2011 H2 Chemistry Paper 3 Suggested Solutions

2011 H2 CHEMISTRY 9647/03 SUGGESTED SOLUTIONS

1

 –

(a)

Let the oxidation number of iodine in ICl4 be x .  –1 = –4 + x  x  = +3

(b)

ClF5: 5 bond pairs; 1 lone pair; square pyramidal  –

ICl4

: 4 bond pairs; 2 lone pairs; square planar +

 –

(c)(i) HClO + H + 2I



 –

Cl + I2 + H2O  –3

(ii) Mass of NCS = 0.5 × 6.0 × 10 + 2(16.0)] = 0.00200 g

× 0.0050 × [4(12.0) + 4(1.0) + 35.5 + 14.0

(d)(i) R –CONH2 + Br2  R –CONHBr + HBr (ii) +1 (iii) 8 electrons (2 inner electrons and 6 valence electrons) 2 –

(iv) The other product is CO 3 . (e)

V: FeBr3 /Br2, room temperature VI: KMnO4(aq)/H2SO4(aq), heat VII: PCl5, room temperature VIII: NH3, room temperature C: 4-bromobenzoic acid D: 4-bromobenzamide

 –

R –N=C=O + 2OH



R –NH2 + CO3

2 –

2

5

 –6

(a)(i) M r = (0.562 × 8.31 × 298) / (1.00 × 10 × 386 × 10 ) = 36.1 (ii) carbon dioxide (iii) M r = 2(36.1 – 44/2) = 28.2 (iv) E: CaC2O4

carbon monoxide

CaC2O4  CaO + CO + CO2  –

(b)(i) HO2CCH2CO2 has an electron-withdrawing  –CH2CO2H which disperses the negative charge, in addition to resonance stability (electron delocalisation) of the carboxylate ion, making it more stable than the ethanoate ion (which only has resonance stability). (ii)

 –

 –

O2CCH2CO2 is more negatively charged (2 –) compared to ethanoate ion (1 –), which makes it more likely to accept a proton back to form  –

HO2CH2CO2 . +2

 –2.85

(iii) [H ] = K 1[HA] = (10

2

 –6

)(0.10) = 0.000141 mol dm

+

pH = –log10[H ] = 1.93 (iv) Key features of titration curve sketch: 3

Initial pH = 1.93 at 0 cm (as calculated in part (iii)) 3

 –

pH at MBC1 = pK 1 = 2.85 at 5 cm ([HO2CCH2CO2 ] = [HO2CCH2CO2H])  –5.70

pH at 1st equiv. point = 0.5 log 10(10

)(0.050) = 3.50 at 10 cm

3

 –

3

 –

 –

pH at MBC2 = pK 2 = 5.70 at 15 cm ([HO2CCH2CO2 ] = [ O2CCH2CO2 ])  –8.30

pH at 2nd equiv. point = 14 + 0.5 log 10(10

)(0.0333) = 9.11 at 20 cm

pH beyond 2nd equiv. point = 14 + log 10(0.001/0.04) = 12.40 at 30 cm

(c)(i) F: O=C=C=C=O

linear with respect to all atoms

(ii) NH3(g): H2NOCCH2CONH2

HCl(g): ClOCCH2COCl

3

3

(d)

BrCH2CH2Br  HO2C –CO2H 1. NaOH(aq), heat 2. K2Cr2O7(aq)/H2SO4(aq), heat BrCH2CH2Br  HO2C –CH2CH2 –CO2H 1. ethanolic KCN, heat 2. H2SO4(aq), heat

3

2+

 –4

(a)(i) [Ca ] = 0.0080/40.1= 2.00 × 10 2+

 –4

[Mg ] = 0.0049/24.3 = 2.00 × 10

 –4

 –

[Cl ] = 0.0071/35.5 = 2.00 × 10

 –3

mol dm

 –3

mol dm

 –3

mol dm

 –4

 –

[HCO3 ] = 0.0366 / [1.0 + 12.0 + 3(16.0)] = 6.00 × 10

 –3

mol dm

CaCl2 : MgCl2 : Ca(HCO3)2 : Mg(HCO3)2 = 1 : 1 : 3 : 3 (ii) calcium carbonate  –

When the sample was partially evaporated, [HCO 3 ] increased. This 2 –

caused the position of the equilibrium CO 2(g) + H2O(l) + CO3 (aq)



2 –

 –

2HCO3 (aq) to shift to the left, forming more CO 3 . Since CaCO3 is more insoluble than MgCO 3, the K sp sp value for CaCO 3 is smaller than that for 2 –

MgCO3. The increased [CO 3 ] caused the ionic product [Ca

2+

2 –

][CO3 ] to 2+

2 –

increase to a value more than K sp sp(CaCO3) first, with [Mg ][CO3 ] remaining less than K sp sp(MgCO3). Thus, CaCO3 is precipitated first. (iii) The rock is probably made up of calcium carbonate and magnesium carbonate. Rainwater contains dissolved carbon dioxide and the following equilibrium is established: 2 –

 –

CO2(g) + H2O(l) + CO3 (aq)  2HCO3 (aq) When rainwater percolates through the rock, carbon dioxide reacts with some of the dissolved CO 3

2 –

from the partial dissolution of calcium and  –

magnesium carbonates to form HCO 3 as present in the mineral water.

The decrease in concentration of CO 3 equilibrium 2+

2 –

2 –

MCO3(s)  M (aq) + CO3 (aq)

causes the position of the next

where M = Ca or Mg

to shift to the right, enabling more carbonate to dissolve to produce the metal cations present in the mineral water. (b)(i) 6Ca5(PO4)3F + SiO2 + H2O  SiF4 + 2HF + 3CaO + 9Ca 3(PO4)2 2+ 3

3 – 2

5

(ii) K sp sp = [Ca ] [PO4 ]  –26

(iii) 1 × 10

3

 –15

units: mol dm

2

5

= (3x ) (2x ) = 108x 

2+

 –6

 –6

[Ca ] = 3x  = 3(2.47 × 10 ) = 7.42 × 10

 –3

mol dm

(c)(i) PCl3 can be made by refluxing chlorine with phosphorus, with continuous

removal of PC l3 as it is formed. PCl5 can be made by reacting PC l3 further with excess chlorine. (ii) G: POCl3

tetrahedral with respect to central P

(iii) H: sulfur dioxide

PCl3 + SO3  POCl3 + SO2 PCl5 + H2O  POCl3 + 2HCl

4

(a)

The secondary structure of proteins is the arrangement of a polypeptide chain in space around a single axis. It is formed and stabilised by the interactions of amino acids that are fairly close to one another on the polypeptide chain, through hydrogen bonding between C=O group of one peptide and N-H group of another. The tertiary structure of proteins refers to the way in which the polypeptide with its primary and secondary structures can be organised in space to form a more complex polypeptide configuration through the formation of interactions between the R-groups of the amino acids, such as hydrogen bonding, disulfide bridges, van der Waals’ forces and ionic linkages. The quaternary structure of proteins consists of more than one polypeptide chain coming together to form the complete protein maintained by the same forces which are responsible for tertiary structure.

(b)

ΔG  should

be negative because the reaction is spontaneous.

ΔH 

should be negative due to the formation of various interactions (e.g. hydrogen bonding, disulfide bridges, van der Waals’ forces and ionic linkages) between the R-groups of the constituent amino acids, which is an exothermic process.

ΔS 

should be negative due to the coalescence of a more ordered haemoglobin molecule, which reduces the number of ways in which the energy can be distributed. 2+

(c)(i) Fe : [Ar] 3d 2+

6

Fe has incompletely filled d-orbitals. In an octahedral environment, the d-orbitals are split into two energy levels. When white light shines, the electron in a d-orbital of the lower energy level will absorb a quantum of light and jump to one of the partially filled d-orbitals of higher energy. The energy gap E between the two levels corresponds to a range of wavelengths in the visible spectrum. The red colour observed is the complementary colour of the light absorbed during this d-d transition.

(ii)

‘high spin’ state

‘low spin’ state

(iii) Electrons prefer to occupy orbitals singly to minimise inter-electronic repulsion. (iv) Oxyhaemoglobin (‘low spin’ state) will contain the larger energy gap, E, between its d-orbitals.

Where the energy gap is larger, the energy required to overcome the energy gap, E, in adding subsequent electrons to the higher energy dorbitals is more than that required to overcome inter-electronic repulsion in adding electrons to the partially filled lower energy d-orbitals. Thus, the lower energy d-orbitals are filled first, before higher energy d-orbitals are used (i.e. ‘low spin’ state). Conversely, where the energy gap is smaller, the energy required to overcome E in adding subsequent electrons to the higher energy dorbitals is less than that required to overcome inter-electronic repulsion in adding electrons to the partially filled lower energy d-orbitals. Thus, the electrons occupy all the d-orbitals singly, before pairing up in the lower energy d-orbitals (i.e. ‘high spin’ state). (d)(i) Acidic/alkaline conditions, heat. +

 –

H  /OH acts as a catalyst during hydrolysis. (ii) met-asp-gly-ser-ala-gly-glu-ser-lys-tyr

5

(a)

Acidity increases from ethanol to water to phenol. The electron-donating ethyl group in ethanol intensifies the negative charge on the ethoxide ion, compared to hydroxide (from water). This destabilises the ethoxide ion, making ethanol a weaker acid than water. The p-orbital of oxygen overlaps with the π -electron -electron cloud of the benzene ring. The negative charge on oxygen is delocalised into the benzene ring. This dispersion of charge stabilises the phenoxide ion, making phenol a stronger acid than water.

(b)

dilute nitric acid, room temperature

(c)

4-nitrophenol is likely to be more acidic than phenol due to the presence of the electron-withdrawing  –NO2 group on the para -position -position which further disperses the charge, stabilising the 4-nitrophenoxide ion even further.

(d)(i) I: Sn, concentrated HC l, heat; then NaOH(aq), room temperature II: excess NaOH(aq), room temperature IV: CH3COCl, room temperature (ii) I: reduction II: deprotonation / acid-base / neutralisation III: nucleophilic (aliphatic) substitution (bimolecular) IV: nucleophilic (acyl) substitution (iii) To make it a better nucleophile (e)

Cl2(aq), room temperature

(f)(i)

3Cl2 + 6NaOH  NaClO3 + 5NaCl + 3H2O 2+

(ii) n (Fe (Fe ) = 11.3 × 10 –3 × 0.500 = 0.00565 mol  –

+

 –

ClOx  + 2x H + 2x e  Cl + x H2O  –

Mole ratio of C lOx  : e = 1 : 2x  0.150 / (39.1 + 35.5 + 16.0 x ) = 0.00565 / 2 x   x  = 2 KClO2 + 4Fe

2+

+

+ 4H



KCl + 4Fe

3+

+ 2H2O