An organic compound ' $A$ ' has the molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$. It…
An organic compound ' $A$ ' has the molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$. It undergoes iodoform test. When staturated with $\mathrm{HCl}$ it gives ' $\mathrm{B}$ ' of molecular formula $\mathrm{C}_{9} \mathrm{H}_{14} \mathrm{O}$. ' $\mathrm{A}$ ' and ' $\mathrm{B}$ ' respectively are
Propanal and mesitylene
Propanone and mesityl oxide
Propanone and 2,6-dimethyl-2, 5-heptadien4 -one
Propanone and mesitylene
Solution
The compound A with formula \(\mathrm{C}_3 \mathrm{H}_6 \mathrm{O}\) gives iodoform test, it is propanone. It forms a compound B having carbon atoms three times to the number of carbo propanone, it is 2, 6-dimethyl-2,5-heptadien-4-one
\(\begin{array}{ll}
\mathrm{CH}_3-\underset{\stackrel{|}{ \atop } \atop \huge \mathrm{CH}_3}{\mathrm{C}}=\mathrm{O}+\mathrm{CH}_3-\underset{\stackrel{||}{ \atop } \atop \huge \mathrm{O}}{\mathrm{C}}-\mathrm{CH}_3+\mathrm{O}=\underset{\stackrel{|}{ \atop } \atop \huge \mathrm{CH}_3}{\mathrm{C}}-\mathrm{CH}_3 \xrightarrow{\text { dil } \mathrm{HCl}}\mathrm{CH}_3-\underset{\stackrel{|}{ \atop } \atop \huge \mathrm{CH}_3}{\mathrm{C}}=\mathrm{CH}-\underset{\stackrel{||}{ \atop } \atop \huge \mathrm{O}}{\mathrm{C}}-\mathrm{CH}=\underset{\stackrel{|}{ \atop } \atop \huge \mathrm{CH}_3}{\mathrm{C}}-\mathrm{CH}_3 \\
\qquad \qquad \qquad \qquad \text{(A)} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{(B)} \\
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{2, 6-Dimethyl-2, 5-heptadien-4-one}
\end{array}\)
/