Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: haiph12 on March 15, 2004, 10:42:33 AM
-
help me to cpmplete this scheme
Edit: edited title for better indexing. Mitch
-
why nobody help me?
-
Just start from formulas of both parent substances: C6H5NO and C5H9NO. As you can see the product A shows sum of atoms from both formula. So addition facilitated by ethoxide anion take place at alpha carbon of pyrolidone
and alcohol group is formed from aldehyde group. Then this OH group is oxidised to carbonyl one (product B)
B can enolise (two beta-carbonyl groups) and can attach one HCl to double bond and hydrolyse to carboxylic acid - amino group joins the next HCl (product C -add H2O over arrow between B and C). Next step is decarboxylation of acid (product D) followed by neutralisation and elimination of HCl
(product E - C6H4N-CO-CH2-CH2-CH2-NH-CH3). Then reduction of CO to CHOH (product F) followed
by exchang of OH by I (product G) and elimination of HI produce C5H4N-CH=CH CH2-CH2-NH-CH3 product N). I doubt if the final product could exist in open ring form - it rather cyclises to N-methyl pyrolidine.
AWK
-
the first thank you AWK for every thing.
You study organic chemistry very much,don't you.
second , final product( N ) ( cyclises to N-methyl pyrolidine is true) because N is Nicotin and is optical active compound.
Can you help me complete the scheme( in topic : do you solve the problem) ?
Thank you very much AWK
-
Though nicotine is an optical active compunds, your total synthesis gives a mixture of enantiomers
-
Yes,
and the scheme in topic : do you solve the problem
can you help me?
help me
Thank.