question_answer

Arrange the following compounds in the increasing order of the property indicated against each. Give reasons for your answer. (i) \[C{{H}_{3}}C{{H}_{2}}OH,\,C{{F}_{3}}C{{H}_{2}}OH,\]\[CC{{l}_{3}}C{{H}_{2}}OH\]-Acid strength (ii) 2-methyl-2-propanol, 1-butanol and 2-butanol?Reactivity towards sodium.

Answer:

                (i) Due to -\[I\]-effect of the halogen, the electron density in the O-H bond decreases. As a result of this electron-deficiency, the O -H bond weakens and thus facilitates the release of a proton as compared to\[C{{H}_{3}}C{{H}_{2}}OH\]. Further, since F has stronger -\[I\]- effect than \[Cl,\] therefore, \[C{{F}_{3}}C{{H}_{2}}OH\] is a stronger acid than \[CC{{l}_{3}}C{{H}_{2}}OH\] while\[C{{H}_{3}}C{{H}_{2}}OH\] is the weakest acid. Thus, acid-strength increases in the order : \[C{{H}_{3}}C{{H}_{2}}OH<CC{{l}_{3}}C{{H}_{2}}OH<C{{F}_{3}}C{{H}_{2}}OH\]. (ii) It is an acid-base reaction since alcohols are acidic in nature and sodium is a strong base. As such, there activity of these alcohols towards sodium increases as the acidic character of alcohols increases. Now since the acidic character of alcohols increases in the order: 3° < 2° < 1°, therefore, the reactivity of Na towards alcohols increases in the same order, i.e., \[\underset{(least\,reactive)}{\mathop{2-methyl-2-propanol\,(3{}^\circ )}}\,\,\,\,\,\,<\]                \[\,\underset{(more\,\,reactive)}{\mathop{2\text{-butanol}\,(2{}^\circ )}}\,\,\,\,<\,\,\,\,\underset{(most\,reactive)}{\mathop{1\text{-butanol}\,\,(1{}^\circ )}}\,\]