Arrannge the followin compounds in the increasing order of the property indicated against each. Givereasons 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.

Arrannge the followin compounds in the increasing order of the propert

1.	Arrannge the followin compounds in the increasing order of the property indicated against each. Givereasons 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.
Answer» Solution :(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 PROTON as cojmpared to `CH_(3)CH_(2)OH`. Further,s ince F has stronger -I-effect than Cl, therefore, `CH_(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. such, 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^(@)))`