* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Week 6 Solutions - Brown University Wiki
Survey
Document related concepts
Woodward–Hoffmann rules wikipedia , lookup
Enantioselective synthesis wikipedia , lookup
Petasis reaction wikipedia , lookup
Wolff–Kishner reduction wikipedia , lookup
Aromatization wikipedia , lookup
Ring-closing metathesis wikipedia , lookup
George S. Hammond wikipedia , lookup
Asymmetric induction wikipedia , lookup
Aromaticity wikipedia , lookup
Strychnine total synthesis wikipedia , lookup
Hydroformylation wikipedia , lookup
Transcript
(1) Russell Shelp Russell Shelp 3. a) Provide the chair structure of the product of the following reaction IMMEDIATELY after the reaction occurs. What is the lowest energy chair conformer? Br2 OH H2O OH + Br Br Br2/R-OH addition occurs through an anti-periplanar stereochemistry, so we know the bromine and OH group must initially be anti-periplanar right after the reaction occurs. In order to be antiperiplanar, they must be axial. Initial conformation: OH Br or OH Br After a while, equilibrium is reached between the other chair conformers and the lowest energy chair conformers prevail: Br Br OH OH OH or OH Br Br b) Label the following structures as conformers, enantiomers, or diastereomers. I find the best way to approach this problem is to first convert both Newmans and chair diagrams to planar cyclohexane rings. This makes it easier to identify stereochemical relationships. i. Cl H H Me Me vs. H H H H Converting to cyclohexane rings gives us: Cl Cl Cl vs. Me Me The left structure shows the methyl equatorial and the right structure shows the chlorine equatorial, but the structures are otherwise identical. These structures are conformers. ii. H Et H H H H Me H Br Me Et vs. Br The problems showed the Ethyl of the right hand molecule in an incorrect axial position, sorry about that! Converting to planar cyclohexanes gives us: Et Et vs. Me Me Br Br These structures are mirror images of one another and are therefore enantiomers. iii. Br Br H H H H H H vs. Me Converting to a planar cyclohexane gives us: Me Me vs. Br Br Only one of the three stereocenters have been switched, making these structures diastereomers. Mary O'Connor 4. 5. Mechanism: As always, let’s start out with numbering the starting material and the product and making a bonds broken/bonds made list. Bonds Broken C4-C5 pi O-H sigma Bonds Made O-C4 sigma C5-H sigma We know that the oxygen has to make a bond to form a quaternary center. The likely candidate for this is carbon number 4 as it already has three bonds to carbon. If the alcohol oxygen made a bond with carbon 4, it would form the ring system in the product. But how do we make a bond with carbon 4? So far, we have seem bonds formed between electrophiles and nucleophiles. Since oxygen has lone pairs, it will be the nucleophilic atom. The question then becomes, how do we make carbon 4 electrophilic? One way to make carbons electrophilic is to turn them into carbocations. That means we will need a way to break the C4-C5 pi bond. Another consideration is to look at the reagent. Our sole reagent is H2SO4, which is a strong acid. We know that alkenes in strong acids can be protonated. Given this, we can start pushing arrows: Remember that the tertiary carbocation forms preferentially over the secondary carbocation. Since the alcohol is the only source of oxygen, it attacks the carbocation. Once deprotonated by the conjugate base, the product is formed. Chance Dunbar 6. 9) H 2SO 4 H 2O HO This mechanism may at first seem daunting – but if you approach it in a systematic manner, you will be able to figure out whats going on. Like in all mechanism problems the first thing we want to do is to number carbons and make a bonds broken/bonds formed table. In this case numbering carbons is actually pretty difficult but lets try it out. We know that we’ve had a ring expansion – thus one of the carbons in the 5 membered ring that was formed has to come from one of the carbons outside the initial 4 membered ring. Furthermore, we know one of the branched carbons in the product has to come from the branched carbon in the starting materal (shown with C in both molecules). a 1 a C 1 C C 1 HO Also note that the carbon that is labeled 1 in the first compound (note that they are the same as the molecule is symmetric) must have been the one to attack the carbon currently labeled a to make a 5 membered ring. If it is attacking carbon a – we know that the other branched carbon has to correspond to the carbon labeled in red in the starting material. This may all seem confusing currently. It takes some practice – especially in a pretty challenging example like this one. The full numbering is shown below: 7 5 6 6 1 1 5 H 2SO 4 2 4 H 2O 3 4 7 3 2 HO If the numbering didn’t make sense before – try to see if you can reason out why the numbering scheme above works. Notice that C1,2, and 3 is part of both rings and is unbranched in both. Furthermore note that C4 still has the connection to C7 and C5 is still connected to C6. Now lets make a bonds broken/bonds formed table: Bonds Broken C5-C6 pi C1-C4 sigma Bonds Formed C1-C5 sigma C4-O sigma C6-H sigma Now lets think about the mechanism – we have an alkene in out starting material and we are subjecting it to a very strong acid. We know that it will get protonated and form the more stable carbocation – in this case secondary. 7 5 7 5 6 H-HSO4 1 4 6 1 4 3 2 3 2 In that step we broke the C5-C6 pi and formed the C6-H sigma. Now we know that to form the 5 membered ring we broke the C1-C4 sigma and made the C1-C5 sigma. This is a case of a rearrangement. 6 7 5 1 5 6 1 2 4 4 3 2 3 7 Now there is a positive charge on C4, which means the remaining water in solution will attack here, forming the C4-O sigma. 6 1 5 2 4 3 7 HO H 2O H And finally, we have to deprotonate the alcohol group. And now you can also check your BB/BF table to see if you did everything you needed to in the mechanism. HO OH H B Full Mechanism 7 5 6 1 4 3 H-HSO4 6 7 5 6 1 2 4 2 1 5 3 4 3 7 2 H 2O HO OH H B Natalie Palaychuk 7. Please provide a mechanism for the following reactions: a) Start this problem by numbering the starting material and the product and creating a “bonds made” and “bonds broken” list. 1 4 1 8 OH 6 3 5 Hg(OAc)2 AcOHg 4 H 2O 7 3 5 2 Bonds Made C3-O C4-HgOAc 2 O 8 7 6 Bonds Broken C3-C4 pi O-H (HgOAc)-OAc This is the mechanism for oxymercuration of a double bond. Like Br2 addition, this mechanism proceeds through a three-membered ring intermediate. The tertiary carbon of the double bond (C3) is more electrophilic since it has more carbocation character than the secondary carbon of the double bond (C4). Once the three-membered ring is formed, it is opened by the OH within the molecule, not water. But, Why doesn’t water attack C3?! Because: intramolecular reactions (electrophile and nucleophile in the same molecule) are much faster than intermolecular reactions (electrophile and nucleophile are on separate molecules) due to the increased proximity. AcO OAc Hg AcO OAc Hg Hg OH OH OAc Hg OH AcOHg AcOHg H O O BH4 NaBH4 is the second of two reagents used during oxymercuration. It removes the Hg(OAc) from C4 and replaces it with a hydrogen, in a process know as “reduction”. OH Hg(OAc)2 H 2O AcOHg O NaBH4 O b) CHALLENGE *Note: D means deuterium, an isotope of hydrogen H 2O HO D H2SO 4 D As with any mechanism, it is best to start by numbering the carbons and constructing a “bonds made” and “bonds broken” list. 3 1 2 4 5 2 H 2O 7 4 8 1 10 6 HO D H2SO 4 9 3 5 6 8 7 9 D 10 If you find the numbering difficult, consider these tips: • You know that carbon 7 had the deuterium, so you can generally assume that it retains the deuterium in the product. While this is not always the case (maybe there was a deuterium shift), it’s a good place to start. • The terminal alkene seems to have been retained from starting material to product, so we can assume that it is still 1. Again, this is not always the case but it’s a good place to start. • The methyl groups in the starting material are 4 and 10. When numbering the product, assign the methyls to 4 or 10, depending on the surrounding numbers. Bonds Made C3-C9 C5-C6 pi C9-OH C7-H C8-H Bonds Broken C3-C5 pi C6-C7 pi C8-C9 pi There are a couple of starting moves we could take. The best way to get used to doing mechanisms is to go down unsuccessful paths and learn to recognize patterns. For example: • Protonating the 3,5 double bond can give us a cyclic product, but the cyclic double bond in the product will end up in the wrong location • • Protonating the 8,9 double bond can also give us a cyclic product, but the cyclic double bond in the product will end up in the wrong location. Also, the resulting carbocation (and OH addition) will be in wrong location. Protonating the 6,7 bond is just right! Notice that there is a hydride shift moving the carbocation downhill from a secondary carbon to a tertiary carbon. This gives up OH addition at the correct location. H+ D D D H H O H D H 2O D B D HO D Mary O’Connor 8. Box$problems!$ $ a)$ OH 1. BH3 2. H2O2, NaOH 1. Hg(OAc)2, H2O 2. NaBH4 H2SO 4, H2O OH HO *Scrambled stereochemistry $ As$I$indicated,$remember$that$the$reduction$step$in$oxymercuration:reduction$ scrambles$stereochemistry!$Also,$remember$that$water$will$always$add$to$the$more$ substituted$side$of$the$alkene.$ $ With$acid$catalyzed$hydration,$water$still$always$adds$to$the$more$substituted$side$ of$the$alkene,$but$thermodynamically$favored$carbocation$rearrangements$will$ occur$whenever$possible!$ $ With$hydroboration:oxidation,$water$adds$to$the$less$substituted$side$of$the$alkene.$ $ It’s$really$useful$to$start$recognizing$the$patterns$of$the$products$for$each$of$these$ reactions.$I$would$also$recommend$starting$a$reaction$chart,$where$you$summarize$ each$reaction$and$any$of$its$important$features$(when$doing$this,$it’s$useful$to$go$to$ the$reaction$summary$at$the$end$of$chapter$6—there’s$one$of$these$summaries$at$ the$end$of$each$chapter$going$forward!).$$ $ $ $ $ $ b)$ O O 1. O3 2. Me2S H2, Pd/C 1. OsO4 2. NaHSO3, H2O These products are diastereomers. OH OH + OH OH + (diastereomers) $ Remember$that$ozonolysis$is$like$molecular$scissors;$you$use$it$to$cut$an$alkene$in$ half,$and$create$a$carbonyl$on$each$side.$It$creates$a$ketone$or$an$aldehyde$ depending$on$the$substitution$of$the$alkene.$ $ $ To$go$from$an$alkene$to$an$alkane,$you$can$use$a$transition$metal$catalyst$(Pd/C$or$ Pt/C)$with$H2!$The$hydrogens$will$add$syn.$The$product$is$a$mix$of$diastereomers$ because$the$chiral$centers$in$the$starting$material$do$not$change,$but$the$hydrogens$ can$both$add$to$either$the$top$or$bottom$face$of$the$ring.$ $ Osmylation$adds$two$alcohols$syn$to$one$another!$In$this$case,$the$product$is$a$mix$ of$diastereomers$because$the$starting$material$has$two$other$chiral$centers$that$ remain$the$same$in$the$product,$while$the$alcohols$can$add$to$either$face$of$the$ring.$ $ Ella Cohen 9. Box Problems. Indicate stereochem when necessary. a) Things to remember: Osmylation and Hydrogenation are both syn additions Halohydrin is an anti addition For the osmylation reaction, it looks like the wrong product. Why are the alcohols on opposite sides of the molecule if osmylation is a syn addition? It’s actually just a conformer of the product. Here’s a different perspective on this: As we can see, the methyl is just rotated so that it is anti to the ethyl group. For the halohydrin reaction, the water will preferentially attack the carbon with the phenyl group as this carbon as better able to stable the partial positive charge in the transition state. A benzylic tertiary carbocation is more stable than just a normal tertiary carbocation due to resonance. This reaction will form a pair of enantiomers. Hydrogenation is a simple syn addition and forms a pair of enantiomers in this reaction. b) Sometimes it’s hard to think about going backwards. However, we can use the 2 equivalents of product to our advantage. Since ozonolysis cleaves double bonds, then the both sides of the double bond must be identical. There must also be two double bonds since there are two carbonyls in the product. c) Hydroboration is a syn addition, which means the alcohol and the hydrogen will add to the same face of the double bond. Since the stereochemistry of the alkene is different between starting materials, each reaction will produce a different pair of enantiomers. The pairs of enantiomers are diastereomers to each other, as shown by the indicated absolute configuration. Chance Dunbar