CH3D3 2010 exam questions_answers_Revised (2).pdf

CH3D3 2010: Answers to exam questions Section B: 1. (a) ΔH can be taken directly from the graph as shown by the arrow in plot (B) above. In this experiment, the change in enthalpy is ~ − 35 kcal mol. -1 [please note this value may not directly correlate with values in the Table below, see discussion...

CH3D3 2010: Answers to exam questions Section B: 1. (a) ΔH can be taken directly from the graph as shown by the arrow in plot (B) above. In this experiment, the change in enthalpy is ~ − 35 kcal mol. -1 [please note this value may not directly correlate with values in the Table below, see discussion later!] (a) Ligand 14 has the largest binding affinity – largest value for K (=[RL]/[R][L]). a (c) Using ΔG° = ΔH ° − T ΔS° and ΔG° = − RT ln K See table below: Temperature (K) ΔH (kJ mol ) -1 ΔG° K (M ) -1 a ΔS° (kJ mol ) -1 (kJ mol ) -1 Ligand 12 280 -10.4 1.00 × 10 -26.80 58.6 298 -12.3 7.69 × 10 -27.87 52.3 313 -14.8 6.25 × 10 -28.74 44.5 280 -35.1 5.00 × 10 -35.91 2.9 298 -38.9 2.50 × 10 -36.50 -8.1 313 -40.3 1.11 × 10 -36.23 -13.0 280 -58.9 1.00 × 10 -37.52 -76.4 298 -61.5 5.00 × 10 -38.22 -78.1 5 4 4 Ligand 13 6 6 6 Ligand 14 7 6 313 -65.7 3.33 × 10 -39.08 6 -85.1 (d) Binding of Ligand 12 seems to be entropically driven as ΔS is large and positive (–TΔS). This may tell us that the binding of the ligand involves restructuring of water, i.e. hydrophobic effect is operating. Binding of Ligand 13 is enthalpically driven as ΔH is the largest term, contributing to a greater Gibbs energy. This may tell us that the binding of the ligand involves hydrogen bond formation, van der Waals interactions or electrostatic interactions not present (or weaker) in the complex with 12. (e) See below: Values estimated from gradient of line of best fit (y=mx+c) of three data points. Taking data from any two temperatures is also acceptable. ΔC (J mol K ) -1 -1 p Ligand 12 -132.4 Ligand 13 -159.3 Ligand 14 -204.0 (f) A negative and large value of change in heat capacity, ΔC indicates there is p structural reorganisation taking place upon ligand binding. Typically this is interpreted as water being displaced from host upon ligand binding i.e. the hydrophobic effect is operating, but it could be indicative of protein movements to accommodate ligand which in this case may be consistent with the larger enthalpy change. How is this typically resolved experimentally – using a structural method such as NMR or X-ray crystallography to compare the protein in the presence and absence of ligand. Further discussion not required in answer: We can see that the effect on entropy is not the same across all three ligands, despite the heat capacity change being large and similar. Ligands 13 and 14 both have much greater association constants than 12 and this may be a consequence of the way the data has been manipulated in order to make the question. (g) If c > 1000, the data exceed the capability of the instrument (see plots below) resulting in the gradient of the ITC curve being increasingly steep. Reliable fitting of the data is thus not possible, preventing reliable estimation of K. In these cases, competition experiments can be used, with the desired ligand displacing a lignd of weak or moderate binding strength. 2(a)-(c) is an unseen problem and is intended to be integrative with CH264, a Year 2 module on shapes and conformations of molecules. (U) 2(d) – (e) builds upon lecture material but a similar problem has NOT been tackled in class or set as an exam question in previous years. It is intended to be a more challenging, open-ended question requiring some thought as well as familiarity with lecture materials. (L/U) 2. (a) butane H CH3 H H H H H H CH3 CH3 anti- H CH3 gauche- 70 30 gas phase / liquid K = 30/70 = 3/7 55 45 aqueous K = 55/45 = 9/11 (b) In liquid or gas phase: ΔG° = – RT ln K eq = –(1.987cal / mol × 4.2)(298)(ln3 / 7) = –2107Jmol −1 In aqueous solution: ΔG° = – RT ln K eq = –(1.987cal / mol × 4.2)(298)(ln9 / 11) = –499Jmol −1 Hence difference in energy and the energy penalty when anti-butane dissolves in water is -499 – (-2100) = 1601 J/mol. (c) Given the hydrophobic energy per unit area is 168 J/mol Å , then the area buried as a result of the change from anti to gauche is: 2 1601J / mol = 9.5Å 2 2 168J / molÅ (d) Without a conformational change, butane would make the system energetically unfavourable because of exposure to water molecules that have fewer degrees of freedom. The ordered water molecules released from the buried non-polar butane surface area thus become the origin of the ‘hydrophobic effect’ responsible for this conformational change. When released to bulk water they are able to gain more translational and rotational freedom (entropic gain) as there are a greater number of conformations available to them in which they can make enthalpically stabilising hydrogen bonds with neighbouring water molecules. In the anti-butane conformation the water molecules are considered to be of lower entropy (more ordered) since they cannot easily make hydrogen bonds with exposed non-polar surface. (e) This is a surface contour plot of a free energy surface showing interactions between functional groups that can be considered hydrogen bond (HB) donors (blue, α, y-axis) and hydrogen bond acceptors (red, β, x-axis). The saddle at ΔG = 0 kJmol -1 is broad and the blue area in the southwest quadrant where solvent-solvent interactions tend to dominate (solvophobic) is relatively compressed. Numerically the value in that quadrant is also small, such that interactions between SOLUTES are relatively weak if the binding is driven by maximising the free energy of those interactions. This indicates that for many HB donors and acceptors it is not the magnitude of interaction that is important, but some other factor that can be manipulated to drive the free energy parameter. Logically this is solvent accessible molecular surface area of the solute(s), as shown in the first part of the question. If DMSO replaced water, the surface plot would change by weakening even further the solvent-solvent (southwest) quadrant and pushing the zero saddle point to the extreme right of the graph, because HB donor properties of DMSO are very poor. As an excellent HB acceptor however it will also outcompete almost every other functional group on the x-axis and hence extend the red ( ve) northwest quadrant across most of the surface plot, pushing the saddle into the lower, southeast corner. This is a free energy description of an excellent HB acceptor solvent; solvophobic interactions are less significant than for water, although solvent accessible surface area (‘solvophobic effect’) will still play a role. + 3(a)-(b) is an unseen problem (U) and we have not covered similar material in lectures. We deal with a number of ‘molecular receptors’ that use hydrogen bonds as a key design element and cover the ideas developed by Jorgensen to explain simply the relative strengths of HB interactions. Calculation (c) is an unseen example (U but also D) based upon methods developed by Hunter and covered in Lectures and a worked examples class as well as appearing in previous examinations, so a ‘question spotter’ might have reasonably prepared for this type of material. Nonetheless it is challenging and there are plenty of places to misunderstand or go awry in a calculation of this type. Part (d) is best answered using the correct results from (c) but could be interpolated from answers to (a) and (b) alone. Part (e) is an untested hypothesis but is I believe a reasonable question in that it is asking the students to make a prediction that builds on their experience during the module. It requires some critical thought and explanation. 3. (a) and (b) Molecule 15 has three positions available for hydrogen bonding that are readily complementary to complex 16 as shown in the figure below, compared with only two for receptor 3. In addition, the urea function displays an H-bond donor, donor (DD) motif which has two positive secondary electrostatic interactions when interacting with the acceptor, acceptor (AA) motif in the guanine nucleobase. According to WL Jorgensen (JACS 1991), this ought to significantly strengthen this interaction relative to, for example a donor-acceptor / acceptor-donor interaction (secondary electrostatic interactions of dipoles). O R 15 O R N O O N O N H 17 N O N H H H H N H O N H H N N OAc H N N AcO N O O O H OAc N N N O O O N N AcO H N N H OAc N H OAc OAc OAc N OAc O OAc 16 OAc OAc 16 Figure 1 Looking at molecule 17 we see there are only two possible hydrogen bonding sites which match those available on the same base pair 16. (see E. Mertz, S. Mattei, S. C. Zimmerman Org. Lett. 2000, 2, 2931-2934 for full details). (c) Treating cytosine NH as an amide-like NH donor (it is effectively a vinylogous amide) works reasonably well; the urea NH donors then partner with an amide carbonyl and treating the guanine nitrogen as an amidine-like N in the following for 15 with 16: 2 Free energy change = amide NH donor /amide carbonyl + urea NH donor/amide carbonyl + urea NH donor/amidine-like guanine N: ΔΔG = −(2.9 − 2.2)(8.3− 0.8) − (3.0 − 2.2)(8.3− 0.8) − (3.0 − 2.2)(8.9 − 0.8) + 6 = −5.25 − 6 − 6.48 + 6 = −11.73kJmol −1 − ΔG Hence using K a = e RT we see that the estimated association constant is a 113 M -1 (experimentally 1016±155 M ). This appears to be an underestimate, suggesting that assumptions made may not be valid (e.g. treating the cytosine amidine as an amide NH) or there are additional molecular interactions that need to be taken into account, for example the possibility of an additional HB to the other NH in the cytosine amidine (see figure, below). This type of bifurcated hydrogen bond is precedented in high resolution crystal structures of nucleobases (see for example, G. A, Jeffery, W. Saenger Hydrogen Bonding in Biological Structures, 1991, p. 20-21) and seems likely in this instance. -1 O O R R N N O O O O N bifurcated hydrogen bond 15 H N N 17 bifurcated hydrogen bond H H H H H N H N O O N N H N N AcO O H N H OAc H N OAc N O O H N H O OAc OAc OAc N N AcO N O N N OAc O OAc OAc OAc 16 N OAc 16 Figure 2 Doing this gives: ΔΔG = −2 × (2.9 − 2.2)(8.3− 0.8) − (3.0 − 2.2)(8.3− 0.8) − (3.0 − 2.2)(8.9 − 0.8) + 6 = −10.5 − 6 − 6.48 + 6 = −16.98kJmol −1 and hence: − ΔG K a = e RT −16.98 K a = e 0.008314 T K a = 950 M −1 which is very now close to the experimental value. Treating the interaction of 17 with G C pair 16 as in Figure 1, the complexation free energy is estimated as amide NH donor /amide carbonyl + amide NH donor/amide carbonyl ΔΔG = −(2.9 − 2.2)(8.3− 0.8) − (2.9 − 2.2)(8.3− 0.8) + 6 = −5.25 − 5.25 + 6 = −4.5kJmol −1 −ΔG RT Using Ka = e gives a value of 6 M for the association constant. Free energy change = amide NH donor /amide carbonyl + amide NH donor/amide carbonyl. -1 However if we also assume there could be a bifurcated hydrogen bond as in Figure 2, the estimate becomes: ΔΔG = −2(2.9 − 2.2)(8.3− 0.8) − (2.9 − 2.2)(8.3− 0.8) + 6 = −10.5 − 5.25 + 6 = −9.75kJmol −1 and hence: − ΔG K a = e RT 9.75 K a = e 0.008314 298 K a = 51M −1 Which is close to the experimentally observed value of 70 M. -1 (d) As noted in answer above to (a) and (b) the DD / AA urea function brings beneficial secondary electrostatic interactions as well as a second primary hydrogen bond to the complex with 15 and 16. When this is replaced by only one amide on the right-hand side of the receptor, we see that we lose more than just a single HB. (e) No there would NOT be expected to be a difference since although molecule 17 looks ideally placed to bind to U A pair 18, the interactions and geometry are more or less the same as for the G C base pair and we can also make the same assumption of a bifurcated hydrogen bond if necessary. However reasonable arguments developed in opposition to this will accrue a proportion of the marks. O R R O N N 15 O 17 O O N H O N H H O N N H N N O OAc O N H N N AcO O O H N H OAc OAc N N OAc O OAc OAc 18 OAc O N OAc O H N N N N AcO H H OAc 16 OAc