39Something wrong with this question. Dialkylcopperlithium(Gilman reagent) reacts only with acid halides. There is no acid halide after all those reactions..besides this is not 1,2 versus 1,4 addition either.
LDA forms a dianion with the carboxylic acid, but H+/heat dene par dianion toh gaya? We will get a six membered ring then(because it is a delta-hydroxy carboxylic acid, so we get a delta-lactone)...or am I missing something..
Gilman reagents do not reduce esters(because the next step involves PCl5, there should be alcohol). They reduce only acid halides like I said..
24sorry forgot to menttion i thing ,,,
in first step only 1 eq grignard is used
sorry again [2]
39
Pritish Chakraborty
·2010-03-19 05:38:51
Last product can be a diketone instead of an acid anhydride as well...mechanism of alkyne ozonolysis is not known. I am assuming that NaNH2 does away with both chlorine atoms. If those last few answers don't match, you know then that you have to eliminate only one chlorine instead of both.
If water would have been present, ring would cleave into two terminal carboxylic acid groups(for the anhydride).
3DONT U THINK c=o bond shud be attacked first as its more polar??
39
Pritish Chakraborty
·2010-03-19 05:59:48
no iitm...the ketone is sterically hindered. Both the alpha carbons of that ketone have a heavy group attached.
