Decreasing order of stability of ions is (i) $\mathrm{CH}_3-\stackrel{+}{\mathrm{C}}…
Decreasing order of stability of ions is
(i) $\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}-\mathrm{CH}_3$
(ii) $\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}-\mathrm{OCH}_3$
(iii) $\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}-\mathrm{COCH}_3$
(i) $>$ (ii) $>$ (iii)
(ii) $>$ (i) $>$ (iii)
(ii) $>$ (iii) $>$ (i)
(iii) $>$ (i) $>$ (ii)
Solution
The dispersal of the charge stabilizes the carbocation. More the number of electron donating groups are present, greater the dispersal of positive charge and therefore more the stability of carbocation can be observed.
So, the order is
$\begin{aligned} & \mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}-\mathrm{OCH}_3>\mathrm{CH}_3-\stackrel{+}{\mathrm{C}}_{\mathrm{H}}^{\mathrm{H}}-\mathrm{CH}_3 \\ &>\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}-\mathrm{COCH}_3\end{aligned}$