[azncuzin202]

For an organic synthesis problem, I start out with 1-methylcyclopent-1-ene and need to convert it to tetrahydro-6-methyl. Lactones are formed via the interaction of a hydroxy group and a carboxylic acid on the same molecule. With that in mind, I plan on forming an epoxide using mCBPA and opening it with NaOH. However, does the methyl group on the 1-carbon block this?

[azncuzin202]

Never mind, that epoxide idea doesn't really make sense now that I look at it.

[azncuzin202]

Will hot KMnO4 cleave the double bond into their carbonyl products without interfering with the methyl group?

[Dan]

need to convert it to tetrahydro-6-methyl.

This is not a complete chemical name - we can't help you if we can't tell what the question is.

Please upload a picture or write the full IUPAC name.

[orgopete]

Question (or hint), are you studying Baeyer-Villager oxidations?

[azncuzin202]

No I can't use that because we haven't learned that. The IUPAC name is 6-methyltetrahydro-2H-pyran-2-one. I feel like the KMnO4 (hot) would open the ring giving me a carbonyl and a carboxylic acid. Does it not?

[Dan]

No I can't use that because we haven't learned that.

You shouldn't be penalised for suggesting something that would work, regardless of whether you saw it in lectures or found out about it through your own initiative. The Baeyer-Villiger route orgopete hinted is the most concise route I can see.

I feel like the KMnO4 (hot) would open the ring giving me a carbonyl and a carboxylic acid. Does it not?

Yeah, should do. What next?

[azncuzin202]

I'll just use water and H+ as solvent to get the final product.

[azncuzin202]

Actually can someone explain why the KMnO4 needs to be hot. I know that if you use KMnO4, H2O, -OH, and acetone forms a glycol but what is happening when KMnO4 is used by itself? To be honest, I'm beginning to have doubts as to whether or not I can use this reagent considering I can't find the exact reaction in the book.... =/

[spirochete]

Actually can someone explain why the KMnO4 needs to be hot. I know that if you use KMnO4, H2O, -OH, and acetone forms a glycol but what is happening when KMnO4 is used by itself? To be honest, I'm beginning to have doubts as to whether or not I can use this reagent considering I can't find the exact reaction in the book.... =/

Two possibilities:

1) Hot, acidic KMnO4 gives cleavage

2) Colder, basic KMnO4 gives dihydroxylation

[azncuzin202]

Thanks  spirochete. After re-looking at the rubric, i can get extra points for doing an alternate synthesis. So I'm going to try an do the baeyer-villiger oxidation, I just won't get the full twenty because we never learned this in class. From what I've gathered, I need to turn that double bond into a carbonyl. After that, would I use MCBPA to form the lactone?

[orgopete]

Let's roll this back a little. KMnO4 will give a carboxylic acid and a ketone. The ketone must be removed to form a lactone.

Obviously, we don't know what has been discussed in class, yet many posters ask questions as though we do know. I was only guessing.

I don't know why permanganate should give ring cleavage. There are precedents. The reaction of ozone and OsO4 with an alkene looks similar. Ozone cleaves the ring and OsO4 does not. A cyclic iodate ester cleaves the ring. This is how rationalize the reaction of permanganate. The cyclic manganate ester is similar to the iodate and heated results in ring cleavage. If kept cold, the cleavage is slowed. In the presence of base, hydrolysis of the manganate ester releases the diol.

[azncuzin202]

Let's roll this back a little. KMnO4 will give a carboxylic acid and a ketone. The ketone must be removed to form a lactone.

Obviously, we don't know what has been discussed in class, yet many posters ask questions as though we do know. I was only guessing.

I don't know why permanganate should give ring cleavage. There are precedents. The reaction of ozone and OsO4 with an alkene looks similar. Ozone cleaves the ring and OsO4 does not. A cyclic iodate ester cleaves the ring. This is how rationalize the reaction of permanganate. The cyclic manganate ester is similar to the iodate and heated results in ring cleavage. If kept cold, the cleavage is slowed. In the presence of base, hydrolysis of the manganate ester releases the diol.

I expect the KMnO4 to give me a structure like       
  ..O                                                               
..||
CH3-C-CH2-CH2-CH2-COOH

I would than convert the ketone into an alcohol and use H2O/H+ to get the final product. Can you or anyone verify if this could work out. As far as converting the ketone to an alcohol is concerned, is there a way to do it without grignard reagents?

[azncuzin202]

Actually NaBH4 would do the trick correct? It would convert the ketone into an alcohol without interfering with the carboxylic acid.

[orgopete]

Correct, NaBH4 would work.