Chemistry - Carbonyl compounds Question with Solution | TestHub

When mixed ethers are used, the alkyl iodide produced depends on the nature of alkyl groups. If one group is Me and the other a pri- or sec-alkyl group, then methyl iodide is produced. Here reaction occurs via ${\mathrm{S}}_{\mathrm{N}}2$ mechanism and because of the steric effect of the larger group, ${\mathrm{I}}^{-}$attacks the smaller (Me) group. ${\mathrm{CH}}_3{\mathrm{OC}}_2{\mathrm{H}}_5+\mathrm{HI}\to {\mathrm{CH}}_3\mathrm{I}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$ When the substrate is a methyl $t$-alkyl ether, the products are $t-R\mathrm{I}$ and $\mathrm{MeOH}$. Here reaction occurs by ${\mathrm{S}}_{\mathrm{N}}1$ mechanism and formation of products is controlled by the stability of carbocation. Since carbocation stability order is $3^{\circ }>2^{\circ }>1^{\circ }>\stackrel{\oplus }{{\mathrm{CH}}_3}$. $\therefore$ Alkyl halide is always derived from tert-alkyl group.