Problem Points Credit. 1. Nomenclature (one structure) 30

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1 alifornia State Polytechnic University, Pomona 1 hem 2010 Midterm #2 Fall, 2018 Beauchamp ame KY (int your name legibly) oblem Points redit 1. omenclature (one structure) xplain relative stabilities based on logic arguments 20 of organic chemistry 3. hair conformations or hain conformations or 25 Stereochemistry or 2/3 structures, hybridization 4 Acid/base equations, draw mechanism details, 24 estimate K eq, explain answer 5. S and mechanisms, including stereochemical 40 details 6. edict products (20) 30 7.opose short syntheses (2) arbocation rearrangement Free adicals, edict products provide mechanism 30 Total 244 This is a long exam. It has been designed so that no one question will make or break you. The best strategy is to work steadily, starting with those problems you understand best. Make sure you show all of your work. raw in any lone pairs of electrons, formal charge and curved arrows to show electron movement in mechanism and explanation problems. If resonance is part of an answer, draw the best resonance structure, plus at least one additional resonance structure to show that resonance is present. nly write answers in the space available. o your best to show me what you know in the time available. Your mission in life is not merely to survive, but to thrive; and to do so with some passion, some compassion, some humor and some style. Maya Angelou

2 alifornia State Polytechnic University, Pomona 2 1. ovide an acceptable name for the following structure. Indicate the absolute configuration of any chiral centers shown in three dimensional form (/S) and any /Z stereogenic centers. (30 pts) 2. a. What is the most stable cation? xplain your reasoning, using structures, if necessary. Include curved arrows, formal charge and lone pairs. int: onsider the resonance of a group and a lone pair of electrons. (10 pts) 2 2 The first example is very destatilizing. Two types of resonance are shown, the resonance of a nitrile group and the resonance of an allyl carbocation. This places 2 cation sites next to one another which would be highly repulsive and destabilizing. 3 3 The second example is very stabilizing. The nitrogen lone pair can share its electron density 3 2 with the carbocation site, making an additional 3 2 bond and completing the octet of the carbocation carbon. This is very good resonance. b. What is the most stable anion? xplain your reasoning, using structures, if necessary. Include lone pairs, formal charge and lone pairs. int: onsider the resonance of a group and a lone pair of electrons. (10 pts) The first example is more stable because resonance puts the negatve charge on more electronegative nitrogen The second example is less stable because anion resonance puts the negatve charge next to another lone pair of electrons which destabilizes it due to greater electron/electron repulsion. The negative charge cannot be pushed onto the nitrogen atom because it is sp 3 and has full octets.

3 alifornia State Polytechnic University, Pomona 3 3. a. Use ewman projections of the 3 4 bond of 4-phenyl-2-methylhexane to show the lowest energy and highest energy conformations and calculate the relative energies. Show the most stable conformation first. alculate a K equilibrium between the least stable and most stable conformations. Assume = 2 cal/(mol-k) and T = 300 K. (20 pts) Approximate clipsing nergy Values (kcal/mole) Some were estimated by me. most stable ewman onformation Me i- t-bu Ph Me i t-bu Ph Approximate Gauche nergy Values (kcal/mole) Some were estimated by me. Me i- t-bu Ph Me i t-bu Ph least stable ewman conformation K eq = T 0.0 Ph i = = 8.3 kcal/mole Ph i K eq = , = Ph i- = = 4-phenyl-2-methylhexane = = 1.3 x / 800,000 b. erivatives of the antitumor steroidal saponin were recently prepared. The are highly potent and selective anticancer compounds. They inhibit a + /a +2 exchange leading to higher a +2 in the cytosol and mitochrondria causing cell death (apotosis) (rg. Lett. ASAP, 2014). ircle all chiral centers and any other stereogenic features in the partial structure below, and calculate the maximum number of stereoisomers possible. (5 pts) antitumor steroidal saponin SW chiral centers 3 sugars There are even more chiral centers in the sugar(s) here, but are not shown. rg. Lett. maximum number of stereoisomers = 2 9 possible stereoisomers = 512

4 alifornia State Polytechnic University, Pomona 4 4. The reactant acids and bases are given in two acid/base equations below. Also given with each equation are two pk a values. omplete each acid/base equation including any formal charge, lone pairs and curved arrows to show how the reactants react. Use the pk a values to calculate a K eq for each reaction. ovide a very brief explanation for which side is favored. (24 pts) a. pk a = 6 K a = 10-6 This is the stronger acid because of the inductive withdrawal of the oxygen atom. This nitrogen is less willing to donate its electrons and is therefore a weaker base, which makes its conjugate acid the stronger acid. K eq = pk a values = 6 and 8 equilibrium favors the right side K a ( 3 +) K a ( 2 3 +) 10 = = pk a = 8 K a = 10-8 This is the weaker acid because of the inductive withdrawal of nitrogen is less than oxygen. This nitrogen is more willing to share its electrons and is therefore a stronger base, which makes its conjugate acid the weaker acid. b. mechanism arrows pk a values = 15 and 20 pk a =15 equilibrium favors resonance K a =10-15 the right side resonance pk a =20 A B K a =10-20 resonance arrows K eq = K a (A) K a (B) 10 = = -20 esonance in the amide anion is much better than the ketone anion resonance because the more electronegative nitrogen atom (than carbon atom) helps the oxygen atom to carry the negative charge. Since the amide conjugate base is much more stable, its acid is much stronger than the ketone acid and the equilibrium will lie far to the right. Acid/base equilibria always lie to the side opposite of the stonger acid and stronger base (right in this reaction). resonance arrows

5 alifornia State Polytechnic University, Pomona 5 5. Use 4S-bromo-5-deuteriooctane to provide a simple, arrow-pushing mechanism for each of the following reaction conditions (show curved arrows, lone pairs & formal charge). Fill in the necessary details to clearly indicate any stereochemical features and/or conformational requirements. If reactants are not drawn in the proper orientation to show how the reaction must proceed, then redraw them in a more informative way that shows this. o not consider carbocation rearrangement possibilities. (40 pts) a. raw a 2 structure and then a 3 structure of the reacting molecule. A 3 structure will be provided for the cost of the points of this part. (3 pts) structure = chiral center structure of (4S,5)-5-deuterio-4-bromooctane b. Show a mechanism for each position and simply draw all other possible reaction products (what kind?). Indicate if, Z or neither. You can abbreviate common branch names if they are not part of your mechanism There may or may not be fewer products than there are numbers. (10 pts) + 4 alkene without 2 mechanism + other possible products - 4Z alkene with 3Z alkene alkene / Z configuration / Z configuration c. Show the S reaction (what kind?), indicate the absolute configuration(s) of the center in the product. (6 pts) S 2 mechanism + 4S 4 configuration

6 alifornia State Polytechnic University, Pomona 6 d. Show all steps of the S reaction (what kind?). You can use one intermediate to show all possible S possibilities. Indicate the absolute configuration(s) of the center in the product. You can abbreviate common branch names if they are not part of your mechanism (9 pts) a 2 b - 2 S 1 + S 1 carbocation mechanism S e. Show a mechanism for two products and simply draw all other possible reaction products (you can use the same intermediate for your two mechanisms). Indicate if, Z or neither. There may or may not be fewer products than numbers. (12 pts) 2 a a 1 + related alkenes a show two 2 b related alkene b b b c here are 2 more (extra) b d d c c mechanism 2 2 Z Z Z

7 alifornia State Polytechnic University, Pomona 7 6. Write in the major product and type of reaction for each set of conditions below. Arrow pushing is not required (1.5 each, 30 pts) a c a b a 2 enantiomers S 2 d a K S 2 2 e g i k m o q s Li rearrangement 1. 3 eqs. a a 3 2. LiAl 4 3. workup,, (cat.) P 3 LiAl 4 a S 1 2 (twice) S 2 (twice) S 2 S 1 achiral 2 addition S 2 racemic f h l j n p r t S S S a a 1. Tsl, py 2. a S 2 S 1 a 2 l l S 2 o eaction (1 o neopentyl ) S 2 S acyl substitution S 2 S substitution acyl substitution

8 alifornia State Polytechnic University, Pomona 8 7. opose a reasonable synthetic sequence to make the given target molecules using the given starting materials. Show each step with an arrow and the necessary reagents to accomplish the indicated transformation. If you make a molecule in one part you can use it in any other part. (30 pts) Allowed starting materials 2 4 l a. b. Ph 2 a sodium hydride used above Ph used above 2 2 a sodium hydroxide, addition, addition a sodium cyanide a 3 sodium azide 1. LA,-78 o 2. S 2 2 Li n-butyl lithium K 2 1. LA,-78 o (made above) 2. Ph 3 S 2 K diisopropylamine 1. a 2. S 2 2 used above 2 a S 2 substitution 3 Al LA given given Li 2 K substitution 2 substitution 8. opose a complete arrow-pushing mechanism for the following transformation. (15 pts) LA 2 a Li n-butyl lithium 4 given given o carbocation 3 o carbocation 3 o carbocation with resonance

9 alifornia State Polytechnic University, Pomona 9 a. ow many different types of sp 3 hydrogen atoms are present in 2S-bromopentane? Show all possible products when 2S-bromopentane is brominated with 2 /? Use Fischer projections. Put a dot by any chiral centers. If stereoisomers form, specify what type of isomerism is present (enantiomers, diastereomers, meso compounds, achiral, etc.). Indicate the approximate relative amounts of each product formed if the relative rates of reaction of a bromine atom with an sp 3 - bond are: primary = 1, secondary = 80, tertiary = 1600 and bromine substituted carbon = (21 pts) (2S) starting structure enantiomers = none diastereomers = (, (,F) meso = () (2) achiral (2S,3) (2S,3S) (2S,3) (2S,3S) (2S) A B F G relative amounts = (# )(rel. reactivity) A = (3)(1) = 3 = = = F = (1)(80) = 80 each G = (3)(1) = 3 B = (1)(2000) = 2000 = main product b. ovide a complete arrow pushing mechanism to explain formation of the major product from the above reaction (show proper curved arrow conventions, lone pairs as two dots and single electrons as one dot). learly label each distinct part of the reaction mechanism. alculate an overall for each step of your mechanism using the given bond energies. (15 pts) o o o o o o verall reaction for main product 1. initiation 2a propagation 2b propagation 3. termination (2S) 3 (2S) 7 = = achiral, sp 2 use half-headed arrows achiral achiral, sp = +46 = -64 = -88 adds to both sides, but achiral repeat step 1 = +46 (energy supplied by light) step 3 = very negative step 2a = = 0 step 2b = = -18 It is in your moments of decision that your destiny is shaped. Tony obbins

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