cutting down petroleum consumption, suggest one additional advantage of using the alternative fuel over using gasoline.

(9 marks)


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Mark Scheme ..............................................................................................................................................

5. (a) (i) Complete combustion of octane: C8H18(l) + 12O2(g) 8CO2(g) + 9H2O(l) Standard enthalpy change = 8Hf

o [CO2(g) + 9Hf o [H2O(l)] - Hf o [C8H18(l)] = 8(-394) + 9(-286) (-250) = -5476 kJ mol-1

1

1

Complete combustion of ethanol: C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) Standard enthalpy change = 2Hf

o [CO2(g)] + 3Hf o [H2O(l)] - Hf o [C2H5OH(l)] = 2(-394) + 3(-286) (-278) = -1368 kJ mol-1

1 1

(ii) Conversion of enthalpy changes of combustion from kJ mol-1 units to kJ g-1 units Mr of octane = 114 ; Mr of ethanol = 46

Heat of combustion of C8H18(l) = 1145476

= 48 kJ g-1

Heat of combustion of C2H5OH(l) = 461368

= 29.7 kJ g-1

1

1

As the alternative fuel contains 990% octane and 10% ethanol, its heat of combustion is 0.9(48) + 0.1(29.7) = 46.2 kJ mol-1 The alternative fuel has a lower energy content.

1 1

(iii) Any ONE of the following: Ethanol is an oxygen-containing compound It is easier for the alternative fuel to achieve

complete combustion / less CO is produced / less particulates are formed / less air pollutants.

Ethanol is a renewable energy source. It can be obtained from agricultural products. The cost for the production of ethanol is low in agricultural countries.

1

[9] -------------------------------------------------- Take a break ----------------------------------------------------- 5. (b) Aspartame is a commonly used artificial sweetener.

(i) Copy the structure of aspartame into your answer book, and mark each chiral carbon atom with an asterisk.

(ii) With reference to its structure, explain why aspartame is unstable in cooking.

(iii) In a coffee shop, a packet of sweetener contains about 5% of aspartame and 95% of silicon dioxide. Suggest a reason for including silicon dioxide in the packet.

(4 marks)


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Mark Scheme .............................................................................................................................................. 5. (b) (i)

1

(ii) Aspartame contains peptide links / ester groups. They can undergo hydrolysis. The high temperatures of cooking process can speed up the hydrolysis.

1

1

(iii) SiO2 acts as a filler / increases the bulk mass of the sweetener. 1 [4]

-------------------------------------------------- Take a break ----------------------------------------------------- 5. (c) Consider the isomeric compounds J and K below:

(i) Name the type of isomerism involved.

(ii) Suggest, with explanation, how J and K can be distinguished from each other by

(I) a physical method,

(II) a chemical method, and

(III) a spectroscopic method. (7 marks)

Mark Scheme .............................................................................................................................................. 5. (c) (i) Functional group isomerism 1 (ii) (I) Compare their boiling points/melting points.

J has a higher boiling point/melting point. or, J is a solid while K is a liquid. Intermolecular force between J molecules is H-bond, while that between K molecules is van der Waals forces.

1

1

(II) Any ONE of the following: - Suspend compounds in a little water, and add NaOH(aq). Only J dissolves. - Dissolve compounds in a little NaOH(aq), then add C6H5N2+(aq). Only J gives a

bright red precipitate / an azo dye. - Treat compounds with neutral FeCl3(aq). Only J gives a purple solution.

1+1

(III) Run IR spectrum of J and of K Only J gives a broad and strong absorption from 3230 to 3670 cm-1 of OH in alcohol

1 1

[7]

-------------------------------------------------- Take a break ----------------------------------------------------- 6. (b) The reaction of 2,3-dimethylbutan-2-ol with concentrated H2SO4 gives two alkenes.

(i) Suggest possible structures for the two alkenes.

*(ii) Suggest how the two alkenes can be distinguished from each other by chemical means.

(5 marks)


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Mark Scheme .............................................................................................................................................. 6. (b) (i) structures of the alkenes

1+1

(ii) Treat compounds with (1) O3 (2) Zn dust, H2O (I) gives only 1 product (CH3COCH3) (II) gives 2 products ((CH3)2CHCOCH3 and HCHO)

1 1

The number of products formed can be found using chromatographic method 1

or, The product from (I) is non-reducing, but one of the products from (II) (HCHO) is reducing (1) or, the carbonyl compounds formed can be identified by converting it to a solid derivative

(2,4-dinitrophenylhydrazone) and meaning the melting point of the derivative (1)

[5]

-------------------------------------------------- Take a break ----------------------------------------------------- 6. (c) Outline a synthetic route, with not more than four steps, to accomplish each of the following transformations.

For each step, give the reagent(s), conditions and the structure of the organic product.

(3 marks) Mark Scheme .............................................................................................................................................. 6. (c) (ii) (1) H2SO4(l), cold

(2) H2O, boil

OH Cr2O72-/H+

heat

O H2NOHNOH

1 1 1

Step 1: hydration of C=C Step 2: oxidation *Step 3: condensation of C=O Alternatively, use catalytic hydration (H2O/H+, heat) see 2003-AL-CHEM 2 Q.7(b)(ii)

[3]

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (a) Compound L, with a relative molecular mass of 108, has the following composition by mass:

C 77.8% , H 7.4% and O 14.8%

L does not react with bromine, but undergoes oxidation with acidified Na2Cr2O7(aq) to give compound M. M reacts with PCl5 to yield compound N. The infra-red spectra of M and N are shown below:


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(i) Calculate the molecular formula of L.

(ii) Based on the above information, deduce the structures of L, M and N. (8 marks)

Mark Scheme ..............................................................................................................................................

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (b) The three-dimensional structures of two 2-chlorobutanes (P and Q) are shown below:

(i) Comment on the difference, if any, in physical properties between P and Q.

(ii) Will P and Q exhibit the same chemical properties ? Explain your answer.

*(iii) The reaction of P with a large excess of 37Cl gives Q initially. Outline a mechanism and draw a labelled energy profile for the reaction.

(7 marks)


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Mark Scheme .............................................................................................................................................. 7.

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (c) Perspex is a plastic consisting of polymethyl methacrylate (PMMA). Part of the structure of PMMA is shown

below:

(i) The monomer of PMMA is methyl methacrylate (MMA). Draw the structure of MMA.

*(ii) Propanone is commonly used as a starting material for synthesising PMMA. Suggest a synthetic route, with not more than five steps, for the transformation of propanone into PMMA.

(5 marks)


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Mark Scheme .............................................................................................................................................. 7.

1

4

Note: *Step 1: Addition of HCN on C=O ; *Step 2: acid hydrolysis of CN to give COOH + dehydration of OH group to give C=C ; Step 3: esterification of COOH ; Step 4: Addition polymerisation

2006-AL-CHEM 1

5. (a) Glutathione is a biologically abundant peptide. When subjected to acid hydrolysis, glutathione gives a mixture of amino acids.

(i) Draw the structures of all amino acids formed from the hydrolysis of glutathione, and mark each chiral centre on the structures with an asterisk.

(ii) Suggest an experimental method to separate these amino acids from one another.

(iii) Glutathione is involved in the de-toxification of heavy metals such as cadmium and mercury. Suggest how glutathione performs this function.

(5 marks) Mark Scheme .............................................................................................................................................. 5. (a)

(i) HO2C CO2H

NH3+

*

H3N CO2H

SH

*+

H3N CO2H+

1 1 1

(ii) paper chromatography 1 (iii) Glutathione removes heavy metals by complex formation. 1 [5]


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6. (c) Answer the following multiple-choice questions:

Mark Scheme .............................................................................................................................................. 6. (c) (i) B 1 (ii) D 1

2006-AL-CHEM 2

5. *(a) -Phenylethylamine (PEA) functions as a neurotransmitter and appears to create excited feelings and moods.

Suggest a route with no more than five steps for the synthesis of PEA from methylbenzene.

(5 marks) Mark Scheme .............................................................................................................................................. 5. (a) CH3 CO2H CH2OH CH2OH

CH2CNCH2CH2NH2

MnO4-/H+

heat

LiAlH4 PCl5

KCNLiAlH4

5


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NOTE: In step 2, (1) LiAlH4/dry ether; (2) H3O+ In step 3, SOCl2 can also be used. In step 5, (1) LiAlH4/dry ether; (2) H2O

-------------------------------------------------- Take a break ----------------------------------------------------- 5. (b) Compound B is a strong stimulant. Its structural formula is as follows:

(i) Give the systematic name of B.

(ii) In fact, the above structural formula can represent two stereoisomers.

(I) Draw three-dimensional structures of the two stereoisomers.

(II) State a physical property which is different for the two stereoisomers.

(iii) It is known that among the two stereoisomers, only B has stimulant activity while the other one does not. Why?

(iii) A person is suspected to have taken stimulant B. A urine sample of the person is sent for analysis. Suggest a method to establish whether B is present in the urine sample.

(7 marks) Mark Scheme .............................................................................................................................................. 5. (b) (i) 2-amino-1-phenylethanol 1

(ii) (I)

C

OH

H2NH2CH

C

OH

HH2NH2C

2

(II) They rotate the plane of polarisation of a beam of plane polarised light to opposite directions.

1

(iii) The neuroreceptor is likely to be chiral. The reaction between 2-amino-1-phenylethanol and the neuroreceptor is stereo-specific.

1

(iv) Conduct a chromatographic study 1 Compare the Rf value of the suspected stimulant with that of an authentic sample of B. 1

[7]

-------------------------------------------------- Take a break ----------------------------------------------------- 5. (c) Deduce the structure of compound D on the basis of the information given below:

(1) Elemental analysis data show that D has the following composition by mass:

C 40.0% H 6.7% and O 53.3%

(2) The relative molecular mass of D is estimated to be in the range of 172 to 182.

(3) The infra-red spectrum of D shows, apart from the absorption of CH stretching near 2900 cm-1, a strong and broad absorption around 3400 cm-1, and no appreciable absorption around 1700 cm-1.

(4) All carbon atoms of D have the same bonding environment.

(5) D is highly soluble in water, and the solution does not decolorise bromine water.

(8 marks)


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Mark Scheme .............................................................................................................................................. 5. (c) C H O

Mass (g) 40.0 6.7 53.3 No. of moles (mol)

0.120.40

=3.33 0.17.6

=6.7 0.163.53

=3.33

Relative no. of moles 1 2 1

1

the empirical formula of D is CH2O. 1

Let the molecular formula of D be (CH2O)n 172 < n(12 + 2 + 16) < 182 n = 6

1

the molecular formula of D is C6H12O6.

Infra-red spectrum: Strong and broad absorption near 2900 cm-1 D has OH groups.

1

No absorption around 1700 cm-1. D does not possesses carbonyl group. 1

D does not decolorise bromine D does not have C=C bond. 1

All carbon atoms of D have the same bonding environment, i.e. they are in the same hybridization state, bonded to the same groups, and have the same configuration.

1

D is

OHOH

OHOH

HO

HO

OHOH

OHOH

HO

HO

1

(Accept any one of the stereoisomers of inositol as well as the above structural formula (right).) [8]

-------------------------------------------------- Take a break ----------------------------------------------------- 6. (b) Organic compounds in waste water can be degraded by oxidation with ozone. In a typical process, phenol is

oxidised to give an acyclic compound E (C6H6O4). The infra-red spectrum of E shows a broad absorption band around 3300 cm-1 and a strong absorption at about 1700 cm-1. Hydrogenation of E in the presence of 10% Pd on charcoal produces an acidic compound F (C6H10O4).

(i) Suggest one structure for E and one for F, and explain your answer.

(ii) Suggest two advantages of using ozone to degrade organic compounds. (7 marks)

Mark Scheme .............................................................................................................................................. 6. (b) (i) From IR spectrum

E possess a OH group (broad absorption peak at 3300 cm-1) and carbonyl group (absorption peak at 1700 cm-1)

1 1

C6H6O4 C6H10O6 E F

E has two C=C bonds or one CC bond. F is acidic. F has the structure HO2C(CH2)4CO2H E is an acyclic compound with formula C6H6O4. It has the structure HO2CCH=CHCH=CHCO2H

1 1

1

(ii) Upon reduction, ozone gives H2O which is harmless. Ozone has a short lifetime. The residual O3 will not remain in the atmosphere for a long time.

1 1

[7]

H2/Pd


21

-------------------------------------------------- Take a break ----------------------------------------------------- 6. (c) Consider the following compounds, G, H, J, K, L and M:

(i) Which compound(s) is/are a colourless crystalline solid ?

(ii) Which compound(s) has/have a fishy smell ?

(iii) Which compound(s) react(s) to give an addition polymer ?

(iv) Which pair of compounds react to give a condensation polymer ?

(4 marks) Mark Scheme .............................................................................................................................................. 6. (c) (i) H 1 (ii) M 1 (iii) K and G 1 (iv) G and H 1

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (a) In the past, people chewed willow bark to ease their pain and to lower fever. Later on, chemists identified

salicin as the active ingredient in willow bark.

Salicin undergoes acid hydrolysis to give glucose and 2-hydroxybenzyl alcohol.

*(i) Suggest a chemical test to distinguish between salicin and glucose. Account for the difference in results of your suggested test.

(ii) 2-Hydroxybenzyl alcohol can be converted in two steps to acetylsalicylic acid, which is the active ingredient of aspirin.

(I) Give the reagents used in Step 1 and in Step 2.

(II) In a typical experiment, 2.0 g of 2-hydroxybenzyl alcohol gives 2.0 g of acetylsalicylic acid. Calculate the percentage yield of the conversion.

(7 marks)


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Mark Scheme .............................................................................................................................................. 7. (a) (i) Fehlings reagent 1 Glucose gives a bright red ppt. while salicin does not. 1

Explanation: Salicin is a glucoside. It does not give free aldehyde group in aqueous solution and therefore will not show properties of a reducing sugar.

1

(ii) (I) Step 1: Cr2O72-/H+, heat 1 Step 2: (CH3CO)2O or CH3COCl 1 (II) molar mass of 2-hydroxybenzyl alcohol = 124 g

molar mass of aspirin = 180 g

% yield =

1242

1802

= 68.8 %

1

1

[7]

-------------------------------------------------- Take a break ----------------------------------------------------- 7. (b) The repeating unit of polystyrene (PS) is shown below:

*(i) Outline a mechanism for the formation of PS from styrene.

(ii) Polymerisation of styrene together with 1,4-diethenylbenzene give another polymer N.

Draw part of the structure of N. Suggest one difference in physical property between N and PS, and explain your answer.

(iii) A proton exchange resin is made by reacting N with concentrated sulphuric(VI) acid.

(I) What functional group does the proton exchange resin possess ?

(II) Explain why the proton exchange resin can be used to remove ions of heavy metals from industrial waste water.

(8 marks)


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Mark Scheme .............................................................................................................................................. 7. (b) (i) Mechanism (free radical polymerisation) 3 Initiation:

C O O C

O O

2 C OO

.

( PhCOO ).

( X ).

.

+

CHCH2OCOPh.CHH2CPhCOO

Propagation:

X . CHH2C

+

CHCH2X.

.( X1 )

Termination: Xm + Xn XmXn

(ii) Part of structure of N:

CH CH2 CH CH2

CH CH2 CH CH2CH2CH

CH CH2CH2CH

......

... ...

......

1

Difference in physical property: N is stronger than PS Explanation: N is a cross-linked polymer but PS is not

1 1

(iii) (I) sulphonic acid 1 (II) The resin contains a large number of sulphonic acid groups which are strongly acidic.

Ions of heavy metals will displace the protons from the resin with the metal ions retained by the sulphonate anions.

1

[8]


24

-------------------------------------------------- Take a break ----------------------------------------------------- 7. *(c) (i) Cooking oils are unsaturated triglycerides. They are susceptible to oxidative rancidity, which involves

the formation of peroxide.

Draw the structure of a cooking oil, and write an equation to show the formation of peroxide.

(ii) The extent of oxidation of an oil can be determined by iodometric analysis.

In a typical experiment, 4.85 g of a sample of oil is treated with excess KI(aq) and H2SO4(aq). The iodine liberated requires 21.20 cm3 of 0.012 mol dm-3 Na2S2O3(aq) for complete reaction.

Given that when treated with acidified KI(aq), 1 mol of peroxide liberates 1 mol of I2, calculate the number of moles of O2 absorbed per kg of the oil.

(5 marks) Mark Scheme .............................................................................................................................................. 7. (c) (i) (The fatty acid in the structure should have an even number of C atoms and contain at least one

C=C bond.)

CH2OCHOCH2O COR'

CORCO(CH2)7CH=CH(CH2)7CH3

1

H

...... + O OOHO ...

...

1

(ii) ROOH + 2I- + 2H+ ROH + H2O + I2 I2 + 2 S2O32- 2I- + S4O62- O2 2 S2O32-

No. of moles of S2O32- used = 0.012 21.2 10-3 = 2.54 10-4 mol 1

No. of moles of O2 absorbed per kg of oil = (2.54 10-4) mol (4.85 10-3 kg) = 26.2 10-2

1 1

[5]

AL-CHEM Chemistry of Carbon Compounds(03-06)

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