Ethanol heated with concentrated $\mathrm{H_2SO_4}$ at $443\,\text{K}$ (170°C) gives mainly

1. Your syllabus: concentrated $\mathrm{H_2SO_4}$ is a strong dehydrating agent. 2. At HIGH TEMPERATURE ($\sim 443\,\text{K}$), the dominant reaction is INTRAMOLECULAR DEHYDRATION: removal of an $\mathrm{H}$ and an $\mathrm{OH}$ from adjacent carbons within ONE molecule of ethanol, giving an ALKENE. 3. $\mathrm{CH_3CH_2OH} \xrightarrow{\text{conc.}\,\mathrm{H_2SO_4}, 443\,\text{K}} \mathrm{CH_2{=}CH_2} + \mathrm{H_2O}$. 4. Product is ETHENE (ethylene). 5. At LOWER temperature ($\sim 413\,\text{K}$ / 140°C), INTERMOLECULAR dehydration dominates: two alcohol molecules lose ONE water giving an ETHER ($\mathrm{CH_3CH_2{-}O{-}CH_2CH_3}$, diethyl ether). So option A is at LOWER T. 6. Options B and C describe unrelated reactions. _Source: syllabus Chemistry Part 2, Ch 7, §7.4 (Dehydration of alcohols — temperature-dependent products), p. 11–12._