Topics in the June 2009 Exam Paper for CHEM1102

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1 June 009 Topics in the June 009 Exam Paper for CEM110 ick on the links for resources on each topic. 009-J-: 009-J-3: 009-J-4: 009-J-5: 009-J-6: 009-J-7: 009-J-8: 009-J-9: 009-J-10: 009-J-1: 009-J-13: Periodic Trends in Aqueous xide Coordination Chemistry Strong Acids and Bases Kinetics - Catalysis Physical States and Phase Diagrams Crystal Structures Coordination Chemistry Weak Acids and Bases Calculations Involving pka Kinetics Stereochemistry Alkenes Alcohols rganic alogen Compounds Carboxylic Acids and Derivatives Carboxylic Acids and Derivatives Stereochemistry Alkenes Synthetic Strategies

2 CEM J- June 009 Explain in terms of their electronic configurations and ionisation energies why the halogens (Group 17) are powerful oxidising agents. xidising agents are themselves reduced (i.e. they gain electrons). The electronic configuration of the halogens (Group 17) is np 5. They are small atoms (atomic size decreases across a period as the nuclear charge increases). In each period, the halogen is the element with the highest number of protons in the nucleus that also has an incomplete shell. As a result, they will readily gain a single electron to form the X ion. Similarly, the high nuclear charge and small size means that they have high ionisation energies. ence halogens are poor reducing agents. Compounds of d-block elements are frequently paramagnetic. Using the box notation to represent atomic orbitals, account for this property in compounds of Co +. A cobalt atom has 9 valence electrons: it is in Group 9 of the Periodic Table. A Co + cation has lost electrons so has (9 ) = 7 valence electrons. The configuration of Co + is 3d 7. These occupy the five d-orbitals to maximize the number of unpaired electrons (to minimize the electron-electron repulsion): 3d Co + (d 7 ) Paramagnetism arises as a result of the presence of 1 or more unpaired electrons. Co + has 3 unpaired electrons and so is paramagnetic.

3 CEM J-3 June 009 BF 3 is a Lewis acid in its reaction with diethyl ether. Explain what is meant by a Lewis acid and draw the product of this reaction. A Lewis acid is an electron pair acceptor. BF 3 possess an empty p-orbital on B. C 3 C C C 3 possess a lone pair on. BF 3 reacts with C 3 C C C 3 by accepting this pair of electrons: BF 3 acts as a Lewis acid and C 3 C C C 3 acts as a Lewis base. F F F B What is a catalyst and, in general terms, how does it work? Make reference to an energy level diagram in your answer. A catalyst is a substance that increases the rate of a reaction without being consumed in the reaction. A catalyst works by providing an alternative reaction pathway of lower activation energy, E a. potential energy E a E a reactants reaction coordinate uncatalysed reaction catalysed reaction products The gas methane, C 4, has a critical point at 8 C and 46 atm. Can methane be liquefied at 5 C? Explain your answer. The critical temperature (T c ) is the temperature above which a substance cannot exist as a liquid. Thus, methane cannot be liquefied above -8 C and so cannot be liquefied at 5 C. Define what is meant by an allotrope. Give an example of a pair of allotropes involving (i) oxygen and (ii) a pair not involving oxygen. Allotropes are different structural forms of the same element. (i) Dioxygen,, and ozone, 3, are allotropes of oxygen. (ii) Examples include (a) white and red phosphorus, (b) graphite, diamond and fullerene for carbon and (c) rhombic, monoclinic and amorphous sulfur.

4 CEM J-4 June 009 Complete the following table. (en = ethylenediamine = N C C N ) Formula K [Co 4 ] Na 3 [FeBr(CN) 5 ] [Zn(en) (N 3 ) ] 9 xidation state of transition metal ion Coordination number of transition metal ion Number of d-electrons in the transition metal ion + (II) +3 (III) + (II) (Co is in Group 9 so Co + has 9 = 7) 5 (Fe is in Group 8 so Fe 3+ has 8 3 = 5) 6 (4 N from en and from N 3 - ) 10 (Zn is in Group 1 Zn + has 1 = 10) Charge of the complex ion 3 0 [Co 4 ] - [FeBr(CN) 5 ] 3- [Zn(en) (N 3 ) ] Geometry of the complex ion List all the ligand donor atoms tetrahedral octahedral octahedral 4 1 Br and 5 C 4 N and

5 CEM J-5 June 009 Solution A consists of a 0.00 M aqueous solution of propionic acid, C 3 6, at 5 C. Calculate the p of Solution A. The pk a of propionic acid is As C 3 6 is a weak acid, [ + ] must be calculated by considering the equilibrium: C 3 6 C initial change -x +x +x final 0.00 x x x The equilibrium constant K a is given by: K a =!!!!!!! [!! ] [!!!!!! ]!! = (!.!"!!!) As pk a = 4.87, K a = K a is very small so 0.00 x ~ 0.00 and hence: x = or x = M = [ + ] ence, the p is given by: p = log 10 [ + ] = log 10 [ ] = 3.8 Answer: p = 3.8 At 5 C, 1.00 L of Solution B consists of.4 g of potassium propionate (KC 3 5 ) dissolved in water. Calculate the p of Solution B. The molar mass of KC 3 5 is: molar mass = (39.10 (K) (C) () ()) g mol -1 = g mol-1 Thus,.4 g corresponds to: number of moles =!"## =!!.!"!!!"#$%!!"##!!".!"!!!!"!!! = mol If this is dissolved in 1.0 L, [C ] initial = M. As C is a weak base, [C ] must be calculated by considering the equilibrium: ANSWER CNTINUES N TE NEXT PAGE

6 CEM J-5 June 009 C C initial large 0 0 change -y negligible +y +y final y large y y The equilibrium constant K b is given by: K b =!!!!!! [!"! ] [!!!!!!! ] =!! (!.!"!!!!) For an acid and its conjugate base: pk a + pk b = pk b = = 9.13 As pk b = 9.13, K b = K b is very small so y ~ and hence: y = or y = M = [ - ] ence, the p is given by: p = -log 10 [ - ] = log 10 [ ] = 5.41 Finally, p + p = so p = = 8.59 Answer: p = 8.59 Solution B (1.00 L) is poured into Solution A (1.00 L) and allowed to equilibrate at 5 C to give Solution C. Calculate the p of Solution C. Combining the two solutions will double the overall volume, to.00 L. As a result, the concentration of both the acid and base will halve: [acid] = M and [base] = M. The solution contains a weak acid and its conjugate base. The p of this buffer solution can be calculated using the enderson-asselbalch equation: p = pk a + log [!"#$] [!"#$] = log!.!"!!!.!"! = 4.87 Answer: p = 4.87 If you wanted to adjust the p of Solution C to be exactly equal to 5.00, which component in the mixture would you need to increase in concentration? More base is needed: add KC 3 5

7 CEM J-6 June 009 Peroxydisulfate and iodide ions react according to the following equation. S 8 (aq) + 3I (aq) S 4 (aq) + I 3 (aq) The following rate data were collected at room temperature. 3 Experiment [S 8 (aq)] 0 (M) [I (aq)] 0 (M) Initial rate (mol L 1 s 1 ) Determine the rate law for the reaction. Between experiments (1) and (), [S 8 - (aq)] 0 is constant and [I - (aq)] 0 is halved from M to M. This halves the initial rate. The rate is directly proportional to [I - (aq)]: the reaction is first order with respect to [I - (aq)]. Between experiments () and (3), [I - (aq)] 0 is constant and [S 8 - (aq)] 0 is doubled from M to M. This doubles the initial rate. The rate is directly proportional to [S 8 - (aq)]: the reaction is first order with respect to [S 8 - (aq)]. ence: rate = k[s 8 - (aq)][i - (aq)] Calculate the value of the rate constant at room temperature. From experiment (1), the rate = M s -1 when [S 8 - (aq)] = M and [I - (aq)] = M. ence: M s -1 = k(0.080 M)(0.034 M) k = M -1 s -1 (Note that the units are worked out by requiring that the units on the two sides of the rate law are the same.) Answer: M -1 s -1

8 CEM J-7 June 009 Consider the following pairs of compounds. Indicate the isomeric relationship that exists between the compounds in each set. 8 constitutional isomers (different connectivity) (A) diastereoisomers (same connectivity but different 3D arrangement) NMe (B) NMe enantiomers (non-superimposable mirror images) C 3 C 3 conformational isomers (can be interconverted by rotation of the central C-C bond) C C Et C C Et diastereoisomers (same connectivity but different 3D arrangement. Not enantiomers) (C) Give the full name of compound (A) that unambiguously describes its stereochemistry. (E)--pentene. (The alkyl groups are on opposite sides of the C=C hence (E)). What is the configuration of the stereogenic centre in compound (B)? 1 4 N Me 3 (B) The priorities are indicated on the figure. With the lowest priority at the back, the sequence 1--3 is in a clockwise direction: (R). Is compound (C) a meso isomer? Give a reason for your answer. No. C does not have a plane of symmetry so is optically active.

9 CEM J-8 June 009 Complete the following table. Make sure you indicate any relevant stereochemistry. STARTING MATERIAL REAGENTS/ CNDITINS CNSTITUTINAL FRMULA(S) F MAJR RGANIC PRDUCT(S) 6 /Pd/C (catalyst) 1. dilute Na. C 3 Br Cr 7 / aqueous S 4 Br 1. Mg / dry ether. C 1. NaB 4. + / C The final product may react further: C which might undergo intramolecular esterif ication to give the cyclic ester

10 CEM J-9 June 009 Give the constitutional formula(s) of the organic product(s) formed when each of the following compounds is treated with 4 M sodium hydroxide. The first three reactions proceed at room temperature; the last two require heating. 7 CMPUND C 3 C C RGANIC PRDUCT(S) C 3 C C C 3 C 3 N 3 C N C 3 C 3 Br C 3 3 C C C 3 C 3 C C C 3 C Na 3 C C 3 C C Na + Na N Na + N

11 CEM J-9 June 009

12 CEM J-10 June 009 1,-Dichloropropane can exist in two enantiomeric forms, compounds I and II. In the boxes below draw structures of the two enantiomers of 1,-dichloropropane clearly showing the stereochemistry at the chiral carbon. compound I compound II 5 C l (S )-en a n tio mer (R)-enantiomer There are three other compounds, III, IV and V with molecular formula C 3 6. In the boxes below, give the constitutional formulas and names of these compounds. Structure Name compound III,-dichloropropane compound IV compound V 1,1-dichloropropane 1,3-dichloropropane

13 CEM J-11 June 009 Compounds I, II, III, IV and V are isomers. From the list enantiomers, diastereomers, conformers, constitutional isomers complete the following table. 4 PAIR F CMPUNDS I and III ISMERIC RELATINSIP BETWEEN PAIR F CMPUNDS constitutional isomers I and IV constitutional isomers II and IV constitutional isomers 1,-Dichloropropane can be synthesised in the laboratory by treatment of propene with chlorine as is shown in the following equation. C 3 C C C 3 C C Which of the following best describes the product: (R)-enantiomer, (S)-enantiomer, racemate? racemate (equal amounts of (R) and (S) will be formed)

14 CEM J-1 June 009 Give the mechanism of the reaction that occurs when 1-methylcyclohexene is converted to 1-bromo-1-methylcyclohexane by the addition of Br. Give the structure of the intermediate carbocation that is formed and indicate (with curly arrows) all the bonding changes that occur. 3 Br + Br Br

15 CEM J-13 June 009 Devise a synthesis of the following compounds from the starting materials indicated. Note that more than one step will be required. Indicate all necessary steps and the constitutional formulas of any intermediate compounds. 6 Me Cr 7 / C 3 S heat 1. C 3 MgBr. / conc. S 4 heat

CHEM J-10 June The structure of ( )-linalool, a commonly occurring natural product, is shown below.

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