Solution:
Statement I
In the nitration of aniline, more amount of m-nitroaniline is formed than expected.
Normally, −NH2 is an activating, ortho/para-directing group.
Hence, we expect mainly o - and p-nitroaniline.
However, nitration is carried out in strongly acidic medium ( H2SO4∕HNO3 ), and under these conditions, aniline gets protonated to form the anilinium ion (C6H5NH3‌+).
The −NH3‌+group is an electron-withdrawing, deactivating, meta-directing group.
So the nitration mixture now contains protonated aniline, which gives meta nitro product.
As a result, the actual nitration product mixture contains a higher proportion of m-nitroaniline than we would expect based on the activating nature of −NH2 alone.
Hence, Statement I is correct.
Statement II
In the presence of a strongly acidic medium, aniline is protonated to form anilinium ion, which is meta-directing.
As discussed above, this is true - in a strongly acidic medium, aniline is converted to the anilinium ion (C6H5NH3‌+).
The anilinium ion is meta-directing because the positively charged nitrogen withdraws electrons from the ring, deactivating it, especially at ortho and para positions.
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