Please give the proper ans

Roni, (B) is absolutely correct.

in Option IV there is SIR effect so lp is not delocalised with ring and available freely.

out of Option I,II and III, option II have +H effect of CH3 group present on para position on aniline so less stable tha. option IV

in case of I and III, I have no groups attached to aniline ring while in III NO2 group is attached on meta position so it have -I effect decrease the availability of lp on N, hence I >III

overall IV>II> I>III

why ortho effect not work for iv

SIR and Orho both work😊. here bulky groups are present on 1,2,3 position so SIR.

in noth the effect ortho as well as in SIR resonance diminished so basicity increase due to lone pair availability

ok sir

which option is given in your book??

option A is correct answer

got roni??

If any doubt then please ask😊

When aniline acting as a base becomes NH3 + (on top of a benzene ring), it is usually stabilised by solvation. but if there is a substituent on the ortho position, it inhibits solvation. Thus the tendency to act like a base is reduced. Also, it can be explained by SIP (steric inhibition to protonation) effect.

Always keep in mind roni...in case of ortho subsituted aniline always SIP occurs which decreases its basicity to the least...if ortho subsituted aniline is not there then SIR occurs which increases its basicity to the maximum beacuse due to this, the group is out of the plane and lone pair is freely available...So option A is correct... check this video for more detail https://youtu.be/K4Jt4kxWF_E

Roni any doubt...please check this video

roni please respond 😊

ok

thanks u sir .

roni....is this video helps u...there is always misconception in this effect...so keep in mind these things and make notes of it

now I clear my all doute

ok sir

thanks for support

welcome roni😊👍