Question

Neeta was experimenting in the lab to study the chemical reactivity of alcohol. She carried out a dehydration reaction of propanol at 140°C to 180°C. Different products were obtained at these two temperatures.

1. Identify the major product formed at 140°C and the mechanism followed in this case.
2. Identify the major product formed at 180°C.

Answer 1

Propanol (C₃H₇OH) can undergo dehydration (removal of water) in the presence of concentrated sulfuric acid (H₂SO₄​) or aluminum oxide (Al₂O₃​) as a catalyst. The temperature at which the reaction is carried out significantly affects the major product formed:

1. Dehydration at 140°C (Ether Formation)
   \[\ce{2C3H7OH \overset{conc. H2SO4, 140°C}{->}C3H7OC3H7 + H2O}\]
   • Major Product: Di-n-propyl ether (Propyl ether, \[\ce{C3H7-O-C3H7}\]
   • Mechanism: Intermolecular Dehydration (Williamson Ether Synthesis)
   • Explanation: At 140°C, two molecules of propanol undergo a condensation reaction, forming di-n-propyl ether through the SN_2​ mechanism. This involves the nucleophilic attack of the alcohol oxygen on another alcohol molecule, followed by the elimination of a water molecule.
2. Dehydration at 180°C (Alkene Formation)
   \[\ce{2C3H7OH \overset{conc. H2SO4, 180°C}{->}C3H6 + H2O}\]
   
   • Major Product: Propene (C₃​H₆)
   • Mechanism: Intramolecular Dehydration (E1 Elimination)
   • Explanation: At 180°C, the reaction favors the elimination of water from a single propanol molecule, resulting in the formation of propene. This is a three-step mechanism: