University of Warwick Chemistry Past Paper PDF, March 2015

CH3CHx UNIVERSITY OF WARWICK THIRD YEAR EXAMINATIONS: March 2015 CHEMISTRY (BSc. Year 3, MChem Year 3) PAPER CH3CHG: Bioorganic Chemistry. Time allowed for candidates offering ONE SECTION: Time allowed for candidates offering TWO SECTIONS: 1½ hours 3 hours Read the rubric carefully. Percentages in s...

CH3CHx UNIVERSITY OF WARWICK THIRD YEAR EXAMINATIONS: March 2015 CHEMISTRY (BSc. Year 3, MChem Year 3) PAPER CH3CHG: Bioorganic Chemistry. Time allowed for candidates offering ONE SECTION: Time allowed for candidates offering TWO SECTIONS: 1½ hours 3 hours Read the rubric carefully. Percentages in square brackets are intended as a guide to the time candidates should spend in answering the corresponding part of the question. Read carefully the instructions given in each section that you attempt. Answers to each section should be written in a separate booklet. SECTION A: BIOORGANIC CHEMISTRY (CH3F5) Answer TWO questions from questions 4, 5 and 6 You should NOT attempt MORE than TWO questions in this section since ONLY the first TWO will be marked. UNSEEN BUT STRAIGHTFORWARD PROBLEM WHICH INTEGRATES Y1, Y2 PHYSICAL ORGANIC AND BIOINORGANIC WORK 5. (a) Explain the following observations concerning the indicated NH. O O NH N H NH O NH 2 N N H 2+ O NH O H N H N N O Zn(II) NH 2 H 2N pK a = 9.9 N pK a = 7.1 H N H pK a < 5 [15%] - pKa of uridine proton relatively acidic – show how this is stabilised by resonance into amide C=O for example. 1 CH3CHx N H N O O NH NH NH - O O O N O NH O N O Sketch pyrimidine over the protonated aza-crown which provides positive charge to further stabilize the uridine anion, lowering pKa. O HN O N NH - N NH 2H 2N Presence of coordinated zinc now lowers pKa even further, probably by directly binding to uridine N– donor. UNSEEN CHALLENGING PROBLEM (b) The Z-conformer of ethanoic acid is calculated to be more stable than E-ethanoic acid. In the gas phase Keq = 24 kJ mol-1, but in aqueous solution this preference is reduced to Keq = 7 kJ mol-1. Deprotonation gives the same carboxylate anion. O O O H O H Z-ethanoic acid E-ethanoic acid Base Base O syn lone pair O anti lone pair + HBase (i) Predict whether the syn or anti lone pair indicated in the carboxylate anion is more basic. 2 marks [10%] Draw equlibrium as below to clarify meaning. Syn lone pair predicted to be more basic (or less acidic in return) because the Z-ethanoic acid product is favoured. 2 CH3CHx H O O O O H possibility of HB dimer O O O H Z-ethanoic acid preferred isomer E-ethanoic acid Base O Base syn lone pair O cannot form HB dimer O H H O Base O anti lone pair H Base (ii) Explain the difference in equilibrium constants and why the conformational preference is smaller in water. 3 marks [15%]. Can be answered in part (i) above. Any of: - antiperiplanar anti-lone pair to carbonyl allows better orbital overlap syn lone pair cannot donate into C=O antibonding orbital anti lone pair can donate into C=O antibonding orbital H O O H H O H - O H H O O H H O In water both the acid and its anion will be hydrogen bonded and this attenuates the energy difference. At pH 7 carboxylate deprotonated. 3 CH3CHx more basic syn lone pair can still be involved in HB less basic anti lone pair involved in HB, but more basic syn lone pair is protonated H O O O H O H O H H O H anti lone pair protonated O H H O O H carboxylate deprotonated in water pH = 7 O H H H O H O O H O H H O STRAIGHTFORWARD – RECALL AND EXPLAIN WHAT WE DID IN CLASS + WORKED EXAMPLES (c) Using a suitable example, explain how the linear free energy relationship (Equation A) can be used to calculate the change in free energy upon complexation of two molecules EACH of which contains ONE hydrogen bond donor (α) and ONE acceptor (β), in a solvent (s). G = –( – S )( – [10%] S ) + 6 kJ mol-1.... (A) Sketch a suitable “molecule” containin HB donors and HB acceptors arranged so they are forming two HBs. HB donor O HB acceptor H-Solvent H-Solvent H N N N H Solvent-H + N H Solvent-H N N N N H O O O +2x Solvent-H Solvent-H ΔG = –(α − α s)(β −β s) –(α − α s)(β −β s) + 6 kJ mol-1 ΝΗ Ν ΝΗ Ν Need values for donor (α) and acceptor (β) in order to numerically evaluate. Important to show role of solvent in above process. 4 CH3CHx 2 marks 5