I am confused
The electron density is decreased , so it cant donate electron to Br. i am fine.
But at the same time the -ve charge is stabilized by the -I effect , No ?
Now I remember...acid chloride and acid anhydride cannot give the haloform test due to the presence of electron withdrawing groups.
(Look through my notes)
The answer is B, if those ketone groups are separated by one carbon atom and not as given in the figure.
I don't think it's correct!
And figure is correct as given!
I am confused
The electron density is decreased , so it cant donate electron to Br. i am fine.
But at the same time the -ve charge is stabilized by the -I effect , No ?
In that case is A the answer? Did I overlook something....perhaps an active methylene group?
@Shubhodip : Confused at which case?