Nitration of aniline in strong acidic medium also gives m-nitroaniline because

Question

Nitration of aniline in strong acidic medium also gives m-nitroaniline because (A) In absence of substituents nitro group always goes to m-position. (B) In electrophilic substitution reactions amino group is meta directive. (C) Inspite of substituents nitro group always goes to only m-position. (D) In acidic (strong) medium aniline is present as anilinium ion.

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/33908a134d8aacccf70ed51964307c5b-.png" /> As - overset⊕ N H3 is m-directing and strongly ring deactivating, hence besides para (51%) and ortho (2%), meta product (47%) is also formed in significant yield here.

Q. Nitration of aniline in strong acidic medium also gives m-nitroaniline because

In absence of substituents nitro group always goes to m-position.

In electrophilic substitution reactions amino group is meta directive.

Inspite of substituents nitro group always goes to only m-position.

In acidic (strong) medium aniline is present as anilinium ion.

As $- N H_{3} \oplus$ is m-directing and strongly ring deactivating, hence besides para (51%) and ortho (2%), meta product (47%) is also formed in significant yield here.

Q. Nitration of aniline in strong acidic medium also gives m-nitroaniline because

In absence of substituents nitro group always goes to m-position.

In electrophilic substitution reactions amino group is meta directive.

Inspite of substituents nitro group always goes to only m-position.

In acidic (strong) medium aniline is present as anilinium ion.

As −NH3​⊕​ is m-directing and strongly ring deactivating, hence besides para (51%) and ortho (2%), meta product (47%) is also formed in significant yield here.

As $- N H_{3} \oplus$ is m-directing and strongly ring deactivating, hence besides para (51%) and ortho (2%), meta product (47%) is also formed in significant yield here.