Chemistry 222--Spring Third Problem Set Due 5/1/02 H H HO OH. H Ether C H

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1 hemistry 222--Spring Third Problem Set Due 5/1/02 nswers Part. For the following reactions, give the predominant product. learly show stereochemistry when appropriate (that is, draw a structure that clearly shows double bond configuration and chiral center configuration of one enantiomer). 1) Excess Lil Ether I have chosen a single configuration for the aldehyde's connection to the ring to the left-- that was not specified in the problem, so you could choose whatever you wanted. The product will be both enantiomers of the compound shown, as drawn 2) 3 3=PPh 3 TF 3) 1) 1 equiv: Ts 2 2 2) 2 N-N 2, K, 3) 3 + 4) (catalyst) excess

2 hemistry 222 Problem Set 3 page 2 5) 1) LD, -78 2) 3) 2 There is some question about whether this reaction will give the actual aldol product or will dehydrate. It does depend on how you treat it at the end of the reaction. The indications from the conditions are that the treatment would be relatively gentle, and the aldol would survive.? 2 Part. For the following pathway 2 3 Na, ether P ) Provide structures for compounds,, and. 3 Na, ether 3 2 P catalytic + 3 catalytic ) Explain how the following spectroscopic observations are consistent with your structural assignments: : IR: 3300 (s, b), 1050 (s) NMR: There is a quartet at about 3.7 ppm in the 1 -NMR in acetone. When DMS was used as the solvent (which enhances the chance of observing the coupling to protons), that quartet changed to a quintet. This was key to getting the chemistry to work out right. The dehyrdrohalogenation of the alkyl halide to give the alkene was very popular, but didn't work out with the NMR, especially. This also helps you choose which oxygen was used to make the cyclic ether: the next to the must have 3 neighbors, not counting the proton. That can only be the carbon drawn at the top. The IR at 3300 comes from the stretch, and serves to contrast with, which doesn't have it. : IR: 1700, This corresponds to = stretching from the ketone, and to the - stretch from the cyclic ether. : IR: 1050, no 1700 or There are a lot of - bonds here (stretching at 1050), but no carbonyl and no.

3 hemistry 222 Problem Set 3 page 3 3) omment on the stereochemical nature of (that is, how many isomers would be expected, and what would be the relationship between them?). ssume that you start with all possible isomers of the starting material. This is actually rather interesting stereochemically. There are three stereogenic centers, labeled 1, 2, 3 in the drawing to the right. The configuration of 1 and 2 are fixed by the use of a single enantiomer of the diol, so they do not lead to the production of isomers. The configuration at center 3 is not fixed; one could have either R or S there, leading to two isomers from this reaction, out of a possible 8, by standard analysis. The two compounds that are formed are diastereomers, not enantiomers, since only 1 of 3 possible stereogenic centers is inverted in going from one isomer to the next. If you are easily confused about stereochemistry, you might want to skip this next part of the explanation, since it is an extension beyond the normal stereochemical analysis done in intro organic. It turns out that the ring juncture carbon (the one that is bonded to 2 oxygens as part of the acetal) is not a stereogenic center in this molecule, but would be if we changed the configuration of either center 1 or 2 (but not both). Why? If you start from the ring juncture and go into the acetal ring as drawn, you come to carbons 1 or 2 and they are the same, because they have the same configuration (they are both S). If one is S and the other R, they will not be the same, and the ring juncture would make a fourth stereogenic center. Therefore, the actual total number of possible stereoisomers here is not 8, as standard analysis would suggest, but is 12. To be perfectly honest, I didn't predict this, and discovered it as I was explaining the problem to others. I wouldn't expect you to see this kind of stereochemical relationship without help. 4) Would the isomers of be separable by ordinary means (for example, G) or would they not? Explain briefly. Yes. Diastereomers frequently have different physical properties (P, MP, solubilities, etc.) and so these could be separated. This is a common method for resolution of enantiomers of chiral ketones: make the acetal with a single enantiomer of a chiral diol, separate the diastereomers produced, then remove the acetal from the individual diastereomers, providing purified single enantiomers of the chiral ketone. Part. Reacting phenol with acetone in the presence of an acid catalyst gives a compound know as bisphenol. 1 3 P isphenol a. What other product is formed in the main reaction to produce bisphenol? It's water ( 2 ). This is intended to be a significant clue about the mechanism. Like: do you know of any reactions with ketones that produce water? b. Propose a mechanism for the formation of bisphenol. You may have to combine chemistry that you know from more than one chapter in the textbook isphenol is used in the production of epoxy resins and polycarbonate resins. It is also one of many compounds being examined as an environmental estrogenic compound: it has weak estrogen-like properties.

4 hemistry 222 Problem Set 3 page E D F 3 isphenol Phenol is a weak nucleophile, so you need to start by protonating acetone. This makes a pretty good electrophile, which is good enough to react with the very activated phenol, giving the standard Wheland intermediate. Note that the elements of the Wheland intermediate are just as they are shown in hpt. 12: a single sp 3 carbon, with a positive charge delocalized around the ring, given extra stabilization by the presence of the oxygen para to the site of attack. The elements of first attack by a nucleophile on a protonated ketone are also in place: sp 3 hybrid carbon where the carbonyl carbon was, and a neutral attached. The formation of is the re-aromatization of by the loss of a proton completing the substitution there on the ring. To continue the reaction, you need to get rid of the hydroxyl group. Under these conditions (acidic) it makes sense to protonate it (D) and allow the formation of the carbocation E, which is tertiary and benzylic, so should be quite well stabilized. This can react exactly as expected for a Friedel-rafts alkylation: reaction with the aromatic ring to make the Wheland intermediate F, followed by rearomatization to produce the product. c. Which reactions did you use as models for the mechanism you proposed? [This is a kind of a hint: you don't need to make up the mechanism completely if you can a) for each reactant find an analogous reaction whose mechanism is described in the book; and b) make a functional substitution--for example, nucleophile for nucleophile--which involves the other functional group in the reaction.] The two reactions I would consider most closely related are the aromatic electrophilic substitution reactions (ES, chapter 12) and the formation of acetals from ketones (chapter 17). There are elements of each reaction in the mechanism above. For example, we tell you that the aromatic ring is a nucleophile in the ES reaction. It reacts as a nucleophile here. The protonated ketone is an electrophile in the formation of acetals--it reacts as an electrophile here, too. Part D: Indicate how the following transformations might be carried out. More than one step will be required. Specify all necessary reagents, conditions, and (where

5 hemistry 222 Problem Set 3 page 5 appropriate) solvents. Draw structural formulas for all isolable intermediate products. Do not include mechanisms. You may use any organic or inorganic reagents necessary to accomplish these transformations , 1 hr RT 1) Mg, ether 2) 3 3 You had better plan to protect the ketone from the forming Grignard reagent: after all, there are lots of other molecules to react with. ase (Na 3 or NaN 2 or...) MMPP, or other R 3 Ph 3 P= 3 3 Mostly this isomer, you would probably have to separate isomers here. Wittig gives mostly the cis isomer. The hardest thing to see here is the aldol product (the enal). People will probably form acceptible answers to this without it (since I didn't put size limits on the reagents) but this one might have been more satisfying.

6 hemistry 222 Problem Set 3 page 6 Part E. The reaction shown below gives primarily the product shown, even though one might expect a more complicated mixture. ase 1. Provide a reasonable mechanism for the transformation of starting material to product. 2. If you start with a single enantiomer of the starting material, will you make a single enantiomer of the product? Explain briefly. Yes. The only stereogenic center in the starting material is the only stereogenic center in the product (it's the quaternary carbon at the ring juncture in the product). No bonds to this carbon are made or broken in going from the starting material to product, so the configuration should remain the same. f course bonds are made and broken elsewhere, otherwise there would be no reaction. 3. Redraw the starting material, and show all of the potential sites for enolate formation (that is, all the 1 potentially active α-hydrogens). The compound is shown to the right. The potential enolates may be formed by removing protons from positions 1, 2, or 3 (that is, there are three possible enolates). Note the dotted line to 4: that can not be an enolate, since there are no hydrogens there to remove! Don't be fooled by the fact that it is - to two carbonyls ( and ). No enolate may be 2 formed if there is not proton to remove. 4. Using your answer to number 3, how many possible 3 aldol products are there? Explain briefly. I have labeled the carbonyls,, for discussion. Note that when enolates 1 or 2 are formed, carbonyl is consumed in the process, so there could be no aldol formed between enolate 1 and carbonyl (sorry if this is too obvious). Enolate 1 could react with carbonyls or ; as could enolate 2. Enolate 3 could react with carbonyls or. ence, there could be 6 possible products. Did I mention that there is no enolate 4? 5. Draw the structure for one of the alternate products, and explain why it is not as likely to be formed as is the product shown above. 4

7 hemistry 222 Problem Set 3 page 7 f the 6 possible products, we know that one is formed to a substantial extent. Three of the others are actually quite bad, since the ring size of the product is wrong. Pairings 2-, 2- and 3- make 4-membered rings, which are too small to be stable. Pairing 3- makes a bicyclic product that would be less stable due to a very strained bridgehead double bond The trickiest argument about stability is why pairing 1- doesn't play a larger role in the product mixture. I'm not sure I have a great answer to this, although it pretty clearly has to have something to do with product stability. The best I can come up with without doing stability calculations is that the product that is mentioned (coupling between 1 and ) has two ketones, while this product has one ketone and one aldehyde. Since ketones are more stable than aldehydes, that might be enough to sway the reaction toward product 1-. Notice how the question didn't make you answer this tricky part? When you have this opportunity to answer an easier question, feel free to take advantage of it. 2-1-

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