Akash FT 1

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

9 Answers

4
UTTARA ·

?????

4
UTTARA ·

Finally I am satisfied with the ans given [159]

THREAD CLOSED

24
eureka123 ·

can u expalin for me.....???plzz

4
UTTARA ·

in 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

24
eureka123 ·

what 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

4
UTTARA ·

Ya u're right thats y p-nitro aniline has less dipole moment than X & Y

24
eureka123 ·

but u said ans is 2) [12]

1
gagar.iitk ·

@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

24
eureka123 ·

thanks sir [1]

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