Team Number: High School Name: Team Round. Please write your full name, ID number, and high school at the top of every page.

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1 Team Number: High School Name: Washington University Chemistry Tournament April 2, 2016 Team Round Please write your full name, ID number, and high school at the top of every page. 50 minutes is allowed for this examination. This exam is 7 questions long, and has 18 numbered pages in total. Please check to make sure that your exam is complete, and report to the exam proctors if any pages are missing. Necessary equations and constants, as well as a periodic table, can be found at the end of the exam. Do not write on the scoring sheet on the last page of the exam. Scratch paper will not be permitted during this exam. If you run out of room on the front of the page, you can continue to work on the back of the page, provided that all answers for parts are clearly labelled. Only write work on the backs of pages that correspond to the question on the front of the page. Make sure to circle or box your final answer if appropriate. Correct answers with appropriate work will receive full credit. If work contains reasoning or justification that is partially correct, partial credit may be awarded. Explanations should fully answer the question and provide supporting evidence and logic. These should also be given in complete sentences. Correct answers without reasonable supporting work will not receive credit. If you cannot answer a question or the entire question, it is advised that you do not spend too much time on that question and proceed onto other parts of the exam. No electronics of any kind are allowed during the exam, with the exception of a non-programmable scientific calculator. Cell phones must be turned off, and watches must be removed. A clock will be projected in the exam room. The scoring policy can be found on the WUCT website. Any appeals must be made in writing in the appeals room, Lab Sciences 400. Cheating will not be tolerated on this exam. The cheating policy that has been listed on the WUCT website will be followed. Violators of this policy will be referred to the directors of the competition. 1

2 1. By some unfortunate turn of events, you have found yourself locked inside of a Menards while picking up some compact fluorescent light bulbs for your home on a late night the store has closed while you were shopping, and the manager has turned off the power to the automatic doors. Luckily, you are able to remember some electrochemistry from your high school chemistry course and think you can create a battery that would be able to create enough power to open the automatic door, bringing you to freedom! The temperature is currently 25 C. Walking around the empty store, you procure some copper wiring, a retro tin kitchen backsplash square, a lead fishing weight, and a 3" x 3" Zinc T-Plate (whatever that is used for). You also find two alligator clip leads to connect your battery, 1 kg of kosher salt, a pack of 24 plastic water bottles, a 12-pack of coca cola, drink glasses, a pack of paper towels, and sulfuric acid (!). The table of standard reduction potentials is shown below. You only have enough materials to make a single battery. Half Reaction E (in Volts) e Cu (s) Cu (aq) e Pb (s) Pb (aq) e Sn (s) Sn (aq) e Zn (s) Zn (aq) a. You plan to make the most powerful standard battery possible using only these materials. What is the maximum voltage that can be produced? Assume all conditions are standard. Zinc and Copper battery is most powerful possible E cell = E reduction + E oxidation E cell = 0.34 V V = 1.10 V (This question continues on the next page) 2

3 b. Propose a plan for this battery. Draw a picture of your battery, making sure to clearly label each of the following parts, and indicate what available materials you would use to make each of those. i. Anode electrode ii. Anode solution iii. Cathode electrode iv. Cathode solution v. Salt bridge vi. Container for anode and cathode vii. Connection from electrodes to door viii. Direction of conventional current flow Zinc T-Plate as anode Anode solution is zinc (II) solution, made by dissolving some zinc in acid Copper wiring as cathode Cathode solution is copper (II) solution, made by dissolving some copper in acid. Drink glasses as containers for anode and cathode and respective solutions Alligator clips used to connect electrodes to door Paper towel soaked in salt water serves as salt bridge Current flows from cathode to anode 3

4 2. Examine the table below. Trouton s rule indicates that most standard entropies of vaporization of liquids are close to 85 J/mol*K. Why are liquids such as water, methanol, ethanol, and ammonia exceptions to Trouton s rule? Briefly justify your answer on the lines provided below. Liquid o ΔS vap (J/mol K) acetone 88.3 ammonia 97.6 argon 74 Benzene 87.2 Ethanol 124 Mercury 94.2 Methane 73 Methanol 105 Water 109 Water, Methanol, ethanol and ammonia are exceptions to Trouton s Rule, because they experience hydrogen bonding. Thus, the liquid phase of such liquids is much more ordered than other liquids, so the change of entropy to reach the gas phase is higher than for other liquids. 4

5 3. Dimethyl Sulfoxide (DMSO) is an extremely useful reagent used in research laboratories when scientists perform certain reactions. DMSO is used to inhibit the formation of secondary structures in DNA primers and templates that may interfere with successful amplification of a DNA sequence. a. The chemical formula of DMSO is (CH3)2SO. The connectivity of the atoms in DMSO has been replicated below four times. Use the connectivity to complete all major Lewis structures for DMSO, indicating all lone pairs and non-zero formal charges. It is possible that not all the diagrams will be used. One of the resonance forms has charge separation between sulfur and oxygen. Circle this form. 2 major structures; one with double bond between S and O, all neutral, 2 lone pairs on oxygen, one on sulfur Other has single bond between S and O, O has negative charge, S has positive, 3 lone pairs on oxygen, one on sulfur, student should circle this form b. Let s suppose that we examined trimethyl sulfoxide, a molecule similar to DMSO. In trimethyl sulfoxide, there were three methyl (-CH3) groups attached to the central sulfur atom. Would this molecule or DMSO have a more polar S-O bond? Explain your reasoning in 2 complete sentences or less, on the lines provided below. (Electronegativity of Oxygen is 3.5; Electronegativity of Sulfur is 2.5). More polar: methyl groups are electron donating, so they help give electron density to the sulfur, meaning that there is less disparity between the electron density on the oxygen and that of the sulfur. So trimethyl sulfoxide is less polar. 5

6 c. Dimethyl ketone has a similar structure to DMSO except that the central sulfur atom is replaced by a carbon atom. Given this information, determine which of the two molecules will have a greater bond angle. Explain your reasoning on the lines provided below. Dimethyl ketone has a greater bond angle. DMSO has a lone pair on the sulfur, giving it a tetrahedral geometry; on the other hand, the central carbon in dimethyl ketone has no lone pair, so it has a trigonal planar geometry. Thus, bond angle in DMSO is close to and in dimethyl ketone it is closer to 120. d. Ka is a measure of acidity of a proton. More acidic protons would have greater Ka s, whereas less acidic protons would have lower Ka s. However, chemists usually express acidity in terms of pka, which is equal to the negative log (Ka). The pka of the methyl groups in DMSO is 35, suggesting that the hydrogens are weakly acidic. Given the following bases and their Kb s, how many reagents could deprotonate the hydrogen atom? Justify your answer on the lines provided below. Base Kb Methylamine 4.4 x10-4 Ethylamine 5.6 x10-4 Caffeine 4.1 x10-4 Pyridine 1.7 x10-9 Potassium Hydroxide LDA The conjugate acid of LDA has a pka of 36, and pka of DMSO methyl protons is 35.This means that LDA is a strong enough base to deprotonate anything with a pka below 36. So LDA is the only reagent here that is strong enough to do this deprotonation. 6

7 4. You decide to investigate the properties of the carbonic acid bicarbonate buffer system in vitro. The most important buffer in maintaining ph in the blood is the carbonic acid hydrogen carbonate buffer system as illustrated below: + H (aq) + HCO 3 (aq) H 2CO 3(aq) H 2 O (l) + CO 2(g) An equation for the ph of blood in terms of the concentration of [CO2] and [HCO3 - ] can be derived. We, the question writing committee of WUCT, have performed this dangerous and exhausting derivation. The result follows: ph = p ( K a K 2 ) log [CO 2] [HCO 3 ] a. Hyperventilation can result from strenuous exercise, anxiety, high fever, or high sodium diet. How does this affect the ph of blood? Justify your answer on the lines provided below using the equation derived in part (a). Hyperventilation dramatically increases both oxygen intake and carbon dioxide output, increasing the exchange between oxygen and carbon dioxide in blood. Carbon dioxide is higher in concentration in the blood, so more will leave as ventilation increases. As a result, concentration of CO2 decreases, so ph increases. b. A common technique we use with hyperventilating patients is to have them breathe in a paper bag. Why does this work? Explain your answer in 3 complete sentences or less on the lines provided below. By breathing in a bag, a patient is essentially breathing in air that they previously breathed out; this air has a relatively high concentration of carbon dioxide. As a result, CO2 reuptake occurs through respiration, decreasing the exchange between CO2 and O2 in blood, and ph increases less (This question continues on the next page) 7

8 c. Consider the following reaction for the dissociation of carbonic acid in water: + H 2 CO 3(aq) + H 2 O (l) H 3 O (aq) + HCO 3 (aq). Is the reaction exothermic or endothermic? Use the following information in your explanation on the lines provided below: Ka (25 o C) = 4.5 x 10-7 Ka (180 o C) = 1.6 x G = H T S RTln(K) = H T S RTln(K) = T S H ln( ) = 298 S H ln( ) = 353 S H Subtracting the two we get = 55 S S = 545 J mol K H = 298 ( ) ln( ) = 126 kj mol exothermic Or also accept explanation that Ka decreases with increasing temperature, so that implies heat acts as a product. This means the dissociation is exothermic. 8

9 5. In organic chemistry, there are two main types of substitution reactions that a compound can undergo. In a substitution reaction, a covalent bond is broken to allow for another group to make a bond there using these methods (and many others), organic chemists can start to build up molecules that can be useful for and in a multitude of things, such as pharmaceuticals, dyes, and fragrances. Shown below are simple examples of the two types of substitution, called bimolecular nucleophilic substitution (SN2) and unimolecular nucleophilic substitution (SN1), respectively (R or R just indicates an alkyl group there is no need to worry about what this alkyl group could be). A compound listed above the reaction arrow acts as a reactant for that reaction. SN2: Alkyl halide Alcohol SN1: Alkyl halide Carbocation Alcohol In both of these reactions, bromine (Br) is getting replaced by a hydroxyl group (OH). The reactions may look identical but they are not. One reaction rate depends on both the concentration of the nucleophile (either - OH or H2O these reactants have the same purpose for both SN2 and SN1 reactions and can be treated as essentially the same) and the concentration of the alkyl halide. The other rate depends only on the concentration of the alkyl halide. (This question continues on the next page.) 9

10 a. Using the fact that the equilibrium reaction in the SN1 reaction is the slow, ratedetermining step, come up with rate equations for both SN2 and SN1 reactions, making sure to include k, the rate constant. Your answer should look something like rate = k[a][b][etc.]. S N 2: rate = k[alkyl halide][ OH] (any reasonable form accepted, i.e. putting structures inside brackets, writing hydroxide, etc.) S N 1: rate = k[alkyl halide] (any reasonable form accepted, i.e. putting structure of alkyl halide inside brackets) b. Classify the rates you have written above as zero order, first order, or second order. S N 2: second order S N 1: first order c. Why would the first equilibrium reaction of the SN1 substitution reaction be the rate-determining step? (Hint: look at the products of this first reaction compared to the reactants). Justify your answer on the lines provided below. The products of the equilibrium reaction are not as stable as the reactants because they consist of a carbocation, which does not have a full octet. This makes the equilibrium very slow, and therefore rate-determining. (reasonable answers accepted, use judgement) (This question continues on the next page.) 10

11 d. For the equilibrium discussed in part (c), is ΔS positive or negative? Given that this equilibrium reaction favors the reactants, is ΔH positive or negative? Justify your answer on the lines provided below. ΔS is positive the products are more disordered than the reactants because there are more molecules. ΔG is also positive, because the reactants are favored. G = H T S, so H must be positive. e. For the equilibrium discussed in part (d), is it possible to make this reaction spontaneous, and if so, are high temperatures or low temperatures better for spontaneity? Justify your answer on the lines provided below. G = H T S, and since H is positive, and S is positive, a high enough temperature would result in G being negative, so the reaction would become spontaneous at high temperature. 11

12 6. A single bond is composed of one ϭ (sigma) bond. A double bond consists of one ϭ bond and one π (pi) bond. A triple bond consists of one ϭ bond and two π bonds. Furthermore, a ϭ bond is significantly stronger than a π bond. Based on the above information and your understanding of ϭ and π bonds, answer the following questions: a. Rank the single, double, and triple bonds in terms of strength from strongest to weakest. Triple > Double > Single b. Rank single, double, and triple bonds in terms of length from longest to shortest. Single > Double > Triple c. Amides are used in the formation of fabrics and nylons. Consider the structure of an amide on the left; the R groups are irrelevant to the following question..... Compare the rotational energy barrier of the C-N bond in the amide to the rotational energy barrier of the C-N bond in the amine on the right. Resonance structures and/or other diagrams may be used in your answer. The rotation energy barrier of the CN bond in the amide is much greater than that of the amine. This is because the amine does not have any significant resonance contributors. The amide can have a resonance structure with a double bond between the carbon and nitrogen (C=N), forming a single bond between the carbon and oxygen. This leads to partial double bond character in the amide CN bond, and double bonds are planar, so the rotational energy barrier is higher. 12

13 d. Many molecules with double and/or triple bonds are linear; that is, all atoms lie in the same plane (examples include acetylene C2H2 and CO2). That being said, are any of the 3 possible isomers of C3H4 linear molecules? Explain why or why not? No, none are linear molecules. For the isomer with two double bonds, the terminal CH2 groups will lie in two perpendicular planes. This is caused by the fact that the two double bonds on the central carbon cannot lie in the same plane. For the isomer with an alkyne, the methyl group is not in the same plane. For the cyclical alkene, the hydrogens do not lie in the same plane. 13

14 7. Consider the following reaction profiles that start with the same reactant. For the following parts, please round the energies of Reactant C, Product A, Product B, and the two reaction profile transition states to the nearest integer. E a (Product A) E a (Product B) a. We can classify the two profiles as thermodynamic control or kinetic control. Kinetic control is when the product forms more quickly, whereas thermodynamic control is when the product forms more slowly but is also more stable. Given this information, which of the reaction profiles corresponds to thermodynamic control? Which of the reaction profiles corresponds to kinetic control? Explain your answers on the lines provided below. The reaction profile that produces product A corresponds to kinetic control, because its activation energy is lower, so it would form faster. But its overall energy is higher so it is less stable in the end. The reaction profile that produces product B corresponds to thermodynamic control because it occurs slower (higher activation energy) and is more stable (lower overall energy). A catalyst b. What can we add to lower the activation energy of the reaction that forms product B so that it is similar to that of the reaction that forms product A? (This question continues on the next page) 14

15 c. Which product would be favored if Reactant C was subjected to high temperatures for a long time? Explain your answer. Thermodynamic control will apply here, because high temperature means a higher amount of reactants will have sufficient energy to overcome any activation energy, making both reactions fast. Since product B is more stable, it will be favored. d. What product would be favored if Reactant C was subjected to lower temperatures for a short period of time? Explain your answer. Kinetic control will apply here, because during a short period of time, with lower temperature, fewer molecules will have enough energy to overcome activation energy barriers, so the magnitude of the barrier becomes more important. Thus, product A will form faster in this situation because it has a lower Ea. e. For the following question, round the energies of Reactant C and Product B to the nearest integer. Assuming that Reactant C is in equilibrium with only Product B, calculate the ratio of B to C at equilibrium. Round your answer to the nearest integer. 1 kcal corresponds to kj. At equilibrium, the activation energy and reaction rates do not matter. The difference in free energy is the only factor to consider. The reaction is as follows: Product C Product B Energy of product B is 2 kcal/mol and energy of product C is 5 kcal/mol. This corresponds to a change in free energy of -3 kcal/mole, or G = kj/mol. K eq = e G RT = e kj mol kj K = [B] [C] = mol K 298 K = (End of Exam) 15