Chapter 22, Problem 22.69AP

Interpretation Introduction

Interpretation:

The reason corresponding to the fact that the reaction between $1,3-\text{diphenyl}-1,3-\text{propanedione}$ with ${\text{I}}_{\text{2}}$ and $\text{NaOH}$ forms a precipitate of iodoform even though it is not a methyl ketone is to be stated.

Concept introduction:

In organic chemistry, the carbonyl compound is a class of functional group which contains a $\text{C}=\text{O}$ group. They are of two types aldehyde $-\text{CHO}$ and ketone $\text{RC}=\text{OR}$. They are the oxidized product of alcohol. Iodoform test used to confirm the presence of the carbonyl group. The compounds which contain methyl ketone ${\text{CH}}_{\text{3}}\text{C}=\text{O}$ or secondary alcohol ${\text{CH}}_{\text{3}}\text{CH}\left(\text{OH}\right)$ group gives positive iodoform test.