Doubt from acidic and basic strength

Q6 again see lone pair availability

But in answer key its saying 3>4>1>2

See lone pair of nitrogen at 4 no will have more resonance structure than other 2
So how it can be possible.
But I am checking once again

I think when we give H to the compound aniline ammonium cation will under go resonance.
And when we give H to amide it will also under go resonance but during resonance positive charge will be at oxygen also
It is less stable than aniline.
So remember aniline is more basic than amide.

Question 12 i have able to predict order of pka in first,third and fourth compound but what about second compound it is a which functional group compound

See second compound Is active emethylene

For 2nd compound the H between the 2 C=O us active H as then resonance will stabilise the resulting -ve charge

So what will be overall order of acidic strength

Question 9 is acidic character order is 3>1>2

According to me it should be 1>2>3
Because of SIR effect 1 will be highest and then +M group will decrease acidic nature hence 2>3

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See due to hydrogen Bonding acidic strength decreases

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What’s the correct answer @abhijit_2020

Answer 3>2>1

As you mentioned 1>2>3

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@Shaquib how can 3>2
+M group will decrease the acidic character and hence 2>3

There won’t be H bonding as due to steric hinderance COOH won’t be in the plane of the ring , and that’s why cross conjugation is inhibited and hence it’s acid character will be maximum

See H of Oh is also non-planar

H of OH is just bent but in the plane , but COOH is somewhat perpendicular to the plane of benzene ring

is this not possible.

No , it isn’t possible because that C=O will be out of the plane

This question is a very famous example of ortho effect in COOH where steric inhibition of resonance takes place