AJC Prelims H2 Chem Paper 3 Answers

AJC H2 Chemistry Prelim 2008 - Paper 3 Suggested Answers 1 (a) - 3d and 4s electrons are of similar energies - lose different numbers of these electrons to form ions of similar stability (b) (i) Empirical formula of Y = MnO2 (ii) Z = MnO4 - Step II is a disproportionation reaction 3 MnO4 2- + 2H2O → 2 MnO4 - + MnO2 + 4OH- (iii) - aq CO2 is acidic / CO2 dissolves in water to form carbonic acid, - removes OH- as HCO3 - - Position of equilibrium shifts to the right, hence aids disproportionation (c) (i) Anode: Na → Na+ + e Cathode: MnO2 + H2O + e → MnO(OH) + OH- (ii) E•(MnO2/MnO(OH)) = - 1.71 V (d) (i) Since t½ is constant at about 1.5 mins, order of reaction w.r.t. H2O2 is 1. (iii) k = 0.478 min-1 (iv) k increases, because from the rate equation, if rate increases even when H 2O2 concentration remain the same, then value of k must also increase. (can explain using Arrhenius equation k = A e(-Ea/RT)) AJC/H2 Chemistry/Prelims2008/P3Answers 1

2 (a) (i) ∆Hsol = – 157 kJ mol–1 (ii) - The temperature of the water will increase. (iii) - ∆Hhyd of Ca2+ will be less exothermic than Cu2+. - From Data Booklet, ionic radius of Ca2+ is 0.099nm while Cu2+ is 0.069nm. - This is because Ca2+ has a larger ionic radius and thus smaller charge density than Cu2+. (b) (i) - In reaction I and II, 2 similar Cu–O bonds are broken and - 2 similar Cu–N bonds are formed . Thus, their ∆Hr { values are similar (ii) - In reaction I, one ethylenediamine replaces 2 H2O ligands. - There is an increase in the number of particles in aqueous solution and - thus increase in disorderness or ways of rearrangement. (iii) 1o r 1o r kJmol 5 . 43 = GΔ : II Reaction kJmol 9 . 60 = GΔ : I Reaction - - - ∆Gr { for reaction I is more negative than II. - Ligand exchange is more spontaneous in reaction I than II. - Thus, complex product for reaction I is more stable than II. (iv) - Ligand exchange with polydentate ligands and its analogous monodentate ligands gives similar ∆Hr { values due to formation of similar Cu−N bonds. - Polydentate ligands increases the disorderness of the system more compared to monodentate ligands. - Thus, the positive entropy is the factor which determines a more negative ∆Gr { value and formation of more stable polydentate complex. * mathematically, ∆H r { (reaction I) ≈ ∆Hr { (reaction II); Increase in disorderness ⇒ ∆Sr { (reaction I) > 0 JK–1mol–1 ∴∆Gr { (reaction I) more negative than ∆Gr { (reaction II) (c) (i) CC C N C N S SHH CC S C N C N SH H cis isomer trans isomer (ii) C atoms in C=C : sp2 hybridisation C atoms in C≡N : sp hybridization AJC/H2 Chemistry/Prelims2008/P3Answers 2

(iii) CC CNNC SS− − (iv) cis isomer The trans isomer is not possible as the S– are pointing oppositely and anionic X cannot form 2 dative bonds to the central metal ion. (v) Dative / coordinate bonds 3 (a) NaCl and MgCl2 are both ionic. Hence, they merely dissolve to give ions. (OR Mg2+(aq) undergoes hydrolysis to a very small extent ) AlCl3 and SiCl4 are simple covalent and hydrolyze in water to give white fumes of HCl and an acidic solution. AlCl3 + 6H2O ´ [Al(H2O)6]3+ + 3Cl- [Al(H2O)6]3+ ⇔ [Al(H2O)5(OH)]2+ + H+ or AlCl3 + 6H2O ⇔ [Al(H2O)5(OH)]2+ + H+ + 3Cl- SiCl4 + 2H2O ´ SiO2 + 4 H C l (b) (i) AlCl 3 + 3NaOH ´ Al(OH)3(s) + 3NaCl or Al 3+(aq) + 3OH-(aq) Æ Al(OH)3 (s) (ii) With MgCl2, a white ppt, insoluble in excess NaOH is observed. Al(OH)3 is amphoteric while Mg(OH)2 is basic. Al(OH)3 reacts with NaOH (to give salt and water), hence dissolves. (c) (i) Al-H bond is weaker than the B-H bond. (ii) H- acts as nucleophile. Alkenes are not attacked by nuclephiles as they are electron-rich. 0 1 2 3 4 5 6 7 NaCl MgCl2 AlCl3 SiCl4 pH AJC/H2 Chemistry/Prelims2008/P3Answers 3

(iii) A B C (iv) reaction II: NaCl, conc. H 2SO4, reflux / PCl5, room temp / SOCl2, room temp. reaction III: KCN in ethanol, reflux reaction IV: dilute H 2SO4 (or dilute HCl), reflux (d) (i) phenyl methanol ethanol (ii) Test: Add I 2(aq) in NaOH(aq) to each compound and warm. Ethanol gives a yellow ppt of CHI3. For phenyl methanol, no yellow ppt is observed. 4 (a) (i) Step I: free radical substitution Step II: redox Step III: nucleophilic substitution (ii) ∆H rxn = -117 kJ mol -1 (iii) Butanol and 4-chlorobutanol are assumed to be gaseous molecules when using bond energies in (a)(i). However, both actually exist as liquids under standard conditions. (iv) Cl C CH 2CH2CH2 H H O - δ+δ− CCl O H H CH2CH2CH2 - O + Cl- (v) The yield is likely to be low. Step 1 produces a mixture of chlorine substituted products , not all will undergo nucleophilic substitution to give THF. OR The nucleophile ClCH 2CH2CH2CH2O- may react with another of itself to give a straight chain product rather than a ring product. OH CN CH2OH CO C H 2CH3 O LiAlH4 CH2OH CH3CH2OH+ AJC/H2 Chemistry/Prelims2008/P3Answers 4

(b) CH=CH2 NH2 G: CH2CH2OH NH2 F: CH2COOH NH2 E: D: N O 5 (a) (i) H C R O C H N H H N C R O C - Shape of α- h e l i x - Show hydrogen bonds between C=O and N-H groups (ii) dilute HCl or dil NaOH, reflux (b) (i) Denaturation of proteins refer to the disruption of the shape of the protein molecule without altering its primary structure but resulting in the loss of biological activity. + (i i) I: Lactic acid dissociates to give H ions, thus lowering the pH. This disrupts the ionic bonds in the protein. -COO- + H+ Æ -COOH II. Ethanol disrupts the intermolecular hydrogen bonds. CH3 CH3H CH2O C H3 C H3 CH2 H O O C H C H3 CH2 O H C H3 CH2 O δ + δ - δ - δ + δ - δ - δ + δ + AJC/H2 Chemistry/Prelims2008/P3Answers 5

AJC/H2 Chemistry/Prelims2008/P3Answers 6 (c) (i) Glycine would be optically inactive but not alanine. Unlike alanine, glycine does not have a chiral carbon as it is bonded to 2 H atoms. (ii) I. Glycine has a much higher melting point than 1-aminobutane because the electrostatic attractions (ionic bond) in the crystal lattice of the zwitterionic form is stronger than the intermolecular hydrogen bonding between 1-aminobutane molecules. II. Propanoic acid has a higher melting point than 1-aminobutane. Both have hydrogen bonds between their molecules and similar Mr (similar strength of VDW) However, as oxygen is more electronegative than nitrogen, the hydrogen bonds between propanoic acid is stronger than those of 1-aminobutane. (d) (i) +2 (ii) No change in oxidation number of iron. This is a ligand exchange reaction. (e) Cu+ 1s22s22p63s23p63d10 Cu2+ 1s22s22p63s23p63d9 No vacant d orbital for Cu(I), no d-d transition For Cu(II), d electron to move from lower energy orbital to vacant higher energy orbital, energy is absorbed from the visible light region which is small in magnitude, hence its complementary colour is observed.