Yeah but problem over here is we can't decide by - M since ortho effect is taking place
Ortho effect will act in q,r due to SIR so p is least acidic .
No carabanion will be formed... Since no2 is EWG so it will stabilize carbanion and och3 is EDG so it will make it not very stable
Why won't ortho effect act in P??
Because oh,cn,nh2 are linear so they are in plane....
They are the exceptions
Due to ortho effect.....
When any other group than -OH, NH2, H, D, T, F, is present at ortho position of benzoic acid -COOH then this -COOH group is no longer present in conjugation with ring and acidic strength is most.....
This q can be solved only by - I effect. In No2, two oxygen is attached to it due to which it shows more - I effect than OH and stabilises the cojugate acid ion more.
But why don't we consider H bonding the conjugate in P is getting stabilised by H bonding
Well how do you know H bonding is happening before removal of H or after removal of H? Cuz both have opposite effect on acidity. Just ask your teacher about this thing.
Yes, ur ryt both do have the opposite effect, I am considering the H bonding after the removal of proton(which increases the acidity). See if we consider H bonding before the removal of proton it is happening in all the three cases so it will decrease acidity of all three compounds.
Yeah so I think and also confirmed with my teacher that D must be correct.
pKa strength of Benzoic Acid Derivatives at 25°C.
Ohh, thanks for looking into it
But according to the post of Pka values he has sent above, it shows ortho nitro benzoic acid to be more acidic than the Oh wala
Maybe I'll have to ask again to my teacher then.
But then I would say only 1 explanation works, the one given by Chirag.
Why don't you ask your teacher once ??
Yeah, I messeged him but he didn't reply to it