I Ibraiz Full Member 10+ Year Member Joined Feb 23, 2010 Messages 209 Reaction score 2 Jul 23, 2010 #1 Members don't see this ad. Please refer to the question in picture Uploaded with ImageShack.us
R rippinitez Full Member 10+ Year Member 5+ Year Member 15+ Year Member Joined Aug 20, 2007 Messages 206 Reaction score 0 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 Joined Sep 10, 2009 Messages 451 Reaction score 0 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 Joined Jan 30, 2009 Messages 4,218 Reaction score 14 Jul 23, 2010 #4 Mechanism : http://www.curvedarrowpress.com/partd/aminefamide.html Upvote 0 Downvote