[Aurelius1219]

Problem:

LiAlH4 + 3 (CH3OH ) = ?

I'm stumped.  I know it's not a good reaction to have in the lab, but beyond that, I know nothing.  My TA told me to treat the Alcohol like Water, Replacing the leading H with the methyl group.  Only problem is, to do a straight across transfer with the LiAlH4 + H2O reaction we have in our notes, I would need 4 moles of alcohol, not 3.

The water reaction we have is LiAlH4 + 4(H2O) = LiOH + Al(OH)3 + 4 H2

Any help would be great

[AWK]

Split the reaction with water into 4 steps, and  proceed the first 3 steps with alcohol.

[FeLiXe]

the 4th H is probably not reactive enough to do something with an alcohol