Question

Propanone reacts with iodine in acid medium according to the following equation:
$C{H}_3-\stackrel{\stackrel{O}{||}}{C}-C{H}_3+I_2\stackrel{H^{+}}{\to }C{H}_3-\stackrel{\stackrel{O}{||}}{C}-C{H}_{2}I+HI$
The following data were obtained when the reaction was studied
$$\begin{array}{cccc}(C{H}_3{)}_{2}CO,M& I_2,M& \left(H^{+}\right),M& \text{Relative rate}\\ {10}^{-2}& {10}^{-2}& {10}^{-2}& 1\\ 2\times {10}^{-2}& {10}^{-2}& {10}^{-2}& 2\\ 2\times {10}^{-2}& 2\times {10}^{-2}& {10}^{-2}& 2\\ 2\times {10}^{-2}& {10}^{-2}& 2\times {10}^{-2}& 4\end{array}$$
For the above reaction the following mechanism have been proposed.

Regarding the given reaction which of the following is / are incorrect:

• A. Overall order of the reaction is two and step - I is the RDS of the reaction.
• B. Overall order of the reaction is two and step - II is the RDS of the reaction
• C. Overall order of the reaction is three and step - I is the RDS of the reaction
• D. Order with respect to $I_2$ is one and step - I is the RDS of the reaction

Answer 1

The correct option is A Overall order of the reaction is two and step - I is the RDS of the reaction.

From the given data
(1) on doubling the conc. of $C{H}_3-\stackrel{\stackrel{O}{||}}{C}-C{H}_3$ rate gets doubled.
$\therefore$ reaction is the first order with respect to $C{H}_3-\stackrel{\stackrel{O}{||}}{C}-C{H}_3$
(2) on doubling the conc. of $I_2$ no change in rate
$\therefore$ rate = $k[I_2{]}^{\circ }$
(3) on doubling the conc. of $H^{+}$ rate gets doubled
$\therefore rate=k\left[H^{+}\right]$
$\therefore$ Over all rate $=k[C{H}_3-\stackrel{\stackrel{O}{||}}{C}-C{H}_3]\left[H^{+}\right]$
Second order reaction
So, step one I is the RDS.