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| 1. |
Account for the following : pK_(b) of aniline is more than that of methylamine. |
| Answer» Solution :In aniline, the LONE pair of electrons on the N-atom are delocalised due to resonance with benzene ring. As a result, electron density on the NITROGEN decreases. In CONTRAST, in `CH_(3)NH_(2)`, +I-effect of `CH_(3)` increases the electron density on the N-atom. Therefore, aniline is a weaker BASE than methylamine and hence its `pK_(b)` value is higher than that of methylamine. (Lower the value of `K_(b)`, higher the value of `pK_(b)`). | |