Secondary and tertiary alcohols are not favoured for the preparation…

The correct answer is Option (1) → elimination reaction competes over substitution reaction.

Ethers can be made by heating an alcohol with concentrated sulphuric acid at about 413 K. One alcohol molecule attacks another protonated alcohol molecule in an S_N2 step, and water is lost:

    \( \displaystyle \ce{CH3CH2-OH + H^{+} -> CH3CH2-OH2^{+}} \)

    \( \displaystyle \ce{CH3CH2-OH + CH3CH2-OH2^{+} -> CH3CH2-O-CH2CH3 + H2O + H^{+}} \)

    \( \displaystyle \text{Net: }\ce{2CH3CH2OH} \xrightarrow[\text{413 K}]{\ce{H2SO4}} \ce{CH3CH2-O-CH2CH3 + H2O} \)

The same acid can instead remove water from a single molecule to give an alkene, an elimination reaction. For ethanol this happens at a higher temperature:

    \( \displaystyle \ce{CH3CH2OH} \xrightarrow[\text{443 K}]{\ce{H2SO4}} \ce{CH2=CH2 + H2O} \)

With secondary and tertiary alcohols, the carbon carrying the –OH group is crowded by alkyl groups, which hinders the S_N2 attack. These alcohols also form stable carbocations that lose \( \ce{H^{+}} \) very easily. So elimination is faster than substitution, and the alkene becomes the main product instead of the ether, for example:

    \( \displaystyle \ce{(CH3)3C-OH} \xrightarrow{\ce{H2SO4}} \ce{(CH3)2C=CH2 + H2O} \)

Why the other options are incorrect

• Option (2) substitution reaction competes over elimination reaction — Incorrect. This is the reverse. For secondary and tertiary alcohols, elimination wins over substitution.

• Option (3) substitution reaction can not occur with secondary and tertiary alcohols. — Incorrect. Substitution does occur with these alcohols. For example, tertiary alcohols react with HCl fastest of all.

• Option (4) an elimination reaction can not occur with secondary or tertiary alcohols. — Incorrect. Elimination is actually easiest for these alcohols; the ease of dehydration is 3° > 2° > 1°.

NCERT (2026-27 edition):
Preparation of Ethers – …