pract_final_key - KEY Massachusetts Institute of Technology...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: KEY Massachusetts Institute of Technology 5.13: Organic Chemistry II December 19, 2005 Final Exam Question 1 __________/10 points Question 2 __________/15 points Question 3 __________/30 points Question 4 __________/10 points Question 5 __________/10 points Question 6 __________/15 points Question 7 __________/10 points Question 8 __________/12 points Question 9 __________/10 points Question 10 __________/12 points Question 11 __________/12 points Question 12 __________/12 points Question 13 __________/12 points Question 14 __________/14 points Question 15 __________/16 points TOTAL _________/200 points Name (printed) ________________________________ Name (signed) __________________________________ T.A _____________________ There are 18 pages (2-19) of questions in this exam. 1 1. (10 points total) Write an arrow-pushing mechanism for the reaction below. Note: Aste risk(*)=13C. H + heat O O O O H Me O O O O N H Me O * O R2N+H2 + O + O * O H OH N H Me O H*H OH H N H CO2 IMINIUM FORMATION +3 * OH2 H N Me CO2 * Aza-cope +4 O N Me O Me OH * H*H HO N Me * CO2 CO2 N Me H2O HYDROLYSIS +3 N O O H H2O * N Me CO2 13 Solution must account for C in formaldyde, otherwise no more than 5 pts should be awarded. 2 Figure by MIT OCW. 2. (15 points total) Compound A is prepared from B and C and has the spectroscopic data listed below. Draw the structure of A in the box provided, and write an arrow-pushing mechanism for its formation from B and C in the space below. 2. points 2. points Ph OH O + N N + O Catalytic H Heat CH3 B O O N N A C N N 1. point OH Ph O H N O Ph N O H O Partial credit for another mechanism leading to the right molecule that doesn't include [3.3] sigmatropic REARR: [0-3 points]. ∆ O N N 7 O 2. points Data for A: 1H NMR (ppm) IR 7.05-7.15, m, 5H 1685 cm-1 5.80, t, J = 6.3, 1H 3.67, t, J = 6.5, 4H Molecular weight 3.47. t , J = 6.5, 4H 3.22, d, J = 6.3, 2H 2.34, t, J = 7.4, 2H 273.17 2.12, t, J = 7.4, 2H 1.71, s, 3H O O O Ph N O Ph O MECHANISM 8 Ph Claisen Rearr O N O Ph N Ph O O 3 Figure by MIT OCW. 3. (30 points total, 1 point per box) For the following 15 structures, write the number of chemically non-equivalent (number of “different types”) of hydrogens and carbons in the appropriate boxes below. (Be careful to put the numbers in the correct boxes – we can’t read your mind, i.e. wrong numbers will receive no credit – no exceptions.) # non-equivalent H # non-equivalent C CH3 a. 2 All or nothing 3 CH3 Cl Br b. 4 c. CH3 CH2 d. CH3 CH2 2 CH2 Me 3 1 Me 2 e. 2 2 CH3 2 2 CH3 6 Me 3 Me Me Me Figure by MIT OCW. 4 # non-equivalent H f. Me Me Me 3 1 Me g. i. 3 Me HO 2 Me 1 6 Me H 7 3 6 7 3 4 5 7 5 7 O Me 2 h. 3 Me 1 2 # non-equivalent C 4 5 3 4 5 6 H 32 4 Me 7 H Me 1 2 1 j. 65 7 3 4 Figure by MIT OCW. 5 # non-equivalent H O k. 2 O N Me H 3 Me Me N 3 (DIASTEREOTOPICITY) O m. 2 restricted rotation O i. (3 was accepted) # non-equivalent C O 9 8 Me Me (Me's are diastereotopic) O n. Me 6 4 1 1 Me o. (10e-, aromatic) Figure by MIT OCW. 6 4. (10 points) An alcohol (R-OH) was treated with sodium hydride and 1-bromo-2-butyne to give compound D (molecular weight = 166.10). Using the 1H NMR data listed below, determine the structure of the product and the starting alcohol. Draw the structures in the boxes provided. 1. NaH, THF 2. 1-bromo-2-butyne 1 chirality 2pts. R-OH OH 5.68, ddd, J= 17.0, 10.5, 8.5, 1H 5.27, dd, J=10.5, 1.5, 1H 5.19, dd, J=17.0, 1.5, 1H 4.14, d, J=15.0, 1H 3.94, d, J=15.0, 1H 3.42, d, J=8.5, 1H 1.86, s, 3H 0.91, s, 9H Me O draw D here draw R-OH here H NMR data for D (ppm) = From ROH = From ALKYNE Correct ether synthesis +2 pts. CH3 Br + Me RO RO HH RO Has : 9 H S 2,(tBu) 2pts. 2pts. 3 H IN Alkene region, All Coupled to each other 2H, S OR 2 1H IF R IS CHIRAL WHICH IT IS IN THIS CASE H H H 2pts. Add'l 1H is coupled to 1 Alkene H H H H H Z Y Y and Z have no H directly attached. 3.42ppm O CH3 ONLY tBr and O Remain Figure by MIT OCW. 7 5. (10 points) At room temperature, compound E is converted to compound F in high yield. Using the data provided, determine the structure of F (and draw the structure in the box provided), and write an arrow-pushing mechanism for its formation from E. H2N N O S NO2 Me O 23 0C Me O Ph or E H N H Ph SO2 N Data for F : +3 Me Ph Ph coupled to * + 4 if not concerted but correct arrow pushing over all O Me IR 1H NMR (ppm) 1955 CM 7.15-7.30, m, 5H 5.99, d, J = 2.0, 1H 5.25, dq, J =7.0, 2.0, 1H 1.59, d, J =7.0, 3H Molecular weight + 2 for allene -1 allene 130.19 CH3 C10H10 = MW 130 + HN=N-SO2Br Me Ph CORRECT ARROW PUSHING AND STRUCTURE =10 pts 1 OF ABOVE CORRECT: 5pts PARTIAL CREDIT FOR PARTIALLY CORRECT MECH AND/OR STRUCTURE BOTH OK. Figure by MIT OCW. 8 6. (15 points) Propose a synthesis of G from H, maleic anhydride, and benzyl bromide (BnBr = PhCH2Br). (All of the substituents on the five-membered ring in G are cis to one another, and your synthesis must establish this relative configuration.) Your synthesis must use H, maleic anhydride, and BnBr. You may use any other reagents in addition to these. Write your synthesis neatly in the forward direction, and for each transformation, write the reagents necessary over the arrow. BnO BnO OBn OBn BnO BnO O O H H BnO BnO OBn OBn O O O O O O +3 pts +3 pts H H H H H H OBn OBn OHC OHC O O +2 pts +2 pts O O O O +2 pts +2 pts H H O O maleic anhydride maleic anhydride +6 pts +6 pts H H endo endo H H 2 H H away from CH2OBn away from CH2OBn OH OH (1) O3 (1) O3 (2) Me2SS (2) Me O O OBn OBn OBn OBn G G (all cis) (all cis) O O BnO BnO LiAlH4 LiAlH4 CHO CHO O O All H's are CIS Au H' s CiS O O OBn OBn HO HO OH OH OH OH OH OH MANY POSSIBLE CORRECT SEQUENCES AFTER MANY POSSIBLE CORRECT SEQUENCES AFTER DIELS DIELS - ALDER +2 pts +2 pts NaH NaH BnBr BnBr G G (x.s. of each) 9 Figure by MIT OCW. (7) (2 points for each box; 10 points total) Please provide the indicated information. If you use a base or an acid, please specify whether a “catalytic amount”, “1 equivalent”, etc. is required. cat. HO (a) Ph Ph CH3 Ph CH3 O (c) N OH (d) H O Ph cat. HO or RO + or H H O Me OH O Ph O CH2N2 O H2SO4 NH H 2O (e) O Ph O O (b) or RO O O O MCPBA O Figure by MIT OCW. 10 (8) (12 points) Please provide an efficient synthesis of the indicated target compound. All of the carbon of the target compound must come from methyl acetate. O O O Me OMe methyl acetate Me Me Target compound Various routes were possible and partial credit was given depending on efficiency. Below are the most common: O 1 eq MeO OMe MeOH H OO + OMe O- LDA (B) Route 2: O Route3: (A) OMe D A O- cat. MeO OMe OMe OMe H O (C) ∆ (-H2O) O- (A) OMe O (C) O (A) H H + O OMe OMe 1 eq MeO 2. H+, H2O ∆ O O + O OMe O (D) O- C OO OO O ∆ (-H2O) H2O ∆ B OMe Route 1: H cat. H+ or OH- O + O + (C) O 1 eq MeO 2. H + cat MeO MeO O OO 1 eq MeO, + 2. H , H2O, ∆ (B) OMe Figure by MIT OCW. 11 (9) (10 points) The Strecker reaction, followed by a hydrolysis reaction, is an excellent method for synthesizing amino acids, which are the building blocks of proteins. Provide the best mechanism for this process. Please show all arrow pushing. Note: You do NOT have to draw the mechanism for the hydrolysis reaction. O R H NH3 HCN H CN O Strecker H2N CN reaction R hydrolysis H2N CO2H H HO OH R NH3 R H R H H2N R H H2O NH2 R CN R H+ transfer H H H NH3 2pts for each step ! NH2 R H CN Figure by MIT OCW. 12 (10) (12 points) Provide the structure of A and the best mechanism for both of the illustrated transformations. Please show all arrow pushing. HO OH H 1) LiAlH4 2) H H A C8H12O + HO OH HO OH2 O OH2 H O O O AlH3Li H workup OH H H2O H OH H 4 POINTS + 2 POINTS A HO H + H OH2 OH2 H + + 4 POINTS H2O * FULL CREDIT FOR ALL STEPS AND CLEAN, THROUGH MECHANISM Figure by MIT OCW. 13 Bu3SnI I Figure by MIT OCW. 14 (12) (12 points) Provide the best mechanism for the illustrated process. Please show all arrow pushing. Your mechanism should rationalize why the reaction proceeds with complete retention of stereochemistry. O O HO HO Me PT NH2 N O HO HO HO (2) H Me N Me Cl NO H2O N O O Me (1) PT NH N O O H+ HO N O O HO NaNO2 HCl NH2 H Na -O N O O Me O HO Me + N2 N OH2 4 inversion inversion Me +4 Cl Overall retention HO O Me + 4 H Cl Neighboring Group Participation O Me + 4 pts formation of HO O + 10 pts if via HO N2 Me NH3 +8 pts if double inversion mechanism involving NO2 as nucleophile Figure by MIT OCW. 15 (13) (10 points) Please provide a detailed mechanism for the illustrated transformation. Show all arrow pushing. (Bn = CH2Ph) Hint #1: Number your carbons! Hint #2: PhSH is catalystic! OBn O + O cat. AIBN, ∆ OMe OBn MeO cat. PhS-H Initiation: NC N ∆ N CN CN + N2 2 CN + CN H PhS H + PhS Propagation: OBn PhS OBn PhS OBn PhS O + MeO OBn PhS OBn PhS O MeO O OMe OBn O MeO + PhS Figure by MIT OCW. 16 (14) (10 points) Compound A is converted to B, C, and D upon heating. The reaction is accelerated by irradiation. Provide the structures of B, C, and D, and provide the mechanisms by which they are formed (please show all arrow pushing). Me O Me Me A Cl ∆ C4H9Cl B C3H6O C + + C7H15ClO D O + Cl O + C O Cl Cl + O B H OH O O Cl Cl OH + O D Figure by MIT OCW. 17 (15) (16 points total) In an amazing process, Nature transforms squalene oxide into steroids (as a single stereoisomer!). For each of the process illustrated below, provide the best mechanism. Please show all arrow pushing. Me squalene oxide Me Me Me Me O Me Me Me H2O H Me Me Me Me Me Intermediate l HO Me Me Me Me Me Me Me Me Ianosterol, a steroid HO Me Me Me (a). (10 points) Squalene oxide into intermediate l: Me Me Me Me Me O Me Me Me H Me Me Me Me Me HO Me Me Me Me Me Me Me Me Me HO Me Me Me Me Me Me Me Me HO Me * or Show Stepwise C Charges Figure by MIT OCW. 18 +2 for each step: Protonation opening of epoxide each cation π − Cyclization * if no errors -2 if show wrong connectivity after cation π − Cyclic. (b) (6 points) Intermediate 1 into lanosterol: Me Me +2 for deprotonation H H Me +1 for every 1,2-shift Me Me HO Me Me Me H2O H Me HO Me H Me Me Me Me Me Me HO Me Me Me Me Me Me Me Me * or show stepwise C charges ! -1 if show deprotonation in same step at alkyl shift. -1 if show formation of double bond s deprotonation, but show arrow -2 for every 1,3 shift -1 for every 2 missing mech. arrows Figure by MIT OCW. 19 ...
View Full Document

Ask a homework question - tutors are online