Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: amcavoy on December 04, 2007, 12:05:48 AM
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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
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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
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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.
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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
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Thanks a lot Rico, that was really helpful. By the way, what did you use to write out that mechanism in a jpeg format? :)
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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
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why not to cyclize pent-4-ene-1-ol to MeTHF under acidic catalysis?
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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?
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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