Dipole moment of p-nitroaniline when compared to nitrobenzene (X) and aniline (Y) will be
1. Smaller than both X and Y
2. Greater than both X and Y
3. Greater than Y but smaller than X
4. Equal to zero
ans is 2
I want explanation
4in aniline net dipole moment downwards coz N donate it's lone pair to benzene
in nitro benzene N withdraws the electron from benzene ring due to -I and -R effect..
and in case of p-nitro aniline....amine group donates the electron and and nitro group withdraws the elctrons..so a net downward dipole moment which will be more than the abv cases
This was explained to me by Govind
24what i am saying is ...
in p-nitroaniline...whatever amine donates is taken by nitro and is not given to teh rest of ring.....
whcih is the case i aniline and nitrobenzene....where distribtuin of electrons is i nthe ring
i may be saying all this in a vague way..but plz correct me
4Ya u're right thats y p-nitro aniline has less dipole moment than X & Y
1@eureka
the net dipole moment vector in case is basically determined by the reso and i effect of the system in case of nitro both gets added and in case of NH2 i effect apposes the reso effect but when NH2 is present at para position r effect of both get increased result in more dipole moment of the system
try compare when both at meta positions