Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: a confused chiral girl on May 30, 2007, 09:12:47 PM
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HI~
I am a bit confused about this question, because NaBH4 reduces aldehydes as well as ketones into alcohols. However, in these choices, there changes all the double bond to O into OH. I am thinking about choice #2, but the O on the aldehyde is not reduced though...please help :)
thanks~~
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unless they specify somethng tricky like substoichometric amounts of NaBH4, reduction will occur at both centres
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The aldehyde would reduces slightly easier but both would go. So you would get (1) first and that would proceed through to (3)
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Heh, this is a bit of a trick question!!
I think you would get 2 first because the beta-ketoaldehyde is almost never in the keto-keto tautomer. It will be almost completely enolized toward the aldehyde!
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Yes, of course. I remember following the prep of a beta-ketoaldehyde oxime once and being surprised that the oxime formed at the ketone almost exclusively.
The final outcome is the same with excess borohydride though right?
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Heh, this is a bit of a trick question!!
I think you would get 2 first because the beta-ketoaldehyde is almost never in the keto-keto tautomer. It will be almost completely enolized toward the aldehyde!
Yes! I thought it was tricky too!! ;D because if we use LiAlH4, then we know for sure that both aldehyde and ketone would react to form alcohol, since LiAlH4 is strong. That's why I was a bit unsure that the ketone would get reduced as well~
so the aldehyde CHO gets reduced to CH2OH right? and the ketone to OH~