A Organic Chem problem

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Well i am having problem prepearing this alcohol from bromobenzene and 1-butene




please help.This question is from Solomons organic chem.
 

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Maybe you should redraw the target, you've got three bonds to oxygen. And I could be mistaken but I think you're missing a carbon.
 
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Well i made a little mistake in the bond to oxygen atom but have updated the figure and corrected it.
please help.
 

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Right off the top I can suggest a nucleophilic attack through a elimination-addition (not addition elimination) mechanism. Look it up in the book, it is the chapter regarding phenols. Adding a hydroxyl group to the double bond of butene results in an nucleophilic compound with the negative charge on the terminal carbon, this terminal carbon compound under the right conditions perhaps can participate in a substitution reaction replacing bromine (NOT sn2 sn1 pathway). Some problems in the solomons book require a bit of the imagination, you need to suggest a possibility without any knowledge of the existence of such pathways and reactions. Although there may be a existing, commonly used reaction to achieve this product.


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Here is another way I thought of to synthesize your desired compound:

1) Turn bromobeneze into a grignard reagent by the usual methods. Put it aside.

2.) Do a hydroboration oxidation on the terminal alkene to create the primary alcohol.

3) Do a swern oxidation on the primary alcohol to create the aldehyde.

4). React the created grignard reagent with the aldehyde and then add some acid. Your result will be your desired product.
 
Are you sure that's the right number of carbons? If so, ozonolysis with reductive work-up of the butene, followed by Grignard addition of the bromobenzene to the aldehyde.
 

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The regiochemistry of the alcohol would be wrong with the hydroboration/oxidation thing.


Ozonolysis would carve out a carbon. The product drawn has 10.
 

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How about epoxidation of the alkene with MCPBA, then make the Gilman reagent of the bromobenzene (nBuLi, then CuI). Mixing that with the epoxide should open the epoxide at the less hindered position and give the right regiochemistry.
 
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The regiochemistry of the alcohol would be wrong with the hydroboration/oxidation thing.

I'm not sure what you are talking about. Hydroboration oxidation of a terminal alkene leads to a primary alcohol. Hydroboration oxidation adds an OH to an alkene in what appears to be an "anti Markovnikov" fashion. You would get 1-butanol with Hydroboration oxidation.
 
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well i tried some but i am not getting desired roduct.
well i got by one of the suggestion of hydroboration oxidation alcohol but it give wrong osition of hydroxyl group.
it should on next carbon but got on first carbon of sidechain.
problem is age 563 12.20 of organic chemistry by craig fryhle and solomons 7th edition.
 
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I'm don't understand what you are saying. With hydroboration oxidation you want the primary alcohol CH3CH2CH2CH2OH. Then when you oxidize it you get the aldehyde CH3CH2CH2COOH. Reacting that with the grignard reagent will give you your product.
 

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Grave, write it out on paper. That method would give the alcohol at the benzylic position (off of the carbon directly attached to the phenyl ring) but the drawing shows (C6H5)CH2CH(OH)CH2CH3.

Amb, have you learned about epoxides and Gilman reagents yet, because I am pretty sure that is the way to do it.
 
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well i have learned about epoxides
but gilman i dont know that. it's not in solomons which i am primarily following in undergraduate chemistry text.
 
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Lithium dialkyl cuprate...thats called Gillman Reagent I think (R2CuLi). A very important reaction indeed. Look it up in Morrison & Boyd and you will find a nice explanation too.

Cheers
Vivek
 

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Yeah, R2CuLi, they aren't always referred to as Gilman reagents, but that's what I learned them as. They are similar to Grignard reagents, but the copper makes them "softer" anions so they behave a bit differently. They are very good at opening epoxides. They are also useful for adding into alpha-beta unsaturated carbonyl systems, since they add 1,4 (Michael addition) instead of 1,2 like Grignards.
 
Yes, ozonolysis cleaves a carbon. Yes, the product has ten carbons. BnBr plus butene makes eleven carbons.

I'm guessing you're confusing BnBr with PhBr.
 
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Ah yes thank you movies I see now. I really need to put on my glasses before doing O Chem. Theoretically you could just do all the steps I listed and dehydrate OH of the benzylic position to give the double bond. Then you could just rehydrate it to move the OH over to the right carbon, of course the yield might not be to good, but who cares, its answers the question.
 

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CSF, the question says bromobenzene, not benzyl bromide.
 
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CSF, the question says bromobenzene, not benzyl bromide.
Whoops, you're right. My mistake.
 

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I apologize, I meant the benzyne mechanism, not elimination addition.



gravenewworld said:
Here is another way I thought of to synthesize your desired compound:

1) Turn bromobeneze into a grignard reagent by the usual methods. Put it aside.

2.) Do a hydroboration oxidation on the terminal alkene to create the primary alcohol.

3) Do a swern oxidation on the primary alcohol to create the aldehyde.

4). React the created grignard reagent with the aldehyde and then add some acid. Your result will be your desired product.
There are several problems with your proposal, right of the bat I don't believe that a grinard reagent can be formed with bromobenzene.


How about epoxidation of the alkene with MCPBA, then make the Gilman reagent of the bromobenzene (nBuLi, then CuI). Mixing that with the epoxide should open the epoxide at the less hindered position and give the right regiochemistry.
Again, not very plausible, similar reasoning as above.

Which chapter is your problem on ambuj?

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Actually, forget what I said above, it just might work.
 

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Don't you have the solution's manual, by referring to the book it seems that the exoxide etc...were what they were looking for...all you had to do was to refer to the solution's manual.
 
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I guarantee that you can make Grignards from bromobenzene. I've done it. You can buy phenyl Grignard from Aldrich.

You used to be able to buy phenyllithium, but I think Aldrich stopped carrying for some reason. Probably because it is so easy to make from butyllithium.
 
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Well actually i have not Purchased solution manual to solomons as my teacher suggested me not to look at solution but now i have left the Private coaching and have no way to solve question excet do myself or Post it in Physicsforums or Purchase solution manual .well the question is from the chaeter on aldehyde and ketone and grignard reagent.
Anyway thanx all of you for suggestion can someone Please solve the question as i have to comPlete it by monday for school.
 
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GCT

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The solution is to form a grinard reagent with the bromobenzene and create an epoxide from the primary butene, the now anionic benezene will attack the partial positive carbon of the epoxide and will create an oxyanion in doing so, all that's left is hydrolysis of this compound to obtain the desired product...all this, I found in the solution's manual.
 

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