CBSE 2025 · Region 5 · Set 3 · Q20 · 2 marks
Write the mechanism of dehydration of ethyl alcohol with conc. $\displaystyle \mathrm{H}_{2} \mathrm{SO}_{4}$ at $\displaystyle 443$ K.
Marking-scheme solution
Step $\displaystyle 1$: Formation of protonated alcohol.
Step $\displaystyle 2$: Formation of carbocation: It is the slowest step and hence, the rate determining step of the reaction.
Step $\displaystyle 3$: Formation of ethene by elimination of a proton.
Step $\displaystyle 2$: Formation of carbocation: It is the slowest step and hence, the rate determining step of the reaction.Step $\displaystyle 3$: Formation of ethene by elimination of a proton.Alcohols, Phenols and EthersAlcohols and PhenolsApplyvery_short_answermedium
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CBSE Class 12 Chemistry past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.