week 11 problem set (solutions)
TRANSCRIPT
Week11ProblemSet(Solutions)4/24,4/25,4/26
ConceptsCovered
AlkynesOxidationAlcohols
Reactions/Reagents
DeprotonationofAlcohols/AlkynesJonesOxidation(Cr2O7)
Dess-MartinPeriodinane(DMP)Corey-SuggsOxidation(PCC)
PinacolRearrangementPeriodicCleavageof1,2Diols
ThionylChloride(SOCl2)PhosphorousTribromide(PBr3)
(4) Synthesis
As with all syntheses, some basic things to check first before you dive in are the number of carbons in the product and starting material and where the starting material ends up in the product. The starting material has 2 carbons and a phenyl group, while the product has 6 carbons and two phenyl groups. The deuterium label helps to show that the starting material was most likely used twice in the synthesis of the target compound. If the starting material were used twice, we would have two carbons leftover that would have to be added via outside reagents. Whenever there is a two-carbon piece missing, one of the first things you should think of is acetylide, as it is a common building block used in 350.
Retrosynthesis (it’s easier backwards!):
First, the trans alkene was most likely formed via sodium metal reduction of the corresponding alkyne. While elimination is also a route to product trans alkenes, an E2 reaction would not be selective in this system since the alkene can be formed between multiple different carbons with the same substitution.
If we make a disconnection between the alkyne and one of the connected carbon chains, the bond was most likely formed via an SN2 reaction. (The disconnection site was arbitrarily chosen, so you could have picked the other side of the alkyne first and it would not matter). While alkynes can be formed from alkenes via a double elimination after Br2 addition, this would not make sense here since we just came from an alkene. Since SN2 reactions cause in inversion of configuration, the electrophile (shown with a good leaving group “GLG”) would have the opposite configuration as demonstrated with the deuterium now point out of the page. This disconnection is helpful because it breaks the larger compound into two smaller ones that can both be further analyzed via retrosynthesis.
Looking at the electrophile (next page):
Chance Dunbar
Following a rotation in the plane of the page, it becomes apparent that the electrophile has the same stereochemistry as the starting material. This is convenient because the starting alcohol can be converted into a GLG in one step via tosyl chloride, which will retain the configuration.
Now looking at the alkyne piece:
Again, since this is an SN2 disconnection, there would be an inversion of configuration at the site of attack, as shown by the change in deuterium from a wedge to a dash.
The final step is to take this electrophile back to the alcohol. Notice that the chiral center has had an inversion of configuration. Remember, this is the only stereoisomer we are working with, so we can only use reactions that have a controllable/predictable stereochemistry. Conveniently, there’s a reagent that will convert an alcohol into a GLG via an SN2-type fashion: PBr3 (or SOCl2).
Full Synthesis (order of electrophiles added can be reversed):
(8) Boxes
These boxes are definitely a bit tricky. Let’s gone through each set of them one at a time. First, some strategy. While it is helpful to draw out all the possible stereoisomers of C7H13OH, it is not necessary to
solve the problem (and might take up valuable time during an exam). However, it is excellent practice to do so on your own and is highly recommended. The process to find all the isomers is identical to Week 5 – Problem 4. Please review this if you need help generating isomers! Here are all of them so you can check to see if you got them all:
If you did not generate the stereoisomers, you should at least write out the constitutional isomers (without stereochem) to help you visualize which isomer(s) would be appropriate for specific boxes.
Looking at the top section, let’s start by filling in with what we know. Only two structures are provided – the rest of the information is given with reagents. This tests out ability to work backwards through problems. The best place to start is probably the reaction of the first boxed isomer with PBr3 and EtO- as most of the other reactions depend on knowing this structure. PBr3 is a reagent that converts alcohols into Br’s via an SN2-type mechanism, so an inversion of configuration must have occurred. EtO- is a strong unhindered base that will perform an E2 elimination towards the more substituted side. Interestingly, we see that the alkene formed is not the most substituted alkene possible. This leaves us with two options: the isomer has a 1,2 relationship between the alcohol and methyl group but no anti hydrogen is available to form the tri-substituted alkene; OR the isomer has a 1,3 relationship between the alcohol and the methyl group. However, the logic of the latter option quickly disintegrates if we think about what the specifics of the reaction. If the relationship between the substituents was actually 1,3, then the elimination could not be controlled and two alkenes would be formed. Since this is not the case, the former option must be true.
The full reaction therefore must be:
The stereoisomer shown above is the only one that can satisfy the reaction scheme. PBr3 will cause inversion of the OH group as it changes it to a Br. The bromo compound has no hydrogens anti to the Br on the tertiary carbon, meaning that elimination cannot occur to form the tri-substituted alkene. Instead, the elimination must occur on the other side as this is the only option available.
The hard part of this set of boxes is completed. With the structure of the first isomer deduced, we can continue with the reaction scheme. When this compound is reacted with TsCl followed by EtO-, the elimination will now occur on the more substituted side. Why? Because the conversion of OH to OTs does not involve an inversion of configuration at the carbon, so a hydrogen anti to the OTs is available on the adjacent tertiary carbon for elimination.
A fork in the road. It’s up to you which one to do first. I will choose to first work backwards from the given ketone compound. The ketone formed via PCC is hopefully an easier transformation to recognize. We know that PCC oxidizes alcohol to ketones, so the alcohol that came before should have been an isomer that has a 1,2 relationship between the alcohol and the methyl substituents. Going forwards from the tri-substituted alkene, there must have been an addition of alcohol to the less substituted side of the double bond. The only reaction that we know that does this is hydroboration. The reaction scheme must therefore look like this:
Since hydroboration add the OH and H syn across the double bond, we must show the proper stereochemistry of the reaction products. Two stereoisomers of C7H13OH are produced and both should be shown. When both are reacted with the oxidizing agent PCC, the alcohol becomes a ketone. Since the methyl group was both up and down in the products of the hydroboration reaction, the methyl group on the cyclohexanone is drawn with a squiggly line to indicate both configurations are present at that site.
Now we can fill in the reagents needed to convert the hydroboration products back to the tri-substituted alkene. The simplest method is to use H3PO4 to do an E1 reaction. The tri-substituted alkene will be predominantly formed as it is more stable. Another option is to do an E2. However, the hydroboration products have the OH syn to the H on the tertiary carbon. If the OH is converted into a good leaving group, it must also undergo an inversion of stereochem so that the LG and the H end up trans to each other. As a result, PBr3 or SOCl2 must be used to change OH into a LG as they invert the configuration. Afterwards, an unhindered base like EtO- will give the desired E2 reaction.
Looking at the osmylation reaction, we know that a syn diol will be produced. The question then becomes, how we convert a diol into a ketone?
This is one of those reactions that you will have to remember based on pattern recognition. We know that one of the OH groups leaves and then the leftover OH is turned into a ketone. The conversion of a diol into a ketone (with the number of carbons intact) is evidence of a Pinacol rearrangement. The Pinacol reaction occurs when a diol is subjected to a strong acid, like H2SO4. The more substituted alcohol will leave first generating the more stable carbocation. A hydride transfer will occur to shift the carbocation directly under the remaining hydroxyl, which will then donate its lone pair to form a double bond with the carbon. The missing reagent is therefore a strong acid (H2SO4).
The first set of boxes is complete. Moving on to the middle set of boxes! The reaction scheme is fairly simple: An alcohol is converted into a LG and undergoes SN2 with NaN3. The azide compound shows us the relationship between the alcohol and the methyl must be 1,3. The question is a matter of stereochemistry. Working backwards, we know that an SN2 with NaN3 will cause an inversion of configuration at the site of attack. If the leaving group was formed via PBr3, then there was a double inversion of configuration or net retention of stereochemistry at the original alcohol carbon center. If the leaving group was formed via TsCl, then there will be a net inversion of configuration.
Chance Dunbar
The last box in the middle section involves converting either alcohol to a compound whose molecular formula is C7H12O. Compared to the C7H13OH molecular formula, there is a net loss of H2. This strongly suggests an oxidation has occurred. If the degrees of unsaturation are calculated for both compounds, it is determined that C7H13OH has 1 degree of unsaturation while C7H12O has a 2 degrees. This also confirms that an oxidation is occurring as an additional pi bond has formed, resulting in the loss of H2. Since the only functional group to start with is an alcohol, we can assume the oxidation is occurring there. To oxidize an alcohol, a number of oxidizing agents can be used: Jones, PCC, or DMP.
Finally, we bring our attention to the bottom set of boxes. We know that ozonolysis cleaves alkenes into two carbonyls. Since the pair of carbonyls is formed on the same linear chain, the reaction must have occurred in a cyclic alkene.
The last step is to connect one of the isomers of C7H13OH to the cyclic alkene shown above. What is the relationship between the substituents? We know that elimination reaction has to occur to convert an alcohol to an alkene, and that the most direct route of eliminating alcohols is through an E1 with H3PO4. If the OH and Me were 1,3 to each other, then the elimination could not be controlled. However, if the relationship is 1,4, then because of the symmetry of the molecule, there will be only one elimination product. Therefore, it must be one of the 1,4 isomers. Only two stereoisomers of the 1,4 constitutional isomer exist: the cis and trans isomers (the beginning of this solution shows all possible stereoisomers). Both isomers when undergoing an elimination will product both enantiomers of the cyclic alkene. This is due to meso nature of the molecule. Therefore, both cis and trans isomers are valid here, which is indicated by the squiggly bond to OH in the answer key.