Match the conversions in List-I with the conditions used for them in…

The correct answer is Option (3) → A-II, B-IV, C-III, D-I

With a protic acid, an alcohol can lose water to give an alkene or, from two molecules, an ether; the conditions decide which. The ease of dehydration to alkenes is tertiary > secondary > primary, because the tertiary carbocation forms most easily; the easier the dehydration, the milder the conditions needed.

(A) Ethanol → Ethene — This matches (II) H₂SO₄, 443 K. At the higher temperature, elimination wins and ethene is the product.

(B) Ethanol → Ethoxyethane — This matches (IV) H₂SO₄, 413 K. At the lower temperature, an ethanol molecule attacks a protonated ethanol molecule (S_N2), and the ether is the main product.

(C) Propan-2-ol → Propene — This matches (III) 85% H₃PO₄, 440 K. A secondary alcohol needs milder conditions than ethanol, but stronger ones than a tertiary alcohol.

(D) 2-Methylpropan-2-ol → 2-Methylpropene — This matches (I) 20% H₃PO₄, 358 K. A tertiary alcohol is the easiest to dehydrate, so dilute acid at a low temperature is enough.

NCERT (2026-27 edition):
7.6.1 Preparation of Ethers – “Alcohols undergo dehydration in the presence of protic acids (H₂SO₄, H₃PO₄). The formation of the reaction product, alkene or ether depends on the reaction conditions. For example, ethanol is dehydrated to ethene in the presence of sulphuric acid at 443 K. At 413 K, ethoxyethane is the main product.”

7.4.4 Chemical Reactions – “Secondary and tertiary alcohols are dehydrated under milder conditions. For example”

    \( \displaystyle \ce{CH3CH(OH)CH3} \xrightarrow[\text{440 K}]{85\%\ \ce{H3PO4}} \ce{CH3-CH=CH2 + H2O} \)

    \( \displaystyle \ce{CH3-C(CH3)2-OH} \xrightarrow[\text{358 K}]{20\%\ \ce{H3PO4}} \ce{CH3-C(=CH2)-CH3 + H2O} \)

“Thus, the relative ease of dehydration of alcohols follows the following order: Tertiary > Secondary > Primary”