Arrange the following compounds in the increasing order of the property indicated against each. Give reasons for your answer.
(i) `CH_(3)CH_(2)OH,CF_(3)CH_(2)OH,C Cl_(3)CH_(2)OH-` acid strength.
(ii) 2-methyl-2-propanol, 1-butanol and 2-butanol `-` Reactivity towards sodium.

Explanation

(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 this facilitates the release of proton from `CF_(3)CH_(2)OH` as compared to `CH_(3)CH_(2)OH`. Further, since `F` has stronger `-I` effect than `Cl`, therefore, `CF_(3)CH_(2)OH` is a stronger acid than `C Cl_(3)CH_(2)OH` while `CH_(3)CH_(2)OH` is the weakest acid. Thus, acid strength increases in the order :
`CH_(3)CH_(2)OH lt C Cl_(3)CH_(2)OH lt CF_(3)CH_(2)OH`.
(ii) It is an acid-base reaction since alcohols are acidic in nature and sodium is a strong base. Thus, the reactivity 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^(@) lt 2^(@) lt 1^(@)`, therefore, the reactivity of `Na` towards alcohols increases in the same order, i.e.,
`underset(("least reactive"))("2-methyl-2-propanol "(3^(@))) lt underset(("more reactive"))("2-butanol "(2^(@))) lt underset(("most reactive"))("1-butanol "(1^(@)))`