[Azazel]

Hello, I am new to this forum.

I am an undergraduate student in my third year, and a step in my synthesis project has been the reduction of the C=C double bond with LiAlH4. Now, as many of us know, this usually doesn't occur. However, in this case, it does occur. This happens due to the fact that the reactant has a Hydroxy group in the allylic position. I've tried to find the proper mechanism online, but failed. If anyone knows the mechanism for this ( esp. with a reference to the proper literature. ), I would be very glad.

I am also not a selfish bastard, and will try to assist others in question they have.

[Mitch]

Have you tried scifinder scholar?

[Azazel]

The paper that I got in the scifinder scholar was exactly what I needed wrt diastereoselectivity of the reaction ( the main issue about that part of the project). No mechanism in that paper though. ( Organic letters ) .

Maybe a more general search is needed.

[movies]

It's an alkene, not an alkyne, right?

[AWK]

LiAlH4 does not reduce unpolarized double or triple bonds

[Azazel]

It's an alkene, not an alkyne, right?

yes.

LiAlH4 does not reduce unpolarized double or triple bonds

I know. I've wrote about it in my opening post. See? :p

This happens due to the fact that the reactant has a Hydroxy group in the allylic position.

I need the mechanism for that, and I couldn't find it...

[movies]

So I'm not familiar with the reaction with alkenes, but with alkynes containing a propargyl hydroxy group, the key is that you can form a five membered ring chelate between the alkyl aluminum and the alkoxy group.  I would suspect that the same thing would be key to the alkene reaction.

The mechanism would probably proceed as follows:

1) deprotonation of the alcohol by LAH (LAH is quite basic)
2) hydroalumination of the alkene via a 4-member transition state
3) chelation of the resultant alkyl aluminum to the pendant alkoxide
4) the alkoxide chelate is then broken down on workup

[Azazel]

Hey, I think I sort of worked it out on paper. I'll recheck on scifinder tomorrow.

Thanks a bundle!