[CHEKAL]

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?

[discodermolide]

I would use sodium borohydride.
LAH and AlCl₃ is too hard for this.

[CHEKAL]

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?

[discodermolide]

Yes the thinking is ok. Have another read at that link I posted.

[CHEKAL]

do you have any suggestions of what form the mechanism will take for the removal of the ketone carbonyl? with LAH and AlCl3?

[discodermolide]

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?

[CHEKAL]

i used equimolar amounts of both.

[CHEKAL]

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?

[discodermolide]

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?

[CHEKAL]

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

[discodermolide]

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.

[CHEKAL]

that would be of great help if thats ok

[discodermolide]

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.
 

[CHEKAL]

How does the chlorine get replaced by hydrogen in the final step?

What drawing package do you use?

Thanks again for your help

[discodermolide]

The reduction of alkyl halides with LAH gives alkanes.
So I presume the reduction with AlClH₂ will do the same.
I use ChemDraw.