MgBr 17 H 16 N 2 O 3 HO N. no partial. 4 each; can be either order -2 if steps unnumbered 1) CH 3 OH/TsOH. 1) NaH 2) CH 3 I

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1 I. (43 points) ame Page EP1 omplete the following reaction schemes as necessary. Sequential experimental steps should be numbered appropriately! omplete structures should be shown. (a) J equiv. Mgr ) 3 3 2) (b) L The product of this reaction is (check one): a single, achiral compound X a racemic mixture a mixture of diastereomers 5 3 (c) J (d) ibid (e) J ) excess 4 each; can be either order -2 if steps unnumbered 1) 3 /Ts 2) l 1) a 2) 3 I 3 3 in pyridine 2) l / acetic acid 2 3 the sole product (with no by-products) resulting from the nucleophilic opening of a lactone draw the expected major anomer l

2 II. (40 points) ame Page EP2 A. Provide the complete, step-wise mechanism for the following observation (J ). 3 3 ( 3 ) 3 ( 3 ) 3 3 ( 3 ) 3 + ( 3 ) 3 3 addition reaction mechanism = 4 tetrahedral intermediate = 4 elimination mechanism I = 4 alkoxide ion = 4 elimination mechanism II = 4 + ( 3 ) (a) Provide the complete IUPA name (including stereochemical configuration) for the following: (S)-2-methoxy-2-methylhexanoic acid class of : stereochem, punct, numbering, prefixes, name 3 3 (b) Draw the structures for the following: the 2 most favorable enol forms of 2,4-dimethyl-1,3-cyclohexanedione

3 ame III. (42 points) Page EP3 omplete the following reaction schemes as necessary. Sequential experimental steps should be numbered appropriately! ull structures must be shown. (a) J ) a 3 3 ( 3 ) 3 2) r 3 ( 3 ) (b) (c) ursin is -cell differentiation hormone; it is a tripeptide: LysisGly; draw the major structure of LysisGly that is expected to be present at low p (for instance, p < 1). Use the ischer projections S 4 (d) The tripeptide ValProPro is a hypotensive agent. omplete the following synthesis of ValProPro; protection and/or deprotection steps should be shown separately; use ischer projections. 2 2) 2 /Pd- 2 1) R could be ( 3 ) 3 anh r Pro- ( 3 ) 2 4 with added ValProPro (at its isoelectric point) D: 4 2 /Pd- 4) 3 2 3) R anh r if ( 3 ) Pro- used 4 3 hydrolysis 3

4 ame IV. (37 points) Page EP4 You recall the acid-catalyzed aldol reaction, right? It has appeared on each of the past two examinations this semester. irst you show the mechanism of formation for the reactive (enol) form, then you show its acid-catalyzed addition reaction to give the!-hydroxy-carbonyl (or aldol) addition reaction product. In the aldol condensation, as you will recall, the!-hydroxy-carbonyl undergoes an acid-catalyzed E2 elimination reaction. Provide the complete, step-wise mechanism for the entire sequence: (a) formation of the enol; (b) addition (aldol) reaction; and (c) elimination of water As always, you may use - and for any ronstead acid and base you need. formation of enol: protonation mech = 3 intermediate = 4 deprotonation mech = 3 formation of aldol: protonation mech = 3 intermediate I = 4 addition of enol = 3 intermediate II = 4 deprotonation = 3 E2 elimination: protonation mech = 3 intermediate = 4 elimination mech =

5 V. (30 points) ame Page EP5 A. When the following epoxide is treated with acetic acid, a 2-step mechanism is proposed to explain the following observation. Provide the complete, step-wise mechanism using only these two reagents - and the information that the process only requires 2 steps protonation of epoxide = 5 intermediate = 5 attack by acetate = Acylating agents differ in reactivity based on the exact combination of atoms that make up the acylating agent. Methoxycarbonyl chloride (see below), for example, is observed to be a less reactive acylating agent than acetyl chloride (see below). Provide words and pictures that explain why the methoxycarbonyl chloride is a less reactive acylating agent than acetyl chloride (J ). l 3 methoxycarbonyl chloride l 3 acetyl chloride pictures = 3; words = 4 l 3. What is the starting material needed for the following enolate intramolecular substitution reaction (J )? l [( 3 ) 2 ] 2 Li 2 2 (LDA) + Lil + [( 3 ) 2 ] l 3 resonance from the 3 group decreases the partial plus of the carbonyl carbon, lowering it reactivity 7

6 VI. (4 points) ame Page EP A. Provide the following structures, as described. (a) The open chain, ischer projection (b) The beta-anomer resulting from the reaction for D-mannose, the -2 epimer of between D-glucopyranose and 2 under D-glucose. acidic conditions. (c) The open chain, ischer projection for the reaction product between D-fructose and (d) The open chain, ischer projection for the reaction product resulting from the a 4 reduction of L-galactose, the -4 epimer of L-glucose.. Years ago, while studying carbohydrate chemistry, someone discovered that treating ordinary table sugar (sucrose) with chlorine (l 2 ) gives the trichloro derivative shown below. This substance is now widely marketed as the non-caloric artificial sweetener called "Splenda" (or sucralose). In animal testing, it is noted that all of the sucralose that is fed to test subjects is excreted as exactly the same structure, unchanged. Sucrose, on the other hand, is rapidly hydrolyzed into its monosaccharide components and considered a source of dietary carbohydrate. What is the simplest explanation for why the "Splenda" molecule does not undergo hydrolysis, even though it is so structurally similar to sucrose? sucrose 2 l 2 2 l l l sucralose ("Splenda") 2 2 enzyme-substrate specificity biochemical reactions require enzymes as catalysts; the "Splenda" molecule must be different enough, structurally, that the enzyme that works on sucrose does not work on sucralose; this might be the lack of -bonding from the missing 's, it might be the sterics of the chlorine atoms groups, etc. basic idea of E-S specificity =