UNIVERSITY OF MANITOBA DEPARTMENT OF CHEMISTRY

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1 PAGE 1 of 11 UNVESTY F MANTBA DEPATMENT F CEMSTY STUCTUAL TANSFMATNS N GANC CEMSTY FNAL EXAMNATN - PAPE # 432 Dr. Phil ultin Friday December 17, NAME: STUDENT NUMBE: 1. (20 Marks) For each of the following terms or concepts, provide a concise definition and a specific illustration (i.e. a complete and accurate chemical structure or reaction). a. A 1,3 strain b. A stereospecific reaction c. Conformational locking

2 PAGE 2 of 11 d. Diastereotopic atoms (in a molecule) e. The haptophilic effect (in hydrogenations)

3 PAGE 3 of (20 Marks total) Concisely explain (using clear three-dimensional drawings to illustrate your explanation) the reason(s) behind each of the following observations. a. (4 Marks) Cis-2,3-dichloro-1,4-dioxane crystallizes at -140 ºC. Why are the two C-Cl bond lengths in the crystal structure very different? (Acta Cryst. 1963, 16, ) Å Cl Cl Å b. (8 Marks) The trans isomer of 1-iodo-2-(p-bromobenzenesulfonyl)cyclohexane is solvolysed very rapidly in acetic acid to give trans-1-iodo-2-acetoxycyclohexane (i.e. retention of stereochemistry). The cis isomer is solvolysed over a millionfold slower, and the product is a mixture of stereoisomers. Bs Ac fast Ac Bs Ac slow Ac + Ac Bs = S Br

4 PAGE 4 of 11 c. (4 Marks) Why does the preferred envelope conformer of 1-methylcyclopentane have the methyl group on the flap of the envelope, rather than on another position? C 3 C 3 d. (4 Marks) n the 1 NM spectrum of the sesquiterpene shown, the vicinal coupling constant J ab = 1.8 z, while J bc = 9.1 z. What are the relative stereochemical orientations of a, b and c? (Chem. Eur. J. 2003, 9, ) a b c

5 PAGE 5 of (20 Marks) Provide the necessary structure or reagent/solvent to correctly complete each of the following reactions. Don t forget to show stereochemistry where appropriate. a. (3 Marks) Ph Ac 1. AgBF 4 toluene/ 2 reflux N 2. LiAl 4, TF 0 o C rg. Lett. 2004, 6, b. (2 Marks) rg. Lett. 2003, 5, c. (3 Marks) 3 C 1. 9-BBN, TF TBS 2. Na, C C 3 C 3 rg. Lett. 2004, 6, d. (2 Marks) 1. C 3 2. Jones' oxidation C 3 Eur. J. rg. Chem., 2004, e. (3 Marks) 3 C 1. NBS, TF/ 2 2. Li, DMF Cl rg. Lett. 2003, 5,

6 PAGE 6 of 11 f. (2 + 2 Marks) Br 1. N 2. N C 3 C 3 C 3 C 3 J. rg. Chem., 1983, 48, g. (3 Marks) N 1. K (s), N 3 (l) tbu/tf 2. Br C 3 Tetrahedron Lett., 2004, 45,

7 PAGE 7 of (20 Marks Total) Provide detailed stepwise mechanisms to explain the reactions shown below. Show all intermediate structures including appropriate stereochemistry. Commentary is not necessary. a. (6 Marks) The following rather surprising reduction reaction leads to complete deoxygenation of the ketone (Tetrahedron Lett. 1995, ). Me NaB 4

8 PAGE 8 of 11 b. (7 Marks) ecall that acid treatment of alkenes often leads to structural rearrangement, especially if it permits the relaxation of ring strain! (J. Chem. Soc. Chem. Commun. 1970, 248) C 3 3 C C 3 C 3 3 C 3 C C 3

9 PAGE 9 of 11 c. (7 Marks) Note that the following reaction is performed in the absence of any base or buffer. mcpba C 2 Cl 2 Ac (Tetrahedron Lett. 1968, 1449)

10 PAGE 10 of (20 Marks) D-soascorbic acid is oxidized with aqueous hydrogen peroxide, and the reaction mixture is then acidified with Cl. The product of this reaction is then heated in acetone with toluenesulfonic acid, to yield a chiral product C The spectra of this product are shown (spectra from D-soascorbic acid C a. What is the structure of the product? (aq), then Cl (aq) 2. acetone, Ts PDUCT C b. Assign the signals in the 1 and 13 C NM spectra to appropriate nuclei in your structure. CDCl 3 solvent peaks

11 PAGE 11 of 11

12 ANSWE KEY PAGE 1 of 11 UNVESTY F MANTBA DEPATMENT F CEMSTY STUCTUAL TANSFMATNS N GANC CEMSTY FNAL EXAMNATN - PAPE # 432 Dr. Phil ultin Friday December 17, NAME: STUDENT NUMBE: 1. (20 Marks) For each of the following terms or concepts, provide a concise definition and a specific illustration (i.e. a complete and accurate chemical structure or reaction). a. A 1,3 strain 3 C 3 C 3 C 3 C van der Waals' conflict between an allylic substituent and the cis substituent on the alkene destabilizes the conformer with a relatively bulky group eclipsing the alkene. This is referred to as A 1,3 strain. b. A stereospecific reaction A stereospecific reaction is one in which stereoisomerically different starting materials form specific stereoisomerically different products under identical reaction conditions. The S N 2 reaction is a good example. K acetone Ts K acetone Ts c. Conformational locking A molecule is conformationally locked if it is physically impossible for it to adopt an alternative low-energy conformation. Generally, this term refers to cyclohexane systems such as the trans decalin below that can only assume one of the possible chair conformations, although they may be able to adopt a high-energy boat or twist geometry.

13 ANSWE KEY PAGE 2 of 11 d. Diastereotopic atoms (in a molecule) Two atoms of thesametypeinamoleculeare diastereotopic if altering each one in turn leads to the formation of products that are diastereomers of one another. These hydrogens are diastereotopic Ts e. The haptophilic effect (in hydrogenations) When an -bond donor group is located in a fixed geometric relationship to an alkene within the same molecule, it directs catalytic hydrogenation of that alkene to occur syn to the donor group. This is the haptophilic effect. 2, Pt (cat.) major product

14 ANSWE KEY PAGE 3 of (20 Marks total) Concisely explain (using clear three-dimensional drawings to illustrate your explanation) the reason(s) behind each of the following observations. a. (4 Marks) Cis-2,3-dichloro-1,4-dioxane crystallizes at -140 ºC. Why are the two C-Cl bond lengths in the crystal structure very different? (Acta Cryst. 1963, 16, ) Å Cl Cl Cl Cl Å Cl The chair conformation of the ring forces one of the two C-Cl bonds to be axial and the other equatorial. The axial bond will be longer because of the anomeric effect. Donation of the adjacent oxygen lone pair into a σ* orbital of the C-Cl bond weakens and lengthens this bond relative to the equatorial C-Cl bond which is not aligned correctly for this donation to occur. Cl b. (8 Marks) The trans isomer of 1-iodo-2-(p-bromobenzenesulfonyl)cyclohexane is solvolysed very rapidly in acetic acid to give trans-1-iodo-2-acetoxycyclohexane (i.e. retention of stereochemistry). The cis isomer is solvolysed over a millionfold slower, and the product is a mixture of stereoisomers. Bs Ac fast Ac Bs Ac slow Ac + Ac Bs = S Br The trans isomer benefits from anchimeric assistance by the adjacent iodine atom. This increases the rate of displacement of the leaving group. t also forces the attack of acetate to occur from the backside of the resulting iodonium ion - that is, a double displacement leads to net retention of stereochemistry. Bs Bs The cis isomer cannot benefit from anchimeric assistance. Acid-catalyzed ionization of the leaving group leads to a simple S N 1 mechanism for nucleophilic displacement, which of course can occur from either face of the planar cation intermediate Ac Ac

15 ANSWE KEY PAGE 4 of 11 c. (4 Marks) Why does the preferred envelope conformer of 1-methylcyclopentane have the methyl group on the flap of the envelope, rather than on another position? C 3 C 3 When the methyl group is on the "flap" of the envelope, it is gauche to the adjacent hydrogens, whereas when it is on one of the other positions, it will be eclipsed. This means that the "flap" conformation is lower in energy. C 3 C 3 d. (4 Marks) n the 1 NM spectrum of the sesquiterpene shown, the vicinal coupling constant J ab = 1.8 z, while J bc = 9.1 z. What are the relative stereochemical orientations of a, b and c? (Chem. Eur. J. 2003, 9, ) b and c must be cis to one another because the fusion of a 3-membered and 6-membered ring can only occur cis. The small a - b coupling constant suggests a dihedral angle of nearly 90 o. Because of the ring fusion, this implies a trans relationship between a and b. c a b a b c

16 ANSWE KEY PAGE 5 of 11 The Woodward- Prévost reaction proceeds through a trans iodoacetate like the starting material here. Step 1 is just the completion of the W-P reaction. Step 2 cleaves the acetate ester off, by reduction. 3. (20 Marks) Provide the necessary structure or reagent/solvent to correctly complete each of the following reactions. Don t forget to show stereochemistry where appropriate. a. (3 Marks) Ph Ac 1. AgBF 4 toluene/ 2 reflux N 2. LiAl 4, TF 0 o C rg. Lett. 2004, 6, Ph N b. (2 Marks) Any oxidizing reagent that can stop at the aldehyde stage is acceptable. Note that Swern conditions are suggested, but other DMS oxidations would work too. PCC/C 2 Cl 2 or PDC/C 2 Cl 2 or DMS/oxalyl chloride C 2 Cl 2, -78 o C then Et 3 N rg. Lett. 2003, 5, ydroboration of the terminal alkene creates a new stereocentre adjacent to the alcohol. The selectivity is a function of the preferred rotamer of the allylic bond, which has the hydrogen eclipsing the alkene. The bulk of the molecule thus blocks approach from the b tt c. (3 Marks) 3 C 3 C C 3 TBS C BBN, TF 2. Na, 2 2 rg. Lett. 2004, 6, C 3 C TBS C 3 C 3 C 3 d. (2 Marks) Dissolving metals reduce alkynes to trans alkenes. n this case the ketone also was reduced (as we noted, isolated ketones are reducible if enough metal is added), and that is why the second step was an oxidation. 1. Li, N 3 (l), Et C 3 2. Jones' oxidation Eur. J. rg. Chem., 2004, C 3 The double bond furthest from the carbonyl is the most reactive towards electrophiles. The top face is least hindered, because of the conformation of the 7-membered ring which puts the methyl in a pseudoequatorial orientation. Thus, the halohydrin intermediate has the Br on top and the on the bottom. The second step simply cyclizes the halohydrin. e. (3 Marks) Cl 3 C 1. NBS, TF/ 2 2. Li, DMF rg. Lett. 2003, 5, Cl 3 C

17 ANSWE KEY PAGE 6 of 11 An epoxide is being formed on the less-hindered side of the molecule. This suggests a peracid reaction of an alkene. We can get the required alkene from the bromide by elimination. Notice that there is no risk of getting the wrong alkene isomer because only one adjacent proton has the required antiperiplanar orientation with the bromine. f. (2 + 2 Marks) N C 3 Br C Et 3 N or other non-nucleophilic base mcpba C 2 Cl 2 J. rg. Chem., 1983, 48, N C 3 C 3 Another dissolving metal reduction. The regioselectivity is controlled by both the electron-donating methoxyl group, and the electron-withdrawing amide. This is an example of trapping the final anion in the reduction with a carbon electrophile, allyl bromide. g. (3 Marks) N C 3 1. K (s), N 3 (l) tbu/tf 2. Br Tetrahedron Lett., 2004, 45, N C 3 Note that all the predict the product reactions were drawn from the papers submitted by students for bonus marks this year. hope that at least some of you found a few of these a bit familiar!

18 ANSWE KEY PAGE 7 of (20 Marks Total) Provide detailed stepwise mechanisms to explain the reactions shown below. Show all intermediate structures including appropriate stereochemistry. Commentary is not necessary. a. (6 Marks) The following rather surprising reduction reaction leads to complete deoxygenation of the ketone (Tetrahedron Lett. 1995, ). Me NaB 4 workup (+ + ) Me Na B + C 2 Me Me NaB3 You should recall that since ketones are routinely reduced to secondary alcohols by NaB 4, the direct reduction of this ketone by two successive hydride attacks is highly unlikely. Also, we know that NaB4 generally only reduces esters sluggishly, so attack on the carbonate is also less probable. The fact that the carbonate group is lost during the reaction should suggest that it is linked to the deoxygenation in some way, and that this most likely involves generation of C 2. Your approach to the mechanism could be guided by this initial analysis of what is happening in the overall reaction.

19 ANSWE KEY PAGE 8 of 11 b. (7 Marks) ecall that acid treatment of alkenes often leads to structural rearrangement, especially if it permits the relaxation of ring strain! (J. Chem. Soc. Chem. Commun. 1970, 248) 3 C C C 3 3 C 3 C C C 3 Note that it is the bond that breaks and migrates in this reaction, and not the 2-10 bond. Since the product is not bridged, we need to create the next cationic centre within the large ring. Moving the 2-10 bond would put the cation outside. 3 C C 3 3 C 1,2-shift C 3 3 C - C 3 C 3 C 3 C 3 C 3 3 C C 3 This rearrangement is similar to one you saw on a problem set. That reaction followed a somewhat different pathway involving migration of the 2-10 bond. Why does this one not do that? Notice that there is no water present in this reaction to trap the cation. Undoubtedly the 2-10 bond does migrate, but with nothing to trap the cation, it simply equilibrates back. The reaction here is completely intramolecular, and obviously represents the preferred outcome in non-aqueous conditions.

20 ANSWE KEY PAGE 9 of 11 c. (7 Marks) Note that the following reaction is performed in the absence of any base or buffer. mcpba C 2 Cl 2 (Tetrahedron Lett. 1968, 1449) Ac Ar + Ac Ar + ArC This reaction was a mechanism practice problem in the textbook!

21 ANSWE KEY PAGE 10 of (20 Marks) D-soascorbic acid is oxidized with aqueous hydrogen peroxide, and the reaction mixture is then acidified with Cl. The product of this reaction is then heated in acetone with toluenesulfonic acid, to yield a chiral product C The spectra of this product are shown (spectra from D-soascorbic acid C a. What is the structure of the product? (aq), then Cl (aq) 2. acetone, Ts PDUCT C b. Assign the signals in the 1 and 13 C NM spectra to appropriate nuclei in your structure. CDCl 3 solvent peaks

22 ANSWE KEY PAGE 11 of 11 The formula shows 3 degrees of unsaturation. t is obvious from the 13 C NM and that there is an ester in the molecule. There must also be 2 aliphatic C 3 groups with no neighbours. You know that acetone was involved in the reaction, so probably these methyls come from acetone. The signal at ppm in the 13 C spectrum might be an alkene or even an alkyne, but since there is only one such signal the alkene would have to be completely symmetrical. Since you see 7 carbon signals from 7 total carbons you know that this cannot be the case there cannot be any symmetry in the molecule (and, you are told the molecule is chiral). Given these results, there must be 2 rings in the molecule, plus the carbonyl. The 1 NM shows four hydrogens with chemical shifts consistent with being adjacent to heteroatoms (i.e. oxygen), and the splitting pattern suggests that one of them has only one neighbour. Likewise, the 2 hydrogens in close chemical shift proximity are probably a C 2 group. You also have 3 carbons with chemical shifts consistent with aliphatics attached to oxygen. The product is 2,3--isopropylidene-D-erythronolactone. You did not have to show stereochemistry for full marks, although you should recall that a 5-5 ring fusion can only be cis. 13 C NM assignment: lactone carbonyl (C1) (C5) 75.4 (C2 or C3 or C4) 74.5 (C2 or C3 or C4) 70.1 (C2 or C3 or C4) 26.7 (C6 or C7) 25.5 (C6 or C7) t is possible to assign these carbons more specifically but since that requires additional information you were not required to do so for full marks. 1 NM assignment: ppm (1, dd) ppm (1, d) (2) 4 and 4 (two distinct signals, an ABM pattern with 3) These can be assigned more specifically, since you notice that one of them has a much larger vicinal coupling constant with 3 than the other one does. From this one can conclude that the upfield signal at around 4.4 ppm is the syn to 3, while the anti hydrogen is nearly orthogonal to the C3-3 bond and hence has a negligible coupling constant. This analysis was not required for full marks, however. Approx 1.5 two distinct 3 singlets for the methyls (diastereotopic).