Explain by writing a reaction, which product will be formed on heating n-butyl alcohol with concentrated sulphuric acid.

Explanation

When pieces of sodium metal are put in n-propyl alcohol, hydrogen is liberated and sodium n-propoxide salt is formed `underset("Sodium")(2Na)+underset("n-propyl alcohol")(2CH_(3)-CH_(2)-CH_(2)OH) to underset("Sodium n-propoxide")(2CH_(3)-CH_(2)-CH_(2)-ONa)+underset("Hydrogen")H_(2) uparrow`