CBSE 2024 · Region 1 · Set 1 · Q26 · 3 marks
(a)Write the mechanism of the following reaction : \[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \xrightarrow[443 \mathrm{~K}]{\mathrm{H}^{+}} \mathrm{CH}_{2}=\mathrm{CH}_{2} \](b)Write the main product in each of the following reactions :
(a)
Write the mechanism of the following reaction : \[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \xrightarrow[443 \mathrm{~K}]{\mathrm{H}^{+}} \mathrm{CH}_{2}=\mathrm{CH}_{2} \]
(b)
Write the main product in each of the following reactions :
Marking-scheme solution
(a) 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.
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CBSE Class 12 Chemistry past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.