I Ibraiz Full Member 10+ Year Member Jul 23, 2010 #1 Advertisement - Members don't see this ad Please refer to the question in picture Uploaded with ImageShack.us
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R rippinitez Full Member 10+ Year Member 5+ Year Member 15+ Year Member Jul 23, 2010 #2 In the first reaction, an amide is being reduced, you lose the double bonded oxygen. That's just what happens in that particular reaction. Don't confuse the reduction of "ketones, aldehydes, esters and acids" with "amides". Upvote 0 Downvote
In the first reaction, an amide is being reduced, you lose the double bonded oxygen. That's just what happens in that particular reaction. Don't confuse the reduction of "ketones, aldehydes, esters and acids" with "amides".
Baylor2011 DDS Apprentice 10+ Year Member Jul 23, 2010 #3 LiAlH4 is a very strong reducing agent...if you had something like maybe sodium amide which is a weaker reducing agent, then yes it would reduce it to a hydroxyl group first. Go look up all the reducing agents/oxidizing agents and compare/contrast Upvote 0 Downvote
LiAlH4 is a very strong reducing agent...if you had something like maybe sodium amide which is a weaker reducing agent, then yes it would reduce it to a hydroxyl group first. Go look up all the reducing agents/oxidizing agents and compare/contrast
L loveoforganic -Account Deactivated- 10+ Year Member Jul 23, 2010 #4 Mechanism : http://www.curvedarrowpress.com/partd/aminefamide.html Upvote 0 Downvote