(a) Arrange the following in the increasing order of boiling points:
`C_(2)H_(5)NH_(2)`, `C_(2)H_(5)OH`, `(CH_(3))_(3)N`
(b) Arrange the following in increasing order of base strength in gas phase:
`(C_(2)H_(5))_(3)N`, `C_(2)H_(5)NH_(2)`, `(C_(2)H_(5))_(2)2NH`
(c) Arrange the following in decreasing order of solubility in water:
`(CH_(3))_(3)N`, `(CH_(3))_(2)NH`, `CH_(3)NH_(2)`

Explanation

Boiling points depend upon the extent of H-bonding which, in turn, depends upon: (i) electronegativity of element carrying the H-atoms and (ii) the number of H-atoms on it. Since O is more electronegative than N, therefore, O forms stronger H-bonds than. In other words, b.p. of `C_(2)H_(5)OH` is higher than those of `C_(2)H_(5)NH_(2)` and `(CH_(3))_(3)N`. further, `C_(2)H_(5)NH_(2)` carries two H-atoms on N but `(CH_(3))_(3)N` does not carry any H-atom on N. therefore, `C_(2)H_(5)NH_(2)` forms H-bonds but `(CH_(3))_(3)N` does not. in other words, b.p. of `C_(2)H_(5)NH_(2)` is higher than that of `(CH_(3))_(3)N`. thus, the overall b.ps of the three compounds increase in the order:
`(CH_(3))_(3)N lt C_(2)H_(5)NH_(2) lt C_(2)H_(5)OH`