Question

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.

Answer 1

Decreasing order of reactivity of sodium metal is:

1° > 2° > 3°

Alcohols react with sodium metal to form alkoxides and hydrogen is liberated:

\[\ce{R - O - H + Na -> RO - Na+ {+} 1/2 H2}\]

The order of reactivity of alcohols is primary > secondary > tertiary. This can be explained on the basis of cleavage of O – 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 As a result, the electrons of O – H bond cannot be withdrawn strongly towards oxygen and O – H remains strong. Therefore, greater is the number, of alkyl groups present, smaller will be reactivity of alcohol.