1 Page of 18
THE UNIVERSITY OF BRITISH COLUMBIA
CHEMISTRY 123 FINAL EXAMINATION This examination consists of 18 numbered pages. PLEASE CHECK THAT YOU HAVE A COMPLETE PAPER 27 April 2010 TIME LIMIT: 2.5 HOURS
SURNAME: ________________________________ GIVEN NAME(S): __________ANSWER KEY__________ (IN INK) (CAPITALS) (IN INK) STUDENT NUMBER: ________________________ (IN INK) SIGNATURE: _______________________________ (IN INK)
The only calculator allowed is the Sharp EL-510R. All other calculators will be confiscated. Cell phones or other electronic communication devices are not permitted. Molecular models are allowed.
Lecture Section (check your section)
___ ___ ___ ___ ___ ___ ___ ___ 201 (MWF 202 (MWF 203 (MWF 210 (MWF 211 (MWF 299 (T,Th 266 (T,Th 222 (T,Th 1:00) Bizzotto/Love 2:00) MacFarlane/Ruddick 3:00) MacFarlane/Ruddick 10:00) Patey/Sherman 11:00) Patey/Sherman 9:30) Monga 11:00) Lekhi/Stewart 2:00) Lekhi/Stewart
Question 1 2 3 4 5 6 7 8 9 10 11 12 13 TOTAL
Maximum 6 10 16 6 2 3 2 3 4 16 20 10 12 110
Obtained
Initials
REGULATIONS FOR EXAMINATIONS
1. Each candidate must be prepared to produce upon request, a Library/AMS card for identification. 2. Candidates are not permitted to ask questions of the invigilators, except in cases of supposed errors or ambiguities in examination questions. 3. No candidates shall be permitted to enter the examination room after the expiration of one half hour from the scheduled starting time, or to leave during the first half hour of the examination. 4. Candidates guilty of any of the following, or similar, dishonest practices shall be immediately dismissed from the examination and shall be liable to disciplinary action: (a) Having at the place of writing any books, papers or memoranda, calculators, audio or visual cassette players or other memory aid devices, other than those authorized by the examiners. (b) Speaking or communicating with other candidates. (c) Purposely exposing written papers to the view of other candidates. The plea of accident or forgetfulness shall not be received. 5. Candidates must not destroy or mutilate any examination material; must hand in all examination papers; and must not take any examination material from the examination room without permission of the invigilator.
*****ADDITIONAL DATA***** Attached (last page) is an Equation Sheet and a Periodic Table. You may carefully remove the last page.
*ANSWER ALL QUESTIONS*
Page 2 of 18 1. [6 marks] Explain the following events using the concept of phase equilibria. You may use point form and include sketches, if you wish. a. An excited Science student explained the "ice drop" demonstration done in his Chemistry 123 class to his biochemistry professor, but forgot to explain why it worked. The Science student explained that in the "ice drop" demonstration a 1 L block of ice was placed on an ice block holder. Two metal weights (~20 lbs each) connected by a fine metal wire were hung over the ice block (see figure 1). After approximately 30 minutes, the weights crashed to the floor having passed through the ice block. The ice block itself was left intact. Not wanting to be outdone by chemists, the biochemistry professor decided to improve on the demo by replacing the usual ice, H2O(s), with dry ice, CO2(s). To the professor's great puzzlement the demo failed again and again!
1 L Ice block
Ice block holder Weights Figure 1: Ice block demonstration
Ice block holder
Recall that the solid-liquid phase equilibrium line for water has a negative slope with respect to pressure, so when ice is compressed it melts even though the temperature does NOT change. In the ice drop demonstation, the pressure of the wire over the ice block melts the ice. Once the wire passes through the ice, the water freezes again as the temperature is still OC, leaving the ice block in- tact. Dry ice on the other hand, does not melt when compressed because of its positive slope of the solidliquid CO2 phase equilibria.
Page 3 of 18 b. A chemistry professor with warm hands carefully removed a sample of ethane in a sealed tube from a cool bench drawer. The professor and students could easily see that the bottom half of the tube contained liquid, and the top half contained vapour. But after the professor had held the tube for a few minutes, the students noticed that any distinction between liquid and vapour had completely disappeared, and the tube was now completely filled with a uniform fluid. Initially, the temperature is below the critical temperature, Tc, and the liquid and vapor phases coexist in equilibrium. The heat from the professor's hands increases the temperature such that the temperature increases beyond Tc. At this point, the two phases no longer co-exist. Instead, there is a uniform supercritical fluid.
2. [10 marks] Consider the nucleophilic substitution reaction:
CH 3 Br ( g ) + Cl - (aq ) CH 3Cl ( g ) + Br - (aq ) (1)
CH3Br (g) CH3Cl (g) CH3I (g) I (aq) Cl (aq) Br (aq)
H0f, 298 K (kJ mol-1) 35.4 81.9 14.4 55.2 167.2 121.6
G0f, 298 K (kJ mol-1) 246.4 234.6 254.1 51.6 56.5 82.4
a. Use the data provided to show that reaction (1) is not spontaneous under standard conditions at 298 K. Assume a constant pressure process. To show this we must calculate G298 K: G 0 298 K = G 0 ( products ) - G 0 (react ) f f = (234.6 kJ mol -1 + (82.4 kJ mol -1 ) - [ (246.4 kJ mol -1 + (56.5 kJ mol -1 )] =14.1 kJ Since G298 K > 0, the reaction is non-spontaneous.
Page 4 of 18 b. Draw a reaction coordinate diagram on the axis below that is consistent with reaction (1). Label the reactants and products. Assume there is a kinetic barrier to the reaction.
Standard Free Energy
Products
14.1 kJ (see (a)) Reactants Reaction Coordinate
c. What is H0 at 298 K of the reaction: 3H 2 ( g ) + Br2 (l ) + 2C ( s, graphite) 2CH 3 Br ( g ) (2)
The H0 for this reaction is 2 Hf0 (CH3Br (g)) i.e. twice the standard enthalpy of formation of CH3Br. From the data table, this is: 2 mol ( 35.4 kJ mol-1) = 70.8 kJ
d. An eager student suggests the following two step synthesis to form the products in reaction (1):
CH 3 Br ( g ) + I - (aq ) CH 3 I ( g ) + Br - (aq ) CH 3 I ( g ) + Cl (aq ) CH 3Cl ( g ) + I (aq )
- -
(3) (4)
From the point of view of thermodynamics, would you expect this two-step process to be an improvement over reaction (1)? Explain your answer. (Hint: This question does not require calculations)
Because G is a state function and the net reaction of (3) + (4) is the same as reaction (1), G for (3) + (4) must be the same as G for reaction (1), which was calculated to be 14.1 kJ in part (a). Thus, this would not be an improvement.
Page 5 of 18 3. [16 marks] 100 litres of a 0.1 M hydrazine solution is prepared to remove dissolved oxygen in a large industrial water boiler system. The Kb of hydrazine (H2N-NH2) at 298 K is 9.810-7. Neglect the possibility that hydrazine may accept a second proton. a. Write down the balanced reaction for hydrazine with water. There are no easily dissociable H in H2N-NH2 but H2N-NH2 can accept a proton:
b. For the reaction in part (a): i. which is the the weaker of the two acids? ___________water_______________________
ii. which is the weaker of the two bases? _______________H2N-NH2______________________
c. Does the equilibrium position of the reaction in part (a) lie to the left or right at 298 K?
______Left____________________
d. Calculate the pH of the hydrazine solution. This is a base hydrolysis reaction. The concentrations at equilibrium are:
[ H N - NH 3 ][OH - ] x2 Kb = 2 = = 9.8 10 -7 [ H 2 N - NH 2 ] 0 .1 - x Assume x << 0.1 M x = [OH-] = 3.13 10-4 M pOH = -log (3.13 10-4) = 3.5 pH = 14 pOH = 10.5
+
Page 6 of 18 e. Is the solution in part (d) a buffer? Explain. No, there is not an appreciable amount of both weak base and its conjugate acid. It only contains an appreciable amount of the weak base.
f. To reduce the pH, 2 M nitric acid (HNO3) is added to the hydrazine solution. Calculate the volume of 2 M nitric acid solution needed to achieve a pH of 8. The addition of 2 M HNO3 (strong acid) to hydrazine (weak base) is a neutralization reaction but there is not so much acid as to completely consume the weak base so the resulting solution is a buffer. Let x be the initial mols of HNO3 added. Since HNO3 is a strong acid, it completely reacts with the weak base:
Solving for x:
[ H 3O + ] = K a
[ Acid ] [ Base] [ Base] pH = pK a + log [ Acid ] x 8 = 7.99 + log 10 - x x = 4.95 mols
The volume of 2M HNO3 used: 4.95 mols 2M = 2.47 L
Page 7 of 18 4. [6 marks] A new type of rechargeable battery is being developed, the ZnCd battery. The battery is operated under constant and high pressure. During operation, the Zn electrode is oxidized to Zn(OH)2(gel) and the Cd(OH)2(gel) is reduced at the Cd electrode in a concentrated KOH electrolyte separated by an ionic exchange membrane: Zn( s ) + 2OH - (aq ) Zn(OH ) 2 ( gel ) + 2e - Cd (OH ) 2 ( gel ) + 2e - Cd ( s ) + 2OH - (aq ) NOTE: A gel is neither a pure liquid nor a pure solid. a. These batteries will be used over a wide temperature range, so the determination of H0rxn is important. This can be easily estimated simple from measurements of the cell potential at different temperatures. The standard cell potentials were 0.434 V at 298 K and 0.440 V at 423 K. Use these values to calculate the H0rxn for the overall cell reaction. Clearly state any approximations you have made. To calculate H0rxn, can use the equation: ln K1 H 0 1 1 = ( - ) K2 R T2 T1
This equation presumes that H0, S0 of the reaction are not dependant on the temperature. To determine the equilibrium constants at the two temperatures, use the relationship: G = RT ln K = - nFE 0 Rewriting this relationship in terms of K and solving for K298: nFE 0 ln K = RT (2mol )(96,500 C mol -1 )(0.434V ) ln K 298 = (8.314 J K -1 mol -1 )(298 K )
= 33.80 K =e
33.80
= 4.78 1014
Similarly, one can calculate K423 to be 3.07 1010. From here H0 can be determined:
K1 H 0 1 1 ln = ( - ) K2 R T2 T1 4.78 1014 H 0 1 1 = ( - ) 10 R 423 298 3.07 10
ln
H 0 = -81.0 kJ
Page 8 of 18 b. Notice that the standard cell potential increases with an increase in temperature. Using your knowledge of thermodynamics, what can you conclude about the sign of S0 for the overall cell reaction? Provide a rationale for your choice of sign. If the cell potential increases with increasing temperature, this means that G is decreasing with increasing temperature since G = nF . At the two temperatures: G298 = H 298 S G423 = H 423 S. Now assuming H and S are independent of temperature, and G298 K > G423, then 298 S > 423 S This can only be true if S is positive. 5. [2 marks] Name the following compound according to IUPAC nomenclature. Designate the stereochemistry where appropriate.
CH3 F
(3Z,5R)-3-fluoro-5-methyl-3-octene
6. [3 marks] Given that the pKa of H2 is 36 and the pKa of CH3OH is 16, explain why bubbles form when NaH (sodium hydride) is added to a flask containing methanol maintained at room temperature. Provide a balanced equation and indicate which way the position of the equilibrium lies to help illustrate your explanation.
7. [2 marks] A carboxylic acid with the molecular formula C3H5O2Br is optically active. Draw the structure of the R-isomer.
O
OH Br
Page 9 of 18 8. [3 marks] Menthol (C10H20O) is one of the most important flavouring chemicals used in pharmaceuticals, cosmetics, toothpaste, chewing gum, etc. A number of possible stereoisomers of menthol exist in nature. Draw the lowest energy chair conformation of the naturally occurring menthol stereoisomer shown below. Lowest Energy conformer:
OH CH3
CH3
H3C
Menthol
9.
[4 marks] Consider the following reactions:
i.
ii.
iii. a. Which of the above reactions would proceed the fastest? Briefly explain your answer.
II HS- is the best nucleophile of the three because S is more polarizable than O.
b. Which of the above reactions would proceed the slowest? Briefly explain your answer III H2O is the poorest nucleophile of the three because it is neutral and O is less polarizable than S.
Page 10 of 18 10. [16 marks] Below are eight pairs of structural formulas. In the box to the right of each pair, place the number that corresponds to the term describing the relationship between the two structures. NOTE: Each term may be used more than once and not all terms need to be used. 1. 2. 3. 4. 5. 6. Conformers Constitutional isomers Diastereomers Enantiomers Identical None of the above relationships
Page 11 of 18 1. 2. 3. 4. 5. 6. Conformers Constitutional isomers Diastereomers Enantiomers Identical None of the above relationships
Page 12 of 18 11. [20 marks] (-)-Paroxetine, also known as Paxil (shown below) is a selective serotonin re-uptake inhibitor that is used as an antidepressant. (-)-Paroxetine has affinity for various protein receptors.
a. Draw the enantiomer of (-)-Paroxetine.
b. Would you expect (-)-Paroxetine and its enantiomer to have the same biological effect? (In other words, would (-)-Paroxetine and its enantiomer have the same interaction with the same protein target?) Briefly justify your answer. No: enantiomers have different shapes and fit into the enzyme site/biological receptor (which are chiral) in a different manner. Hence their interaction with the same protein target will probably be different.
c. Would you expect (-)-Paroxetine and its enantiomer to have the same boiling point? Briefly justify your answer. Yes, enantiomers have the same chemical and physical properties except optical activity.
Page 13 of 18 d. (-)-Paroxetine has a specific rotation of -24.3 at 298 K. What is the specific rotation of its enantiomer under the same conditions? +24.3 e. What would be the specific rotation of a sample that contains 39% of (-)-Paroxetine and 61% of its enantiomer? Observed specific rotation = (0.39) (-24.3) + (0.61) (24.3) = -9.477 + 14.823 = 5.346 f. What is the free energy difference (Go) between a 1 M solution of (-)-Paroxetine and a 1 M solution of its enantiomer at 298 K in the identical achiral solvent. Since the two compounds are enantiomers, they have the same chemical and physical properties. Hence, G0(+) = G0(-) Therefore, G0 = 0 g. Assuming racemization is possible, calculate G at 298 K for the process of bringing the solution in part (e) to thermodynamic equilibrium in the same achiral solvent as in part (f). In part (e), the solution contains 39% of (-) enantiomer and 61% (+) enantiomer. At equilibrium, the solution will contain 50% of both (-) and (+) enantiomer. Using 1 mole of compound, the process can be described as: 0.39 (-) + 0.61 (+) 0.50 (-) + 0.50 (+)
Where the stoichiometric coefficients are chosen to keep the number of moles consistent with the initial mole fractions and those of the racemate. Q for this reaction can be written as:
Q=
[(+)]0.50 [(-)]0.50 (0.50) 0.50 (0.50) 0.50 = = 0.97589 [(+ )]0.61[(-)]0.39 (0.61) 0.61 (0.39) 0.39
G = G 0 + RT ln Q = 0 + (8.314 J K -1mol -1 )(298 K ) ln(0.97589) = -60.4 J
Page 14 of 18 12. [10 marks] For each reaction shown below, draw the structure of the major organic product(s). Show stereochemistry where appropriate. a.
b.
OH OH
concentrated HCl O
The product has a chemical formula of C8H16O
c.
Br NaCN DMF (solvent) CN
d.
CH3OH Br OCH3
*
OCH3
Page 15 of 18 13. [12 marks] The optically active compound shown below was found to react to give different products, depending on whether or not sodium iodide (NaI) was added to the solution. Without added NaI, the major product is A. Note that NaCl is not soluble in acetone. Note that the pKa of a protonated alcohol [ROH2]+ is -2 and the pKa of HCl is -7.
a. Draw the mechanism for the formation of A. Do not draw transition states.
HO Cl
CH3
O
H
HCO3
O CH3
CH3
Page 16 of 18 b. When a small amount of NaI is added, B forms in addition to A. Draw the mechanism for the formation of B. Do not draw transition states.
Page 17 of 18 c. If a sufficient quantity of NaI is used, the resulting solution is optically inactive. Explain using a mechanism to illustrate your rationale. The resulting solution of A and B is a racemic mixture.
Page 18 of 18
Equation Sheet
1. E = q + w or U = q + w 2. H = E + PV or H = U + PV 3. G = H - TS 4. Greaction = Greaction + RT ln Q
5. Greaction = - RT ln K
9.
=
RT ln K nF
10. R = 8.314 J K-1 mol-1 = 1.987 cal K-1 mol-1 = 0.0821 L atm K-1 mol-1 11. 1 L atm = 101.3 J = 24.2 cal 12. F = 96,500 coulombs mol-1 13. 1 J = 1 volt coulomb 14. Kw = 1.00 10-14 at 25C (298.15 K) 15. Kelvins = degrees Celsius + 273.15 16. 1 atm = 760 mmHg
K H 1 1 - 6. ln 1 = K R T2 T1 2
7. G = wel = - nF 8. = -
RT ln Q nF
PERIODIC TABLE OF THE ELEMENTS
Group 1
1
17
1
18 2
H
1.008
2
4
13
5
14
6
15
7
16
8
H
1.008
He
4.003
3
9
10
Li
6.941
Be
9.012
B
10. 811
C
12.011
N
1 4.007
O
15.999
F
18.998
Ne
20.17 9
11
22.99
12
13
14
15
16
17
18
Na Mg
24.305
3
21
4
22
5
23
6
24
7
25
54.938
8
26
5 5.847
9
27
10
28
11
29
12
30
Al
26. 982
Si
28.086
P
3 0.974
S
32.064
Cl
35.453
Ar
39.94 8
19
20
31
32
33
34
35
36
K
3 9.098
Ca
40.08
Sc
44.956
Ti
47.9
V
50.941
Cr
51. 996
Mn Fe
43
"(98)"
Co
58.933
Ni
58.7
Cu
63.54 6
Zn
65.38
Ga
69 .72
Ge
72.59
As
7 4.922
Se
78.96
Br
79.904
Kr
83.8
37
38
39
40
41
92.906
42
95 .94
44
45
46
47
107.8 7
48
112.41
49
50
51
52
53
54
Rb
8 5.468
Sr
87.62
Y
88.906
Zr
91.22
Nb Mo Tc
73 74 75
Ru
1 01.07
Rh
102.9
Pd
106.4
Ag Cd
79
196.9 7
In
114.82
Sn
118.69
Sb
1 21.75
Te
127.6
I
126.9
Xe
131.3
55
56
137.33
57
138.91
72
178.4 9
76
77
78
80
200.59
81
82
83
84
85
86
Cs
132.9
Ba La* Hf
88
226.03
Ta
180.95
W
183 .85
Re
186.21
Os
190.2
Ir
192.22
Pt
195.09
Au Hg
Tl
204 .37
Pb
207.2
Bi
2 08.98
Po
"(209)"
At
"(210)'
Rn
"(222 )"
87
89
227.03
104
1 05
106
107
108
109
Fr
223
Ra Ac# Rf
58
Db
{261}
Sg
60
144 .24
Bh
61
145
Hs
62
150.4
Mt
63
151.96
59
64
65
66
162.5
67
164 .93
68
69
1 68.93
70
173.04
71
* #
Ce
140.1 2
Pr
140.91
Nd Pm Sm Eu
92 93 94
244
Gd
157.25
Tb
158.9 2
Dy Ho
98 99
Er
167.26
Tm Yb
101
258
Lu
174.97
90
91
95
243
96
247
97
247
100
257
102
259
103
Th
232.0 4
Pa
231.04
U
238 .03
Np
237.05
Pu Am Cm Bk
Cf
251
Es
2 52
Fm Md No
Lr
260