Midterm II: Answer Key

Transcription

1 hemistry 7 A (Kahn, Summer 22) Midterm II: Answer Key

2 . Predict if the molecules shown below are chiral or not. If the molecule is not chiral, indicate what symmetry element is responsible for the lack of chirality. Total of 6 points. A) F I l 2 pts Achiral. This molecule has no chirality centers, and has a plane of symmetry. (Even if it were in a s-gauche conformation, the two s-gauche forms can rapidly interconvert via s-trans conformer and would be present in equal amounts). B) 3 2 pts hiral. (two chirality centers with different substituents) ) hiral. (the molecule does not have any plane or point of symmetry. It does have 3- fold axis of rotation, but axes of rotation does not prevent chirality ) ) N 3. N 3 Achiral. The molecule has a point of symmetry in the center. This was one of the examples shown in the class. 3 3 Achiral The molecule has a plane of symmetry. E) F) 3 l 3 hiral. (The molecule has two chirality centers and no planes of symmetry. Note that plane through halogens and carbons is not a symmetry plane). 2

3 2. Reactant B adds to reactant A via two parallel reactions to give two products, and : Reaction : A B Reaction 2: A B The following thermodynamic parameters were determined for these two reactions: Reaction : Reaction enthalpy: º = -. kcal/mol; reaction entropy: Sº = -5 cal mol - K - Reaction 2: Reaction enthalpy: º 2 = -.35 kcal/mol; reaction entropy: Sº 2 = -6 cal mol - K - A) Propose a hypothesis why Sº is negative in both reactions (4 points). In both reactions two reactant molecules combine to form one product molecule. Each molecule has three translational degrees of freedom and also three rotational degrees of freedom. Each of these degrees give a positive contribution to the entropy of the system (you will learn details in Phem, or see huffy.bioeng.washington.edu/~hyre/docs/_thermo.pdf) Because the number of molecules decreases in the reaction, the number of translational and rotational degrees of freedom is decreased, and thus entropy is decreased. This is always true for reactions in the gas phase on in nonpolar solvents, where the entropy of solvent does not change significantly. In strongly associating solvents, such as water, the solvent entropy may dominate, and Sº may be even positive. B) Assuming that the product distribution is entirely under thermodynamic control, calculate temperature at which products and form at equal amounts (6 points). Use: Molar gas constant: R =.987 cal mol - K - = 8.34 J mol - K - Temperature scale: T(K) = T(º) If and form at equal amounts, this means that they have identical stabilities. Identical stabilities of and imply that reaction free energies for reaction and 2 are equal T = = = 35 K S S = G rx _ = G r _ 2 T S = 2 T S 2 ) If your goal is to maximize the yield of product, would you use higher or lower temperatures? Explain why. Assume that A, B,, and do not decompose at extremely cold or extremely hot temperatures and that the activation barriers for both reactions are low (4 points). If we want to maximize yield of, we must make G more negative than G 2. There are several ways to get the right answer. ne simple way is to consider two extreme temperatures. At very low temperature ( K), G =, and reaction 2 is favored (it has more negative ). At 35 K reactions and 2 are equally favored. At very high temperatures, T*( S ) becomes larger than S 2 2 and a reaction with smaller entropy decrease is favored (remember, entropy does not want to decrease). The entropy decrease is smaller in reaction, thus reaction is more favorable at high temperature. 3

4 ) alculate the ratio of to at 5 º (6 points). K K eq eq [ ] = = e [ A][ B] [ ] = = e [ A][ B] G G2 } [ ] e = [ ] e G G = e G G2 5 º = K, =.84 kcal/mol = *.5 = kcal/mol G G [. 84 ] = e [ ] = *.6 = -7.8 kcal/mol r, expressed as a percent: 52.% ; 47.8%. Now you also see that the higher temperature favors formation of. =.9 3. Give a full systematic name specifying appropriate stereochemistry. A) (4 points) (S)-2-methyl--butanol (S)-2-methylbutan--ol (S)-2-methyl-n-butanol (Z)--chloro-3-ethyl-4-methyl-3-hexene B) l (4 points) ) (4 points) (S)-3-bromo-3-methyl--hexene (R)-5-ethyl-,5-dimethyl-,3-cyclopentadiene ) 3 (4 points) 4. 2-butene with unknown stereochemistry was reacted with deuterium gas in the presence of platinum catalyst. In reactions with alkenes, the deuterium gas ( 2 ) behaves the same way as 4

5 the normal hydrogen would. The reaction was carried out in the sample tube of a polarimeter. No change in the polarization of the light was observed during the reaction. a. escribe how polarimeters work (5 points) Polarimeter consists of a light source (typically producing monochromatic light at 589 nm), a polarizer that creates plane-polarized light, a sample tube, and an analyzer. The analyzer could be another polarizer that can be rotated to minimize the amount of light passing through. Light source Polarizer Sample tube Analyzer etector Light that enters the sample tube is plane-polarized. If the sample is not optically active, no light passes through the polarizer if analyzer is at 9 degree angle relative to the polarizer. If the sample is optically active, the plane of polarization is rotated by an angle α = [ α] l c, where [α] is the specific rotation, characteristic of the compound. Some light will pass through the polarizer. Then the analyzer is rotated by α degrees until no light passes through. b. Would such an experiment reveal if the starting 2-butene was cis or trans isomer? Justify your answer by predicting outcomes from these two possibilities. iscuss the chirality of products in both cases. ( points) No, such an experiment cannot distinguish between cis and trans 2-butane because the products are optically inactive in both cases. The addition of deuterium is syn, thus the cis isomer would yield an erythro pair. Because 2- butene is symmetric alkene, the erythro enantiomers will have plane of symmetry, and we really have a meso compound. Meso compounds are optically inactive Pt 2 The trans isomer would yield threo pair of isomers, because the addition can occur from below or above the plane of the molecule, equal amounts of two enantiomers will form. Thus, the syn addition to the trans isomer will give a racemic mixture. Racemic mixtures are optically inactive Pt c. Propose another experiment that would allow to distinguish between the cis and trans 2- butene. (5 points) ) Measure heats of hydrogenation: trans-2-butene is more stable than cis-2-butene thus the heat of hydrogenation from trans-2-butene will be less. a) Measure rates of hydrogenation, cis should be faster because it is less stable. 2) Measure a suitable physical property of 2-butene. We could easily determine boiling points (cis boils at 3.5 º, trans boils at º) or densities, or the refractive index. 2a) The meso product and threo pair of enantiomers have different physical properties. 3) No reaction with achiral reagent will yield optically active products. Some chiral reagents might work. an you think one that gives optically active products with trans butane only? 5

6 5. What product is formed in the acid-catalyzed reaction between methylpropene and isopropanol? (8 points) Give the reactants that would be required to carry out the following syntheses (5 points each): 2 -= = 2 ) B 3, TF 2) 2 2, 2, - N N 3 = mixture of stereoisomers: Ring expansion occurs via carbocation mechanism. annot use 2 2 here because no carbocation is formed in the radical mechanism 6