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Explain the following: (a) Ethylamine is soluble in water, whereas aniline is not. (b) Although amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. (c) Aniline does not undergo Friedel Crafts reaction. |
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Answer» Solution : (a) Ethylamine DISSOLVES in water DUE to intermolecular hydrogen bonding as shown below: `H-underset(C_(2)H_(5))underset(|)overset(H)overset(|delta-)(N).......overset(delta+)(H)-underset(H)underset(|)overset(delta-)(O)......overset(delta+)(H)-underset(C_(2)H_(5))underset(|)overset(H)overset(|delta-)(N)........overset(delta+)(H)-underset(H)underset(|)(O).......` However, because of large hydrophobic part (i.e., hydrocarbon part) of aniline, the extent of hydrogen bonding is less and therefore, aniline is insoluble in water. (b) Under strongly ACIDIC conditions of nitration (in the presence of a mixture of conc. `HNO_(3) + H_(2)SO_(4)` ), aniline gets protonated and is converted into anilinium ion having `-NH_(3)^(+)` group. This group is deactivating group and is m-directing. So, the nitration of aniline gives o, p-nitroaniline (mainly p-product) while the nitration of anilinium ion gives m-nitroaniline.  Thus, nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of aniline. (c) Aniline being a Lewis base reacts with Lewis acid such as `AlCl_(3)` to form a salt. `underset("Lewis base")(C_(6)H_(5)NH_(2)) + underset("Lewis acid")(AlCl_(3)) to underset("Salt")(C_(6)H_(5)overset(+)(NH_(2))AlCl_(3)^(-))` As a result, N of aniline acquires +ve charge and hence it acts as a STRONG deactivating group for electrophilic substitution reaction. Hence aniline does not undergo Friedel Crafts reaction. |
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