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Arrange the compounds shown below in order of decreasing basicity (most basic first)

 

$ \begin{array}{1 1}(a)\;(I) > (II) > (III) > (IV)\\(b)\;(IV) > (III) > (II) > (I)\\(c)\;(IV) > (II) > (III) > (I)\\(d)\;(I) > (III) > (II) > (IV)\end{array}$

1 Answer

(IV) is the least basic ,it is an amide,not an amine,(I) is more basic aliphatic amines are usually much more basic than aromatic once.(III) is more basic than (II) because the ring of II holds an $e^{\ominus}$ with drawing group making the unshared pair of $e^{\ominus}$ on the amine nitrogen less available for sharing.
Hence (d) is the correct answer.
answered Mar 3, 2014 by sreemathi.v
 
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