Diethyl ether is prepared by:

Solution

Chemistry ma’am · 3 steps

Two ethanol molecules lose one water between them, and at 413 K that water loss gives you the ether.

1. Protonate the alcohol. Conc. H₂SO₄ protonates the -OH of ethanol to give C₂H₅OH₂⁺. Now water is a good leaving group.
2. Second ethanol attacks. A second ethanol molecule attacks the carbon of C₂H₅OH₂⁺ by S_N2 and pushes out water. Losing H⁺ then leaves diethyl ether.

   2C₂H₅OH →(conc. H₂SO₄, 413 K) C₂H₅-O-C₂H₅ + H₂O

   Option	What ethanol or the reagent actually gives
   (a) conc. H₂SO₄, 413 K	diethyl ether (bimolecular dehydration)
   (b) Oxidation	ethanal, then ethanoic acid
   (c) Na	sodium ethoxide and H₂ gas
   (d) Hydrolysis of C₂H₅Cl	ethanol

3. So. Only (a) gives diethyl ether. At the lower temperature of 413 K, two molecules join and lose one water between them. At 443 K the same acid gives ethene instead.

The idea to keep. With conc. H₂SO₄, temperature decides the product: about 413 K gives the ether (two molecules lose one water), about 443 K gives the alkene (one molecule loses water).

Answer from the NTA final answer key. Solution by PrepWiser.