Organic Synthesis Help

[KCM92]

I have an assignment in my organic class (First semester of organic for chemistry majors) to show how to perform the following synthesis.

 

 

We can only use reactions that we have covered in class this semester which include: alkene additions (hydroboration, oxymercuration, etc.), reactions of alcohols (oxidation, converting to tosylates/mesylates, reduction), epoxide formation and breaking reactions, nucleophilic substitution and elimination, metathesis reaction, Grignard reaction, cleavage of glycols. We are graded on using the most efficient method, and get bonus points if we include another method.

 

I started and am stuck on how to get the starting material to a ring. I'll post what I have so far and I would appreciate any advice on where to go from here.

 

Step 1 (working backwards):

Either step would work, I assume the peroxyacid one would be more efficient though.

 

 

Step 2: (t-butoxide is supposed to be over the reaction arrow)

 

 

This is where I am stuck, I'm not sure how to get the starting material into a cyclobutane/ene ring. I have one possible method figured out, but I'm not sure how well it would work.

 

 

 

Any advice/help would be much appreciated. Thanks in advance.

[hypervalent_iodine]

Hi KCM92,

 

Apologies it's taken so long to get a reply. I've been (and still am) on a trip away and it seems that your assignment hasn't piqued the interest of our other members.

 

Firstly, your initial strategy of 'working backwards' is a very good way to start things like this. Though you probably haven't covered this yet, the approach you took actually has a name, that being retrosynthesis, and is used frequently to work out synthetic routes to complicated structures.

 

You are right to assume that the peroxyacid is more efficient than your second route. It's not so much because the reaction itself is better, but because the synthesis of the 1,2-dimethylcyclobutene ring versus the 1,2-dimethyl-2-bromo-1-hydroxycyclobutane is much easier.

 

I'm not sure I agree with your step 2. Making the ring installed with a bromo-substituent would be quite hard. If you imagine a disconnection at the bond between the two methylated ring protons, you will note that one of the carbons would need to act as a nucleophile and the other as an electrophile. Doing this to end up with the starting material listed in Scheme two would not be able to occur in one step. Since this is an assignment, I don't think it would be fair to simply tell you what would be better. Can think of what other substituents you might be able to include in place of the bromine, which can then be eliminated to give a double bond?

 

Once you've worked that out, you'll need to work out where to make a bond disconnection to make the ring. Feel free to come back with ideas on that and I can help you more. There is one name reaction in particular that you mentioned in your OP that would apply here.

 

Also, I'm curious to know if you've copied the assignment down correctly? It seems odd that they would add two completely superfluous carbons.

[KCM92]

Are you thinking of dehydration of an alcohol for the second step? I could do an ozonolyzis on the starting material to both ends, then a Grignard addition with methylmagnesium bromide to get the two hydroxyl groups and appropriate amount of carbons. From there I'm stuck on getting the ring to form.

 

The problem is written down correctly. I think my professor just wanted to make it slightly more difficult.

 

Also, am I correct in assuming that I don't have to worry about the stereochemistry because there is only one possible configuration for the final product (the two bonds attached to the oxygen are always going to be on the same plane)?

 

 

Thank you for the reply.

[hypervalent_iodine]

Are you thinking of dehydration of an alcohol for the second step? I could do an ozonolyzis on the starting material to both ends, then a Grignard addition with methylmagnesium bromide to get the two hydroxyl groups and appropriate amount of carbons. From there I'm stuck on getting the ring to form.

 

I am indeed. Upon reflection though, I realise that this route might not work since it relies on being able to install different functionality at each end of the molecule, which I realise now would be difficult. My idea to close the ring was to install a ketone moiety on what would be the first ring carbon and a bromine on the second last carbon of that chain, make a Grignard reagent at that centre and cyclise by reacting the intramolecular Grignard reagent with the ketone:

 

 

Another way you could set up your ring closure, using reactions known to you, is to brominate in a Markovnikov fashion on each of the double bonds, then eliminate them both under kinetic control to give you the terminal double bonds. From there you could cleave them with ozonolysis as you have in your method to give you the dione.

 

I also had a quick check on SciFinder to see if I was missing any methods in preparing your ring. The one reaction that predominated is a photoisomerism reaction and involves a radical pathway. I'm not sure how much you know of those types of reactions, but I thought I might mention them.

 

I think what might be best for you is if you realise that cyclisation reactions are typically just an intermolecular reaction applied to an intramolecular system; the reaction itself is the same, it's just that you're now using it on functional groups in the same molecule (for example, the Dieckmann cyclisation, which is an intramolecular Claisen condensation). So from here it is really just a problem of looking at your 1,2-dimethylcyclobut-1-ene, which you indicated as you preferred precursor to the final product, and figuring out ways you know to make double bonds. It might help you to make a list of the ones you mentioned in your OP and go from there.

 

I think my professor just wanted to make it slightly more difficult.

 

Oh, of course. The compound you listed is something I could envisage being made in a two-step procedure, so it's definitely an exercise in getting you to apply yourself to a difficult and purely theoretical situation.

 

Also, am I correct in assuming that I don't have to worry about the stereochemistry because there is only one possible configuration for the final product (the two bonds attached to the oxygen are always going to be on the same plane?

 

Well, you wouldn't be able to distinguish between the (R, S) and (S, R) enantiomers and definitely you would only get that particular configuration, since the (S, S) and (R, R) diastereomers would not be able to be formed given how constricted they would be. With that in mind then, I would say no, you don't need to consider any asymmetric syntheses, although it I am not the one marking you.

[KCM92]

That helps a lot, thank you very much. I'll talk to my professor about the chirality to make sure.

[hypervalent_iodine]

I shouldn't make posts at 2am. The photoisomerism reaction is actually a pericyclic, photoinduced [2+2] addition rather than a radial reaction, although a radical pathway would also result in your product. You mightn't be too aware of many pericyclic reactions out side of the Diels Alder reaction, though, so I wouldn't worry too much about that pathway.

[Horza2002]

If you actually wanted to do this reaction, I would say the best way to make a cyclobutane/ene ring would be a photoinduced [2+2] reaction....trying to get that Grignard to cyclise before it polymerise might be a little tricky.

[hypervalent_iodine]

Well, yeah, but undergrad is rarely concerned with making things easy, now is it?

 

Also, you must be really bored.

[KCM92]

Here's what I got for the final answer

 

[Horza2002]

In theory, most of that looks ok....however in practise this would not work at all; selectivly reducing one of those ketones to the alcohol is going to be extremely difficult.

 

With that said, the step converting your dibromide to the diterminal alkenes (step two going from the start material) will not work at all. Yes in theory, you could eliminate form the primary methyl groups to give you the product you want, and maybe this will be the kinetic product...but the more substituted alkenes are more stable (i.e. four groups is more stable than 3 groups which is more stable thn 2 groups which is more stable than 1 group). This arises from hyperconjugation from the adjoining carbon-carbon or carbon-hydrogen bonds. Electron density from these bonding orbitals is able to overlap with the pi* antibonding orbital of the alkene thereofr making the system more stable.

 

In addition to being more subsituted, your starting material is more stable as the two double bonds are in conjugation with each other.

[mississippichem]

I shouldn't make posts at 2am. The photoisomerism reaction is actually a pericyclic, photoinduced [2+2] addition rather than a radial reaction, although a radical pathway would also result in your product. You mightn't be too aware of many pericyclic reactions out side of the Diels Alder reaction, though, so I wouldn't worry too much about that pathway.

 

Agreed. To add a bit, I don't believe that the radical pathway is symmetry allowed here. Thats definitely proceeding [2+2].

[KCM92]

We don't necessarily need it to work in practice, it just needs to be the most efficient theoretical synthesis and an alternative synthesis for bonus points.

 

With that said, the step converting your dibromide to the diterminal alkenes (step two going from the start material) will not work at all. Yes in theory, you could eliminate form the primary methyl groups to give you the product you want, and maybe this will be the kinetic product...but the more substituted alkenes are more stable (i.e. four groups is more stable than 3 groups which is more stable thn 2 groups which is more stable than 1 group).

 

If I use a bulky base such as t-butoxide the anti-Zaitsev product will form, that's what I intended for elimination of the bromine atoms. I didn't include the reagents on my post.

[Horza2002]

 

If I use a bulky base such as t-butoxide the anti-Zaitsev product will form, that's what I intended for elimination of the bromine atoms. I didn't include the reagents on my post.

 

You mean SHOULD form. As any of the organic chemists here will tell you, there is a huge difference between what happens and what should happen. Your reasoning is good, i agree that a bulky base should indeed give you the intended product. But, discrimination between a primary and secondary hydrogen is not always as clear cut as it would seem. Usually only between primary and tertiary is it clear cut.

 

Sorry if you think I'm being picky, but I'm just trying to point out all the different angles that can occur from this