[Raitei]

Explain mechanistically how sodium borohydride is able to reduce both the C=O and C=C double bond in 2-cyclopentenone, but cannot reduce the C=C double bond in 2-cyclopentenol.

[macman104]

I have not heard of NaBH4 reducing conjugated ketones like that, can you provide a reference?

[Raitei]

I'm trying to reason mechanistically why such a reduction of a c=c bond in 2-cyclopentenol CANNOT occur, i'm just a little confused applying basic NaBH4 reduction principles mechanistically to 2-cyclopentenol

[macman104]

The first act of NaBH₄ is the activation of the C=O bond by sodium.  There is no such highly polarized bond (or something to attack the sodium) in the 2-cyclopentenol system.  That would be my assumption...

[Raitei]

This might be a lot to ask, but i'm kind of a visual learner, I saw on other posts that people were able to post mechnisms on the actual forum, is there anyone that could take a stab at it in a drawn-out mechanism form?

[macman104]

Which mechanism are you looking for?  Just a general NaBH4 mechanism?

[Raitei]

Yes, the general visual mechanism of why NaBH4 cannot  work with the c=c bond in 2-cyclopentenol.

[Raitei]

Sorry if i'm not being clear, i'm a little confused myself.

[aldoxime_amine]

Do you want to say that the conjugated system in cyclopentenone makes the c c double bond polar(albeit weakly) and hence is also reduced?

[Raitei]

Kind of, i think i'm figuring this out with your help....

Let me try asking this question a little differently to make more sense to you guys:
Why will NaBh4 reduce a c=c double bond adjacent to a carbonyl, but won't reduce any other kind of c=c bond?

I think that's the root of what i'm not getting.

Thanks!

[ARGOS++]

[kiwi]

Kind of, i think i'm figuring this out with your help....

Let me try asking this question a little differently to make more sense to you guys:
Why will NaBh4 reduce a c=c double bond adjacent to a carbonyl, but won't reduce any other kind of c=c bond?

I think that's the root of what i'm not getting.

Thanks!

start by thinking about how the presence of two bonds to an oxygen atom in a ketone affects the polarity of the carbon directly connected to it. if you have other groups in conjugation with a ketone (or other decent electron-withdrawing group) their reactivity will be affected - something you can sketch out an explanation for quite readily if you understand resonance. if you then attempt to sketch similar resonance forms for an isolated alkene you'll find you cannot - the electrons are localised, causing different/normal reactivity. the reaction you have given is a general example of conjugate addition:

http://en.wikipedia.org/wiki/Conjugate_addition