90 o. (Note This could be worked out by analogy with N-chlorosuccinimide in (c), where chlorine clearly has an oxidation number of +1.

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1 1 Victoria Junior ollege Suggested Answers November /03 (ree Response) 1(a) +3 (b) I 90 o I 90 o Since I atom has 4 bond pairs and 2 lone pairs, the shape of I 4 - will be square planar. 90 o 90 o Since atom has 4 bond pairs and 1 lone pair, the shape of 5 will be square pyramidal. (c)(i) H + H + + 2I I 2 + H 2 (ii) Amount of S formed = 6.0 / = Amount of I 2 formed = 1 / 2 ( ) = Mol ratio of N-chlorosuccinimide : H : I 2 = 1 : 1 : 1 M r of N-chlorosuccinimide ( 4 H 4 N 2 ) = Mass of N-chlorosuccinimide formed = = g (d)(i) R-NH 2 + Br 2 RNHBr + HBr (ii) +1 (iii) 6 (Note This could be worked out by analogy with N-chlorosuccinimide in (c), where chlorine clearly has an oidation number of +1.)

2 2 (iv) 3 2-2H - + R-N== R-NH (Note 2 cannot be formed because in the presence of H -, 2 would form H 3 - or 3 2-.) (e) H 3 H 3 2 H NH 2 V VI VII VIII Br 2, anhydrous AlBr 3 Br KMn 4, H 2 S 4 (aq) Br P 5, room temperature Br NH 3, room temperature Br D 2(a)(i) pv = nrt ( ) ( ) = n(8.31)(298) n = = ( ) / M r M r = 36.1 (ii) 2 (iii) Volume of 2 and the other gas is the same. Hence, by Avogadro s hypothesis, mol ratio of 2 to other gas is 1 : = ( ) / 2 = 28.0 Gas is. (iv) E : a 2 4 a 2 4 a (b)(i) Malonic acid has an electron-withdrawing - 2 H group that helps to disperse negative charge on the conjugate base which is the monoanion, and stabilise it (or the monoanion is stabilised by hydrogen bonding with the unionised - 2 H group). Hence acid dissociation is favoured, leading to a higher K a and lower pk a.

3 3 H 2 (ii) δ + H - δ - Hydrogen bonds The removal of H + from a species that already carries a negative charge would be electrostatically unfavourable (or the stabilising hydrogen bonding would be destroyed by the ionisation of the second 2 H) and cause the pk 2 to be higher than pk 1 of ethanoic acid. (iii) H + = [(0.10)( )] ½ = moldm -3, ph = -lg (0.0119) = 1.93 (iv) ph Volume of NaH (cm 3 ) Initial ph nly malonic acid present. Hence, ph = -lg(k a c) ½ = 1.93 irst Equivalence Point 10cm 3 of 0.1 moldm -3 NaH required to neutralise 10cm 3 of 0.1 moldm -3 malonic acid irst Maimum Buffering apacity Half the equivalence point i.e 5cm 3. ph = pk 1 = 2.85

4 4 Second Equivalence Point Another 10 cm 3 of NaH required for second neutralisation. Second Maimum Buffering apacity i.e 15cm 3, ph = pk 2 = 5.70 Addition of Ecess NaH(aq) After NaH (aq) reacts completely with malonic acid, there will be ecess of NaH (aq). The ph of the solution approaches 13. (c)(i) ====, linear (ii) H 2 H 2 H 2 N NH 2 (Note The hint given in the question was that the δ+ end of the nucleophilic water adds onto the central carbon, and the δ- H adds onto the = group giving malonic acid.) (d) BrH 2 H 2 Br NaH(aq) NaN(alc) H + /K 2 r 2 7 HH 2 H 2 H NH 2 H 2 N 2 N H 2 S 4 (aq) H 2-2 H H 2 -H 2 H 2 2 H (Note NH 2 H 2 N represents an iso-nitrile which is incorrect.)

5 5 3(a)(i) Ion oncentration /moldm -3 a Mg Salt oncentration /moldm -3 (Alternatively) oncentration /moldm -3 Mg a a(h 3 ) H Mg(H 3 ) Note 1. The ratio of M 2+ to - present is 1:1. If M 2+ is present in two salts, then M 2+ that forms M 2 is half the concentration of - (i.e. [ - ] = , [M 2+ ] = ) because of the mole ratio. (M 2 M , M = Mg or a) 2. The remaining moldm -3 of M 2+ will be involved in forming the salt M(H 3 ) 2 (i.e. [H 3 - ] = , [M 2+ ] = ) with concentration of (ii) a 3 will precipitate first. As the sample was partially evaporated, water was removed. This shifts the position of equilibrium to the left, producing more 3 2- ions. Since salts of Group II becomes more insoluble down the group, K sp of a 3 < K sp of Mg 3, hence the ionic product of a 3 will eceed K sp first, and precipitate out earlier. (iii) omposition of the rock : Mg 2, Mg 3 and a 3 The presence of atmospheric 2 dissolved in rainwater will shift the - equilibrium to the right and this caused more H 3 to be formed in the mineral water. The chloride rocks would merely dissolve. (b)(i) a = 1, b = 1, c = 1, d = 2, e = 3, f = 9 (ii) K sp = [a 2+ ] 3 [P 3-4 ] 2 mol 5 dm -15

6 6 (iii) (3) 3 (2) 2 = = moldm -3 [a + ] = = mol/dm 3 (c) (i) P 4 (s) (g) 4P 3 (l) Small amounts of chlorine gas passed over white phosphorus. Phosphorus burns and P 3 is distilled over. P 4 (s) (g) 4P 5 (s) Ecess chlorine gas is passed over solid P 4 to produce P 5. (ii) ompound G : P 3 Tetrahedral (iii) By-product H and S 2 P 3 + S 3 P 3 + S 2 H P 5 + H 2 P 3 + 2H 4(a) The secondary structure refers to the regular arrangement of the polypeptide chain, stabilised by hydrogen bonds formed between the N-H and = groups of the peptide groups in the polypeptide backbone. In the tertiary structure of protein, a polypeptide chain folds through the interactions between the side chains of the amino acids. The types of side chain interactions in tertiary structures are ionic interaction between 2 - and NH 3 + groups, hydrogen bonding or dipole-dipole interactions between polar groups, van der Waals interactions between non-polar side chains, and disulfide bridges (-S S-) between groups containing thiol (-SH). The quaternary structure refers to the precise arrangement of 2 or more polypeptide chains (also known as sub-units) held together by -R group interactions such as Van der Waals forces, ionic bonds and hydrogen bonds. (b) Since the folding process is spontaneous, ΔG is negative. As bonds are being formed during the formation of secondary, tertiary and quaternary structures, ΔH is negative. The degree of disorder is decreasing as the chains coil or come together, therefore ΔS is also negative.

7 7 (c)(i) Presence of partially filled 3d orbitals allows an electron in the lower energy 3d orbital to absorb energy from the visible light spectrum, and be ecited to the higher energy d orbital. The colour observed (red) is the complement of the colour absorbed (green). (ii) High Spin State Low Spin State (iii) Electrons are negatively charged, and will eperience electronic repulsion when paired together. ccupying the orbitals singly helps to minimise repulsion. (iv) yhaemoglobin will contain a larger energy gap, E as electrons would only pair up in the lower orbital if the energy gap was greater than the repulsion energy. (d) (i) H (aq) or NaH(aq), for several hours (ii) met-asp-gly-ser-ala-gly-glu-ser-lys-tyr Hydrolysis of the third, sith and ninth peptide bond will give rise to met-aspgly, ser-ala-gly and glu-ser-lys respectively. Hydrolysis of the first and fourth peptide bond will give rise to asp-gly-ser. Hydrolysis of the fifth and eighth peptide bond will give rise to gly-glu-ser. Hydrolysis of the seventh peptide bond will give rise to ser-lys-tyr. 5 (a) HA + H 2 A - + H 3 + The greater the ability of the anion to disperse the negative charge, the more stable the anion. The forward acid dissociation reaction will be favored hence resulting in a stronger acid. or H 3 H 2 H, -H 3 H 2 eerts an electron-donating inductive effect and intensifies the negative charge on - and destabilises the anion, H 3 H 2 -.

8 8 or phenol, the p orbital of the oygen atom overlaps with the π-electron cloud of the benzene ring. This decreases the intensity of the negative charge on - and stabilises the anion, 6 H 5 -. Strength of acids : Phenol > water > ethanol (b) Reagent : Dilute HN 3 (aq) ondition : Room temperature (c) The presence of electron-withdrawing N 2 group will decrease the intensity of the negative charge on -, stabilise the anion and favour the acid dissociation, resulting in 4-nitrophenol being a stronger acid than phenol. (d) (i) H H - 2 H 5 2 H 5 N 2 I II III IV Sn, conc. H, followed by NaH(aq) NH 2 NaH(aq) room temperature NH H 5 Br NH 2 H 3 room temperature NHH 3 (ii) Step I : Reduction Step II : Acid-base Step III : Nucleophilic substitution Step IV : Nucleophilic substitution (iii) The formation of the phenoide anion would make K a better nucleophile than J. (e) Reagents : 2 (aq) onditions : Room temperature (f) (i) NaH 5Na + Na 3 + 3H 2 (ii) Amount of e 2+ = 11.3/ = mol Amount of electrons lost by e 2+ = (since e 3+ + e - e 2+ ) K K 2 K 3 Amount of electrons 2 4 6

9 9 gained per to form - Amount of compound required or 0.150g of K, molecular mass of K Mass of present Therefore = 2, 4e H e H 2 +