CEM 852 Exam LDA, THF, 0 C, 15 min; then

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1 CEM 85 Exam- April, 005 This exam consists of 5 pages. Please write ALL your answers in the answer books. Please write legibly and draw all structures clearly. Good luck. 1. Provide examples of the following name reactions: (1 pts) a. ozaki-iyama-kishi coupling b. Glaser coupling c. aza-prins cyclization d. Paterno-üchi reaction. Provide a brief written description of the role played by each of the following reagents typically employed in the Sharpless asymmetric dihydroxylation: (3 pts) a. K s () 4 b. K 3 Fe(C) 6 c. S 3. Provide the product or products of the reactions outlined below. Show all intermediate compounds and be sure to indicate the product s relative or absolute stereochemistry. For reactions where multiple products are possible, indicate the major and minor species. (8 pts) (a) 1. LDA, TF, 0 C, 15 min; then. 3, C Cl, -78 C, 30 s 3. 3 P=CC, C Cl, C, 3 h 4. Li, - (:1), ˇC, 3 h I (b) 1. (Z)-1,4-dichloro--butene (1.0 eq.), a ( eq.), DMF. Cy, TF, 0 C; then, aac, 3. Cr 3, S 4, Et, 0 C 4. Zn, rc C Et, Et (c) Tf 1. ( 3 Sn), LiCl, cat. Pd(P 3 ) 4, TF, 60 C, 1 h. IS, C Cl, rt, h 3. t-uli, Et, -78 C; then CeCl 3, TF -78 C; then -78 C, h 1

2 (d) TMS 1. Li; then Se Se. Cu(C)Li TF, -78 C; then 4 Cl 3. LiAl 4, Et 4. Et Zn, C I, benzene, 60 C (e) 1. 3 Al, Cp ZrCl, (C Cl), 3 C, 4 h; then TMS.5 mol % of Pd (dba) 3, r ZnCl (1 eq), TF, 10 mol % tri(-furyl)phosphine]. K C 3,, 3 C, 3 h 3. uli, TF, -78 C; then I (f) n 1. S, a, /t-u, Δ.,-dimethoxypropane, cat. PCl 3 3. mcpa, -78 C 4. Ac, aac, Δ (g) Ac r 8 mol % Pd(P 3 ) 4 Et 3, C, 50 C A 6 mol % Pd(0) n-u 4 Ac 4. Provide the reagents necessary to convert the starting material to the product. Most of these transformations will require more than one step. e sure to consider the product s relative or absolute stereochemistry. To maximize your chances at partial credit, it is recommended that you show all intermediate compounds. (8 pts) (a) n (b) Si 3 Et

3 (c) Pr Pr r (d) C Ac C (e) n n n (f) TS g) C 5. eactions of A and with mcpa are complementary with regards to the direction of epoxidation. This can be explained by considering A-strain combined with the putative 10 olefin-hydroxyl bite angle. Thus, compound A prefers reactive conformation I while reacts via conformation II during its mcpa epoxidation. An examination of I and II can also explain why the mcpa epoxidation of is more selective that that of A. Please explain. (5 pts) A m-cpa C Cl 0 C I m-cpa C Cl 0 C II 3

4 6. As shown in Scheme 1, allyl or (E)-crotylboration with the, tartrate derived reagents react at the Si face of simple aliphatic or aryl aldehydes to give the corresponding homoallylic alcohols. In a related fashion the S,S tartrate derived boronates react at the e face (not shown). Scheme 1. C ipr n-c 8 17 C () () C ipr n-c 8 17 = (86% ee) = (88% ee) Similarly the (,)-(E)-crotylboronate reacts at the Si face of the optically active aldehyde shown in Scheme. In this case, Si face attack also corresponds to a Felkin approach and affords a high degree of stereocontrol (97:3). As expected the (S,S)-(E)-crotylboronate preferentially attacks the e face, but as this approach is anti-felkin a loss of selectivity is seen. Scheme. z 3,4-anti-4,5-anti diastereomer major (85:15) anti-felkin addition with respect to C C ipr C ipr z C C ipr () C ipr () z 3,4-anti-4,5-syn diastereomer major (97:3) Felkin addition with respect to C C ipr C ipr z C ipr () C ipr () z In contrast to the examples above, reaction of the alanine derived aldehyde with the (,)-(E)- crotylboronate proceeds with e face attack to give as the major product the anti-felkin product shown in Scheme 3. Explain this somewhat surprising result. (8 pts). Scheme 3. C ipr oc C () C ipr () oc oc (75:5) 4

5 7. Treatment of halide A with u 3 Sn results in a 5-exo-dig radical cyclization to give intermediate, which is capable of undergoing a 10-endo-trig/6-exo-trig radical macrocyclizationtransannulation process to afford C. 1 -M of C reveals that this compound has no vinyl protons. Provide the structures of ( pts) and C (3 pts). E E r A; E = C (Z/E 9:1) u 3 Sn, AI, 80 C 10-endo/ 6-exo radical macrocyclizationtransannulation C 8. The Colvin rearrangement is a method to convert aldehydes to alkynes. Provide a complete arrow (electron) pushing mechanism for this transformation. (ote: A 1 -M of TMSC shows two peaks one of which is at. ppm and the other at ca. 0 ppm.) (6 pts) TS C 1.4 eq. LDA, 1.4 eq. TMSC, TF, -78 C, 1 h; then 5 C, h TS 9. While the Colvin rearrangement above works well, other substrates are known to give unwanted side products. For example benzaldehyde does not rearrange to phenylethyne but instead gives benzyl alcohol and α-diazoacetophenone (Scheme 4). In another example, benzil undergoes the desired transformation, but the product is accompanied by significant amounts of pyrazole (Scheme 5). Provide a complete arrow (electron) pushing mechanism for the formation of either the pyrazole or the benzyl alcohol. (5 pts) Scheme 4. uli, TMSC, C TF, 0 C, h; C then Scheme eq. uli, 1.5 equiv. TMSC, Et, 0 C, 10 min ()C C() TMS C() onus Question: In a few hours MSU will take to the court for a semi final game in this year s CAA men s basketball championship. For a 3-point play name one organic chemistry faculty member from each of the other three final four schools: Illinois (Urbana-Champaign), Louisville, and orth Carolina (Chapel ill). 5