Final exam for CHEM6112, March 16th, Time allowed hours. Please write your name at the top of every answer sheet.

Transcription

1 Final exam for CHEM6112, March 16th, 1995 Time allowed hours Please write your name at the top of every answer sheet. Show all working and reasoning associated with any numerical problems Accounts for 40 % of the total course mark The numbers in bold are the points per section 1) Suggest reagents (and any missing intermediates) for each of the steps in the following reaction schemes. a) Mn2(CO) > Mn(CO) > Mn(CO)5Me i) Reductive cleavage using say a suspension of sodium metal ii) Alkylation using say MeI or Me3O + BF4 - b) Mo(CO) >? > Mo(=CMe)(CO)5 i) Treatment with MeLi, giving Mo(=CMe)(CO)5 ii) Alkylation with Me3O + BF4 -

2 2) Wilkinson s catalyst [Rh(PPh3)3Cl] is a widely used hydrogenation catalyst. Using the hydrogenation of cyclohexene with Wilkinson s catalyst as an example catalytic cycle, write chemical equations involving: a) ligand substitution [Rh(PPh3)3Cl] + sol > [Rh(PPh3)3sol] + + Cl - b) oxidative addition [Rh(PPh3)3sol] + + H > [RhH2(PPh3)3sol] + c) migratory insertion - RhH2(PPh3)3 ] + + sol > - RhH(sol)(PPh3)3] + d) reductive ellimination - RhH(sol)(PPh3)3] > [Rh(sol)(PPh3)3] + + cyclohexane 3) a) Which of the following compounds obey the 18 e rule? (you will loose marks for each incorrect answer) V(CO)6, Fe(Cp)(CO)2Me, Ir(CO)3Cl, Fe3(CO)12 Fe(Cp)(CO)2Me, Fe3(CO)12 are 18 e compounds V(CO)6 is a 17 e compound, Ir(CO)3Cl is a 16e compound.

3 b) Which of the following cations is likely to be the most stable? Fe(Cp)2 +, Co(Cp)2 + Co(Cp)2 + is an 18e compound and is likely to be very stable be Fe(Cp)2 +, is a 17e compound and is likely to be oxidizing (can prepared). c) What is meant by hapticity? If the ligand cyclooctatetraene is bound to a metal center through it s π electrons what hapticities are possible? Hapticty is the number of atoms on a ligand that are coordinated to the metal center. COT can have hapticities of 2,4,6 and 8. 4) Reduction of the two cobalt(iii) complexes shown below by Cr(H 2 O) 6 2+ occur in acidic solution at significantly different rates as indicated by the rate constants for the two reactions. (H 3 N) 5 Co-N (H 3 N) 5 Co-N k = 4.0 x 10-3 M -1 s -1 k = 17.4 M -1 s -1 a) What is the most likely reason for the large difference in rate for the two reactions? One of the compounds has a ligand that can act as a bridging ligand and the other does not. The compound with the bridging ligand probably reacts via an innersphere mechanism whereas the other compound probably reacts via an outer sphere mechanism. b) Explain what products can be expected from the reactions and how a product analysis might help you confirm the reaction mechanism. In the case of an outer sphere electron transfer the products will be aquo ions. However, in the case of an innersphere reaction, the bridging ligand will be present in the coordination sphere of the Cr(III) product. The presence of a nonaquoligand in the coordination shell of the Cr(III) product provides

4 strong evidence for an innersphere mechanism as substitution into the coordination shell of Cr after the oxdiation is very slow.

5 5) the a) Using the Tanabe Sugano diagrams supplied explain why solutions of FeF 3-6 are essentially colorless, solutions of CoF 3-6 are colored with only one absorption in visible region but Co(NH 3 ) 3+ 6 is colored with two absorptions in the visible region. Assign the spectroscopic transtions and explain the observations. Note most octahedral cobalt (III) complexes have two absorptions in the visible region. FeF 6 3- is high spin d 5. All the d-d transtions for this complex are spin forbidden and hence they will be very weak. CoF 3-6 is a high spin d 6 ion. The ground state is 5 T 2. There is only one spin allowed d-d transition 5 E <-- 5 T 2 and hence one stong band in the visible region. Co(NH 3 ) 3+ 6 like most coblat (III) complexes is low spin d 6. The ground state is 1 A 1. There are a number of spin allowed d-d transtions, two of which lie at relatively low energy and are likely to be observable in the visible; 1 T 1 <-- 1 A 1, 1 T 2 <-- 1 A 1. b) Why is CoF 6 3- different from Co(NH 3 ) 6 3+? CoF 3-6 is different from most cobalt (III) complexes because it is high spin rather than low spin. F - is a weak field ligand that is not very covalent. Both of these factors favor high spin over low spin complexes. The former because the ligand field splitting is small and the latter because the electron pairing energy is high. In most cobalt complexes the ligand field splitting energy and the electron pairing energy are such that low spin is favored. 6) a) What are ORD and CD (explain what the experiments are)? ORD optical rotatory dispersion is the measurement of the optical rotation of a compound as a function of the wavelength of the light used. CD circular dichroism is the measurement of the difference in extinction coeffecient for left and right handed circularly polarized light as a function of wavelength. b) Why are they useful? They allow the assignment of relative configuration. The shape of an ORD or CD spectrum for a species can be compared to those for other chemically

6 related compounds. If the shape of the curves are probably have the same absolute configuration. similar then the compounds

7 7) a) Explain qualitatively the data presented in the following graph. The graph shows ionic ocnductivity versus temperature for LiI, AgI and Na β alumina. Ionic motion in the solid is an activated process, consequently Arrhenius type behavior is to be expected. In Na β alumina a line with a single slope is seen and the conductivity is high. In this material there are many vacant sites in the solid due its sturtcure and ionic hoping between vacant sites is relatively easy. In LiI two regions with different slopes are present. At lower temperatures the slope of the graph relates to the activation barrier for motion of an extrinsic defect in the solid. At high temperatures the activatyion energy has a component from the formation of intrinsic point defects and the mobility of these defects. In AgI the discontinuity is a consequence of a phase transtion to a high temperature material that has many available ion sites and low energy paths between the sites. b) What structural features of Na β alumina make it such a good fast ion conductor? Na β alumina has a structure that can be though of as spinel like blocks help appart so as to give gallaries between the blocks. The Na ions sit in the gallaries and can easily hop from site to site. It is the availability of a large number of vacnat ion sites and the low barrier to motion between site sthat is responsible for the materials high ionic conductivity.

8 8) a) Sketch a diagram showing the iteraction of the metal d-orbitals on a transition metal with the frontier orbitals on a CO ligand. b) Why is CO described as a π acceptor ligand? CO is described as a π acceptor ligand because it s vacant π* orbitals can overlap with the metal d orbitals and the metal d electrons can then be delocallized in part onto the CO. c) In what way does the π acceptor behavior of the ligand influence the CO stretch frequency? As the π acceptor behavior involves putting electrons into what was originally an antibonding orbital the C-O bond is weakend and the frequency is lowered. stretch d) Infrared spectroscopy is a useful tool for studying the structures of metal carbonyls. Explain what factors influence the number of bands observed in the IR spectrum of a metal carbonyl and the frequencies of those bands. The number of bands in the IR spectrum of a carbonyl depends upon it s symmetry. The number that can be expected for a given geometry can be calculated using group theory and compared to those observed. If more are observed than predicted the symmetry of the carbonyl must be lower that used in the group theory calculation. The frequencies of the carbonyl stretching bands depend upon how the CO is bound. Terminal Cos have higher strech frequencies that bridging ones. The frequencies also depend upon how electron rich the metal center is. Electron rich metal centers lead to low stretch frequencies.

9 9) a) Which of the following compunds are likely to contain metal-metal bonds? (Incorrect answers will loose marks) ZrBr, ReCl 3, PdF 2, K 2 ReCl 4, IrO 2 ZrBr, ReCl 3, K 2 ReCl 4 b) In what way is the electrical conductivity of TiO different from that of NiO and why are they different? TiO is a metallic conductor but NiO is hoping semiconductor. In TiO the metal d-orbitals are large enough to overlap and give wide bands. These partially filled wide bands lead to metallic conductivity. In NiO The poor d- orbital overlap and the resulting narrow bands lead to localized electronic behavior. 10) a) Explain what is meant by heterogeneous catalyst and homogeneous catalyst A homnogeneous catalyst is catalyst that exists in the same phase as reactants and products. A heterogeneous catalyst is catalyst that is in a different phase to the reactants and products. the b) Why might the (100) surface of a metal catalysts have a different catalytic activity from the (111) surface? The arrangement of atoms on a (111) surface is different from that on a (100) surface and consequently the available binding sites and catalytically active sites on those surfaces will be different. c) Zeolites, like ZSM5, are widely used industrial catalysts. Which of the following types of reaction might be effectively catalysed using the hydrogen form of an alumino silicate zeolite. (Hydrogen form means that H+ is the charge balancing extraframework cation). Incorrect answers loose marks. i) Hydrogenation of alkenes ii) Isomerization of alkenes iii) Alkylation of benzene using methanol iv) Oxidation of propene to allylalcohol.

10 ii) Isomerization of alkenes iii) Alkylation of benzene using methanol