Explain the dehydration reaction of $2$-methylbutan-$1$-ol with concentrated $H_2SO_4$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The dehydration of $2$-methylbutan-$1$-ol with concentrated $H_2SO_4$ at $443 \ K$ proceeds via the formation of a carbocation intermediate.
$1$. Protonation of the alcohol: The hydroxyl group is protonated to form a good leaving group $(-OH_2^+)$.
$2$. Formation of carbocation: Loss of water molecule generates a $1^{\circ}$ carbocation.
$3$. Rearrangement: The $1^{\circ}$ carbocation undergoes a $1,2$-hydride shift to form a more stable $3^{\circ}$ carbocation.
$4$. Elimination: Loss of a proton from the $3^{\circ}$ carbocation yields $2$-methylbut-$2$-ene as the major product,following Saytzeff's rule.

Explore More

Similar Questions

Alcohols $(i) \ CH_3CH_2CH_2OH$,$(ii) \ CH_3-CH(OH)-CH_3$ and $(iii) \ CH_3-C(CH_3)(OH)-CH_3$ were treated with Lucas reagent (Conc. $HCl + ZnCl_2$). What results do you expect at room temperature?

Difficult
View Solution

Ethylene may be obtained by dehydration of which of the following with concentrated $H_2SO_4$ at $160 - 170\,^{\circ}C$?

Find the major product for the following reaction.

Which of the following will not produce $CH_4$ gas when reacted with $CH_3MgBr$?

Difficult
View Solution

Absolute alcohol is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo