2012 Y6 H2 Chem T3 CT Answers _with Comments

7/27/2019 2012 Y6 H2 Chem T3 CT Answers _with Comments

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RAFFLES INSTITUTION2012 Year 6 H2 Chemistry Term 3 Common Test

Suggested Solutions

Section AQuestion No. 1 2 3 4 5 6 7 8 9 10

Answer B D A C C D C D C B

Question No. 11 12 13 14 15 16 17 18 19 20Answer A D C A B D A B B C

1 angle of deflectioncharge

mass

ion Br+ 1Br+ 1Br + Br +

charge

mass

1

79

1

81

2

81

2

82

angle of deflection: 81Br+ < 79Br+ < 82Br2+ < 81Br2+

least most

2 SF4 is polar.

Both I3 and N3

are linear, i.e. their bond angles are 180o.

Allene (or propadiene) is not planar. The two hydrogen atoms on the left lie on anequatorial plane while the two hydrogen atoms on the right lie on an axial plane.

These two dipoles

do not cancel out.

:SF

F

F

F

These two dipoles

cancel out.

CH

HC

H

H

C

In addition to instantaneous dipole induceddipole (id-id) and permanent dipole permanentdipole (pd-pd) interactions between propanone

and water molecules, the carbonyl oxygen ofpropanone can also form hydrogen bonds withwater. Hence, propanone is soluble in water, i.e.these two liquids are miscible.

+

+

:

O

HH

O


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2

3 Cu+ + e Cu E = +0.52 V

Cu2+ + e Cu+ E = +0.15 V

reduction: Cu+ + e Cu

oxidation: Cu+ Cu2+ + e

2Cu+

Cu2+

+ Cu E

cell = (+0.52) (+0.15) = +0.37 VSince Ecell > 0 V, the disproportionation of Cu+ is energetically feasible under

standard conditions.

2Sn2+ Sn4+ + Sn Ecell = (0.14) (+0.15) = 0.29 VSince Ecell < 0 V, the disproportionation of Sn

2+ is not energetically feasible understandard conditions.

4 Amt of Cr(s) formed = 1.04 / 52.0 = 0.02 molCr3+ + 3e CrAmt of electrons required to reduce 0.02 mol of Cr3+ to Cr = 3 x 0.02 mol = 0.06 molAmt of charge transferred = 0.06 mol x 96 500 C mol1 = 5790 CNote: 1 A = 1 C s1

Time taken = Amt of charge transferred / Current= (5790 C) / (5 C s1)= 1158 s

5 Let the solubility be z mol dm3.AlY3(s) Al

3+(aq) + 3Y(aq)

z 3zKsp = K= [Al

3+(aq)][Y(aq)]3= (z)(3z)3= 27z4

6 CH2ClCOOH + KOH CH2ClCOOK + H2OLet vol. of CH2ClCOOH added be V cm

3.Amt of KOH left = initial amt of KOH amt of acid added

= (5.0 x 103 x 0.10) (V x 103 x 0.05)= (5.0 x 105)(10 V), where 0 V 10

Amt of excess acid = (V 10) x 103 x 0.05= (5.0 x 105)(V 10), where V > 10

V / cm3Amt of KOH

left / molAmt of CH2ClCOOK

formed / molAmt of excess

CH2ClCOOH / molRemarks

0 5 x 10 0 0 only KOH(aq) present

A 5 2.5 x 104 2.5 x 104 0half-equivalence point

reachedB 10 0 5 x 104 0 equivalence point reached

C 15 0 5 x 104 2.5 x 104 buffer formed, but[CH2ClCOOH 50 cm3),there is only one equivalence point on the titration curve.

Based on the information, we could make the following deductions. The weak base has one pKb value, i.e. 1 mol of base reacts with 1 mol of HCl.

Hence, the weak base cannot be H2NCH2CH2NH2, which has two pKb values and1 mol of it requires 2 mol of HCl for complete neutralisation.

Due to the hydrolysis of the conjugate acid of the weak base, the solution isacidic (pH < 7) at equivalence point. Methyl orange is a suitable indicator for thetitration as its working range coincides with the region of rapid pH change (i.e. theregion from just before to just after the equivalence point).

At least 25 cm3 of HCl(aq) is required to reach half-equivalence point. When25 cm3 of HCl(aq) is added to the weak base, the acid reacts completely and

neutralises about half the total amount of weak base. At this point, the resultingsolution contains unreacted weak base and the conjugate acid of the weak base.Hence, a buffer solution is formed.

16 The highest oxides (i.e. an oxide in which the element exhibits its highest possibleoxidation number) are Na2O, MgO, Al2O3, SiO2, P4O10, SO3 and Cl2O7.

Basic oxides (ionic): Na2O, MgOAmphoteric oxide (ionic): Al2O3Acidic oxides (covalent): SiO2, P4O10, SO3, Cl2O7

Na2O + H2O 2NaOH (pH > 12)MgO + H2O Mg(OH)2 (pH 9) Note: MgO is sparingly soluble in water.

Al2O3 is insoluble in water because the heat generated from its dissolution will not besufficient to overcome the large amount of energy (i.e. highly exothermic latticeenergy) required to break its giant ionic lattice. Thus, a solution containing solidaluminium oxide has a pH value of 7 (at 25 oC).

SiO2 is insoluble in water because the heat generated from its dissolution will not besufficient to overcome the large amount of energy required to break its giantmolecular structure. Thus, a solution containing solid silicon dioxide has a pH valueof 7 (at 25 oC).

P4O10 + 6H2O 4H3PO4 Danger! Beware of highly exothermic reaction.SO3 + H2O H2SO4 Danger! Beware of hot, acidic mist.

Cl2O7 + H2O 2HClO4 Note: Chloric(VII) acid is very acidic.

From Data Booklet:Na+ + e Na E = 2.71 V

Mg2+ + 2e Mg E = 2.38 V

Al 3+ + 3e Al E = 1.66 V

Cl2 + 2e 2Cl E = +1.36 V Note: Cl2 is oxidising.

reducing power: Na > Mg > Al > Cl2 (no E data for Si, P4 and S8)

Electrical conductivity: Na < Mg < AlAs the number of valence electrons which can

delocalise in the metallic lattice increases, the

electrical conductivity also increases.


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17 MCO3(s) heat MO(s

MgCO3 decomposes atMg2+ polarises, and henextent than Ba2+ doedecomposed to give CO2

Gdecomp = Hdecomp TThe decomposition reaheated to a certain tempAt low temperatures, GAt high temperatures,

18

progesterone

Only progesterone reactOnly aldosterone reactsNeither of the two comp

19

aspirin

The phenol group of saligroup in aspirin. EthanoyOnly salicylic acid forms3RCOOH + PCl3 3RBoth aspirin and salicylic

20 At pH 7, the guanidino gcannot undergo nucleopIn addition, with the exuncharged. Thus, at pH

OHO

O

O

7

) + CO2(g) Hdecomp > 0; Sdecomp > 0 (M =

a lower temperature because the higher charce weakens, the CO bond in carbonate ion. Depending on the temperature, MgCO, but BaCO3 may not have begun decomposing

decomption does not occur spontaneously until therature.

ecomp > 0 because |Hdecomp| is greater than |T

decomp < 0 because |TSdecomp| is greater than |

aldosterone

with alkaline aqueous iodine (see rectangle).ith Fehlings solution (see oval).

unds reacts with sodium hydrogen carbonate (

salicylic acid

cylic acid does not react with ethanoic acid tol chloride should be used instead.a violet complex with neutral FeCl3(aq).OCl + H3PO3acid react with the same amt of PCl3.

oup will be protonated (i.e. positively charged)ilic acyl substitution with ethanoyl chloride.eption of the guanidino group, the rest of th, leupeptin migrates to the negative electrode (

OHO

OH

NH is isoelectr

and NH2 is isoel

O. Hence, this

isoelectronic with

can donate a lone

ion. At pH < 12.5,

C(NH)(NH2)2+.

guanidino group

Mg or Ba)

ge density ofto a greater

may have.

carbonate is

Sdecomp|.| Hdecomp|.

a weak base).

ive the ester

, so leupeptin

molecule iscathode).

nic with O,

ectronic with

moiety is

COO, i.e. it

pair to an H+

it exists as


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8

Section B

B1 (a) Prediction: Solubility of cerium(IV) sulfate in sulfuric acid decreases whenconcentration of sulfuric acid increases.

Explanation: As [H2SO4] increases, [SO42(aq)] also increases. By Le

Chateliers Principle, the position of equilibrium of the reactionCe(SO4)2(s) + aq Ce

4+(aq) + 2SO42(aq) shifts to the left and the solubility

of Ce(SO4)2 decreases.ORH2SO4(aq) is a strong acid which dissociates completely in aqueous solution

to give H+ and SO42 ions. The SO4

2 ion is a common ion to both H2SO4 andCe(SO4)2. Due to common ion effect, the solubility of Ce(SO4)2 decreaseswith increasing [H2SO4(aq)].

solubility ofcerium(IV)

sulfate

concentration of sulfuric acid(slight curve is also acceptable)

(b) Temperature

Examiners CommentsB1(a)

The inverse relationship between the solubility of Ce(SO4)2 and the concentration of sulfuric acid must be clearlyshown in your answer. Vague answers such as solubility decreases, which do not mention how solubility of

Ce(SO4)2 varies with the acid concentration, are not given credit. The common ion, SO42, must be clearly

identified in the explanation.

Some students were able to give the expected prediction, but were unable to correctly display their prediction inthe form of a graph. Many of these students did not seem to understand what they were doing.

B1(b) Most students managed to identify temperature as the variable to be controlled since it is mentioned in the

definition of solubility given at the beginning of the question. It is the temperature of water which should be keptconstant, not the temperature of the surroundings.


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(c) Part 1 solubility of cerium(IV) sulfate in water

Method 1:1. Weigh a clean and dry conical flask. Let its

mass be m1. Place 60 cm3 of distilled water

into the conical flask and obtain the new mass,

m2. Mass of distilled water, m

w= m

2m

1.

2. Place the conical flask with the distilled waterin a water bath which is maintained at thetemperature at which the solubility of Ce(SO4)2is to be found.

3. Add a little Ce(SO4)2 to the distilled water andstir the mixture with a glass rod. Continueadding Ce(SO4)2 , with stirring, until some solidremains undissolved. Allow time (e.g. 10 min)for the mixture to reach equilibrium.

4. Weigh the conical flask with the saturatedsolution and undissolved Ce(SO4)2. Let themass be m3. Filter this mixture. Collect the

residue and leave it to dry. Weigh the dryresidue. Let its mass be m4. Mass of saturatedsolution, ms = m3 m1 m4

5. The mass of Ce(SO4)2 which is soluble in

60 cm3 of distilled water, mc = ms mw

6. Solubility of Ce(SO4)2 in water =mc

mwx100

Method 2:1. Pour 60 cm3 of distilled water into a conical

flask. Place the conical flask with the distilledwater in a water bath which is maintained atthe temperature at which the solubility ofCe(SO4)2 is to be found.

2. Add a little Ce(SO4)2 to the distilled water andstir the mixture with a glass rod. Continueadding Ce(SO4)2, with stirring, until someremained undissolved. Allow time (e.g. 10 min)for the mixture to reach equilibrium.

3. Weigh a clean and dry evaporating dish. Let itsmass be m1.

4. Filter the mixture in the conical flask. Collectthe filtrate and place it in the weighedevaporating dish. Weigh the filtrate andevaporating dish. Let the mass be m2.

5. Heat the evaporating dish to drive off all thewater. Weigh the evaporating dish and theresidue. Let the mass be m3.

6. Mass of Ce(SO4)2 in filtrate, mc

= mass of residue = m3 m1

Mass of distilled water in filtrate, mw = m2 m37. Solubility of Ce(SO4)2 in water =

mc

mwx100

Key points:

Description of a sequential method for preparing the saturated solution using 60 cm3 ofdistilled water and filtering off the excess solid

Mention stirring and waiting time for equilibrium to be established

Description of how temperature is controlled, e.g. using a water bath during thepreparation of the saturated solution Description of how the mass of Ce(SO4)2 and mass of water could be found

Examiners Comments for Part 1

Procedure lacks essential details, e.g. suitable apparatus, appropriate volume or mass, what measurements totake, when to stop adding the solid, how to maintain constant temperature, the need for equilibrium to beachieved before filtration, etc.

Procedure lacks clarity, e.g. not labelling the steps, poor choice of words such as pour solid, illogical sequencesuch as not weighing the dry beaker before adding in water, etc.

Ice-baths (you cannot dissolve a solid in ice!) and sand-baths (too hot!) are unsuitable for this experiment. Athermostat is not the same as a temperature-controlled (or thermostatically controlled) water bath. There iscertainly no need for a thermodynamically controlled water bath in this experiment.

Many students failed to recognise the need to either dry the residue or evaporate the filtrate to find the mass ofsolid dissolved. Some students dry the residue and then wash it with cold water, which defeats the purpose ofdrying the residue. Drying agents such as anhydrous calcium chloride should not be used as they contaminatethe residue. The liquid obtained from filtration is a filtrate, not a distillate (which is obtained from distillation).

Many students assumed the mass of 60 cm 3 of water to be 60 g. Although this is a good approximation (sincedensity of water is about1 g/cm3), students should obtain the mass of water used by weighing.

Although the definition of solubility is given in the question, it was generally ignored. Some students confusedsolubility with solubility product, Ksp.

A filter funnel is not the same as a separating (or separatory) funnel. You do not use a separating funnel toseparate a solid from a solution; a separating funnel is used to separate two immiscible layers.


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Part 2 solubility of cerium(IV) sulfate in sulfuric acid

1. Using a burette, transfer 20.00 cm3, 40.00 cm3, 60.00 cm3 and 80.00 cm3 of the acidprovided into four different 100 cm3 graduated volumetric flask labelled as S1, S2, S3and S4.

2. For each graduated flask, top up to the 100 cm3 mark with distilled water. Stopperand shake each flask to ensure that the dilute sulfuric acid prepared is homogeneous.

Concentrations of the sulfuric acid are 1.00, 2.00, 3.00 and 4.00 mol dm3 in theflasks S1, S2, S3 and S4 respectively.

Key points: An appropriate method for diluting the sulfuric acid described (means of measuring

volume of acid and of water; sensible choice of volumes) Burettes/pipettes and graduated flasks (or other suitable precision apparatus) used to

measure volumes Four solutions with different concentrations (to be specified) prepared and the range

should cover at least a fourfold increase in concentration, e.g. 1 mol dm3 to 4 mol dm3,and at least one should be greater than 2.5 mol dm3. 5 mol dm3 should not be one of

these because a decent scatter plot should contain 5 data points The volumes to be used must correspond to the desired concentrations to be attained

Alternative method: Serial dilution

Examiners Comments for Part 2

Language of procedure writing to include quantity and instrument used.o The key to doing well is to read as many laboratory procedures as possible.o Note that each step ideally should contain three parts:

1. Start with a verb (an action word).

2. Describe the variablewith suitable quantities to be measured & recorded.

3. List the instrument to be used.

o E.g. Weigh and record about 2.5 g of Na2O (s) using an electronic mass

balance . [not required in this expt but for those that require weighing]

E.g. Transfer 25.0 cm3 of FA1 using a pipette to a 250-ml

graduated volumetric flask .

o Many students left the measuring instrument out! For those that stated the apparatus used, the

apparatus used were not precise enough. E.g. Students suggested using measuring cylinders and

beakers to measure volumes. Instead, burettes or pipettes should be used. To prepare a dilute solution

of known concentration (standard solution), we have to use a standard volumetric flask (see comment

below) instead of just placing the solution in a beaker.

Adding up volumeso Generally, the total volume of a mixture of miscible solutions of different densities is not equal to the

individual volumes. E.g. 10 cm3 of 5 mol dm3 H2SO4 + 10 cm3 of H2O 20 cm3 of combined solutiono Hence, there is a need to use graduated standard volumetric flask to make standard solutions of known

concentration, i.e. topping up to the mark to prepare a solution of a particular volume.

Presentationo Students should try to tabulate their desired volumes to be used instead of writing all over the place.

Calculation of concentrationo A significant number of students do not know how to calculate the appropriate volumes of acid and water

to make their desired concentrations of acid.[diluted solution] = volume of 5 mol dm3 H2SO4 x 5 mol dm3

total vol. of solution in standard flask


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(d) Corrosive nature of concentrated sulfuric acidOR toxicity of cerium compounds

B2 (a) (i) Lead metal

(ii) Ecell = +1.69 (0.36) = +2.05 V

(iii) 2PbSO4(s) + 2H2O(l) PbO2(s) + 4H+(aq) + 2SO4

2(aq) + Pb(s)

(b) (i) Anode: Zn(s) + 4OH(aq) [Zn(OH)4]2(aq) + 2e

Cathode: O2(g) + 2H2O(l) + 4e 4OH(aq)

Overall: 2Zn(s) + 4OH(aq) + O2(g)+ 2H2O(l) 2[Zn(OH)4]2(aq)

(ii) +1.62 = +0.40 E

anodeEanode = 1.22 V

(iii)

(iv) Mass of Zn used = 0.60 x 40 = 24 gAmt of Zn used = 24/65.4 = 0.3670 molAmt of e = 2 x 0.3670 = 0.7339 molQuantity of charge = 0.7339 x 96500 = 7.08 x 104 CTime needed = 7.08 x 104 / 13.7 x 103 = 5.17 x106 s

salt bridge

K+

NO3

[H+(aq)] = 1 mol dm3

H+

H2(1 atm)

Pt

V

T = 298 K

[Zn(OH)4]2

= 1 mol dm3

[Zn(OH)4]

Zn

Examiners Comments for B2

Question a(iii) seemed to be a problem for many students. These students failed to understand that the process

of charging required electrical power and thus the overall equation is in the opposite direction to the reactionwhich took place to give the positive Ecell in a(ii). A few students define rather than determine the Ecell in a(ii).Please read the question carefully.

For Questions b(i) and (ii), many students did not analyse the information given in the question. It is analkaline/neutral medium and thus the half equation O2 + 4H+ + 4e 2H2O should not be chosen for the cathodereaction. Some students who chose the correct half-equation made careless mistakes and did not balance theoverall equation correctly. The arrows used in the equations for the anodic, cathodic and overall processesshould point in one direction only (oxidation at anode and reduction at cathode). Reversible arrows should not beused.

Things to include:

Voltmeter

Salt bridge

Labelled electrodesStandard conditions

(T = 298K, P = 1atm andconcentrations= 1 mol dm-3)

Examiners Comments for B1(d)

Students need to differentiate risk from precautionary measures.o Risk: potentially health-threateningo Precautionary measure: methods to reduce the risk


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B3 (a) (i) A titration curve with three vertical sections at 10, 20 and 30 cm3 andthree pKa values of 2.1, 4.1 and 9.5 at 5, 15 and 25 cm

3 respectively asshown below:

Region A B C D E F G

pH 1.5 2.1 (pKa1) 3.1 4.1 (pKa2) 7.8 9.5 (pKa3) 11

Dominantspecies

glu+Equal amt ofglu+ and glu

gluEqual amt of

glu and gluglu

Equal amt of

glu and

glu2glu2

Hence, at pH 2.5, as required by question (a)(ii), the species present would be glu+ and glu.

(ii)

HO

O

NH 3+

OH

O

HO

O

NH 3+

O

O

cationic form (glu+) zwitterionic form (glu)

Examiners Comments for B2 (continued)

Students who revised their work answered Question b(iii) well. The diagram should be constructed using a ruler.Please note that the standard hydrogen electrode is used as a reference electrode and has a voltage of 0.00 V.The electrode to be used for this half-cell should be platinised platinum. Using a graphite electrode in this half-cellwill lead to a potential difference.

Question b(iv) was generally well-answered. Some students did not use 40 g of Zn as given on pg 15 for their

calculations. Quite a number of students forgot to calculate the quantity of electricity passing through the cell andcalculated only the length of time. Some students incorrectly thought that the number of moles of electrons

transferred was 4 times that of zinc. Many students did not know that 13.7 mA = 13.7 x 103 A. Almost allstudents omitted the units for quantity of electricity, which was C (coulombs). A number of students used theAr ofCu instead of Zn. In many cases, rounding-off errors were spotted. Please do not copy the entire string ofnumbers from your calculator. Round-off intermediate and final answers to 4-5 and 3-4 significant figuresrespectively.

0 5 10 15 20 25 30

Volume of aq NaOH/ cm3

pH9.5

4.1

2.1

A

Additional information:(1) pKa associated with glu are as shown:

(2) As pH increases, glu is progressively deprotonated, forming 4 species:

Note that acid with lower pKa is deprotonated first.

(3) The species present at different parts of the titration curve are shown below:

H

3

N C

C H

2

C O

2

H

H

C H

2

C O

2

H

+

p K

a

= 2 . 1

p K

a

= 4 . 1

p K

a

= 9 . 5

H

3

N C C O

2

H

H

C H

2

C H

2

C O

2

H

H

3

N C C O

2

H

C H

2

C H

2

C O

2

H

H

3

N C C O

2

H

C H

2

C H

2

C O

2

H

2

N C C O

2

H

C H

2

C H

2

C O

2

+

+ +

g l u

+

g l u

g l u

g l u

2

-COOH has a smaller pKathan side-chain COOH as itis closer to the electronwithdrawing NH3+ groupwhich stabilises its anion.

1

2

3


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Using Henderson-Hasselbalch equation,2.5 = 2.1 + lg {[zwitterionic form]/[cationic form]}

Hence, the ratio of zwitterionic form to cationic form is 2.51.(OR cationic form : zwitterionic form = 0.398)

(b)

H2N O

N

O

NH

OH

O

OH (zwitterionic form also accepted)

(c) Any one of the following:van der Waals interaction (ala with ala, pro with pro, or ala with pro)hydrogen bonding (ser with ser, ser with glu, or ser with lys)

ionic bonding (glu with lys)

Examiners Comments for B3(a)(c)

Most students did not give a complete sketch of the pH titration curve. Please show 3 vertical (or near vertical)regions and label their corresponding equivalence volumes. More often, the half-equivalence points and theircorresponding pKa values and volumes were not well-labelled.

General formation of amide bond is as shown: RCOOH + RNH2 RCONHR + H2O. The N of amine grouploses an H. Since proline is a secondary amine, it has no more H attached to it after amide bond formation.

Note that if C is written in the structure, all H atoms attached to it should be shown. Hence, the followingstructure would be wrong. The correct structure is shown on the right.

For(c), marks are awarded for the type of R group interaction only if the amino acid residues that give rise to itare correctly given.o For van der Waals (or id-id) interactions, the R group should be non-polar. Glycine is not acceptable as its R

group (which is an H atom) is too small to result in significant vdw interactions.

o Hydrogen bonding can occur between 2 ser residues (OH - - - OH), glu and ser (COO- - - HO) and lys

and ser (NH3+ - - - OH). Note that R groups of lys and glu exist as (CH2)4NH3+ and (CH2)2COO

respectively. Hence, the interaction between lys and glu is ionic interactions and not hydrogen bonding.o

Some students thought that electrostatic attraction is the same as ionic bonding. Do note that electrostaticattraction is an attraction that arises from unequal distribution of charges. Hence, ionic bonds, ion-dipoleinteractions, hydrogen bonding, pd-pd and id-id interactions are all electrostatic in nature and electrostaticattraction is not the same as ionic bonding.

C

N

C

C

C

O

H

2

N

O

N H

O H

O

O H

N

C

O

H

2

N

O

N H

O H

O

O H

C H

2

C H

2

C H

2

H

w r o n g

c o r r e c t


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14

(d) quaternary structure

(e) Glycine is small so it can fit into the space within the helix and still allows thechains to come close together.

(f) Papaya juice contains enzymes that can digestor break down the covalent bonds in the protein.

Boiled papaya juice is ineffective as the heat has denatured the enzyme.

(g) Choose a sensible method such as using an acidic/basic medium (such aslemon juice or bicarbonate powder) or mechanically breaking the bondsthrough hammering/hitting the meat or gentle heating. Methods that will resultin the meat becoming inedible, e.g. using mineral acids or heavy metal ions,are not accepted.

(h)

2 + [O]

O

NH2

SH OH

O

NH2

S

OH

O

NH2

S

OH

+ H2O


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15

B4 (a) Mg(NO3)2 decomposed first.

The ionic size of Mg2+ is smaller than that of Ba2+. As a result, the charge density of Mg2+ is higher than that of Ba2+. Mg2+ has a higher polarising power compared to Ba2+ and is able to

polarise the electron cloud of the NO3 ion to a greater extent.

The NO bond in the NO3 ion is weakened to a greater extent in

Mg(NO3)2. Thus, Mg(NO3)2 is thermally less stable.

Mg(NO3)2(s) MgO(s) + 2NO2(g) + O2(g)

(b) (i) X to Y: Volume increase is due only to increase in temperature of thegases. VT since P is constant.

(ii) W to X: From W to X, the volume is affected by not only an increase inthe temperature but also the increase in the no. of moles of gasresulting from the decomposition of Mg(NO3)2. Since both n and T arechanging, the volume increases exponentially instead of linearly.

Examiners Comments for B44(a)

Some students had the misconception that thermal decomposition involves the breaking of the ionic bondsbetween the cations and the anions. As a result, some students thought that Ba(NO 3)2 will decompose first sinceit has a less exothermic lattice energy.Please note that in thermal decomposition, it is the covalent bondin the anion (in this case, the N-O bond) thatwill be broken.

Students were not very careful / specific with the terms used. For example, there were flawed statements like: Mghas a higher charge density than Ba. (Note: Mg does NOThave a charge!) orMagnesium nitrate polarises theanion to a greater extent (Note: It is the cation, Mg2+, that polarises the anion, not magnesium nitrate or Mg).

Some students did not compare Mg2+ and Ba2+ specifically. Instead, they either gave general trends down thegroup without clearly stating the relative positions of Mg and Ba, or just simply stated that Mg 2+ has a high charge

density and was able to polarise the electron cloud of the anion.

4(b)(i)

Many students wrongly explained that the volume increase from X to Y was due to the thermal decomposition ofbarium nitrate. They failed to see that the question hinted that they should make use of the ideal gas equation todeduce the relationship between V and T. Also, some students did not read the question stem properly andhence missed out the fact that P was kept constant.

4(b)(ii)

Very few students could see that the volume increase from W to X results from a combination of 2 factors:temperature increase as well as an increase in the no. of moles of gas in the system (due to thermaldecomposition of magnesium nitrate).


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16

Section C

C1 (a) (i) A nucleophile is an electron-rich species that donates a pair ofelectrons (to an electron deficient species) to form a covalent bond.

(ii) I - nucleophilic additionII - eliminationIV - acid-base reaction/proton transfer

(iii) LiAlH4 reacts violently with water (to form H2 gas).

Note: LiAlH4 + 4H2O LiOH + Al(OH)3 + 4H2

dry ether (or any non-protic solvent such as CCl4, followed by wateras proton source)

Note: A non-protic solvent is a solvent that cannot donate H+ ions.

(iv) NaH is a strong base / H is least polarisable / H is a poor nucleophile

(b) (i) C6H7NO2

(ii) x = 120o ; y = 104.5o (allow 104o105o for y)

(iii) 1 = nitrile; 2 = alkene; 3 = ester

(c) H2O

(d) Q CH2=C(CH2NH2)CH2OHR CH3CH2OHS CH2=C(CH2NHCOCH3)CH2OCOCH3T CH3CH2OCOCH3U CH3CH(CH2NH2)CO2CH2CH3

- Reactions of Q and R with ethanoyl chloride: 1:2 ratio with Q as both NH2 andOH are acylated

- 1:1 ratio with R as only one OH group is acylated- No hydrogenation of alkene with LiAlH4- No reduction of ester with hydrogen/Ni

Examiners CommentsC1(a)(i)

The majority of the students are unable to give the full definition of a nucleophile. Many students thought thatnucleophiles must be negatively charged and they failed to recognise that what a nucleophile must possess is alone pair of electrons for donation to form a covalent bond with an electron deficient species.

C1(a)(ii) Students should understand the difference between state and describe so that they do not waste time by

giving too much detail.

When asked to state type of reaction, it is not sufficient to just say addition/substitution. The word nucleophilicor electrophilic is crucial.

The term neutralisation should be avoided here as it involves the formation of a salt.

The reaction in step IV is NOT hydrolysis. Hydrolysis usually involves the rupturing of a chemical bond by theaddition of water.


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C2 (a)

(i)

(ii) React with 2,4-DNPH and then determining the melting points of the

hydrazones.

(iii) Reagents:I2(aq) / NaOH(aq)Condition: warm

Observation: Pentan-2-one gave a yellow ppt, but not pentan-3-one.

Organic products: CHI3 and CH3CH2CH2COONa+

Examiners CommentsC1(a)(iii)

Many students had the wrong impression that water is a stronger nucleophile and hence will hydrolyse the ester.They failed to recall that the hydrolysis of esters does not take place so readily and it requires heating.

Some students suggested methane or ethane as another suitable solvent without realising that these compoundsexist as gases at rtp and hence cannot be used as solvents.

C1(a)(iv)

Many students failed to recognise that NaH is an ionic compound and there is no such thing as Na H bond inthis context.

Some students also have the serious misconception that H+ from NaH is a poor nucleophile. They failed torealise that H+ can never serve as a nucleophile as it is electron deficient and has NO lone pairs of electrons.

It should be the H ion from NaH that is a poor nucleophile because its electron cloud is less polarisable.

C1(b)(i)

About half the cohort did not answer to the question. They gave the structural formula instead of the molecularformula. This is shocking!

C1(b)(ii)

When asked to state the bond angle, please state a value. Signs like are not acceptable. Do not state arange of values too.

Students failed to link 1(d) to 1(a) and thought that the reaction of LiAlH4 with an ester is similar to the hydrolysis

of an ester.

Students also need to learn the correct names of functional groups (e.g. cyanide is not acceptable for the CNgroup). When students encounter a C=O group, they should always check if the C=O is part of an ester or amidelinkage before concluding the functional group is a ketone.

The explanations for1(d) are also unclear be explicit and explain how the mole ratios of ethanoyl chloride to Qand R help to identify the structures (i.e. say Q has two groups (alcohol and amine) that will react with CH3COCl

and not Q may have alcohol or amine functional groups).

Students also need to remember which functional groups are affected by the various reducing agents. Manyfailed to recognise that LiAlH4 does not reduce alkenes and hydrogen gas reduces nitriles.

Just because hydrogen gas does not reduce esters does not mean it is a weak reducing agent!


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(iv) Type of reaction: Nucleophilic Addition

Condition: trace amounts of NaCN or NaOH(aq)

Mechanism:

B1

Examiners Comments for C2(a)

This question was generally high-scoring. However, there are also many students who did not take care whenreading the question and as a result, lost marks unnecessarily.

(a)(i) The question asked for a balanced equation. Many students did not balance the equation with H2O.Please take note that the structure of the product has to be clear. CH3CH2CH2 cannot be replaced withC3H7 because it represents two possible structures, CH3CH2CH2 and (CH3)2CH.

(a)(iii) While many students stated correctly the use of the iodoform test to distinguish the ketones, very fewstudents gave the correct products. CH3CH2CH2CO2Na+ was usually just given as the anion without thecounter ion. Note that the sodium ion is required.

(a)(iv) Please note the use ofcurved/curly arrows in the writing of mechanisms and their significance. To manystudents, arrows show the movement of attacking species. However, arrows show movement of electrons,with a half arrowhead representing one electron and a full arrowhead representing a pair of electrons. Andthe arrows must be curved! Dont be flippant when drawing mechanisms. Respect the convention.

Arrows should also start from the electron pair and end close to where the pair of electrons goes. Dontleave your arrow hanging in midair. And please DONT draw arrows starting from negative charges andending them at positive charges. These charges are drawn in by humans; they dont physically exist.

Note that the intermediate must also show correct structure so the condensed C 3H7 is not allowed andsuch answers were penalised again.

The question also did not ask you to address the stereochemistry of the product so there is no need toshow the attack of the carbonyl carbon by the cyanide ion from either side of the plane around the carbonylgroup.


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(b)

C C

C

H

O

O H

+

cis

(hydrogen bonds

In the cis isome

intramolecular hy

intermolecular hy

OR The cis isom

can, and thus, th

(c) (i) Li+, Na+ a

increases

which is

experienc

(ii)

(iii) The crowthe organiinstantansolvent.

(iv) HOOCCH

(v) In the organd the

aqueoushydration

(vi) AlBr3

(vii) AlBr3 will

AlBr3 + 3

19

C CHOOC

H COOH

H

C

H

OH

trans

not required)

r, the COOH groups are oriented close en

drogen bonding. Hence, the cis isomer has l

rogen bonding and a lower melting point.

er cannot pack as efficiently or closely as the

cis isomer has a lower melting point.

nd K+ belong to Group I. Down the group, n

but electrons are added to a higher principal

urther away from the nucleus. Hence, vale

weaker attractive force and ionic radius incre

ether complexes with the K+ ions and carriesc solvent since the CH2CH2 chains of the eous dipole-induced dipole interactions with

2CH2CH2CH2COOH

nic solvent, the K+ ions are complexed with thnO4 ions are unsolvated or naked (as

edium, where the MnO4

ions are surroundeshell.)

ydrolyse water instead.

2O Al(OH)3 + 3HBr

dative bonds

+

ugh to form

ss extensive

trans isomer

clear charge

uantum shell

ce electrons

ses.

the ions intoher can formthe organic

crown ether,compared to

by a dense


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20

C3 (a) (i) Titanium dioxide exists as oppositely charged ions held together bystrong ionic bonds / electrostatic forces.

Octocrylene exists as molecules held together by relatively weakervan der Waals forces.

More energy is required to overcome the stronger electrostatic forces

of attraction in titanium dioxide and thus it has a higher boiling point.

(ii) Phenylbenzimidazole sulfonic acid has polar OH and NH groups

that can form hydrogen bonds with water molecules. Thus, it is water-

soluble and is not a suitable sunscreen.

Octocrylenehas a large hydrophobic hydrocarbon chain that makes it

water-resistant, and thus, is a suitable sunscreen.

Examiners CommentsC2(b)

Many students did not draw diagrams to illustrate the proximity of COOH groups in the cis isomer or the moresymmetrical arrangement in the trans isomer.

A number of students mixed up the cis and trans isomers.

Most of the answers were lengthy and did not have the key points.

C2(c)(i)

For trends down a group, please do not explain using effective nuclear charge, which is for comparing trendsacross a period.

Some students mistook effective nuclear charge for nuclear charge, which is determined by the number ofprotons.

C2(c)(ii)

Many students could not draw the correct complex with six dative bonds. The cavity of the crown ether is notlarge enough to accommodate multiple K+ ions!

C2(c)(iii)

Most students wrongly explained the increased solubility solely based on overcoming the ionic bonds in KMnO 4.

The more important point here is the id-id interaction between the alkyl chains of the crown ether and thebenzene ring as the question asks why it is soluble in benzene.

Many students used the term hydrophobic, which is not accepted. Please describe the molecules as non-polarand the interactions as id-id (not hydrophobic interactions).

C2(c)(v)

Most students gave incorrect responses for this question. Many students suggested that the reactants did notdissolve as well in water, which does not explain the enhanced oxidising strength of purple benzene.

C2(d)(i)

Some students suggested CH3CH2+ as the Lewis acid, but the question requires identification of a reagent.

C2(d)(ii)

Most students described the hydrolysis reaction as AlBr3 dissociates in water. AlBr3 is a covalent compound, sohydrolysis describes the breaking of its covalent bonds more accurately. Dissociate is a term often used forionic compounds.


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21

(iii)

COO-Na+

O-Na+

O

HO

(iv) correct reagents and conditions + correct observations

Any one of the following tests.

Test 1:Reagents and conditions: 2,4-dinitrophenylhydrazineObservations: Orange ppt formed with oxybenzone and no orangeppt obtained with octocrylene.

Test 2:Reagents and conditions: SodiumObservations: With oxybenzone, effervescence of H2 was observed.Gas evolved gave a pop sound with a lighted splint.No effervescence of H2 gas observed with octocrylene.

Test 3:Reagents and conditions: KMnO4(aq), dilute H2SO4(aq), heatObservations: Decolourisation of purple MnO4

with octocrylene andno decolourisation of purple MnO4

with oxybenzone

Test 4:

Reagents and conditions: neutral FeCl3Observations: Purple colouration observed with oxybenzone and nopurple colouration observed with octocrylene

Examiners CommentsC3(a)(i)

Ionic bonding involves electrostatic attraction between ions, NOT attraction between molecules or betweenatoms. Confusing terms like intermolecular ionic bonding also surfaced.

Octocrylene is incapable of hydrogen bonding between its molecules.

C3(a)(ii)

Many students only focus on the solubility of phenylbenzimidazole sulfonic acid, but ignored octocrylene.

Some merely said that octocrylene is incapable of forming hydrogen bonding with water and hence it is insoluble.However, the focus should be on the bulky hydrophobic alkyl groups and the benzene ring. In fact, the nitrile andester groups are capable of hydrogen bonding with water molecules!

C3(a)(iv)

Some students used Br2(aq) with the intention of reacting it with the phenol group in oxybenzone. However, thepresence of an alkene in octocrylene would also bring about decolourisation of Br2.


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(b) (i) sp2

(ii)

(c) (i) 9Note: no. of orbitals in a sub-shell = 2l + 1, where l = angularmomentum quantum number. For s orbital, l = 0; p orbital, l = 1;dorbital, l= 2; forbital, l= 3; gorbital, l= 4. Hence, (2 x 4) + 1 = 9.

(ii) 18Note: Each orbital can accommodate 2 electrons, so 9 orbitals canaccommodate up to 18 electrons. Thus, there are 18 elements.

(iii) 4p 5s 4d 5p 6s 4f 5d(Use Aufbau Principle. You may refer to Understanding Advanced

Physical Inorganic Chemistry by Jeanne Tan & Kim Seng Chan.)

(iv) 6d 7p 8s 5g

(v) Two gelectrons

(d) 1 ppm = 1 m3 of gas in 106 m3 of air

100 ppm = 100 m3 of gas in 106 m3 of air

= 100/106 x 25.5

= 0.00255 m3 of gas in 25.5 m3 of air in laundry room

pV = nRT = mRT /Mr

m = pV (Mr) / RT = 101 x 103 x 0.00255 x 166 / (8.31 x 303) = 17.0 g

Examiners Comments3(d)

Some students cannot answer this question is because they did not convert ppm to m3 correctly.

Examiners CommentsC3(b)

The two molecules must be orientated in such a way that three H-bonds can be drawn.

Students often left out the lone pair of electrons on the donor atom.

The partial charges + and must be placed, respectively, on the H and N atoms of the H-N bond.

+

+

Documents

2012 Y6 H2 Chem T3 CT Answers _with Comments