CBSE 2025 · Region 5 · Set 2 · Q20 · 2 marks
Write the mechanism of dehydration of ethyl alcohol with conc. $\displaystyle \mathrm{H}_{2} \mathrm{SO}_{4}$ at $\displaystyle 413$ K.
Marking-scheme solution
(i)
$\displaystyle \mathrm{CH_3-CH_2-\overset{..}{\underset{..}{O}}-H \;+\; H^{+} \;\longrightarrow\; CH_3-CH_2-\overset{H}{\underset{..}{\overset{+}{O}}}-H}$
(ii)
$\displaystyle \mathrm{CH_3CH_2-\overset{..}{\underset{|}{O}}{:}\;\underset{H}{}\;+\;CH_3-CH_2-\overset{+}{O}\!\!\big<^{H}_{H} \;\longrightarrow\; CH_3CH_2-\overset{+}{\underset{|}{O}}-CH_2CH_3 \;+\; H_2O}$
(with the lone pair on the first oxygen attacking the carbon bearing the $\displaystyle \mathrm{-\overset{+}{O}H_2}$ group; the O of the product carries an H below it)
(iii)
$\displaystyle \mathrm{CH_3CH_2-\overset{+}{\underset{|}{O}}-CH_2CH_3 \;\longrightarrow\; CH_3CH_2-O-CH_2CH_3 \;+\; \overset{+}{H}}$
(the O of the reactant carries an H below it)
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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.