Give the reaction of $CH_3-C(CH_3)_2-OCH_3$ with the following:
$(i)$ Anhydrous $HI$ in ether
$(ii)$ Concentrated $HI$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The reaction of tert-butyl methyl ether with $HI$ proceeds via the protonation of the ether oxygen atom.
$(i)$ With anhydrous $HI$ (or $HI$ in ether),the reaction follows an $S_N2$ mechanism. The nucleophile $I^-$ attacks the less sterically hindered methyl group,resulting in the formation of tert-butyl alcohol and methyl iodide $(CH_3I)$.
$(ii)$ With concentrated $HI$,the reaction follows an $S_N1$ mechanism. The protonated ether undergoes cleavage to form a stable tert-butyl carbocation,which then reacts with $I^-$ to form $2-$iodo$-2-$methylpropane,while the methyl group forms methanol $(CH_3OH)$.

Explore More

Similar Questions

In the given reaction
$C_6H_5 - O - CH_2 - CH_3 \xrightarrow{\text{Excess } HI / \Delta} [X] + [Y]$
$[X]$ and $[Y]$ will respectively be

When $CH_2 = CH - O - CH_2 - CH_3$ reacts with $1 \ mol$ of $HI$,one of the products formed is

Ethoxybenzene on reaction with hot and concentrated $HI$ forms:

The ether that undergoes electrophilic substitution reactions is

The product in the reaction (figure) $+ HBr$ 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