W10.215.Exam2_KEY - l. (20 points) There are two possible...

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Unformatted text preview: l. (20 points) There are two possible enolates (B and C) that can result from the deprotonation of (S)-2-ethylcyclohexanone (A). An energy diagram for these two pathways is shown below. i) Which of these enolates is expected to form if Compound A is added to a cold solution of lithium diisopropyl amide (LDA)? Briefly state the reason, and be sure to include structure-related issues in your reply. Which enolate forms? circle one C equal Briefly state the reason: Under irreverable base conditions, the deprotonation favors the less substituted protons when the pKa values are about Name W10.215.Exam2.p1 reaction pathway ii.) If enolate C is combined with bromoethane, an achiral product is formed. Draw it: iii.) Intermediate C is referred to as the thermodynamic enolate. As you can see, although it is more stable than intermediate B, it involves a higher activation energy to get there. Given the fact that intermediate B (called the kinetic enolate) always forms faster than intermediate C, briefly explain how it is that intermediate C can EVER form at all? Conditions such as a higher temperature and/or substiometric amount of base allow for an equilibrium of the enolate formed, which favors the more stable thermodynamic enolate. NOTE: Answers that only refer to the relative stability of B versus C are not worth points because they restate the given information. iv.) Complete the following reaction by providing the products. Indicate stereochemistry for the reaction. Do not use abbreviations. G) 09 Li NOTE: the products are stereoisomers ||. (20 points) Name W10.215.Exam2.p2 A.) The following questions refer to the shown protons on molecules A, B, C, and D. i.) Which molecule has the most acidic protons? ii.) Which molecule has the least acidic protons? iii.) Which molecule is more acidic, B or D? Circle the more acidic molecule. B.) J. Org. Chem, 1969, 34, 2724. Three membered nitrogen rings (aziridines) are found in the mitomycin family of natural products, which have shown antitumor acitivity. Provide a mechanism for the following reaction, which constructs an aziridine ring. Note that carbon nitrogen TE-bOl'ldS can undergo addition reactions of enolates in exactly the same way as carbon oxygen n-bonds. N0 0 KGB 90k ; l + BrdL OCH2CH3 CH3OCH2CHZOCH3 OCH20H3 (solvent) 0 C—enolate is also acceptable +2 for arrows O 0 K63 90k 0' E: t B N r\7]\OCH2CHS e k +2 for arrows Q 0 +1 for intermediates Q a. —Dlv OCHZCH3 OCHZCHg 0 +2 for arrows +1 for intermediates 0 C.) Provide the IUPAC name or structure for the following compounds. (S)-ethyl 4-hydroxyhexanoate (S)-3-bromohexanoic acid -1 pt for each error -1 pt for each error Ill. (27 points) Name W10.215.Exam2.p3 A.) Liquid crystalline compounds may have a wide range of uses such as media for optical data storage. (Po/ym. Prepr. 1996, 37, 125). The product of the following set of transformations demonstrates liquid crystalline properties when heated. Complete the following reactions by providing starting material, reactants (number experimental steps as necessary), or products. 1. Mg, or2 equiv Li C02 H3CO@Br —> 2. H3O" NOTE: carboxylic aci formation can be A O SOCI2 o H3C04©—< —> H3C04©—< OH Cl < O o H O Hsco©—<O O 1/2 equiv H @0043 + N<CH20H3)3 O HO OH C23H1807 IT B.) Treatment of 2,5-diethylcyclohexanone with sodium methoxide results in the conversion from the cis- isomer to predominantly the trans- isomer. / I...~ N aOC H3 / 1,," / HOCH3 . I O O - CIS- Isomer trans- Isomer For the following questions refer to the above isomerization. i.) Draw the most stable chair ii.) Draw the most stable chair iii.) Explain why the product is conformation of the starting conformation of the product. favored. matenal' H - the base is used to enolize the or ketone and thus go from axial- H H equatorial subsitution pattern to O H W a equatorial-equatorial H O subsitution pattern, which is O H lower In energy F 3 3 iv.) Draw the step-wise mechanism for the conversion of the cis- isomer to the trans- isomer. / (‘ NaOCH3 o ., H CH3 OCH (0 H 3 o- / + 2 for arrows —> + 2 for intermediate 0 + 2 for arrows IV. (28 points) Name W10.215.Exam2.p4 A.) Complete the following reactions by providing starting material, reactants (number experimental steps as necessary), or products. Indicate stereochemistry for the reactions and if more than one stereoisomer is formed, draw one structure and write "+enantiomer" or "+diastereomer." Specify if reagents are used catalytically. Do not use abbreviations. i.) Chem. Comm, 1967, 753. Na+ O O" O 1) excess LiAlH4 2 equiv AOMOA /\o / 2) H3O+ Br / /\O O 3) 2 equiv F>Br3 Br Cel‘HoBQ O Ho / H2O, cat. HCl, heat CH3OH 0 cat. H2804 + 2 C02 0 OH 4 + 4 4 O Hsceo H3CO / / 1.) NaH 0 —> 2) H 0+ 0 + t- HsCO O - 3 enan Iomer C11H1603 ii.) Tetrahedron: Asymmetry, 1996, 7, 1005. cat. HCl —> ,. ‘,.- / \ W ,CH3 0 O C16H23NO2 F iii.) J. Amer. Chem. 800., 2002, 124, 12078. 1) 0504 0 cat. HCl —> \/\)J\OCH3 2) H3C3k 0ng o o ox \‘ W CH3 0 + enantiomer C9H1604 4 3) H30+ (workup) H3C + enantiomer V. (25 points) Name W10.215.Exam2.p5 A.) Draw the product in which all enolizable protons are exchanged with deuterium by the treatment of 2-(3-oxopentyl)cyclohexanone with D20. You do not need to worry about stereochemistry. D D 0 D20 0 D DD DD (excess) 0 O 4 B.) Treatment of 2-(3-oxopentyl)cyclohexanone with KOH results in an intramolecularAIdoI condensation reaction to form a product containing two six-membered rings as shown below: 0 KOH O —’ 90% product A 2-(3-oxopenz‘yl)cyc/ohexanone O i.) 2-(3-Oxopenty|)cyclohexanone has multiple enolizable sites that are all energetically similar. How many total intramolecular aldol condensation products could be formed (including the one given)? ii.) Draw only one of the other possible aldol condensation products. 03% «as iii.) In 15 words or less explain why only one aldol condensation product is formed. The other aldol condensation products form strained rings or the product is more stable C.) Draw the stepwise mechanism for the above reaction going through the aldol intermediate. Use 'OH as your base and H20 as your conjugate acid. You do not need to worry about stereochemistry. /\H H r?) 90 (I) ( —> O H C O C-enolate is also acceptable 6 9 + H20 F @OH + 3 points for each step LOH H O 0 Draw the aldol intermediate. ( 90H -E2 mechanism also allowed OH 7 (b r O 9 0%} ...
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W10.215.Exam2_KEY - l. (20 points) There are two possible...

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