the mistake in ur logic is that anion on B is not stabalised ..whereas the anion on A and C is stabalised....and in A we have both inductive and resonance effect but in C we have only inductive...and i quoted that values just to show that Hydrogen Bonding doesnt make a huge difference in the case of NO2
15 Answers
Ans 1..It is A only..see B cant be the answer coz the anion formed is not resonance stabalised...so we have only 2 options left A and c..actually in case of C.. NO2 grp is meta to it..so that will not be much aciidc..and for that hydrogen bonding case u can see the table given in NCERT..the pKa values of p-Nitrophenol and o-nitrophenol differ by 0.1 only...whereas m-nitrophenol differ by 1.something ...
Ans 2..Ortho effect - due to which COOH grp changes the plane and COO- anion is stabalised as the ring do not donate electrons to it//bcoz it is now in different plane
Ans 3.. 2nd one is more acidic..
arre yaar....table s se value mat dekho...mere logic mein galti batao...
ok..lets leave b..
par a toh bilkul nahin ho sakta naa ???becoz Hydrogen isnt free for donation
answer should be a becoz due to Mesomeric effect of NO2 a positive charge appears only on the carbon bonded to OH at position a so the electrons of oxygen will be towards this carbon and hence the ydrogen is not bonded to oxygen very strongly so it can be removed easily thus it is most acidic......
eureka we discussed this. a) is the most acidic due to ortho effect. Same reason in Q2 :)
In Q3 we shall not decide acidic order by resonance stabilisation. Meta position is not affected by resonance. Now both methoxy and chlorine groups have -I effects. But chlorine being a halogen has the strongest -I here, so the second compound is most acidic.
noooo final verdict : wait for asish and his pKa table :D:D:D
tables dont decide acidity in exam hall [1]
i feel convinced by the discussions done
mere bolne ki kya jaroorat thi?? I had uploaded my table naa?? [3]
btw ans is correct