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Arrange the following in increasing order of their boiling points. CH_(3)CHO,CH_(3)COOH,CH_(3)CH_(2)OH |
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Answer» Solution :`underset("Acetaldehyde")(CH_(3)-overset(O)overset(||)(C)-H)""underset("Acetic ACID")(CH_(3)-overset(O)overset(||)(C)-OH)""underset("Ethyl alcohol")(CH_(3)-CH_(2)-OH)` Acetaldehyde `(CH_(3)CHO)` does not contain a H atom attached to O atom but acetic acid `(CH_(3)COOH)` and ethyl alcohol `(CH_(3)CH_(2)OH)` have. As a RESULT, `CH_(3)CHO` does not form H-bonds but `CH_(3)COOH` and `CH_(3)CH_(2)OH` form H-bonds. consequently, the b.p. of `CH_(3)CHO` is much LOWER than those of `CH_(3)COOH` and `CH_(3)CH_(2)OH`.  But acetic acid due to its dimeric structure forms stronger H-bonds, than ethyl alcohol and HENCE b.p. of ethyl alcohol is lower than that of acetic acid. Combining both the trends the b.ps. of these three compounds increase in the order: `CH_(3)CHO(294K) lt CH_(3)CH_(2)OH (351K) lt CH_(3)COOH (393K)` |
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