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Account for the following observations : (i) PK_(b) for aniline is more than that for methylamine. (ii) Methylamine solution in water reacts with ferric chloride solution to give a precipitate of ferric hydroxide. (iii) Aniline does not undergo Friedel Crafts reaction. |
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Answer» Solution :(i) Lone pair of electrons on nitrogen in aniline is delocalised due to resonance with benzene ring. Thus, electron density on the amino group in aniline decreases. Consequently, its basic strength also decreases. On the other hand, electron density on amino group in methylamine increases due to `+I`-effect of `CH_(3)` group. Thus, it has a greater basic strength `(K_(b)). pK_(b)" and "K_(b)` are inversely related. Therefore, `pK_(b)` for aniline is more than that for methylamine. (ii) `CH_(3)NH_(2)+H_(2)O rarr CH_(3)overset(+)(N)H_(3)+OH^(-)` `FeCl_(3)+3OH^(-) rarr UNDERSET(underset("hydroxide")("Ferric"))(Fe(OH)_(3))+3Cl^(-)` It is due to the formation of `OH^(-)`, that the precipitation of ferric hydroxide takes place. (iii) `AlCl_(3)` is used as a catalyst in Friedel-Crafts REACTION. `AlCl_(3)` is a Lewis acid (electron deficient). `AlCl_(3)` gets attached to the lone pair of electrons in aniline. Amino group is thus not able to ACTIVATE the benzene ring for electrophilic SUBSTITUTION. Moreover electrophiles like `overset(+)(C)H_(3)" or "CH_(3)overset(+)(C)O` are not formed as the catalyst is consumed in association with amino group. Therefore, aniline does not undergo Friedel Crafts reaction. |
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