AJC H2 Chemistry 9647 N2012 P1 Suggested Solutions

©2013AndersonJC/CHEM 1 H2 Chemistry 9647 N2012 P1 Suggested Solutions 1 Molecular formula of glucose = C6H12O6 (from the diagram given, you should be able to observe that there are 4 hydroxyl groups, which will be replaced by chlorine atoms) C6H8O2(OH)4 C6H8O2Cl4 (hence, empirical formula = C3H4OCl2) C 2 Fe + 2Fe 3+ 3Fe2+ initial moles 1 2 0 change in moles –1 –2 +3 moles left 0 0 3 Given that reacti on goes to completion, you may comp lete the above table using each of the options to come up with the values below. n(Fe) n(Fe 3+) n(Fe 2+) A 0 0 3 B 0 1 3 C 0 3 3 D 0.5 0 4.5 Alternatively, for n(Fe2+) formed = 3 (start with the smallest whole no.) = n(Fe3+) left initial n(Fe) : n(Fe 3+) 1 2 + 3 1 5 (option C) C 3 Cr [Ar] 3d 5 4s1 Ge [Ar] 3d 10 4s2 4p2 S [Ne] 3s 2 3p4 S– [Ne] 3s 2 3p5 Sc [Ar] 3d 1 4s2 D 4 From the large increase in IE when the 3rd electron is removed, it can be deduced that there are only 2 val ence electrons present in X. Hence it is in Group II of the Periodic Table. B 5 Factors affecting strength of hydrogen bonds  extensiveness of HB  magnitude of polarity (i.e. di fference in electronegativ ity between F/O/N–H bond) Each water molecule can form, on average, 2 hydrogen bonds as compared with 1 hydrogen bond per HF molecule. C ANDERSON JC

©2013AndersonJC/CHEM 2 6 Due to the formation of hydrogen bonds between CO 2 and H 2O molecules, heat is evolved and H is < 0. As the forward reaction results in a decrease in no. of gaseous molecules, there will be less ways of arranging the particles and S is < 0. A 7 Electrolysis of molten CuCl2 Anode: 2Cl– Cl2 + 2e Electrolysis of aq. H2SO4 Anode: 2H2O O2 + 4H+ + 4e (preferentially discharged over SO42–) Since 2 1 V V 2 2 O Cl  for the same duration, 8 2 n n 2 2 O for passed e Cl for passed e  (4H2O 2O2 + 8H+ + 8e) and hence the current used in electrolysis 2 will be 4I (n  I as nF = It). D 8 2H 2(g) + CO(g) CH3OH(g) initial moles 2.0 1.0 0 change in moles –x – ଵ ଶ x + ଵ ଶ x moles left 2.0 – x 1.0 – ૚ ૛ x ૚ ૛ x C 9 Pyruvic acid is a weak acid wh ile NaOH is a strong alkali. The equivalence point of the titration will be above pH 7 because the salt formed CH3COCO2– hydrolyses in water to give a basic solution. CH3COCO2– + H2O CH3COCO2H + OH– A 10 When aspirin (HA) enter s the stomach where [H +] is very high (10–1 mol dm –3), it will remain effectively as HA due to the suppression of its ionisation in the presence of high [H +]. Hence, [HA] > [A–]. HA H+ + A– D 11 no. of moles of excess HCl = 0.0040 – 0.0025 = 0.0015 mol [H+] = [HCl] = 0.0015 1 = 0.00150 mol dm–3 pH = 2.82 D 12 When molar proportion of uranium to lead is 1 : 3, it can be deduced that the [uranium] has decreased to ¼ of its initia l concentration. Hence 2 half–lives have occurred. C ANDERSON JC

©2013AndersonJC/CHEM 3 13 rate = k [H2O2] (1st order reaction) If [H2O2] is doubled, rate is doubled. Hence, for 0.1M H2O2 0.01 mol dm –3 decomposed in 5 min (given that 10% decomposed) for 0.2M H 2O2 0.02 mol dm –3 decomposed in 5 min (as rate doubled) i.e. 0.02 0.2 x 100% = 10% decomposed Alternatively, Using the concept that “half–life is independent of the initial concentration”, you should be able to conclude that the time taken for H2O2 to decrease to the same extent (i.e. 10% in this case) will be the same regardless [H2O2] is 0.1 mol dm–3 or 0.2 mol dm–3. B 14 From the mechanism proposed, t he catalysed reaction should be a 2–step reaction (Option A or B). And since catalyst does not alter  H of reaction, option A is the answer. A 15 [Cu(H2O)6]2+ + 4Cl– CuCl42– + 6H2O This is a ligand exchange reaction. Hence the no. of d–electrons around copper remains the same but the energy gap between the d–orbitals changes as H 2O and C l– ligands give different extent of the d–orbital splitting. C 16 MgO dissolves in water to give a white suspension not aqueous solution, hence the reaction giving Mg(OH) 2(aq) does not take place readily. D 17 Since there is no large increase in IE when the 2nd or 3rd electron is removed, the element cannot be from Group I or II. The high melting point suggests the presence of gian t lattice structure, hence the element is not likely to be from Group VII too. All the above and the high density co nfirms that the element is likely to be a transition element. D 18 Under hot conditions, compound X is sodium chlorate(V). Recall: 3Cl2 + hot 6OH– ClO3– + 5Cl– + 3H2O C ANDERSON JC

©2013AndersonJC/CHEM 4 19 C CC C C H H HH H H 2sp 2sp 2sp3type of hybridisation 2sp2 2sp2 present in each C atom C 20 This is another way of asking how many different monochlorinated products will be obtained. The 3 are: CH2CH2C(CH3)3 CH3CHC(CH3)3 CH3CH2C(CH3)2CH2 C 21 PGE2 is converted to PGE 2 when the ketone in PGE 2 is reduced to the 2 o alcohol in PGE 2. However, not all reducing agents are suitable for use as there are ot her functional groups that can be reduced (e.g. alkenes and acids). Hence only reducing agent specific for reduction of aldehyde and ketone (NaBH4) can be used. D 22 When cholesterol reacts with cold, dilute KMnO4, the likely product is HO H H3C H H CH3 H CH3 OH OH W: Correct X: Wrong Y: Wrong Z: Correct B 23 In this case, the transition stat e will have an overall negative charge (RX is neutral while nucleophile is anionic ). Hence th e charges on both N and L should be – while that on the reactive carbon is +. B 24 Presence of C–Cl or C–Br bonds in compound will result in formation of halogen radical which destroy the ozone and are not suitable replacement. C–H and C–F are relatively inert! D A D B * ** ** *** * * ANDERSON JC

©2013AndersonJC/CHEM 5 25 Hydrolysis of rosmarinic acid will give O O O O O O O HO O Na NaNa Na Na Na C 26 A Addition of Br 2(aq) produce either (CH3)2C(OH)CH(Br)CH2CH2COCH3 (major) or (CH3)2C(Br)CH(Br)CH2CH2COCH3. Both contain a chiral centre. B Prolonged heating with conc. KMnO 4/H+ produces (CH 3)2CO and HO2CCH2CH2COCH3, NOT HO2CCH2CO2H. C Reduction by NaBH 4/methanol produces (CH3)2C=CHCH2CH2CH(OH)CH3 (C8H16O), NOT C8H18O. D Warming with I2/OH– produces (CH3)2C=CHCH2CH2CO2–, NOT CH3CO2H. A 27 –helix is a type of secondary structure protein and is stabilised by intramolecular hydrogen bonding between C=O and N–H of the peptide bonds. C 28 NO2 NH3 + NH2 Sn, c.HCl NaOH the basic phenylamine will react with the H+ B 29 RC O NH2 RC O NH2 C NH2 O C NH2 O The lone pair of e on the N atom is delocalised into the carbonyl group and hence it is less available in accepting a H+ (less basic) (Option C is a correct statement but it does not explain why amide is less basic.) B ANDERSON JC

©2013AndersonJC/CHEM 6 30 NH2 CH3 NH2 CH3 HBr Br NH2 CH3 Br Br Br + HBr The substitution occurs at 2 nd and 4 th position with respect to –NH 2 group rather than –CH 3 group because –NH 2 group is strongly activating and more reactive. A 31 Each carbon atom in the graphite lattice is bonded to 3 other carbon atoms. The unpaired electron presen t in each of the unhybridised p orbital of the carbon atom s allows the delocalisation of electrons in the lattice (along the plane). The 3  and 1  bond formed by each carbon atom in the lattice give it a valency of 4. B 32 Factors affecting mass of copper deposited  time  magnitude of current D 33 By observation comparing expt 1 and 2, order of reaction wrt 1–bromoethane can be deduced to be 1. (when [1–bromoethane] is doubled, rate of reaction doubles). By observation comparing expt 1 and 4, order of reaction wrt HS– can be deduced to be 1. (when [HS –] is doubled, rate of reaction doubles). Hence both 1–bromoethane and HS – will be involved in the rate– determining step i