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
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
See due to hydrogen Bonding acidic strength decreases
What’s the correct answer @abhijit_2020
As you mentioned 1>2>3
@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
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