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1 NAME: STUDENT NUME: Page 1 of 7 CEM 3390 Midterm Test Monday ctober 26 and Tuesday ctober 27, 2008 This test is graded out of 40 Marks. You must complete all work within TW US. Put all answers in the spaces provided. You may also use the backs of the sheets, if necessary. 1) (20 MAKS TTAL) Stereochemistry and conformation. a) (4 MAKS) riefly explain the difference between stereoselectivity and stereospecificity. b) (8 MAKS) The drug ketamine has the structure shown. In acidic solutions, it predominantly adopts a conformation in which the aromatic ring is axial, while in neutral or alkaline solutions it prefers a conformation in which this ring is equatorial. Draw clear perspective views of the two conformers of ketamine and provide a brief explanation of why the axial conformation is preferred in acid solutions. ketamine N Cl c) (2 MAKS) The 1,2,3,4,5,6-hexahydroxycyclohexanes are known as inositols. There are 9 possible stereoisomers of inositol. f these, two form a pair of enantiomers while the remaining 7 are achiral. Draw the two enantiomeric forms of inositol (a wedge-and-dash representation is fine, no need for perspective or conformation in this question).

2 d) (4 MAKS) Draw a clear perspective view of the preferred conformation of cis,cis,trans-perhydro-9b-phenalenol (Compound A). Page 2 of 7 e) (2 MAKS) The faces of the C=C bond in fumaric acid (trans-butenedioic acid) are enantiotopic. The enzyme fumarase catalyzes the formation of S-malate ((S)-2-hydroxybutanedioic acid) from fumaric acid. i) Write out a balanced reaction equation showing the stereochemistry of the product (Lewis structures using wedge-and-dash notation as needed). ii) Which face of the alkene (re or si) does the enzyme hydroxylate?

3 2) (11 MAKS TTAL) eactions Page 3 of 7 a) (5 MAKS) When the furan derivative A was treated with 1.2 molar equivalents of mercuric triflate in a mixture of nitromethane and dichloromethane solvents, the major product (53%) was the bicyclic organomercury compound, but a smaller amount (10%) of the tricyclic compound C was also formed. Provide a mechanism showing how C was formed in this reaction.

4 Page 4 of 7 b) We noted in class that hydroboration is usually worked up with alkaline hydrogen peroxide, but that other oxidants can afford different products. i) (2 MAKS) What would be the product of the following hydroboration/oxidation reaction worked up by heating with sodium hydroxylamine--sulfonate? ii) (4 MAKS) Write a mechanism for the oxidation of a generic organoborane ( 3 ) using this reagent.

5 Page 5 of 7 3) (9 MAKS) Compound A (C 7 14 ) was stirred in aqueous 2 S 4 for several hours. Two major products and C were formed. The 1 and 13 C NM and infrared spectra of and C are shown on the following pages. What are the two products and C and what was the starting compound A?

6 Product Page 6 of

7 Product C Page 7 of 7

8 ANSWE KEY Page 1 of 8 CEM 3390 Midterm Test ANSWES Monday ctober 26 and Tuesday ctober 27, 2008 This test is graded out of 40 Marks. You must complete all work within TW US. Put all answers in the spaces provided. You may also use the backs of the sheets, if necessary. 1) (20 MAKS TTAL) Stereochemistry and conformation. a) (4 MAKS) riefly explain the difference between stereoselectivity and stereospecificity. NTE: the fact that a reaction produces only one stereoisomeric product does not necessarily make it stereospecific. The key concept in A reaction that can afford more than one stereoisomeric product but which favours one possible product over others is stereoselective. A reaction that is mechanistically constrained such that a given starting material must form one and only one stereochemically defined product is stereospecific. stereospecificity is that mechanism demands a specific stereochemical outcome. b) (8 MAKS) The drug ketamine has the structure shown. In acidic solutions, it predominantly adopts a conformation in which the aromatic ring is axial, while in neutral or alkaline solutions it prefers a conformation in which this ring is equatorial. Draw clear perspective views of the two conformers of ketamine and provide a brief explanation of why the axial conformation is preferred in acid solutions. ketamine N Cl 3 3 The conformation with the aryl ring axial permits a 5-membered cyclic hydrogen bond between the ammonium ion and the ketone. Evidently, the protonated amine is a better -bond donor than the neutral amine, and this overwhelms the steric penalty for putting the aryl ring axial. c) (2 MAKS) The 1,2,3,4,5,6-hexahydroxycyclohexanes are known as inositols. There are 9 possible stereoisomers of inositol. f these, two form a pair of enantiomers while the remaining 7 are achiral. Draw the two enantiomeric forms of inositol (a wedge-and-dash representation is fine, no need for perspective or conformation in this question). (S) (S) () () () () (S) (S) (S) (S) () () Note: you did not have to label the stereocentres - shown here for information only.

9 d) (4 MAKS) Draw a clear perspective view of the preferred conformation of cis,cis,trans-perhydro-9b-phenalenol (Compound A). Page 2 of 8 Notice that this conformation can be regarded as merging a trans decalin with a pair of cis decalin substructures. This question appears in the textbook at the end of Chapter 3. e) (2 MAKS) The faces of the C=C bond in fumaric acid (trans-butenedioic acid) are enantiotopic. The enzyme fumarase catalyzes the formation of S-malate ((S)-2-hydroxybutanedioic acid) from fumaric acid. i) Write out a balanced reaction equation showing the stereochemistry of the product (Lewis structures using wedge-and-dash notation as needed). ii) Which face of the alkene (re or si) does the enzyme hydroxylate? To form the S enantiomer of malic acid, hydroxylation must occur on the si face of fumaric acid (the back face as drawn above).

10 Page 3 of 8 2) (11 MAKS TTAL) eactions a) (5 MAKS) When the furan derivative A was treated with 1.2 molar equivalents of mercuric triflate in a mixture of nitromethane and dichloromethane solvents, the major product (53%) was the bicyclic organomercury compound, but a smaller amount (10%) of the tricyclic compound C was also formed. Provide a mechanism showing how C was formed in this reaction. b) 1. g(tf) 2, MeN 2 /C 2 Cl aq. NaCl Clg Clg A C N: Tf is triflate, the trifluoromethanesulfonate group. This is a very nonnucleophilic anion, the conjugate base of trifluoromethanesulfonic acid. g Tf Tf g Tf g Tf NaCl Clg g Tf g Tf C This is an example of C-C bond formation via mercuration. ecause the cyclic mercurinium ion is equivalent to a carbocation, it can be attacked by relatively electron-rich alkenes and aryl systems in a mechanism similar to a Friedel-Crafts reaction.

11 Page 4 of 8 We noted in class that hydroboration is usually worked up with alkaline hydrogen peroxide, but that other oxidants can afford different products. i) (2 MAKS) What would be the product of the following hydroboration/oxidation reaction worked up by heating with sodium hydroxylamine--sulfonate? ii) (4 MAKS) Write a mechanism for the oxidation of a generic organoborane ( 3 ) using this reagent. S + Na 2 N Na S N The sulfonic acid portion of the reagent is strongly acidic and will be deprotonated by the sodium hydroxide. S N 2 S N N S N + N S N This process repeated for two remaining - bonds N N N N N N N ydrolysis repeated for two remaining -N bonds. N 2 + N N

12 Page 5 of 8 Some students thought that the anionic sulfate group would act as the nucleophile towards boron. While this might seem reasonable, remember that as the conjugate base of a very strong acid, the sulfate group is an extremely poor nucleophile. This is why we use sulfuric acid or tosic acid as non-nucleophilic proton sources! (Note that hydroxylamine--sulfonic acid is not as strong as sulfuric acid) The question that arises if you made this assumption is what would be the product. A sulfate ester was a common choice, but you have to ask how this would form in mechanistic terms. When the group migrates from boron to oxygen, what is the weak bond that breaks in analogy to the - bond in the hydroperoxy situation? From a strictly arrow-pushing viewpoint, you could write a mechanism for the formation of an alcohol via sulfate attack, although it is not favoured in energetic terms because of the poor leaving groups involved. S 2 N pka ~ 1 3 S 2 N alternative 2 N + S + pka ~ N S pka ~ 7

13 Page 6 of 8 3) (9 MAKS) Compound A (C 7 14 ) was stirred in aqueous 2 S 4 for several hours. Two major products and C were formed. The 1 and 13 C NM and infrared spectra of and C are shown on the following pages. What are the two products and C and what was the starting compound A? The formula given for A has NE degree of unsaturation. The nature of the chemistry suggests an acid-catalyzed hydration reaction and the I spectra of and C confirm that they are both alcohols. Consider compound. The 13 C shows 5 types of carbons. The 1 shows a 9 singlet that suggests a t-butyl group and a 3 triplet suggesting a methyl with a C 2 next to it. A broad singlet for 1 at about 1.9 ppm combined with the I says, and the 1 double doublet at 3.08 ppm has to be C- with two different nearest neighbours. We also see two complex multiplets each integrating for 1. Put this together we have identified a total of 7 carbons (tu = 4, C 2 C 3 = 2, C = 1) of 5 types. That s all the carbons. nly one possible structure fits these data. Now look at compound C. The 13 C spectrum shows 7 distinct types of carbons, so the molecule lacks any internal symmetry. The 1 spectrum is more complicated than the previous one. We can, however, see the proton of the alcohol, two 1 multiplets, two singlets each integrating for 3, and a complex upfield signal group integrating for a total of 7. The expansion shows that this group contains a doublet, a complex multiplet and a small multiplet probably amounting to two methyl groups plus a lone hydrogen. Notice as well that there is no 1 signal downfield of 2 ppm which tells us that the alcohol group must be tertiary (i.e. no C proton!). What do we have? Two C 3 groups with no neighbours, one C 3 group with 1 neighbour, and a C 3 group with several neighbours plus a tertiary alcohol. In addition, 3 signals each for 1 proton. The tertiary alcohol and the two C 3 singlets suggest. The C3 with 1 neighbour suggests CC 3. The remaining C 3 group has multiple neighbours, but counting carbons we have only two more carbons to assign. This has to be C 2 C 3. The structure therefore is: C 3 3 C C 3 3 C CMPUND C Now, looking back at the original reaction, the starting compound A is an alkene with 7 carbon atoms and obviously it has several methyls. n acid treatment, it adds 2 across the double bond forming compound, but it also rearranges to some extent leading to compound C. This means that compound A is probably: 3 C C 3 3 C C 3 CMPUND A

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15 Product C Page 8 of 8