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Unformatted text preview: CHEM 171Q, 2004 Preliminary Exam #1, Answers 1. (25 points) Complete each of the following: (a) (10 points) Methyleneimine, which has molecular formula CNH3 , is of current interest in astrophysics and has been observed in interstellar dust clouds. Draw a Kekul structure for methyleneimine. Use lines for all covalent bonds and show all nonbonding electrons. Specify the hybridization you expect for the carbon and nitrogen atoms in your structure.
H C H N H Methyleneimine is isoelectronic with ethene,therefore, one expects a similar geometry, i.e planar with bond angles of 120. The hybridization that fits this geometry for both C and N is sp2. (b) (5 points) The octane rating of gasoline is related to the content of the C8 H18 hydrocarbon with the trivial name isooctane, which has the condensed formula (CH3 )3 CCH2 CH(CH3 )2 . Write the IUPAC name for isooctane. ______________________________________________________ (c) (10 points) It is a common misconception that the pKa of H2 O is 14.0 (deduced from Kw = [H3 O+][HO ] = 10-14; pKw = 14.0). The actual pKa for H2 O is 15.75. For a generic acid HA in aqueous solution, recall that K a(HA) is defined as follows:
HA + H2O [H3O H3O ] [A ] + A 2,2,4-Trimethylpentane Ka = [HA] -1- Write the expression for Ka(H2 O) in aqueous solution in terms of the concentrations of all relevant species.
H2O + H2O [H3O ] [A ] H3O + HO Keq = [H2O] [H2O] recall Ka = Keq [H2O] Ka = [H3O ] [A [H2O] From your expression above, show how pKa(H2 O) is related to pKw.
pKa (H2O) = Kw /[H2O] = pKw + log[H2O] since [H2O] = 55 M for neat water pKa (H2O) = pKw + log(55) = 14 + 1.75 = 15.75 not required for this problem 2. (10 points) Alcohols typically show broad characteristic infrared (IR) absorptions at 3400 cm-1 due to the O-H stretching vibration. However, if the OH group is intramolecularly hydrogen bonded to a Lewis basic group within the molecule, the IR absorption is typically shifted to slightly higher energy. The IR spectra of the two stereoisomers of butane-2,3-diol (CH3 CHOHCHOHCH3 ) both show two IR absorptions in the OH stretching region. Interestingly, the intensity of IR resonance assigned to the intramolecular, hydrogen-bonded OH group is significantly larger in one of the stereoisomers, indicating a greater fraction of the isomer exists in the hydrogen-bonded state. (a) Using Newman projections, draw the most stable conformers of the two stereoisomers of butane-2,3-diol that would be expected to show an intramolecular, hydrogen bonded OH group. It may be helpful to show the hydrogen bond in each structure with a dashed line (i.e. O-----H). Hydrogen bond can only occur between the hydrogen of one OH group and the oxygen of the other OH group if the two groups are proximal. It should be clear that this can only occur in staggered conformations in which the OH groups are gauche. Of the various staggered conformations in which the OH groups are gauche, the most stable ones will be those that minimize torsional strain. These are shown below. -2- CH3 H H CH3 O H O H CH3 H 3C H H O H O H A (b) B Which of the two isomeric conformers drawn above (A or B) would you expect to be more stable? (Hint: A CC/CC gauche interaction is more destabilizing than a CC/CO gauche interaction.) Explain your reasoning. A contains two CC/CO gauche interactions. B contains one CC/CO gauche interaction and one CC/CC gauche interaction. Since the CC/CC gauche interaction is more destabilizing than a CC/CO gauche interaction, A contains less torsional strain and is, therefore, more stable. (c) Which stereoisomer of butane-2,3-diol (A or B) would you expect to show the larger fraction of the IR resonance due to the intramolecular, hydrogen-bonded OH group? Briefly explain your reasoning. Since A and B are the most stable hydrogen bonded conformer for each stereoisomer, their relative population will be related to their relative stability. Since A is more stable than B, the equilibrium population of A will be greater than that of B, consequently the IR resonance due to the hydrogen-bonded OH group will contribute a greater fraction to the total OH stretching absorptions in the IR spectrum. 3. (25 points) Predict whether Keq is > 1 or < 1 for each of the following reactions. Provide a brief but clear explanation for each of your predictions. (This question is about thermodynamics, therefore, you do not have to worry about how a reaction occurs or whether it occurs at a significant rate.) For part (a), estimate the free energy change for the reaction (G) at 25 C. (a) (11 points)
pKa = 5 + OCH3 NH2 + HOCH3
pKa = 15 Keq G > 1 13.6 kcal/mol
-3- Methanol (HOCH3 ) has the higher pKa, thus the higher proton affinity. Therefore, the proton should be more strongly bound to the oxygen in methanol than the anilinium ion (PhNH3 + ), resulting in K > 1. Alternatively, one could recognize that Keq = 10^(pKa) = 1010 > 1. (b) (7 points)
Keq H3C O Keq < 1 The four-membered ring heterocycle (called an oxacyclobutane) is less stable due to greater bond angle strain. (c) (7 points)
H H3C H CH3
Keq H CH3 H CH3 Keq < 1 The conformer on the right contains more torsional strain: 4 gauche interactions between the methyl groups and ring carbons as well as one 1,3-eclipsing interaction between the two axial methyl groups. 4. (10 points) Predict the products of each of the reactions and use the electron-pushing formalism to show the flow of electrons as reactants are being converted to products. Make sure your arrows clearly show which bonds are being made and/or broken. Also, indicate any formal atomic charges in the reaction products. (a) (5 points)
H2SO4 + CH3OCH3 -4- H3C O H3C H O O S O OH H3C O H3C H O O S O OH (b) (5 points)
H3C C H3C H + HOCH3 H3C C H3C H HOCH3 H3C H3C C H O H CH3 5. (20 points) Propose structures for A-D that are consistent with the experimental observations. Your observation/deduction reasoning will be considered for partial credit. Compounds A, B, and C have formula C 6 H12. A and C add one equivalent of H2 in the presence of a catalyst to give D (C 6 H14). The infrared (IR) spectrum of C has an absorption at 1646 cm-1. The 13C NMR spectra of A and D each show two resonances (i.e. two kinds of carbons). The 13C NMR spectrum of B shows one resonance. Observation A/B/C, C6 H12 A/C + H2 D D, C6 H14 C, IR 1646 cm -1
13C Deduction A, B, and C are isomers, each with 1 degree of unsaturation A and C have the same carbon connectivity 0 degrees of unsaturation C=C both A and D must have some symmetry because of 6 carbons there are only two types all 6 carbons in B are equivalent -5- NMR A/D, 2 resonances 13C NMR B, 1 resonance A fl B fl C fl D fl -6- ...
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