From my perspective, this compounds can be retro synthesized by cutting the left C-O bond first through Williamson Synthesis. Then, we break the C-S bond, forming alkyl halide (ex. -Br). Then, we cut the right C-O bond through Esterification of carboxylic acid. Finally, the FGI of carboxylic acid is done through hydrolysis of nitrile.
So, in my opinion, the synthetic approach is:
1. Acid hydrolysis (H2O, H+)
2. Ethanol, in acid solvent (H2O, H+)
3. (NH2)2CS, in base solution (NaOH, H2O)
4. NaI followed with CH3Br
Is the order of my retrosynthesis/synthesis approach corrected or not? Or is there any other efficient ways to safely do the retrosynthesis, based on the reactivity of each functional groups. Any suggestions or ideas are welcomed. Thanks!
I doubt for the last step, since the acidity of alcohol is less than thiol, probably -SMe will form, protection of thiol maybe necessary.
Is this a theoretical exercise? Your starting material is probably even more complicated than the product.
Step 4 should be NaH? and as Yehezkiel stated the thiol will likely react first. Another problem with this sequence is that the alcohol will then react with ester to cyclise to form the 7-membered lactone.
The first 2 steps can be combined into a single step.
As Eugene said, step 1 and 2 can be combined, I don´t think it´s a good idea to generate the thiol group before methylation step; this given chemoselectivity problems that you will have.
If you do the methylation step before introduce thiol group, and you make a deprotonation of hydroxi group (as classic Williamson ether synthesis) a cyclization reaction would happen to produce a tetrahydropyran (Highly stable); I would recommend the use of exhaustive alkylation agent, as dimethyl sulfate or methyl iodide in polar aprotic solvent with high dielectric constant (maybe DMF), without using base.
I hope this be useful.
sorry, but I totally agree with Eugene...this only makes sense if theorical...probably you should concern about how to get the starting material...