Alcohols react with active metals e.g., Na, K etc., to give corresponding alkoxides. Write down the decreasing order of reactivity of sodium metal towards primary, secondary and tertiary alcohols.

Explanation

Decreasing order of reactivity of sodium metal is
`1^(@) gt 2^(@) gt 3^(@)`
Alcohols react with sodium metal to form alkoxides and hydrogen is liberated:
`R-O-H+Na to RO^(-)Na^(+)+(1)/(2)H_(2)`
he order of reactivity of alcohola is primary > secondary > tertiary. This can be explained on the basis of cleavage of 0-H bond. The alkyl groups are electron releasing groups (+I effect) and they increase the electron density around the oxygen. As a result, the electrons of OCH bond cannot be withdrawn strongly towards oxygen and 0-H remains strong. Therefore, greater is the number of alkyl groups present, smaller will be reactivity of alcohol.