Chapter 14, Problem 64P

Interpretation Introduction

(a)

Interpretation:

The oxidized product of the following alcohol when oxidized with ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}\text{O}$ should be determined:

  

Concept Introduction:

A chemical reaction is the symbolic representation of the conversion of substances to new substances.

  $\begin{array}{l}\text{ A + B }\stackrel{}{\to }\text{ C}\\ \text{Reactant }\stackrel{}{\to }\text{ Product}\end{array}$

In a chemical reaction, the substance which is involved in conversion is said to be reactant whereas the newly formed substance is known as a product. Both reactants and products must be separated by an arrow.

The oxidation reaction is the reaction that involves the addition of O atom in the presence of certain oxidizing agents such as ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}\text{O}$ and ${\text{KMnO}}_{\text{4}}$.

Answer to Problem 64P

  

Explanation of Solution

To get the oxidized product of any alcohol, three steps must be followed:

• Locate the C atom in the parent chain that is bonded with −OH group.
• Convert that C atom to carbonyl C atom or carboxylic acid as it is overall removal of H atoms.
• Primary alcohols are oxidized to aldehyde which further oxidized to a carboxylic acid.
• Secondary alcohols are oxidized to ketone (R₂CO).
• Tertiary alcohols are not oxidized as they do not have H atom on the C with the −OH group.

Hence, the oxidized of 2-methylcyclopnetanol will form 2-methylcyclopentanone molecule as 2-methylcyclopnetanol is a secondary alcohol.

  

Interpretation Introduction

(b)

Interpretation:

The oxidized product of the following alcohol when oxidized with ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}\text{O}$ should be determined:

  

Concept Introduction:

A chemical reaction is the symbolic representation of the conversion of substances to new substances.

  $\begin{array}{l}\text{ A + B }\stackrel{}{\to }\text{ C}\\ \text{Reactant }\stackrel{}{\to }\text{ Product}\end{array}$

In a chemical reaction, the substance which is involved in conversion is said to be reactant, whereas the newly formed substance is known as a product. Both reactants and products must be separated by an arrow.

The oxidation reaction is the reaction that involves the addition of O atom in the presence of certain oxidizing agents such as ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}\text{O}$ and ${\text{KMnO}}_{\text{4}}$.

Answer to Problem 64P

  

Explanation of Solution

To get the oxidized product of any alcohol, three steps must be followed:

• Locate the C atom in the parent chain that is bonded with −OH group.
• Convert that C atom to carbonyl C atom or carboxylic acid as it is overall removal of H atoms.
• Primary alcohols are oxidized to aldehyde which further oxidized to the carboxylic acid.
• Secondary alcohols are oxidized to ketone (R₂CO).
• Tertiary alcohols are not oxidized as they do not have H atom on the C with the −OH group.

Hence, the oxidized of 1-decanol will form decanoic acid molecule as 1-decanol is a primary alcohol.

  

Interpretation Introduction

(c)

Interpretation:

The oxidized product of following alcohol when oxidized with ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}\text{O}$ should be determined:

  

Concept Introduction:

A chemical reaction is the symbolic representation of the conversion of substances to new substances.

  $\begin{array}{l}\text{ A + B }\stackrel{}{\to }\text{ C}\\ \text{Reactant }\stackrel{}{\to }\text{ Product}\end{array}$

In a chemical reaction; the substance which is involved in conversion is said to be reactant whereas the newly formed substance is known as a product. Both reactants and products must be separated by an arrow.

The oxidation reaction is the reaction that involves the addition of O atom in the presence of certain oxidizing agents such as ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}\text{O}$ and ${\text{KMnO}}_{\text{4}}$.

Answer to Problem 64P

  

Explanation of Solution

To get the oxidized product of any alcohol, three steps must be followed:

• Locate the C atom in the parent chain that is bonded with −OH group.
• Convert that C atom to carbonyl C atom or carboxylic acid as it is overall removal of H atoms.
• Primary alcohols are oxidized to aldehyde which further oxidized to the carboxylic acid.
• Secondary alcohols are oxidized to a ketone (R₂CO).
• Tertiary alcohols are not oxidized as they do not have H atom on the C with the −OH group.

Hence, the oxidized of cyclopentylmethanol will form cyclopentanecarboxylic acid molecule as cyclopentylmethanol is a primary alcohol.

  

Interpretation Introduction

(d)

Interpretation:

The oxidized product of following alcohol when oxidized with ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}\text{O}$ should be determined:

  

Concept Introduction:

A chemical reaction is the symbolic representation of the conversion of substances to new substances.

  $\begin{array}{l}\text{ A + B }\stackrel{}{\to }\text{ C}\\ \text{Reactant }\stackrel{}{\to }\text{ Product}\end{array}$

In a chemical reaction; the substance which is involved in conversion is said to be reactant whereas the newly formed substance is known as a product. Both reactants and products must be separated by an arrow.

Oxidation reaction is the reaction that involves the addition of O atom in the presence of certain oxidizing agents such as ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}\text{O}$ and ${\text{KMnO}}_{\text{4}}$.

Answer to Problem 64P

2-ethyl-3-pentanol cannot oxidize as it is a tertiary alcohol.

Explanation of Solution

To get the oxidized product of any alcohol, three steps must be followed;

• Locate the C atom in the parent chain that is bonded with −OH group.
• Convert that C atom to carbonyl C atom or carboxylic acid as it is the overall removal of H atoms.
• Primary alcohols are oxidized to aldehyde which further oxidized to the carboxylic acid.
• Secondary alcohol is oxidized to a ketone (R₂CO).
• Tertiary alcohols are not oxidized as they do not have H atom on the C with the −OH group.

Hence 2-ethyl-3-pentanol cannot oxidize as it is a tertiary alcohol.