Question

A compound A $\left( {{{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{C}}{{\text{l}}_{\text{2}}}} \right)$ on hydrolysis gives ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{O}}$ which reacts with ${\text{N}}{{\text{H}}_{\text{2}}}{\text{OH}}$, forms iodoform but does not give Fehling test A is

Answer 1

Hint: Fehling test is a characteristic test used to detect aldehydes. Thus, compounds other than aldehydes do not give Fehling test. The intermediate formed in the reaction does not give Fehling test and thus, it is not an aldehyde. Thus, the product formed in the reaction is a ketone. The intermediate reaction with ${\text{N}}{{\text{H}}_{\text{2}}}{\text{OH}}$ gives iodoform. Formation of iodoform is a characteristic test of methyl ketone $\left( {{\text{C}}{{\text{H}}_{\text{3}}} - {\text{C}} = {\text{O}}} \right)$ group. Thus, the intermediate form is a ketone with methyl $\left( { - {\text{C}}{{\text{H}}_3}} \right)$ group attached to it.

 Complete step by step answer:
-A compound (A) has a molecular formula ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{C}}{{\text{l}}_{\text{2}}}$. This compound undergoes hydrolysis. Hydrolysis means that it reacts with ${{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}$ and forms a compound with molecular formula ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{O}}$ which reacts with ${\text{N}}{{\text{H}}_{\text{2}}}{\text{OH}}$ to give iodoform. The reaction is as follows:

The compound with molecular formula ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{O}}$ can be an aldehyde or a ketone. But it does not give the Fehling test. Fehling test is a characteristic test for aldehydes. Thus, the compound with molecular formula ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{O}}$ is not an aldehyde. Thus, it is a ketone.
Also, ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{O}}$ produces iodoform which is a characteristic of methyl ketone group. Thus, ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{O}}$ is a ketone with methyl group attached to it. Thus, the structure is,

${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{O}}$ is produced by the hydrolysis of ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{C}}{{\text{l}}_{\text{2}}}$. Thus, in ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{O}}$, the $ = {\text{O}}$ atom is replaced by two $ - {\text{Cl}}$ atoms. Thus, the structure of compound A $\left( {{{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}{\text{C}}{{\text{l}}_{\text{2}}}} \right)$ is,

So, the correct answer is “Option A”.

Note: The compound gives Fehling test negative. Thus, it cannot be an aldehyde but can be a ketone. The compound on reaction with ${\text{N}}{{\text{H}}_{\text{2}}}{\text{OH}}$ gives iodoform. Iodoform is a characteristic test for methyl ketone groups. Thus, compound is a ketone containing methyl group.