Organic Chemistry 5.13, Exam 3 Solutions

First Three Letters of Last Name: TA Name: Hour Exam #3 5.13 Fall 2006 KEY Organic Chemistry II November 15, 2006 Name______________________________________________________________ Signature___________________________________________________________ ID#________________________________________________________________ 1. Make sure your exam has 9 numbered pages plus a periodic table. 2. Write your initials on each page. 3. Look over the entire exam before you begin to familiarize yourself with its length. Do what you know first, then attempt the harder problems. 4. Show all of your work. Partial credit receives points! 1. (4 pts) When N,N-dimethylaminopyridine reacts with one equivalent of acid, the sp2 nitrogen becomes protonated.Why don't you see protonation at the sp3 nitrogen when you know that the more p character an orbital has, the more stable it is with a positive charge? H H3C N CH3 H3C N CH3 H3C H+ N CH3 H+ N N N H conjugate acid not formed The Sp3 Nitrogen is actually Sp2. The lone pair on nitrogen is in a P orbital &can delocalize into the pyridine ring. The lone pair of electrons on the Nitrogen atom in the ring are orthogonal to the ring & cannot delocalize .Those electrons are more available for bonding and more basic. 2. (4 pts) Rank the following molecules in order of electrophilicty(1= most electrophilic ) O CH3 Me N CH3 4 O Me O Cl 1 Me S 2 CH3 O Me O CH3 3 Figure by MIT OCW. 3. (18 pts) Provide the missing products for each reaction. Indicate no reaction with N.R. H3C (a) (b) CH3 N CH3 H3C CH3 O H3C NH2 H2O2 D NaBH4 NR 1.CH3l(excess) 2.Ag2O 3. D CH3 1.CH3l(excess) 2.Ag2O 3. D N (c) H3C (d) O H3C (e) H3C H O Cl + HNMe2 + NaBH3CN H+ 1. HNMe2,pyridine 2.LiAlH4 3. H2O N + :N(Me)3 CH3 Me N H3C H3 C Me H N CH3 CH3 Figure by MIT OCW. 4. (21 pts) Provide the missing reagents for each reaction. Several steps may be needed for some transformations. 1.LiAlH4 2. H2O (a) H C C N 3 H 3C (b) H3C H3C HO 1. H3O+/H2O, D 2. EtLi (2 eq.) Me Me 1.H2N OH,H+ 2. LiAlH4 3. H2O O NH2 H 3C or 1. POCl3 2.EtMgBr 1.SOCl2 2. NH3 3. NaOH, Br2 O H Pr OEt O (d) H3C MeMgBr 2 eq. Pr 1.LiAlH4 2. PCC O OH O (c) or 1. SOCl2 2.LiAl(OtBu)3H O NH2 CH3 NH2 H3C CH3 3. H3O+/H2O NH2 OH (e) 1. NaN3 (f) H3C Br 2. LiAlH4 3. H2O or 1. 2. H3C O N K H 2N NH2 + O NH2 Figure by MIT OCW. (1 2 pts) Consider the labeling experiment outlined below. What level of 18O incorporation do you expect in the recovered anhydride (high or low)? Your answer should include a mechanism of hydrolysis and a detailed explanation. 5. O O O Me O O O + Me k1 OH * stop at 50% conversion of acetic anhydride * OH/H * 2O O OH * O Pt O k-1 O O O + k-1 O O * O k1 k2 k2 O O * O HO O * OH * OH + O O O 18 O Incorporation is expected to be low in recovered anhydride. O OH. There fore, k2 >> k-1, or elimination of O O is a better leaving group than faster than revesal to starting material. Most of the 18 O will be found in the acetic acid product. Figure by MIT OCW. 6. (10 pts) Provide a mechanism for the following transformation. O MeO O MeO Me + OH Me O+/H OH H3 2O Me O Me Me O H H OH2 MeO Me Me O Me OH Me MeO MeO H OH Me PT H Me Me O H Me OH O O Me + OH2 O O OH2 Me O + H3O Me Figure by MIT OCW. (10 pts) Under basic hydrolysis conditions, a nitrile goes through a primary amide intermediate before becoming a carboxylate. Show the mechanism for this reaction and explain why it is NOT a facile method for converting nitriles into carboxylates. 7. R R C C N - R O + NH3 N H N N R OH C + H OH R O H + OH OH N H O R O OH/H2O HO H N H NH2 R R O O HO O R OH NH2 O O R O H + NH2 R O + NH3 Base catalyzed hydydrolysis of nitriles is not a facile process because NH2 is not good leaving group( -OH is better!) There, fore, the reaction is slow. Figure by MIT OCW. 8. (9 pts) Provide a synthesis for the following compound. OH Cl HNO3 NO2 H2SO4 NO2 Cl2 AlCl3 Cl H2, Pd OH Cl H2SO4 N N NH2 NaNO2 HCl Cl Cl Cl Figure by MIT OCW. 9. (12 pts) Provide a selective synthesis for ONE of the following compounds. Circle the molecule that you want graded. All of the carbon atoms of the product should come from either ethanol or compounds that contain just one carbon atom. A B O Me N Me Me OH Me NH2 B OH Mg0 PBr3 MgBr Br OH 1. CO2 2. H2O O SOCl2 Cl LiAl(OtBu)3H H HCN CN O O OH 1. LiAlH4 2. H2O OH NH2 O OR 1. H MgBr H OH O 2. H2O O OR OH A Jones O O MeOH EtOH H PCC PBr3 PBr3 EtBr MeBr 1. NaN3 2. LiAlH4 3. H2O OR MeOH PBr3 MeBr SOCl2 OH 1. Mg 2. CO2 3. H2O 0 Et 1. NaN3 2. LiAlH4 3. H2O OH SOCl2 Cl O pyridine N O NH2 1. H H ,H+ 2. LiAlH4 3. H2O N H O Me NH2 + Cl 8 pyridine H 1. LiAlH4 N O 2. H2O H N Figure by MIT OCW. EXTRA CREDIT (5 pts) Synthesize methamphetamine (crystal meth) from benzene and any other reagents. All the carbon atoms in the product should come from reagents that only contain one carbon atom. H N Me 1. Mg 2.CO2 Br2 FeBr3 Me O OH 3.H2O 1.LiAlH4 OH 2.H2O PBr3 OH O 1.MeLi OH O 2.H2O 1. Mg 2.CO2 3.H2O Br H2NMe H + OH N 1.LiAlH4 2.H2O H N Figure by MIT OCW.