[a confused chiral girl]

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~~

[kiwi]

unless they specify somethng tricky like substoichometric amounts of NaBH4, reduction will occur at both centres

[Custos]

The aldehyde would reduces slightly easier but both would go. So you would get (1) first and that would proceed through to (3)

[movies]

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!

[Custos]

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?

[a confused chiral girl]

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!!  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~