Chapter 23, Problem 107IL

(a)

Interpretation Introduction

Interpretation:

To identify the carbonyl compound having molecular formula ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\text{O}$. To draw the structures of two possible carbonyl compounds.

Concept introduction:

The hydrocarbons which have general formula ${\text{C}}_{\text{n}}{\text{H}}_{\text{2n}}\text{O}$ are come in the catagory of carbonyl compounds. The carbonyl compounds have ${\text{R}}_{\text{2}}\text{CO}$ and $\text{RCHO}$ functional groups. So, carbonyl compounds are of two types (1) Aldehydes $\left(\text{RCHO}\right)$ (2) Ketones $\left({\text{R}}_{\text{2}}\text{CO}\right)$.

These compounds have same molecular formula but different structural formula. Therfore their systematic names are different. For aldehydes the suffix written in the end of parent carbon chain is $\text{-al}$, while for ketone $\text{-one}$. For example, $\text{propanal}$ is an aldehyde and $\text{propanone}$ is a ketone.

The reaction of an aldehyde with ${\text{KMnO}}_{\text{4}}$ gives a carboxylic acid. The conversion of an aldehyde to a carboxylic acid is an oxidation reaction, so ${\text{KMnO}}_{\text{4}}$ is acting as a oxidizing agent. However, a ketone does not undergo oxidation in the presence of oxidizing agent.

(b)

Interpretation Introduction

Interpretation:

To identify the carbonyl compound having molecular formula ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\text{O}$. To decide which of the structure is correct.

(c) To write the name of product obtained after oxidation.

Concept introduction:

The hydrocarbons which have general formula ${\text{C}}_{\text{n}}{\text{H}}_{\text{2n}}\text{O}$ are come in the catagory of carbonyl compounds. The carbonyl compounds have ${\text{R}}_{\text{2}}\text{CO}$ and $\text{RCHO}$ functional groups. So, carbonyl compounds are of two types (1) Aldehydes $\left(\text{RCHO}\right)$ (2) Ketones $\left({\text{R}}_{\text{2}}\text{CO}\right)$.

These compounds have same molecular formula but different structural formula. Therfore their systematic names are different. For aldehydes the suffix written in the end of parent carbon chain is $\text{-al}$, while for ketone $\text{-one}$. For example, $\text{propanal}$ is an aldehyde and $\text{propanone}$ is a ketone.

The reaction of an aldehyde with ${\text{KMnO}}_{\text{4}}$ gives a carboxylic acid. The conversion of an aldehyde to a carboxylic acid is an oxidation reaction, so ${\text{KMnO}}_{\text{4}}$ is acting as a oxidizing agent. However, a ketone does not undergo oxidation in the presence of oxidizing agent.

(c)

Interpretation Introduction

Interpretation:

To write the name of product obtained after oxidation.

Concept introduction:

The hydrocarbons which have general formula ${\text{C}}_{\text{n}}{\text{H}}_{\text{2n}}\text{O}$ are come in the catagory of carbonyl compounds. The carbonyl compounds have ${\text{R}}_{\text{2}}\text{CO}$ and $\text{RCHO}$ functional groups. So, carbonyl compounds are of two types (1) Aldehydes $\left(\text{RCHO}\right)$ (2) Ketones $\left({\text{R}}_{\text{2}}\text{CO}\right)$.

These compounds have same molecular formula but different structural formula. Therfore their systematic names are different. For aldehydes the suffix written in the end of parent carbon chain is $\text{-al}$, while for ketone $\text{-one}$. For example, $\text{propanal}$ is an aldehyde and $\text{propanone}$ is a ketone.

The reaction of an aldehyde with ${\text{KMnO}}_{\text{4}}$ gives a carboxylic acid. The conversion of an aldehyde to a carboxylic acid is an oxidation reaction, so ${\text{KMnO}}_{\text{4}}$ is acting as a oxidizing agent. However, a ketone does not undergo oxidation in the presence of oxidizing agent.