[amcavoy]

The problem is "design a synthesis of 2-methyl-tetrahydrofuran starting with 5-bromo-1-pentene."

Could someone tell me if my outline is correct?

Br-(CH₂)₃-CH=CH₂ + CH₃OH + NaH -> HO-(CH₂)₃-CH=CH₂

HO-(CH₂)₃-CH=CH₂ + H₂ (Pd/C) -> HO-(CH₂)₃-CH₂-CH₃ -> H₂O⁺-(CH₂)₃-CH^--CH₃ -> 2-methyl-tetrahydrofuran

[Rico]

Hey amcavoy

I'm sorry to say this, but in my opinion the outline you have given is totally nonsense. I would start looking into some real organic reactions for designing a synthesis of 2-methyltetrahydrofuran, that is consult an organic chemistry textbook! To guide you in the right direction i would look for some kind of hydration of the alkene and afterwards, look for a way to cyclize the product. When you have come up with some suggestions please post them and i will be more than happy to tell you if they are plausible.

Rico   

[amcavoy]

OK one of the reactions I found was 5-hydroxy-1-pentene + Br₂ +H2O -> 2-(1-bromomethyl)tetrahydrofuran.  So to start, I would add H₂O to 5-bromo-1-pentene to get 5-hydroxy-1-pentene.  Then treat this with Br2 and H2O to get 2-(1-bromomethyl)tetrahydrofuran.

How do you replace a halogen with a hydrogen?  I think this is the last step needed.

Also, do you know where I could find a mechanism for the first reaction I have listed? (5-hydroxy-1-pentene + Br₂ +H2O -> 2-(1-bromomethyl)tetrahydrofuran).

Thank you.

[Rico]

Hey amcavoy

Now we are getting somewhere! First of all if you want to substitute the bromine for a hydroxy-group in 5-bromo-1-pentene you should probably use hydroxide-ions instead of water, because water might not be nucleophilic enough to kick out the bromide-ion.
The mechanism for the reaction you have written is quite simple. It starts by a formation of a bromonium-ion, then an intramolecular attack of  the alcohol on one end of the bromonium-ion and then deprotonation by water giving the product (see jpeg attached).
To my knowledge its not possible to replace the bromide with H, in 2-(1-bromomethyl)tetrahydrofuran, by treating it with HBr. Instead this transformation can by done using radical chemistry. Threating 2-(1-bromomethyl)tetrahydrofuran with tributyltin hydride in the presence of a radical initiator would give the wanted product (see jpeg attached).
The way i would synthesize 2-methyltetrahydrofuran was to start of with a hydration of the double bond (either using acid catalysed hydration or by using oxymercuration, since they give the right regiochemistry), then deprotonate the formed alcohol to an alkoxy-anion. This alkoxy-anion will cyclize kicking out a bromide-ion giving the wanted product (see jpeg attached) .

Rico

[amcavoy]

Thanks a lot Rico, that was really helpful.  By the way, what did you use to write out that mechanism in a jpeg format? 

[Rico]

Hey amcavoy.

I'm just glad that i could help    For these drawings i used ISIS draw 2.5 (which is available for free at www.mdl.com, you just need to register and create an account) and then i simply copy pasted it into the windows drawing tool; paint and saved it as a jpeg-file.

Rico

[wintermute]

why not to cyclize pent-4-ene-1-ol to MeTHF under acidic catalysis?

[agrobert]

why not to cyclize pent-4-ene-1-ol to MeTHF under acidic catalysis?

What problems can you see with this reaction?

Brominating a double bond is controlled and so is the displacement.

Acid catalysis of pent-4-ene-1-ol will give two products.  What are they?

[Rico]

I just searched the literature and found an article by Normant describing exactly that reaction (Bull. Soc. Chim. Fr. 1944,5,11,365-366). 4-pentenol was treated with sulfuric acid giving 2-methyltetrahydrofuran as the only product. So that is actually also a possible route to 2-Me-THF.

Rico