[thescepticalchymist]

I have to synthesize a made-up molecule from a simple starting compound (preferably 3 carbons or less).  My professor will know if I copy-paste a synthesis from online, so that definitely wouldn't fly.  In order for me to get the maximum number of points, the synthesis must meet the following criteria:

1) There must be 4-8 steps (protonations and deprotonations not counting)

2) I must use reagents that we learned in both Organic 1 and 2.  We can’t use anything we haven’t learned yet.  NO DIELS-ALDER ALLOWED FOR CYCLIZATION, THE ONLY WAY WE HAVE LEARNED IS INTRAMOLECULAR SN2.)
Here’s everything I have learned so far this semester: https://i.imgur.com/NoRIlI3.jpg, https://i.imgur.com/A6FXLfF.jpg, and https://i.imgur.com/xBTaWC6.png

3) chemoselectivity

4) regioselectivity

5) a cyclization (ring formation)

6) stereoselectivity

7) annotations explaining why a certain intermediate/product was formed (wherever relevant)


The project is due tomorrow.

Here is an example (https://i.imgur.com/CdS6BCW.jpg) of a synthesis that would get the max # of points. And here is mine (https://i.imgur.com/DOch3eT.png) so far.

What I need help with is making sure my synthesis contains the aforementioned 7 things, and that every step is accurate. The help would be much appreciated.

[rolnor]

If you treat the tertiary alcohol with HBr you will get the tertiary bromide so thats a problem.

[thescepticalchymist]

Can that be avoided by protecting the tertiary alcohol before adding HBr, and then deprotecting it after reacting the compound with Br₂?

[clarkstill]

A couple of thoughts:

There is also a stereochemical issue which you don't address: when you form the bromonium ion you can form two diastereoisomers (relative to the adjacent methyl substituent). This will have knock-on effects in the regioselectivity of the final elimination step, since one diastereomer will allow a different antiperiplanar CH to eliminate.

There's another regiochemical issue with the opening of the bromonium ion. Why do you think it would form the 6- rather than the 5-membered ring?

[wildfyr]

Most alcohol protecting groups are acid sensitive.

[thescepticalchymist]

There is also a stereochemical issue which you don't address: when you form the bromonium ion you can form two diastereoisomers (relative to the adjacent methyl substituent). This will have knock-on effects in the regioselectivity of the final elimination step, since one diastereomer will allow a different antiperiplanar CH to eliminate.

Is the other diastereomer the one where the wedges are dashes? I’m just kind of confused.  Also, what exactly is a knock-on effect?

There's another regiochemical issue with the opening of the bromonium ion. Why do you think it would form the 6- rather than the 5-membered ring?

I asked someone about this, and they said that this ring formation is probably actually under equilibrium, because the bromine could act as a nucleophile, albeit a weak one.  That would apparently make the 6-membered ring the major product.  Also, in the example (https://i.imgur.com/CdS6BCW.jpg) that my prof. gave, the 6-membered ring is favored.

[OrganicDan96]

why do you start with an aldehyde, then oxidise it later, it would be better to start with an acid chloride  as you would go directly to the ketone. another problem may be that the ketone is out of conjugation with the alkene as it could go into conjugation in the presence of acid or base.

[thescepticalchymist]

Most alcohol protecting groups are acid sensitive.

Meaning they will be removed by any strong acid?  Wow, we didn’t learn that.

[thescepticalchymist]

why do you start with an aldehyde, then oxidise it later, it would be better to start with an acid chloride  as you would go directly to the ketone.

Didn’t think of that...that does make more sense.

another problem may be that the ketone is out of conjugation with the alkene as it could go into conjugation in the presence of acid or base.

We didn't go over conjugation formally so I don't think he expects us to know that.

[wildfyr]

The two most common and simplest to alcohol handle protecting groups, silyl ethers and acetals, are acid labile. There are other options, but they are not very common, and tend to be difficult to remove and have their own compatibility issues with other functionalities on a molecule.

[clarkstill]

There is also a stereochemical issue which you don't address: when you form the bromonium ion you can form two diastereoisomers (relative to the adjacent methyl substituent). This will have knock-on effects in the regioselectivity of the final elimination step, since one diastereomer will allow a different antiperiplanar CH to eliminate.

Is the other diastereomer the one where the wedges are dashes? I’m just kind of confused.  Also, what exactly is a knock-on effect?

and

If you follow through the stereochemistry to your cyclic product, you will find that for one of the stereoisomers there are two different antiperiplanar CH's available to eliminate, which will give two different regioisomers of the product.

There's another regiochemical issue with the opening of the bromonium ion. Why do you think it would form the 6- rather than the 5-membered ring?

I asked someone about this, and they said that this ring formation is probably actually under equilibrium, because the bromine could act as a nucleophile, albeit a weak one.  That would apparently make the 6-membered ring the major product.  Also, in the example (https://i.imgur.com/CdS6BCW.jpg) that my prof. gave, the 6-membered ring is favored.

Hmm. What do other people think? I'm not convinced the product will re-form the bromonium ion and break the ether C-O bond.. seems pretty uphill to me, although i guess the oxygen could be protonated? But if your professor has made this argument, he's unlikely to mark you down for saying it!

Also - be careful going directly from acid chloride to ketone - generally this will over-alkylate since the ketone is more reactive than the ester. You can use Weinreb Amides though.

[rolnor]

If you use a cuprate you can make ketone from acid chloride, no problem.