CH 233H. Name KEY. Please show your work for partial credit.

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Berenice Dalton
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1 CH 233H Final Exam Tuesday, June 9, 2015 Name KEY Please show your work for partial credit. You may not use notes or other materials with chemical information without the instructor s approval; necessary information is provided on pages at the back of the exam. Please do not use ipods or other music players.

5 points each Addition OR Reduction (Either OK) Oxidation Substitution Elimination

2 1. (25 points) Classify each of the following reactions as one of: substitution, addition, elimination, condensation, hydrolysis, oxidation, reduction. 5 points each Addition OR Reduction (Either OK) Oxidation Substitution Elimination Condensation 2. (20 points) Identify the atomic orbital hybridization for each carbon in the following molecule. 2 points per carbon

3. (50 points) The heats of formation and standard entropies for cyclohexane, cyclohexene, 1,3-cyclohexadiene, H2 and benzene are given. All values are for liquid compounds, except H 2 as a gas.

3 3. (50 points) The heats of formation and standard entropies for cyclohexane, cyclohexene, 1,3-cyclohexadiene, H2 and benzene are given. All values are for liquid compounds, except H 2 as a gas. DGf (298K): kj/mol A. Calculate the following free energies of reaction at 298K for: -Hydrogenation of benzene to cyclohexane -Hydrogenation of 1,3-cyclohexadiene to cyclohexane -Hydrogenation of cyclohexene to cyclohexane The easiest way is to calculate DGf (298K) for each compound noted above. Then calculate DGr (298K) for each reaction (products reactants): Benzene to cyclohexane: { (-38.9)} = kj/mol 1,3-cyclohexadiene to cyclohexane: { (-38.9)} = kj/mol Cyclohexene to cyclohexane: { } = kj/mol You may, of course, calculate DHr and DSr for each reaction before calculating each free energy change: Benzene to cyclohexane: DHr = kj/mol; DSr = J/(mol-K) 1,3-cyclohexadiene to cyclohexane: DHr = kj/mol; DSr = J/(mol-K) Cyclohexene to cyclohexane: DHr = kj/mol; DSr = J/(mol-K) 10 points for each free energy (enthalpies and entropies do not have to be calculated).

B. What do your results tell you about the interaction of 2 or more multiple bonds in these systems? (Be sure to consider implications of an MO description of bonding.

This is because of aromatic stabilization arising from the cyclic array of p orbitals in benzene.

4 B. What do your results tell you about the interaction of 2 or more multiple bonds in these systems? (Be sure to consider implications of an MO description of bonding.) You'd expect a pretty much linear relationship as we react 1, 2 and then 3 moles of H2. However it's not; hydrogenation of benzene releases much LESS energy than you'd expect. This is because of aromatic stabilization arising from the cyclic array of p orbitals in benzene. If you calculated enthalpies (not required) you'd observe cyclohexadiene has about twice the hydrogenation enthalpy as cyclohexene, but that benzene is much, much lower than 3 x cyclohexene. (The entropies are largely in line: 120 J/mol or so for each mole of H2, give or take). 20 points 4. (24 points) Taxol is a significant new treatment for a number of cancers. It is formed in the bark of the yew tree from a terpene precursor called taxadiene. Identify (by tracing out) the isoprene units in taxadiene to prove that these compounds originate from the general terpene biosynthesis. 6 points per unit (highlighted in black)

5. (36 points) The three most common hexoses are D-glucose, D-fructose and D-mannose.

Draw a Fischer projection of D-fructose. 12 points; -3 for any stereochemical change at C3-5; -8 for making the ketose anywhere but C2.

Chemical isomerization of fructose gives approximately equal amounts of glucose and mannose.

5 5. (36 points) The three most common hexoses are D-glucose, D-fructose and D-mannose. Two structures of glucose (Fischer projection and chair form of the pyranose) are shown. A. Fructose is a ketohexose that can be made by isomerization of glucose. Draw a Fischer projection of D-fructose. 12 points; -3 for any stereochemical change at C3-5; -8 for making the ketose anywhere but C2. -5 for anything other than a Fischer projection (if otherwise correct) B. Chemical isomerization of fructose gives approximately equal amounts of glucose and mannose. Draw (in any form, correctly illustrating stereochemistry) a structure for Dmannose. 12 points (any form possible) C. Interconversion of these three uses a reaction called keto-enol tautomerization. Propose a structure for the intermediate responsible for this process. (Hint: bear in mind what bond(s) must be broken and formed in converting either glucose or mannose to fructose.) 12 points. Deprotonated forms OK.

6 6. (45 points) Cytochrome c is a protein that is involved in a wide array of electron transfer processes, including reduction of nitrite ion to ammonia by way of NO. A. Calculate the standard cell potential for reduction of nitrite to nitrogen monoxide by cytochrome c and write a balanced equation for the reaction. Useful half-cell potentials vs. SHE are: Cytochrome c(ox) + 2 e- Cytochrome c(red) NO2- + 2H+ + e- NO + H2O V V 2NO2- + 4H+ + Cytochrome c(red) 2NO + 2H2O + Cytochrome c(ox) E cell = V 10 points B. If a (biological) cell has a concentration of cytochrome c of 30.5 mm and ph 6.8, at what concentration of nitrite will reduction to NO become spontaneous? Assume the cell maintains equal amounts of oxidized and reduced forms of cytochrome c, and that [NO] = 0.1 pm (10-13 M). DG=0=DG +RTln[Cytochrome c(ox)][no]2 ; DG = -nfe cell =-2(96485 C/mol)(0.121 V) [NO2-]2[H+]4[Cytochrome c(red)] = kj/mol Solving, J/mol = RTln(1)(10-13 M)2 X2(10-6.8) = ln(10-26)(1027.4)/x2 12,141 = 101.4/X2 X = M 10 points: 7 for setting up the Nernst Equation correctly C. The primary structure of human cytochrome c is as follows (N-terminus to C-terminus): CYS ALA ASP SER ASN TYR MET ASP VAL GLN GLN LEU LEU CYS PRO ALA PHE HIS ASN PRO ASN VAL ALA LEU PRO TYR VAL HIS TYR TRP LEU SER LEU TYR TYR PHE PHE ASP ALA ALA PRO MET ARG ALA LEU ASN HIS ASN MET ARG The digestive enzyme trypsin cleaves peptides on the carboxyl end of lysine () and arginine (ARG). List the first four fragments from the N-terminus that would be generated by tryptic digestion. -ASP-VAL-- - -PHE--MET- 16 points (4 points per unit). -6 if cleaved on wrong side of. -10 if cleaved from C-terminus.

D. A picture of the structure of cytochrome c is shown; the heme unit (Fe + macrocyclic ligand) is highlighted, and the backbone is illustrated using the cartoon convention.

7 D. A picture of the structure of cytochrome c is shown; the heme unit (Fe + macrocyclic ligand) is highlighted, and the backbone is illustrated using the cartoon convention. All other bonds are present in wireframe. Describe the elements of secondary structure that can be identified from this structure, and explain what holds each in place. Use additional diagrams as necessary. Alpha helices, represented by the curliqueue cartoons. These are held together by H-bonds from on amide N-H to an amide C-O several (3.8, to be exact) residues away in the primary structure. 9 points.

EXTRA CREDIT (20 points) The nominal structure of the amino acid lysine is shown

Draw the structure of the major form of the molecule in solution at ph = 2.0. B.

that of the nextmost-abundant form (conjugate acid or conjugate base of what you

8 EXTRA CREDIT (20 points) The nominal structure of the amino acid lysine is shown below, along with the pka of the conjugate acid form of each functional group. A. Draw the structure of the major form of the molecule in solution at ph = 2.0. B. Show the calculation for the concentration ratio of the form you drew vs. that of the nextmost-abundant form (conjugate acid or conjugate base of what you drew for A above). Ka = [H+][A-] [HA] Ka = [A-] [H+] [HA] = = = 0.708

Selected data that may be of use: Physical constants: g = 9.8 m/s2 ε0 = 8.85419 10-12 C2/(Nm2) c = 2.

381 10-23 m2kg/(k-s2) h = 6.626 10-34 m2kg/s F = 96485 C/mol π = 3.14159 e = 2.

9 Selected data that may be of use: Physical constants: g = 9.8 m/s2 ε0 = C2/(Nm2) c = cm/s R = L-atm/(mol-K) = J/(mol-K) N = k = m2kg/(k-s2) h = m2kg/s F = C/mol π = e = Gravitational Constant Electric susceptibility of a vacuum Speed of light Gas constant Avogadro s Number Boltzmann constant Planck s constant Faraday s constant

10 Properties of State Species CH4 (g) H f kj/mol S J/(mol-K) O2 (g) 0 kj/mol J/(mol-K) CH3OH (l) kj/mol J/(mol-K) CH3OH (g) -205 kj/mol J/(mol-K) H2O (l) kj/mol J/(mol-K) H2O (g) kj/mol J/(mol-K) CO2 (g) kj/mol J/mol-K Ag (s) 0 kj/mol J/(mol-K) Ag+ (aq) kj/mol J/(mol-K) K+(aq) kj/mol J/(mol-K) Zn (s) 0 kj/mol J/(mol-K) Zn+2 (aq) kj/mol J/(mol-K) Cu (s) 0 kj/mol J/(mol-K) Cu+2 (aq) kj/mol J/(mol-K)

11 Electromotive series: Atomic Weights H C O Ca Mn Fe Cu Zn 65.39

Please show your work for partial credit. If you need more spacefor an answer, use the back of the page and indicate where we should look.

CH 233H First Midterm Exam Friday, April 28, 217 Name KEY Please show your work for partial credit. If you need more spacefor an answer, use the back of the page and indicate where we should look. You