Question

Consider the following transformation.
$C{H}_{3}COOK\stackrel{CaC{O}_3}{\to }\begin{array}{c}A\end{array}\stackrel{\text{heat}}{\to }\begin{array}{c}B\end{array}\underset{NaOH}{\overset{I_2}{\to }}\begin{array}{c}C\end{array}$
The molecular formula of $\begin{array}{c}C\end{array}$ is:

• A. $C{H}_3-\underset{I}{\underset{|}{\stackrel{OH}{\stackrel{|}{C}}}}-C{H}_3$
• B. $IC{H}_2-COC{H}_3$
• C. $CH{I}_3$
• D. $C{H}_{3}I$

Answer 1

The correct option is C $CH{I}_3$

$C{H}_{3}COOK\stackrel{CaC{O}_3}{\to }\begin{array}{c}\underset{\left(A\right)}{(C{H}_{3}COO{)}_{2}Ca}\end{array}\stackrel{\text{heat}}{\to }\begin{array}{c}\underset{\left(B\right)}{C{H}_3\stackrel{O}{\stackrel{||}{C}}C{H}_3}\end{array}$

$B\underset{NaOH}{\overset{I_2}{\to }}\begin{array}{c}\underset{\left(C\right)}{CH{I}_3}\end{array}+NaI+C{H}_{3}COONa+H_{2}O$