Answer:-

i. Ethanol on reaction with a very strong base like alkali metal Na or K gives sodium or potassium ethoxide with the liberation of hydrogen gas.

\[\ce{2CH3CH2 – OH + 2Na -> 2CH3CH2O^-Na+ + H2_{(g)}↑}\]

ii. Ethyl bromide on reaction with sodium metal does not liberate hydrogen gas.

\[\ce{2CH3CH2Br + 2Na -> CH3CH2 – CH2CH3 + 2NaBr}\]