CH3BO 2010 Exam Questions PDF

1. Answer ALL parts. MW CH3B) OLD COURSE 2010 exam questions. (a) Illustrate the frontier orbitals, in each molecule, which are involved in the cycloaddition reactions shown below. (i) A [2+2] cycloaddition. Ph Ph Ph irradiation (hυ) + Ph Ph LUMO (ground state) Ph Ph Ph Ph HOMO Ph (excited state) Ph...

1. Answer ALL parts. MW CH3B) OLD COURSE 2010 exam questions. (a) Illustrate the frontier orbitals, in each molecule, which are involved in the cycloaddition reactions shown below. (i) A [2+2] cycloaddition. Ph Ph Ph irradiation (hυ) + Ph Ph LUMO (ground state) Ph Ph Ph Ph HOMO Ph (excited state) Ph Ph 1 mark per molecule [10%] (ii) A [4+2] cycloaddition. O O heat + LUMO HOMO O (or alternative HOMO/LUMO combination) 1 mark per molecule [10%] (b) The conversion of 1 into 2 may be achieved in a two step process. The use of photochemical irradiation results in formation of a reactive intermediate 3, which subsequently rearranges to 2 under non-photochemical conditions. OH Me H OH nonphotochemical Photochemical irradiation (hυ) 3 HO HO 1 (i) 2 Identify the structure of intermediate 3. Me OH HO 3 2 marks [10%] (ii) Write a mechanism for each step of the reaction above, and describe the type of pericyclic reaction that is taking place. OH Me H OH nonphotochemical Photochemical irradiation (hυ) 3 HO HO 1 2 OH Me Photochemical irradiation (hυ) H H antarafacial 6 e process HO 1 OH HO 3 nonphotochemical antarafacial, 8 e process OH HO 2 1 mark for each mechanism and 1 for each process description. [20%] (c) Provide a mechanism for the following reaction and identify the product 4. H Cl i) Et3N (a base) Cl Cl 4 ii) O Et3N H Cl Cl Cl H Cl Cl O C O Cl Cl H O Cl Cl C O [4+2] cyclisation favoured because two pi systems on the ketene are involved. (covered in detail in lectures) 1 mark for step 1, 2 marks for step 2, 2 marks for explanation. [25%] (d) Account for the stereochemistry of the product 5 formed in the sigmatropic rearrangement shown below. Your answer should include an illustration of the transition state of the reaction. Ph O Ph Ph heat O Ph 5 Ph Ph O Ph Ph O Ph Ph 5 Ph Ph O via Ph O O Ph [25%] 2. Answer ALL parts. (a) The synthesis of the alkaloid Pulimiotoxin C may be achieved through the reaction sequence illustrated below, which proceeds via the formation of intermediate 6. O O + H heat O steps Pumiliotoxin C. NH N H (i) OBn 6 BnO O Account for the control of the regiochemistry of 6 in the cycloaddition. δ+ δ+ O δ− O δ− N H OBn The electron donating N group and electron withdrawing C=O create complimentary areas of positive and negative electron density. 2 marks for diagram , 1 mark for explanation. [15%] (ii) Draw a transition state for the cycloaddition reaction, and thus predict the relatively configurations of the three chiral centres in 6. O NH O LUMO of dienophile aligns with HOMO of diene. Endo transition state due NHCO2Bn to secondary orbital overlap. CHO BnO 3 marks for diagram, 2 for explanation. [25%] (iii) The use of a Lewis acid increase the rate of the cycloaddition reaction. With aid of appropriate energy level diagrams, explain why? LUMO+LA LA diene HOMO O LUMO no LA Lewis acid co-ordinates to C=O, lowering LUMO energy and giving better overlap with diene HOMO 2 marks for energy level diagram, 2 marks for explanation. [20%] (b) The reaction of hydroxylamine 7 with aldehyde 8 results in the formation of a reactive intermediate 9. Intermediate 9 subsequently undergoes a reaction with styrene 10 to give the product 11. Ph Ph N H OH + Ph 7 (i) O Ph 9 Ph 8 10 N O 11 Ph Propose a structure for 9, and provide a mechanism for its formation. O Ph N N attacks C=O and then elimination of water. Ph 1 for product and 2 for mechanism. [15%] (ii) Illustrate the mechanism of the reaction of 9 with 10 to form 11. Ph Ph (iii) N O Ph Ph Ph N O 11 Ph 2 mark for mechanism. [10%] (c) Explain why the reaction below fails under basic conditions, but gives a product in 100% yield under acidic conditions. Provide a mechanism for the successful cyclisation. O O HO Using NaOH; 0% yield. O Using HCl; 100% yield. OMe OMe The base catalysed process requires a disfavoured 5-endo-trig reaction (2 marks) whilst the acid catalysed reaction involves protonation of the C=O and rearrangement to structure below, which cyclises via a favoured 5-exo-trig cyclisation (3 marks), the mechanism of which should be illustrated. OH HO OMe [25%] 3. Answer ALL parts. (a) Propose a synthesis of trisaccharide 12 from the three reagents 13, 14 and 15, paying particular attention to the stereocontrol of formation of each anomeric centre. Reagents should be given for each step, together with a mechanism for each step. [80%] Ph O O HO O O OH HO OH O TBDPS=Si(tBu)Ph2 TBDMS=Si(tBu)Me2 Bn=CH2Ph OMe O 12 O OH OH O O Ph Br Ph Bn O O O 13 HO O Br O BnO OBn O TBDMSO O OMe OBn O O O O TBDPS 14 15 Ph Neighbouring group participation. Ph Bn Ag O O O O Br O Ph Ag Bn O O O O Ph Bn Ph Bn O O O O O O 1 mark O 2 marks 13 Ph O Bn O O O HO O BnO O O OBn O TBDMSO 14 O 2 marks (mechanism) 2 marks for explanation of anomeric control. O remove TBDMS Ph O O O with TBAF (1 mark) O O O OBn Bn OBn OAc OAc O O BnO OMe BnO OMe Ag O TBDMSO 2 marks Br 2 marks (mechanism) Anomeric control OBn Ag O OBn O O O O O O O Ph OBn O O Ph OAc O TBDPS 15 BnO OMe TBDPS 2 marks for explanation of anomeric control O O O TBAF to remove TBDPS then hydrogenation to remove Bn groups 2 marks 12 OBn O O Ph TBDPS (b) Predict the structure of the product X from the cycloaddition reaction shown below, and give a mechanism for its formation. [20%] + EtO2C N heat CO2Et N X N H EtO2C OMe CO2Et EtO2C 2 marks for product and 2 for mechanism. H CO2Et X ( -yfrtya^ O M. >q / ow/i-7 OS tHJ 5-((.'Al ' r~\ A £ O ir f-7 V) I? a 1} \ s M ri I T \M V) Vj I 1 3V c. CT °^, S>; ^ r 0rts 5- (A3 -^) i-K- 0 9* ov A -S \ ~~\s 0 v ^sf f I NJ v o O- >- Ox, C) c (0 r\ C / ^ - ; h * /O ^"^\* Kvf 0 rw ^ 0 k^ -^>