Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: CHEKAL on January 02, 2013, 08:28:35 AM
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right, ive done an alternative reaction to the wolff-kishner reduction because it contains hydrazine.
ive reacted a ketone with lithium aluminium hydride and aluminium chloride to remove the carbonyl group and convert it into a methylene group, im really suck of where to even start a mechanism for the reaction.
does anyone have any idea of what the mechanism would be like?
thanks and happy new year
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Do you not mean alkane?
Or is it actually alkene?
Have a look here:
http://en.wikipedia.org/wiki/Wolff–Kishner_reduction (http://en.wikipedia.org/wiki/Wolff–Kishner_reduction)
p.s. there is nothing wrong with hydrazine as long as you take the relevant precautions. It is probably safer than LiAlH4/AlCl3.
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yes i mean alkane sorry! ive already done the reaction, i need a mechanism for it, ive not used the wolff-kishner, how would the carbonyl be removed?
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Why don't you suggest a mechanism?
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i have no idea where to begin, hence why im on here :-\
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i have no idea where to begin, hence why im on here :-\
No need to show an attempt. Hint what do you get when LAH reacts with AlCl3?
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you get AlH3 + LiCl correct?
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If you do it with the correct molar ratios, yes.
See http://books.google.ch/books?id=ti7yMYYW7CMC&pg=PA473&lpg=PA473&dq=reaction+of+LAH+and+aluminium+chloride&source=bl&ots=HiHZP3fRi4&sig=DPDrcWEIYTP8izaNiBZa0bcYkUw&hl=en&sa=X&ei=idvlULHWO4HL4ASM44GQAQ&sqi=2&ved=0CF4Q6AEwBg#v=onepage&q=reaction%20of%20LAH%20and%20aluminium%20chloride&f=false (http://books.google.ch/books?id=ti7yMYYW7CMC&pg=PA473&lpg=PA473&dq=reaction+of+LAH+and+aluminium+chloride&source=bl&ots=HiHZP3fRi4&sig=DPDrcWEIYTP8izaNiBZa0bcYkUw&hl=en&sa=X&ei=idvlULHWO4HL4ASM44GQAQ&sqi=2&ved=0CF4Q6AEwBg#v=onepage&q=reaction%20of%20LAH%20and%20aluminium%20chloride&f=false)
Is that what you did?
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yes, a solution of aluminium trichloride and diethyl ether was made, this was then added dropwise to a LAH/THF solution.
just another additional question to go with it: if i changed the amount of aluminium trichloride, what effect would this have on the reduction reaction? would it change the power of the reduction?
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Yes, it says that in the link I sent you.
It modifies the reductive strength of LAH making it more selective.
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So if I added more aluminium chloride to change the stoichiometry, would it make the reduction less powerful as to only reduce to a secondary alcohol? Would it have an effect at all if I'm wrong?
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would this have an effect?
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Yes t would probably have an effect.
You usually do this with t-Butanol. By doing this you modify the reactivity and hence the selectivity of the LAH reduction.
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would this mean dibenzosuberenone would be reduced to an alcohol if i added more AlCl3? as the stoichiometry i used in my procedure reduced the ketone to a methylene group.
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A methylene group is a CH2 do you mean it got reduced to a Methyl group?
Having looked up the structure of your compound, which you finally supplied, yes it got reduced to a methylene group.
Just a suggestion: If you can supply the structure or a part of it it makes the answering and understanding of the question much easier, instead of trying to fathom out vague questions, which leads to mis-understanding.
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the removal of the ketone group in my procedure resulted in the methylene. would i be able to achieve its alcohol equivalent (cannot find structure online) if i used a recuding agent procedure containing more AlCl3? would more ALCl3 give a weaker reducing agent hence only reduced to secondary alcohol?
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I would use sodium borohydride.
LAH and AlCl3 is too hard for this.
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ive already done the initial procedure, i have to report to my supervisor what effect this would have if i did increase the AlCl3 amount theoretically, does my thinking have any baring to an acceptable answer?
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Yes the thinking is ok. Have another read at that link I posted.
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do you have any suggestions of what form the mechanism will take for the removal of the ketone carbonyl? with LAH and AlCl3?
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Well you might start by considering what reducing agent you actually used. How many equivalents of AlCl3 did you use?
Knowing that Al has a great affinity for oxygen and that you must deliver 2 hydrogen atoms to the carbonyl carbon and thereby remove the carbonyl oxygen can you come up with a mechanism?
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i used equimolar amounts of both.
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could the H2LA-H bond attack the carbonyl from behind adding a hydrogen, then the AlH2Li coordinates to the oxygen? i dont know how i could go further from there?
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So you used AlH2Cl this was the actual reagent used. Did you use an excess of this reagent to reduce your ketone?
This reagent will coordinate to the oxygen and deliver hydrogen to the carbonyl.
By the way how did you work this reaction up?
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yes the reagent was in excess
after the step of coordinating to the oxygen and adding a hydrogen, how will the step towards driving off the carbonyl and adding a further hydrogen happen?
AlCl3 solution was made up, ketone was added to a LAH solution and bother were added dropwise to eachtoher
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I think you dod the reduction as usual, then the AlCl3 displaced the R-OAlHCl with Cl which was then reduced to the alkane.
I can draw what I mean if you want.
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that would be of great help if thats ok
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Now this is a proposal. The reduction step to the alcohol is ok, the bit I'm not sure about is the formation of the chloride. But as far as I can see this is the only way to get to the alkane?
Perhaps someone else will comment.
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How does the chlorine get replaced by hydrogen in the final step?
What drawing package do you use?
Thanks again for your help
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The reduction of alkyl halides with LAH gives alkanes.
So I presume the reduction with AlClH2 will do the same.
I use ChemDraw.
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In what way does the AlCl3 play a role in the reaction? Does it mediate the amount of reagent that actually conducts the reduction?
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I told you this before, see the link I posted on page 1