2.222 Introductory Organic Chemistry II Midterm Exam

Transcription

1 NAME: STUDENT NUMBE: Page 1 of 7 University of Manitoba Department of hemistry Introductory rganic hemistry II Midterm Exam Wednesday February 20, 2002 Put all answers in the spaces provided. If more space is required you may use the backs of the exam pages but be sure to indicate that you have done so. A table of spectroscopic data is attached at the end of the exam. 1. (5 MAKS) iefly explain the differing outcomes of the following set of hydroboration reactions. In each case, the starting ratio of alkene:diborane was >> 3:1. + B 2 6 Ether 3 B + B 2 6 Ether B 2 + B 2 6 Ether B 2

2 Page 2 of 7 2. (5 MAKS) Provide a detailed stepwise mechanism for the following reaction. No explanations are required, just push the arrows. 2, 25 o (racemic) 3. (5 MAKS) Provide a detailed stepwise mechanism for the industrial synthesis of bisphenol A, shown below. No explanations are required, just push the arrows. 2 + anhydrous l + 2

3 Page 3 of 7 4. (14 MAKS) Provide the necessary product, reagent/solvent or starting material to correctly complete SEVEN (7) of the following reactions. Please indicate which ones you want marked! N N a. N 2 N LiAl 4 ether, 35 o b. 3 c N 2 NaN 2 l (aq.) KN, u 2 (N) 2 2, 50 o d. N 2 Fe 3 2 l 2 e.

4 Page 4 of 7 N f. 2 hν 2 l 2 g. fuming 2 S 4 h. NaB 4 3 i. ( 3 ) 2 N l pyridine/ 2 l 2 0 o j. + N N 2 N 2 trace of l Ethanol, k. N 2

5 Page 5 of 7 5. (6 MAKS) You have been assigned to make 1,2-bis(bromomethyl)- 4,5-dihydroxycyclohexane. Propose an efficient synthetic route based on appropriate organic reactions. You have only a small collection of organic materials, shown in the box below, plus any inorganic reagents you might need. This assignment can be achieved in between 3 and 6 chemical steps. Show all necessary reagents, solvents and conditions for each step. Mechanisms and discussion are not required. Methanol, Ethanol, iso-propanol, Acetone Diethyl ether, tetrahydrofuran (TF) Dichloromethane Acetic acid 3

6 Page 6 of 7 6. (5 MAKS) An organic compound (A) of molecular weight 164 (as determined by mass spectrometry) was heated with 1 M aqueous Na until it all dissolved. n cooling to room temperature, the solution remained clear, but when it was neutralized, a solid material precipitated. The solid was identified as phenol. Na acidify A clear solution 2, to p 7.0 Spectra of A are shown below. What was the structure of compound A? doublet triplet Double doublet doublet septet Structure of ompound A:

7 Spectroscopy rib Sheet for Introductory rganic hemistry II 1 NM Typical hemical Shift anges Type of Proton hemical Shift (δ) Type of Proton hemical Shift (δ) (solvent dependent) (solvent dependent) Aryl l Aryl I Aromatic, heteroaromatic X X =, N, S, halide 3 Aliphatic, alicyclic Y =, N, S Y Y Y =, N, S Low Field δ 13 NM Typical hemical Shift anges igh Field Alkene Aryl Ketone, Aldehyde Ester Amide x -Y Y =, N x -= I Typical Functional Group Absorption Bands Group Frequency Frequency (cm -1 Intensity Group ) (cm -1 ) Intensity Medium Strong, broad = Medium Strong = Medium = Strong Strong 2 N Medium, broad Medium ( ) N 1230, 1030 Medium Aryl Medium N Medium Aryl = 1600, 1500 Strong N Strong δ

8 ANSWE KEY Page 1 of 7 University of Manitoba Department of hemistry Introductory rganic hemistry II Midterm Exam Wednesday February 20, 2002 Put all answers in the spaces provided. If more space is required you may use the backs of the exam pages but be sure to indicate that you have done so. A table of spectroscopic data is attached at the end of the exam. 1. (5 MAKS) iefly explain the differing outcomes of the following set of hydroboration reactions. In each case, the starting ratio of alkene:diborane was >> 3:1. + B 2 6 Ether 3 B + B 2 6 Ether B 2 + B 2 6 Ether B 2 In class we noted that all three B bonds in borane are in principle reactive, and so one only needs 0.33 molar equivalents of B 3 to completely consume 1 molar equivalent of an alkene. + B 3 B 2 B owever, we also noted that if a very bulky alkene were used, it might not be possible to get three groups of that size around the boron atom. That is what is happening here. ecall that hydroboration occurs in an anti-markovnikov fashion. The terminal alkene leads to a primary alkyl group, and three of these easily fit around boron. When the alkene is internal, and trisubstituted, the resulting alkyl group is secondary, and thus more sterically bulky. nly two such groups can fit around boron. The tetrasubstituted alkene can only add once, because the carbon bonded to boron is tertiary. B

9 ANSWE KEY 2. (5 MAKS) Provide a detailed stepwise mechanism for the following reaction. No explanations are required, just push the arrows. 2, 25 o (racemic) + concerted cycloaddition N.B. epoxide can be either "up" or "down", and attack by water can occur at either carbon position. owever, this is an S N 2-like process, so the resulting product will be trans. - + Note: the epoxide ring opening requires protonation. Water by itself will not open most epoxides at 25 o. Also note that the fact that a trans diol is formed rules out a mechanism in which the epoxide opens spontaneously to form a carbocation, because water attack at a cation will occur unselectively from both faces. 3. (5 MAKS) Provide a detailed stepwise mechanism for the industrial synthesis of bisphenol A, shown below. No explanations are required, just push the arrows. 2 + anhydrous l + 2 l - + ( - 2 ) l - + This is simply a pair of Friedel-rafts alkylations. Note that alkylation of aromatic rings requires very potent electrophiles. Thus, all of the reagents we have seen that can add to aromatic rings have required activation. The carbonyl is no exception. Protonation increases its electrophilicity dramatically. The second alkylation proceeds via the tertiary, benzylic carbocation that forms on loss of water from the initial adduct.

10 ANSWE KEY 4. (14 MAKS) Provide the necessary product, reagent/solvent or starting material to correctly complete SEVEN (7) of the following reactions. Please indicate which ones you want marked! N N 3 / 2 S 4 N a. N 2 N LiAl 4 ether, 35 o N b. 3 c N 2 NaN 2 l (aq.) KN, u 2 (N) 2 2, 50 o N d. N 2 Fe 3 2 l 2 N 2 e.

11 ANSWE KEY N 3 +, heat (any strong acid in water) f. 2 hν 2 l 2 g. fuming 2 S 4 S h. NaB 4 3 i. l ( 3 ) 2 N pyridine/ 2 l 2 0 o N 3 j. 3 N N 2 + N 2 trace of l Ethanol, N 2 N k. N 2 N N 2

12 ANSWE KEY 5. (6 MAKS) You have been assigned to make 1,2-bis(bromomethyl)- 4,5-dihydroxycyclohexane. Propose an efficient synthetic route based on appropriate organic reactions. You have only a small collection of organic materials, shown in the box below, plus any inorganic reagents you might need. This assignment can be achieved in between 3 and 6 chemical steps. Show all necessary reagents, solvents and conditions for each step. Mechanisms and discussion are not required. Methanol, Ethanol, iso-propanol, Acetone Diethyl ether, tetrahydrofuran (TF) Dichloromethane Acetic acid 3 + heat LiAl 4 ether 35 o P 3, 2 l 2 s 4 acetone/water The key point in this problem was to recognize that a Diels-Alder strategy was the only possible route, given the starting materials available. A tip-off should have been the syn diol we only know one kind of process to make a syn diol, and that is the oxidation of an alkene using s 4 or cold, alkaline KMn 4. (Either method would be fine). nce you see that a cyclohexene must be involved, the Diels-Alder approach should be obvious. The only diene/dienophile combination that could give the necessary number of carbon atoms was 1,3-butadiene/maleic anhydride. The product has a pair of primary bromides. These can be formed from primary alcohols using S N 2 chemistry (the only reaction from needed on this exam!), so you must first reduce the carbonyls. You know that carboxylic acids and esters are reduced by LiAl 4 to primary alcohols, and the analogy with the acid anhydride should be clear. The introduction of bromide can be done in several ways. The use of P 3 is the most straightforward, but you could also form a sulfonate ester derivative of the alcohols and then displace with from K in acetone or some other polar aprotic solvent. When doing a problem of this kind, it is best to think backwards from the product. Since you know what starting materials you have available in this particular case, you can work from both ends towards the middle move forward from the Diels-Alder and backwards from the product to see where the two paths converge. 3 l S S 2 3 K, acetone Et 3 N, 2 l 2 S 3 see textbook, pp for a reminder about this chemistry

13 ANSWE KEY 6. (5 MAKS) An organic compound (A) of molecular weight 164 (as determined by mass spectrometry) was heated with 1 M aqueous Na until it all dissolved. n cooling to room temperature, the solution remained clear, but when it was neutralized, a solid material precipitated. The solid was identified as phenol. Na acidify A clear solution 2, to p 7.0 Spectra of A are shown below. What was the structure of compound A? doublet triplet Double doublet doublet septet Structure of ompound A: 3 3

14 Spectroscopy rib Sheet for Introductory rganic hemistry II 1 NM Typical hemical Shift anges Type of Proton hemical Shift (δ) Type of Proton hemical Shift (δ) (solvent dependent) (solvent dependent) Aryl l Aryl I Aromatic, heteroaromatic X X =, N, S, halide 3 Aliphatic, alicyclic Y =, N, S Y Y Y =, N, S Low Field δ 13 NM Typical hemical Shift anges igh Field Alkene Aryl Ketone, Aldehyde Ester Amide x -Y Y =, N x -= I Typical Functional Group Absorption Bands Group Frequency Frequency (cm -1 Intensity Group ) (cm -1 ) Intensity Medium Strong, broad = Medium Strong = Medium = Strong Strong 2 N Medium, broad Medium ( ) N 1230, 1030 Medium Aryl Medium N Medium Aryl = 1600, 1500 Strong N Strong δ