What is the first step in acid catalyzed dehydration of alcohol?

The correct answer is Option (4) → Protonation of alcohol

The –OH group is a poor leaving group, because \( \ce{OH-} \) is a strong base. In acid-catalysed dehydration the acid first protonates the oxygen atom of the alcohol. The –\( \ce{OH2^{+}} \) group formed can leave as a neutral water molecule, which is a good leaving group. The dehydration of ethanol by conc. \( \ce{H2SO4} \) at 443 K runs in three stages.

First, the alcohol is protonated (fast):

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

Next, the protonated alcohol loses water and a carbocation is formed (slow, rate determining):

    \( \displaystyle \ce{CH3CH2-OH2+ -> CH3CH2+ + H2O} \)

Finally, a proton is lost from the carbon next to the positive carbon, and the alkene is formed:

    \( \displaystyle \ce{CH3CH2+ -> CH2=CH2 + H+} \)

The \( \ce{H+} \) used in the first stage is given back in the last, so the acid acts as a catalyst. The first step is the protonation of the alcohol.

Why the other options are incorrect

• Option (1) Removal of OH — Incorrect. \( \ce{OH-} \) is too strong a base to leave on its own; the group leaves only after protonation, and then as water, not as \( \ce{OH-} \).

• Option (2) Removal of water — Incorrect. Loss of water is the second, slow step; it can happen only after the oxygen has been protonated.

• Option (3) Elimination of proton — Incorrect. Loss of \( \ce{H+} \) from the carbocation is the last step, which forms the C=C bond and regenerates the catalyst.

NCERT (2026-27 edition):
Dehydration – …