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Arrange the following in increasing order of their boiling points: C_(2)H_(5)NH_(2),C_(2)H_(5)OH,(CH_(3))_(3)N |
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Answer» Solution :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` |
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