Department of Chemistry Core Outline Answers. All the answers involve information from the course applied to unseen problems.

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1 Department of Chemistry Core Outline Answers Question 1 All the answers involve information from the course applied to unseen problems. (a) (i) 1 is attached to Wang resin via ester formation (DCC- students are familiar with this in reality other coupling reagents would be used) (6) Boc removed by treating with TFA (20-40% in DCM) must be less than the conc of TFA used for cleavage from Wang. Alternatively the PG strategy could be changed before attaching to resin Boc could be replaced by Fmoc (then removed on resin with piperidine). Need a coupling reagent (DCC) and Boc protected alanine to make resin-bound tripeptide. 90% TFA in DCM will cleave from resin and deprotect peptide. (if Fmoc used need separate piperidine treatment). Correct stereochemistry and N on LHS needed 1

2 Doesn t require problematic HF-corrosive, reacts with calcium. +M on Wang labile with TFA. (b) (i) TBDMS sterically demanding so goes on 6-OH (students have only been exposed to trityl in this course but know TBDMS from another Y2 course) BnCl, base-pyridine or NaH Deprotect silyl ether with TBAF (F - ) or mild acid (acetic) 2

3 (4) In the first reaction there is no NGP so anomer stereochemistry driven by the anomeric effect giving mainly the -anomer In the second reaction the glycosyl cation reacts with acetonitrile to give a nitrilium ion which is then displaced to give the product. 3

4 Question 2 (a) Bragg equation n = 2 dhkl sinθ should be used here. The main issue is to allocate the right 2θ angle to dhkl for the (0 1 0) set of crystal planes. Hence the 2θ angle that corresponds to the reflection from b crystal planes should be selected. Also, care should be taken to half the given 2θ before its use in Bragg equation. After that simple: = 2 x 2.82 Å x sin (28 o /2) = 1.36 Å Practical application of lecture material in problem solving. (b) (i) Component B is the filter in this setup, which stops all continuous (Bremsstrahlung) radiation that comes from X-ray source A. It also cuts out any other characteristic radiation that is typical for a particular anode of the sealed tube X-ray lamp. (Alternative answers may include mirrors or diffraction gratings.) Goniostat D assures extensive rotation of the crystal in all necessary orientations that expose most (all in theory) of the crystal planes against the incoming X-ray beam, which fulfil Bragg diffraction conditions. (1) (1) Two answers here (both required): -good collimator C that maximises the intensity of the X-ray beam, and focuses it on a very small area of the crystal (this also reduces scattering of the X-ray beam by air) -better, more sensitive detector E with low noise background that increases the ratio of the strength of the registered reflection to its surrounding/background area. Moving detector E closer to the crystal will allow registration of reflection with higher 2θ (or simply θ) angles. Maximum registered θ angle is one of the definitions of the resolution, i.e. an increase of registered θ increases the resolution. Alternatively, higher θ angles will be coming from crystal planes with smaller dhkl, which is an alternative definition of the resolution; i.e. registering reflections from smaller dhkl, increases the resolution as well. Either answer will be acceptable. Application of lectures material; unseen probing of the understanding of the relationship between Bragg equation and its experimental application. (c) (i) Structural constraint imposes strict, rigid maths/stereochemical rules on geometry of some chemical groups and bonds. Here the planarity of the aromatic C1 -C6 can be constrained, and C-H bond distances as well. Structural restraint imposes much softer geometrical restrictions within experimentally observed ranges. Here the amide C15O-N14H- torsion angle may vary ( o ) hence it can be restrained. 4

5 The only methods applicable here are direct methods. There is no heavy atom in the structure (anomalous scattering is not taught here so no use for S atom) hence only purely statistical methods that exploit the relationship among phases of triad, quintets etc. of special sets of reflections. The initial phases for those are estimated in a trial and error process. The most satisfying phase relationships are expanded on the whole data sets and electron density maps are evaluated. Hopefully they contain some atoms that serve model building, better estimation of phases, and, subsequently a better, fuller model. Two alternative approaches to control of the quality of the model and the correctness of refinement may be given (one required): - calculation of difference electron density maps, in which maxima and minima should help in correction of the model, and its geometry, and should also indicate still missing atoms; maxima (for example) would indicate heavier (or missing) atoms in certain positions, while minima would indicate less electron rich atoms (or wrongly positioned ones). At the end of the refinement the difference electron density map should be flat. - a statistical approach by monitoring of Rwork factor: (4) The Rwork factor compares the convergence of the observed/experimental Fobs s with Fcalc s that are calculated from the model. The differences between them should get smaller if the model construction, and its refinement, are correct. In theory this should be close to zero, but it never is due to experimental errors, crystal disorder, mobility of the structures etc. In any case, the final stages of refinement should not affect the value of Rwork, indicating completion of the model and its correctness. Lecture material applied in a novel approach for solving of the practical problems of X-ray crystal structure analysis. 5

6 Question 3 (a) Suitable set of canonical structures showing attack by an electrophile at position 2-, 3- and 4-. (4) (b) Bookwork (i) (1) Application of knowledge/bookwork (5) Application of knowledge a good discriminator (1) Application of knowledge 6

7 (c) (i) Application of knowledge Oxazole is less reactive to electrophilic substitution at carbon because it contains a pyridine-like nitrogen which withdraws electron density from within the ring. However, the pyridine-like nitrogen in oxazole is most susceptible towards electrophiles (e.g. H + ). Deductive reasoning Correct structure (1) Me2N- donates electron density (+M) thus promoting electrophilic attack. Deductive reasoning good discriminator 7

8 Question 4 (a) The examples are taken from Y. Wang and G. H. Robinson, Inorg. Chem., 2014, 53, (i) In the case of GaCl2Ph, the phenyl group is not sufficiently bulky to inhibit the formation of multiple Ga-Ga bonds and so oligomerisation occurs. (1 mark) In the case of 8 there is sufficient stabilisation by the Mes groups that trimeric 9 is formed, whereas in the case of 10, the increased bulk of the aryl substituent means that dimeric 11 is obtained. (2 marks). The addition of a second aryl substituent in 12 significantly increases the bulk further and so it is no longer possible for two gallium atoms to approach closely (1 mark). Applications of the principles taught in the course to unseen examples (5) In heavier main group elements there is extensive mixing between and * orbitals (1 mark) which results in the trans-bent geometry and lowering of the energy of the resulting non-bonding n + orbitals (1 mark). Bookwork (b) (i) i. TVEC = P + C + Bu t + B3 + H5 = = 24 electrons (1 marks) SEP = TVEC 2n (n = 5) (1 marks) = = 14. Therefore 7 pairs. (1 mark) SEP = n+2 therefore a nido cluster. (1 mark) (4) Applications of the principles taught in the course to an unseen example A nido cluster involves removing (the most connected) vertex from the parent closo with the same SEP (1 mark). For SEP = 7 the parent closo is an octahedron, so the nido is a square-based pyramid. (1 mark) Bookwork (0.5 marks each) Applications of the principles taught in the course to an unseen example 8

9 (iv) 11 B{ 1 H} NMR -8.1 (2B, B1) and (1B, B2) 1 H NMR 2.89 (2H, Ha), 2.28 (1H, Hb) 1.00 (9 H, Hc) and (2H, Hd) (5) 3 marks for structure, 2 for assignments Applications of the principles taught in the course to an unseen example 9

10 log k Question 5 Data is from DOI: /poc.3571, however the mechanism suggested in the paper is not correct. The whole question is application / problem solving (a) (i) The calculations and the Hammett plot are shown below (6) X sigma k log(k) 4-F Cl F Br H Me Me OMe Hammett plot σ -0.5 y = x y = x The V-shape of the Hammett plot is indicative of the co-existence of two competing mechanisms for this reaction, with different mechanisms operating for different substituents. The values of ρ (see graph above) for electron-donating and electronwithdrawing substituents are -7.2 and 2.6, respectively. The negative sign of the first value (corresponding to the electron-donating substituents) is consistent with the development of a +ve charge in the reaction, which is observed in step A. Formation of this +ve charge right next to the aromatic ring is consistent with the large magnitude of this ρ value. Electron-donating substituents stabilise the radical cation formed in step A, and hence facilitate this step. At the same time, they destabilise the anion formed in step C, and hence slow down step C. Thus the reaction proceeds via step A (which is rate-determining in this case). (5) 10

11 (iv) The positive sign of the second value of ρ (corresponding to the electronwithdrawing substituents) is consistent with the development of a -ve charge in the reaction, which is observed in step C. Formation of this -ve charge two atoms away from the aromatic ring is consistent with the smaller magnitude of this ρ value (although the value is still surprisingly large). Electron-withdrawing substituents stabilise the anion formed in step C, and hence facilitate this step. At the same time, they destabilise the radical cation formed in step A, and hence slow down step A. Thus the reaction proceeds via step C, which is ratedetermining in this case. σ Values for 3-OMe and 3-NMe2 substituents are 0.11 and -0.15, respectively. By noting the position of these substituents on the Hammett plot, it is easy to see that the compound with the 3-NMe2 substituent will react faster. (b) For the electron donating substituent 4-Me, the reaction proceeds via mechanism A. Water is not involved in this step, and hence the rate of reaction does not strongly depend on the concentration of water. (c) For the electron-withdrawing substituent 4-Cl, the reaction proceeds via mechanism C. Water is involved in this step, and hence the rate of reaction should be linearly proportional to the concentration of water. Stronger than expected dependence on water concentration observed in this case is likely due to the solvent effect (see part c) Reactions of both compounds start from a neutral starting material and lead to the formation of charged intermediates. Hence both reactions will be facilitated by polar solvents. The rates of both reactions will thus be faster in DMSO than in toluene. 11

12 Question 6 All problems expect part (a) (a) The Cp and Cp* ligands bind through their π-systems in an η 5 - fashion and are 6-electron donors while the carbene uses the orbitals shown below and is a net 2-electron donor. (b) (c) (d) (e) The oxidation state of the Fe centre in 16 is 2, its d-electron count 6 and valence electron count 18. The oxidation state of the Fe centre in 17 is 2, its d-electron count is 6 and valence electron count is unusual because it is stable and has just 16 electrons. This suggests that the steric bulk of the Cp* and IMes ligands protect the metal centre with the Cl ligand potentially donating some electron density via its lone pairs. The increase in infrared stretching frequency for the PPh3 analogue is due to the metal centre being less electron rich and hence the reduction in back bonding to CO leads to a move to higher wavenumber. Hence PPh3 must be a poorer donor than IMes. The two stretches seen for the dicarbonyl are being replaced by the single stretch of a mono carbonyl but the IR stretch indicates strong back-bonding, so the complex is electron rich and 18-electron, hence [CpFe(CO)(IMes)I] is formed. Bubbles suggest a gas is being released, in this case CH4. This requires a CH bond activation reaction, which involves one of the IMes ligands methyl groups. The resulting Fe(IV) intermediate then undergoes CH3-H reductive elimination leaving behind a product with an Fe-C bond. The reaction is aided by the entropic gain of releasing the gas CH4. The product is still 16 electron and reacts further with CO to form an 18-electron product. (5) (4) (5) 12