Download CH 233H Final Exam Tuesday, June 9, 2015

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.
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.
DGf° (298K): -218.5
-101.7
12.6
-38.9
-2.64 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: -218.5 - {-2.64 + 3(-38.9)} = -99.1 kJ/mol
1,3-cyclohexadiene to cyclohexane: -218.5 - {12.6 + 2(-38.9)} = -153.3 kJ/mol
Cyclohexene to cyclohexane: -218.5 - {-101.7 + -38.9} = -77.9 kJ/mol
You may, of course, calculate DHr° and DSr° for each reaction before calculating each
free energy change:
Benzene to cyclohexane: DHr° = -206.7 kJ/mol; DSr° = -361.5 J/(mol-K)
1,3-cyclohexadiene to cyclohexane: DHr° = -229.1 kJ/mol; DSr° = -254.8 J/(mol-K)
Cyclohexene to cyclohexane: DHr° = -119.9 kJ/mol; DSr° = -141.4 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.)
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.
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. (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
+0.254 V
+0.375 V
2NO2- + 4H+ + Cytochrome c(red) → 2NO + 2H2O + Cytochrome c(ox)
E°cell = + 0.121 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)]
= -23.3 kJ/mol
Solving,
23300 J/mol = RTln(1)(10-13 M)2
X2(10-6.8)4
9.40 = ln(10-26)(1027.4)/X2
12,141 = 101.4/X2
X = 0.045 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):
GLY
CYS
LYS
THR
LYS
GLU
LYS
ALA
ASP
SER
THR
GLY
ASN
TYR
MET
ASP
VAL
GLN
GLY
GLN
LYS
LEU
ILE
LEU
GLU
CYS
PRO
ALA
GLY
GLU
PHE
ILE
LYS
HIS
ASN
PRO
ILE
ASN
VAL
ALA
GLY
THR
LEU
GLY
ILE
PRO
GLY
TYR
LYS
VAL
HIS
TYR
TRP
LYS
ILE
LEU
LYS
GLU
GLY
SER
GLY
LYS
LYS
LYS
ILE
LYS
LEU
TYR
GLU
TYR
LYS
LYS
PHE
GLY
PHE
THR
ASP
ILE
LYS
ALA
ILE
GLY
GLY
ALA
THR
PRO
GLU
THR
MET
LYS
ARG
ALA
LEU
GLY
GLU
ASN
LYS
HIS
LYS
ASN
MET
THR
ARG
GLU
The digestive enzyme trypsin cleaves peptides on the carboxyl end of lysine (LYS) and
arginine (ARG). List the first four fragments from the N-terminus that would be generated by
tryptic digestion.
GLY-ASP-VAL-GLU-LYS
GLY-LYS
LYS
ILE-PHE-ILE-MET-LYS
16 points (4 points per unit). -6 if cleaved on wrong side of LYS. -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. 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 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]
= 10-2.15
10-2
= 10-0.15 = 0.708
Selected data that may be of use:
Physical constants:
g = 9.8 m/s2
ε0 = 8.85419 × 10-12 C2/(Nm2)
c = 2.99792458 × 1010 cm/s
R = 0.08206 L-atm/(mol-K) = 8.314 J/(mol-K)
N = 6.022 × 1023
k = 1.381 × 10-23 m2kg/(K-s2)
h = 6.626 × 10-34 m2kg/s
F = 96485 C/mol
π= 3.14159
e = 2.71828
Gravitational Constant
Electric susceptibility of a vacuum
Speed of light
Gas constant
Avogadro’s Number
Boltzmann constant
Planck’s constant
Faraday’s constant
Properties of State
Species
CH4 (g)
H°f
-74.6 kJ/mol
S°
188.7 J/(mol-K)
O2 (g)
0 kJ/mol
205.1 J/(mol-K)
CH3OH (l)
-238.4 kJ/mol
127.2 J/(mol-K)
CH3OH (g)
-205 kJ/mol
239.9 J/(mol-K)
H2O (l)
-285.8 kJ/mol
69.91 J/(mol-K)
H2O (g)
-241.8 kJ/mol
188.8 J/(mol-K)
CO2 (g)
-393.5 kJ/mol
213.7 J/mol-K
Ag (s)
0 kJ/mol
42.55 J/(mol-K)
Ag+ (aq)
105.6 kJ/mol
72.68 J/(mol-K)
K+(aq)
-254.4 kJ/mol
102.5 J/(mol-K)
Zn (s)
0 kJ/mol
41.63 J/(mol-K)
Zn+2 (aq)
-153.9 kJ/mol
112.1 J/(mol-K)
Cu (s)
0 kJ/mol
33.15 J/(mol-K)
Cu+2 (aq)
64.77 kJ/mol
-99.6 J/(mol-K)
Electromotive series:
Atomic Weights
H
1.008
C
12.011
O
15.998
Ca
40.070
Mn
54.938
Fe
55.845
Cu
63.546
Zn
65.39