tert-butyl methyl ether on heating with anhydrous HI in ether gives

Question

tert-butyl methyl ether on heating with anhydrous HI in ether gives (A)  CH3OH+(CH3)3Cl  (B)  CH3I+(CH3)3COH  (C)  CH3I+(CH3)3Cl  (D) None of the above

Tardigrade Admin · Accepted Answer

Here, ether is the solvent. Being less polar, it favours textS textN text2 reaction and the nucleophile textI- attacks the text1 texto carbon of textC textH3.

Q. tert-butyl methyl ether on heating with anhydrous HI in ether gives

$C H_{3} O H + (C H_{3})_{3} Cl$

$C H_{3} I + (C H_{3})_{3} CO H$

$C H_{3} I + (C H_{3})_{3} Cl$

None of the above

Here, ether is the solvent. Being less polar, it favours $S_{N} 2$ reaction and the nucleophile $I^{-}$ attacks the $1^{o}$ carbon of $C H_{3} .$

Q. tert-butyl methyl ether on heating with anhydrous HI in ether gives

CH3​OH+(CH3​)3​Cl

CH3​I+(CH3​)3​COH

CH3​I+(CH3​)3​Cl

None of the above

Here, ether is the solvent. Being less polar, it favours SN​2 reaction and the nucleophile I− attacks the 1o carbon of CH3​.

$C H_{3} O H + (C H_{3})_{3} Cl$

$C H_{3} I + (C H_{3})_{3} CO H$

$C H_{3} I + (C H_{3})_{3} Cl$

Here, ether is the solvent. Being less polar, it favours $S_{N} 2$ reaction and the nucleophile $I^{-}$ attacks the $1^{o}$ carbon of $C H_{3} .$