CH251Midterm2AU2009csc

CH251Midterm2AU2009csc - Callam Chemistry 251 Autumn...

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Unformatted text preview: Callam Chemistry 251 Autumn 2009 Midterm 2 200 points IH 100 – Tuesday, November 17, 2009 Name (PRINT)_____________________________________________________________ I have neither given or received aid on this exam(SIGN)___________________________________________________________ Place an X in the column next to your recitation section. Failure to mark your correct recitation section will result in –5 points from your exam total. Day Time Room Teaching Assistant Tuesday Tuesday Thursday Thursday Tuesday Tuesday Thursday Thursday Tuesday Tuesday Thursday Thursday 1:30 pm – 2:18 pm 1:30 pm – 2:18 pm 1:30 pm – 2:18 pm 1:30 pm – 2:18 pm 2:30 pm – 3:18 pm 2:30 pm – 3:18 pm 2:30 pm – 3:18 pm 2:30 pm – 3:18 pm 3:30 pm – 4:18 pm 3:30 pm – 4:18 pm 3:30 pm – 4:18 pm 3:30 pm – 4:18 pm AV214 MP1008 SM2186 MP1008 SM2186 MP1008 SM2186 MP1008 KL330 AV214 KL330 AV214 Christopher McDaniel Chad Eichman Christopher McDaniel Chad Eichman Christopher McDaniel Chad Eichman Christopher McDaniel Chad Eichman Chad Eichman Christopher McDaniel Chad Eichman Christopher McDaniel Section 1 2 3 4 5 6 7 Extra Credit Total Points 15 15 30 20 40 40 40 10 200 Score Please read all of the directions for each section of the exam carefully. Place an X in the lower left hand corner of page 3 for 3 points extra credit. There are 12 pages including this cover page for Midterm #2, please check to make sure you have all of the pages. Make sure your answers are clear. Any structures or answers that are unclear or ambiguous are considered wrong. 1/12 Points Earned = Callam Chemistry 251 Autumn 2009 Part 1. Ranking. Rank the following compounds as described in each section. (3 points each for a total of 15 points) (a) Rank the following alkyl halides in order of increasing SN2 reactivity. (1 = least reactive, 3 = most reactive) Br Br Br (b) Rank the following alkyl halides in order of increasing SN1 reactivity. (1 = least reactive, 3 = most reactive) O CH3 Br Br Br (c) Rank the following in order of increasing nucleophilic strength in a polar protic solvent. (1 = weakest nucleophile, 3 = strongest nucleophile) CH3 H3C H3C O H3C O HC S 3 (d) Rank the following in order of leaving group ability. (1 = worst leaving group, 3 = best leaving group) OO OO O S S O O F 3C O F 3C (e) Rank the following in order of increasing nucleophilic strength in a polar protic solvent. (1 = weakest nucleophile, 3 = strongest nucleophile) H N H H H P H H H S H 2/12 Points Earned = Callam Chemistry 251 Autumn 2009 Part 2. Nomenclature. (3 points each for a total of 15 points) (a) For each of the following structures provide the IUPAC name. F Cl Br (b) For each of the following IUPAC names draw the skeletal structure in the space provided below. (R) ­2 ­bromo ­octane (2R,3S) ­2 ­ethyl ­3 ­methylhexane 3/12 Points Earned = Callam Chemistry 251 Autumn 2009 Part 3. Stereochemistry. For each of the following pairs of molecules identify them as one of the following: constitutional isomers (I), same molecule (S), enantiomers (E), diastereomers (D), or none of the above (N) by placing the appropriate letter in the space provided. (3 points each for a total of 30 points) CH3 Cl (a) H3C Cl F CH3 F (f) Br CH3 Br OH (b) OH (g) Cl H D O CH3 H D O CH3 Cl (c) H HO OH O OH H OH HO H OH H OH O OH H 3C O (h ) H3C OH CH3 O OH H 3C O Br (d) H 3C Cl CH3 H 3C Cl Br CH3 O (i ) OH HO O OH HO O H (e) H 3C Br O CH3 Br H OCH3 CH3 (j) O H H CH3 NH2 OH CH3 H 3C O NH2 CH3 OH 4/12 Points Earned = Callam Chemistry 251 Autumn 2009 Part 4. Stereochemistry. Answer the following questions base on the information given below. When optically pure amygdalin is subjected to acid and water it forms three organic products (mandelic acid, D ­glucose, and D ­ galactose) (20 points total) OH HO HO HO HO OH O O HO OH NC O O OH HCl, H2O HO HO 6 5 O 1 OH 43 2 6 5 O 1 OH 4 32 O Mandelic Acid HO HO OH OH OH OH D-glucose Amygdalin D-galactose (a) How many stereocenters are present in amygdalin? What is the total number of stereoisomers for amygdalin? (6 points) (b) Draw the (R) enantiomer of mandelic acid. (3 points) (c) D ­glucose has the R configuration at C ­5 and all of the other groups occupy the equatorial position. Draw the chair conformation for D ­glucose with the correct stereochemistry. (3 points) (d) Pure (R) ­mandelic acid has a specific rotation of  ­154°. If a sample contains 60% of the (R) ­mandelic acid and 40% of (S) ­mandelic acid, what is the e.e. and observed rotation of the solution? (4 points) (e) Calculate the e.e. of a solution of mandelic acid having an observed rotation of +65°. What is the percentage of each enantiomer present? (4 points) 5/12 Points Earned = Callam Chemistry 251 Autumn 2009 Part 5. Predict the Products. Predict the organic product(s) of the following reactions. When appropriate, be sure to indicate stereochemistry. If more than one product is formed be sure to indicate the major product, if stereoisomers are produced in the reaction be sure to indicate the relationship between them. Draw all answers in skeletal form. You will be graded on the product you draw from the reaction no other information is needed for this question. (4 points each for a total of 40 points) O Cl (a) Br NaCN, H 3C N H CH3 H (b) H O O CF3 S O O LiN3, (c ) H H3CO CH3 Br H CH3 1. NaI, acetone O 2. NaCN, H N Product for (c) must be drawn as a Fischer projection (d) Br2, hv , CCl4 Br (e) OH Cl NaOH, H2O, THF 6/12 Points Earned = Callam Chemistry 251 Autumn 2009 O OH (f ) OH 1. NaH (excess), 2. Br (excess) O NaSeCH2CH3, (g ) Cl ( h) 2 O O S CF3 O N aO ONa Br H2NCH2CH2NH2 (i ) Br NaHCO3 (2 equiv) CH3CH2OH (j ) Br D OH 7/12 Points Earned = Callam Chemistry 251 Autumn 2009 Part 6. Mechanisms and Energy Diagrams. (30 points) (a) Provide the mechanism for the reaction shown below. Be sure to include all intermediates, formal charges and arrows representing electron movement. (10 points) CH3 CH3 CH3 H 2O Cl OH OH + (b) Draw the energy diagram for the first step of the reaction above. Label the axes, starting material, product, transition state, Ea, and ΔH°. (5 points) (c) Draw the transition state for first step of the above reaction. Be sure to include relative lengths of bonds breaking and forming, hybridization, and partial formal charges. (5 points) 8/12 Points Earned = Callam Chemistry 251 Autumn 2009 Part 6. Mechanisms and Energy Diagrams. CONTINUED (d) Provide the mechanism for the reaction shown below. Be sure to include all intermediates, formal charges and arrows representing electron movement. Be sure to include all of the initiation, propagation and termination steps. (10 points) Br2, hv, CCl4 Br (e) Using Hammond’s postulate, and the above reaction mechanism explain why chlorination would yield multiple organic products with 3 ­methylpentane while bromination yields 3 ­bromo ­3 ­methylpentane almost exclusively. Use a minimum of words. (10 points) 9/12 Points Earned = Callam Chemistry 251 Autumn 2009 Part 7. Explain. (a) When (S) ­2 ­bromopropanoic acid (1) reacts with concentrated sodium hydroxide, the product formed (after acidification) is (R) ­2 ­hydroxypropanoic acid (2 R). When the same reaction is carried out with a low concentration of hydroxide ion in the presence of Ag2O (where Ag+ acts as a Lewis acid), it takes place with overall retention of configuration to produce (S) ­2 ­hydroxypropanoic acid (2 S). Write detailed mechanisms that account for the formation of these different products. (10 points) O O O 1. Ag2O, NaOH 1. NaOH (excess) H H H O O O 2. HCl, H2O 2. HCl, H2O Br OH OH (b) Reaction of (R) ­sec ­butylamine with a racemic mixture of 2 ­phenylpropanioc acid forms two products having different melting points. Draw the structures of these two products Assign R and S to any stereogenic centers in the products. How are the two products related? (10 points) 2-S 1 2-R OH + O 2-phenylpropanoic acid NH2 (R)-sec-butylamine 10/12 Points Earned = Callam Chemistry 251 Autumn 2009 (c) A mixture of two diastereomers of compound 3 were reacted with sodium bicarbonate (NaHCO3) to produce two organic products (4 and 5). Compounds 4 and 5 have different molecular formula. Predict the structures of compounds 4 and 5 and explain their formation with a mechanism. (The squiggly bond indicates the undefined stereochemistry in the starting material.) (10 points) NaHCO3 + I O 3 Different Molecular Formulas OH 4 5 (d) Answer the question based on the information presented below. (i) Compound 6 is a recently discovered superelectrophile. Compound 6 can be prepared from compound 5 by using reagent X. Propose the structure of reagent X. I Reagent X CH3 CH3 N N N N CH3 I CH3 (ii) When compound 6 is reacted with triethylphosphine (7), two organic products are formed propose their structures. P I CH3 N I 6 11/12 5 6 N 7 CH3 Points Earned = Callam Chemistry 251 Autumn 2009 Extra Credit. (10 points) (a) A limited number of chiral compounds having no stereogenic centers exist. For example X is achiral, constitutional isomer Y is chiral. Explain. (5 points) H3C H3C X H CCC H H CCC H3C Y H CH3 (b) Would you expect the A value for the t ­butyl group on molecule B to be more or less than the A value for tert ­ butylcyclohexane? Explain your answer using pictures and words. (5 points) O O tert-butylcyclohexane B 1 2 3 4 I H Li Na K II Be Mg CS Partial Periodic Table III IV B C Al Si Ga Ge V N P Ar VI O S Se VII F Cl Br VIII He Ne Ar Kr 12/12 Points Earned = ...
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