Ethanol is dehydrated with concentrated $\mathrm{H_2SO_4}$ at $443$ K (170 °C). What is the major product?

**Temperature determines the product.** At ~170 °C with concentrated H₂SO₄, intramolecular dehydration (E1) of ethanol gives **ethene** plus water — the classic preparation of ethene from ethanol. At lower temperatures (≈ 140 °C, less acid) the reaction takes the *intermolecular* path: two ethanol molecules condense via loss of water to give **diethyl ether** instead. Option B is therefore the answer for a *different* temperature regime — a common JEE trap. Mechanism (E1, three steps): (i) protonation of the –OH, (ii) loss of H₂O to give a carbocation, (iii) loss of a β-proton to give the alkene.