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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Aurelius1219 on February 09, 2006, 02:10:45 AM

Title: Alcohol and LiAlH4
Post by: Aurelius1219 on February 09, 2006, 02:10:45 AM

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

Title: Re:Alcohol and LiAlH4
Post by: AWK on February 09, 2006, 03:11:18 AM

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

Title: Re:Alcohol and LiAlH4
Post by: FeLiXe on February 09, 2006, 10:11:13 AM

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