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Question: Nitration of aniline in the strong acidic medium also gives m- nitroaniline because _______ A) In ...

Nitration of aniline in the strong acidic medium also gives m- nitroaniline because _______
A) In absence of substituents nitro group always goes to m- positions
B) In spite of substituents nitro group always goes to only m- positions
C) In strong acidic medium aniline is present as anilinium ion
D) In electrophilic substitution reactions aminos group is meta directive

Explanation

Solution

Aniline is a yellow colour compound which has a brownish shade oily liquid having musty and fishy odour. The smell of aniline is just like a rotten fish. It is a flammable liquid. It is soluble in water and even freely soluble in chemicals such as ether and alcohol. It gets dark in colour when it is exposed to air and light.

Complete step-by-step answer:
It is toxic in nature when someone inhales it through air or when it gets absorbed in skin as it tends to produce nitrogen oxides which are harmful. When the nitration of aniline is done in the strong acidic medium it gives m- nitroaniline because in strong medium the aniline is present as the anilinium ion.as we know that the nitro NO2N{{O}_{2}} group is a strong electron withdrawing group so it will go towards that atom which has more electron density.

In acidic medium there will be the presence of hydrogen ion which attack the high electron density area which is the NH2N{{H}_{2}} and so the formation of NH3+N{{H}_{3}}^{+} ion will take place on the ring and now the ammonium ion is considered as the electron withdrawing group. So the shifting of bonds will take place and the electron density will be the highest at the meta position so the nitro group will be attached to the meta position. This is due to the presence of anilinium ions. And as we know that the amino NH2-N{{H}_{2}} group is ortho para directing but the NH3+-N{{H}_{3}}^{+} is a meta directing group.
Then the reaction for them is the following:
C6H5NH2+H+strongacidC6H5NH3+HNO3mnitroaniline{{C}_{6}}{{H}_{5}}-N{{H}_{2}}+{{H}^{+}}\xrightarrow{strong acid}{{C}_{6}}{{H}_{5}}-N{{H}_{3}}^{+}\xrightarrow{HN{{O}_{3}}}m-nitroaniline

So the correct answer is option ‘C’.

Note: When aniline is oxidised the formation of carbon nitrogen bonds takes place. Aniline is a weak base so it forms an anilinium ion on the reaction with strong acids. When aniline is reacted with carboxylic acid the formation of amides takes place.