
- A$iii > i > ii$
- B$iii > ii > i$
- C$i > iii > ii$
- ✓$i > ii > iii$

The ease of carbocation towards $S_{N} 1$ reaction is shown below. $3^{\circ}<2^{\circ}<1^{\circ}$.
So, the correct order is $i > ii >$ $iii$
Therefore, option $(D)$ is the correct.
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$S{O_2}C{l_2}\left( g \right) \rightleftharpoons S{O_2}\left( g \right) + C{l_2}\left( g \right)$
$CO\left( g \right) + C{l_2}\left( g \right) \rightleftharpoons COC{l_2}\left( g \right)$
on adding more $SO_2$ at equilibrium what will happen?
Statement $(I)$ : The $\mathrm{NH}_2$ group in Aniline is ortho and para directing and a powerful activating group.
Statement $(II)$ : Aniline does not undergo FriedelCraft's reaction (alkylation and acylation).
In the light of the above statements, choose the most appropriate answer from the options given below :
$xBrO_3\,^-+ yCr^{+3} + zH_2O \to Br_2 + CrO_4 ^{-2} + H^+$
the coefficients $x, y, z$ are