🚀 go-pugleaf

"Bob" <bbx107.XYZ@excite.XYZ.com> wrote in message 
news:1lnid2djuk34mnv43hrmq2tergs2u9tpbt@4ax.com...
> On 8 Aug 2006 15:02:15 -0700, "Vladimirr Saigon"
> <kantabusse@lortemail.dk> wrote:
>
>>Does LiAlH4 also reduce double bond carbon-carbon?
>
> No.  The mechanism depends on the high polarity of the C=O bond.

That's generally true, but it's a bit of an oversimplification.

1)  There are several reports of LiAlH4 reducing isolated C=C with 
transition metal all across the first row.  Reactions stoichiometric in MCln 
(M = Ti, V, Cr, Mn, Fe, Co and Ni) give quantitative reduction to the 
alkane.  Catalytic amounts of Ti, Ni and Co also work well.  Of course it's 
really the transition metal hydride that's doing the reduction, but the fact 
is that the reducting equivalents comes from LiAlH4.  It's tantamount to 
saying that H2 can be used to reduce olefins, when in reality it's the TM 
hydride that's doing the chemistry.

2)  There are cases of uncatalyzed reduction that depend on polarity similar 
to what you describe.  For example, cinnamyl alcohol is reduced to 
phenylpropanol by LiAlH4.  Here, the alcohol (alkoxide) helps deliver the 
AlH4- to the olefin, where it can do its job.

3)  I would swear I remember reading that, under the right conditions, 
LiAlH4 will add 1,4- to enones to give the ketone upon hydrolysis, but I 
can't find a reference.  If this does happen, it probably involves a Cu 
catalyst of some sort, and again, it's the Cu-H doing the addition.

In all these cases, LiAlH4 is the reducing agent.  The answer to his 
question depends on perspective.  If he's a synthetic chemist wanting to use 
a reagent for a certain transformation, then the answer is yes, LiAlH4 will 
reduce a C=C bond under the right conditions.  If he's a mechanstic chemist 
trying to understand how a reaction occurred, then no, the LiAlH4 doesn't 
itself reduce a C=C bond.

Eric Lucas