Question

An organic compound A having molecular formula ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{6}}}$ on treatment with aq. ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ gives compound B which on treatment with ${\text{HCl/ZnC}}{{\text{l}}_{\text{2}}}$ gives the compound C on treatment with ethanoic ${\text{KOH}}$ gives back the compound A. Identify the compound A,B,C.

Answer 1

Hint: In the above question, we have to find the organic compound A, B and C. Since the molecular formula ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{6}}}$ which is a representation of alkene is given at the beginning, we can start solving the reaction occurred from the beginning. The first reaction is an additional reaction, followed by substitution of chloride ion against OH.

Complete step-by-step answer:In the above question, the compound A has molecular formula ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{6}}}$ which is of the form ${{\text{C}}_{\text{n}}}{{\text{H}}_{{\text{2n}}}}$ which indicates that it is a propene.
So, A is propene.
When an alkene reacts with aq. ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$, ${{\text{H}}_{\text{2}}}{\text{O}}$ is added to the alkene. In this process, ${\text{OH}}$ group gets attached to the carbon atom containing less number of H atom and H atom from ${{\text{H}}_{\text{2}}}{\text{O}}$ gets attached to carbon atom having more H atom.
So, propene reacts with aq. ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ to form propan-2-ol. So, compound B is propan-2-ol.
The mixture of ${\text{HCl}}$ and ${\text{ZnC}}{{\text{l}}_{\text{2}}}$ form Lucas reagent which replace ${\text{OH}}$ atom with ${\text{Cl}}$. Hence, compound C is 2-chloropropane.
When ethanoic ${\text{KOH}}$ is added to 2-chloropropane it leads to elimination reaction and hence, propene is formed.
The whole reaction can be illustrated as:

Note:In these types of questions, we have to look where the compound molecular formula is given. If it is given in the beginning, we can start solving it as usual. If the molecular formula or molecule is given at the end, we must start solving the equation backward.