Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: sharbeldam on May 02, 2020, 09:15:17 AM
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I tried, but I got stuck. I cant do a wolff kishner, And I think I also started wrong since the major product didnt really help me.
Any tips?
problem attached.
(the first step is lindlar)
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How about use Lindlar, then Pd(PPh4) to isomerize/move the double bond into the ring as a start?
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I appreciate that rolnor, but This reaction is not included in the syllabus and I am as a tutor still not familiar with it. Id be happy for a link that explains it.
Any other ways? in my idea, it does make the double bond to the left but the problem is that it wouldnt be the major product :/
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Here is a link:
https://pubs.rsc.org/en/content/articlelanding/2019/cc/c9cc00223e#!divAbstract
The double bond in this case should migrate into the ring I think. Then the next step would be ozonolysis.
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Ok this is what I did, Id get aldehyde and ketone in the chain tho. I want to turn the aldehyde into halide without touching the ketone . ill keep thinking of it :D
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OK, I see now that you have the ozone-step. To reduce aldehyde in presence of ketone should be possible, maybe NaBH3CN att the right pH?
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I thought it reduces imines but not ketones and aldehydes, so its more selective to aldehydes?
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If you have a little more acidic conditions it will reduce aldehydes selective but I am not sure about the pH. Also diisobutylaluminiumhydride at -78 could work.
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Here is a better option;
https://en.wikipedia.org/wiki/Sodium_triacetoxyborohydride
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Will BH3-THF reduce the ketone? It will reduce an acid - so if you cleave 1-ethylcyclopentene with permanaganate instead of ozone to get the ketone and acid, you can then reduce the acid with BH3-THF and make the alcohol. The mechanism of reduction is complex, but it works for acids but not esters or ketones or aldehydes. (While you can't reduce a ketone by itself with BH3, I am not certain that if you have both a ketone and an acid that reduction will be selective - the triacetoxyborane that rolnor suggests is a lot like the intermediate in the acid reduction.)
But I think they teach it at the sophomore level as being selective for acids.