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

(ii) Mass of NCS = 0.5 × 6.0 × 10–3

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

+ 2(16.0)] = 0.00200 g

(d)(i) R–CONH2 + Br2 R–CONHBr + HBr

(ii) +1

(iii) 8 electrons (2 inner electrons and 6 valence electrons)

(iv) The other product is CO32–

. R–N=C=O + 2OH– R–NH2 + CO3

2–

(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

2 (a)(i) Mr = (0.562 × 8.31 × 298) / (1.00 × 105 × 386 × 10

–6) = 36.1

(ii) carbon dioxide

(iii) Mr = 2(36.1 – 44/2) = 28.2 carbon monoxide

(iv) E: CaC2O4 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–.

(iii) [H+]2 = K1[HA] = (10

–2.85)(0.10) = 0.000141 mol

2 dm

–6

pH = –log10[H+] = 1.93

(iv) Key features of titration curve sketch:

Initial pH = 1.93 at 0 cm3 (as calculated in part (iii))

pH at MBC1 = pK1 = 2.85 at 5 cm3 ([HO2CCH2CO2

–] = [HO2CCH2CO2H])

pH at 1st equiv. point = 0.5 log10(10–5.70

)(0.050) = 3.50 at 10 cm3

pH at MBC2 = pK2 = 5.70 at 15 cm

3 ([HO2CCH2CO2

–] = [

–O2CCH2CO2

–])

pH at 2nd equiv. point = 14 + 0.5 log10(10–8.30

)(0.0333) = 9.11 at 20 cm3

pH beyond 2nd equiv. point = 14 + log10(0.001/0.04) = 12.40 at 30 cm3

(c)(i) F: O=C=C=C=O linear with respect to all atoms

(ii) NH3(g): H2NOCCH2CONH2

HCl(g): ClOCCH2COCl

(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 (a)(i) [Ca2+

] = 0.0080/40.1= 2.00 × 10–4

mol dm–3

[Mg2+

] = 0.0049/24.3 = 2.00 × 10–4

mol dm–3

[Cl–] = 0.0071/35.5 = 2.00 × 10

–4 mol dm

–3

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

–4 mol dm

–3

CaCl2 : MgCl2 : Ca(HCO3)2 : Mg(HCO3)2 = 1 : 1 : 3 : 3

(ii) calcium carbonate

When the sample was partially evaporated, [HCO3–] increased. This

caused the position of the equilibrium CO2(g) + H2O(l) + CO32–

(aq)

2HCO3–(aq) to shift to the left, forming more CO3

2–. Since CaCO3 is more

insoluble than MgCO3, the Ksp value for CaCO3 is smaller than that for

MgCO3. The increased [CO32–

] caused the ionic product [Ca2+

][CO32–

] to

increase to a value more than Ksp(CaCO3) first, with [Mg2+

][CO32–

]

remaining less than Ksp(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:

CO2(g) + H2O(l) + CO32–

(aq) 2HCO3–(aq)

When rainwater percolates through the rock, carbon dioxide reacts with

some of the dissolved CO32–

from the partial dissolution of calcium and

magnesium carbonates to form HCO3– as present in the mineral water.

The decrease in concentration of CO32–

causes the position of the next

equilibrium

MCO3(s) M2+

(aq) + CO32–

(aq) 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 + 9Ca3(PO4)2

(ii) Ksp = [Ca2+

]3[PO4

3–]2 units: mol

5 dm

–15

(iii) 1 × 10–26

= (3x)3(2x)

2 = 108x

5

[Ca2+

] = 3x = 3(2.47 × 10–6

) = 7.42 × 10–6

mol dm–3

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

removal of PCl3 as it is formed.

PCl5 can be made by reacting PCl3 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.

(c)(i) Fe2+

: [Ar] 3d6

Fe2+

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 d-

orbitals 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 d-

orbitals 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 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 which further

disperses the charge, stabilising the 4-nitrophenoxide ion even further.

(d)(i) I: Sn, concentrated HCl, 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

(ii) n(Fe2+

) = 11.3 × 10–3 × 0.500 = 0.00565 mol

ClOx– + 2xH

+ + 2xe Cl

– + xH2O

Mole ratio of ClOx– : e = 1 : 2x

0.150 / (39.1 + 35.5 + 16.0x) = 0.00565 / 2x x = 2

KClO2 + 4Fe2+

+ 4H+ KCl + 4Fe

3+ + 2H2O