CH3-CH2-O-C=O + LiAlH4 -> CH3- CH2-OH + CH3OH
Note: there is obviously a mistake in the formula: it should be a formula of an ether: CH3-CH2-O-C(O)R, where R= H or alkyl group. Reduction of ethers leads to formation of alcohols.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment