Chapter 19, Problem 33P

Interpretation Introduction

(a)

Interpretation:

The skeletal structure of the given fatty acid should be drawn and cis-double bond is to be indicated.

Concept Introduction:

Fatty acids are long-chain carboxylic acid, which may or may not have unsaturation in the molecule. They react with glycerol to form triglycerides. The reaction is known as esterification as a carboxylic acid group reacts with the alcoholic group to form an ester group.

Unsaturated fatty acids can be defined as the long-chain fatty acids which have a long hydrocarbon chain with −COOH group. In unsaturated fatty acids, the carbon chain must have at least one double bond.

Answer to Problem 33P

  

Explanation of Solution

Given:

  

Fatty acids are long-chain carboxylic acids, which consist of two parts; long hydrocarbon chains and polar −COOH group. The long hydrocarbon chain consists of carbon and H atoms, so it has C-C and C-H bonds only, whereas, the polar −COOH group has polar C=O and C-O bonds. The unsaturated fatty acids have cis and trans-double bonds in the hydrocarbon chain.

In the ball-and-stick model, the red ball represents O atom, black ball represents C atom and white ball represents H atoms. Thus, the skeletal structure is:

  

Interpretation Introduction

(b)

Interpretation:

The omega-n designation for the ${\text{CH}}_{\text{3}}\text{-CH}\text{-}{\text{CH=CH-CH}}_{\text{2}}\text{-CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{COOH}$ fatty acid should be determined.

Concept Introduction:

Fatty acids are long-chain carboxylic acid, which may or may not have unsaturation in the molecule. They react with glycerol to form triglycerides. The reaction is called esterification as a carboxylic acid group reacts with the alcoholic group to form an ester group.

Unsaturated fatty acids can be defined as the long-chain fatty acids which have a long hydrocarbon chain with −COOH group. In unsaturated fatty acids, the carbon chain must have at least one double bond.

Answer to Problem 33P

The ${\text{CH}}_{\text{3}}\text{-CH}\text{-}{\text{CH=CH-CH}}_{\text{2}}\text{-CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{COOH}$ fatty acid is an omega-3 fatty acid.

Explanation of Solution

Fatty acids are long-chain carboxylic acids, which consist of two parts; long hydrocarbon chains and polar −COOH group. If there is at least one double bond in the long hydrocarbon chain of the fatty acid it is said to be an unsaturated fatty acid.

The unsaturated fatty acids can also classify as omega-n acids. Here, 'n' represents the position of the first $\text{C=C}$ bond in the molecule from the end of ${\text{–CH}}_{\text{3}}$ group of molecule.

In the given fatty acid, the first $\text{C=C}$ bond in the molecule from the end of ${\text{–CH}}_{\text{3}}$ group of molecule is at C-3 position.Hence,the designation must be omega-3 fatty acid.

  

Interpretation Introduction

(c)

Interpretation:

The stereoisomer of the ${\text{CH}}_{\text{3}}\text{-CH}\text{-}{\text{CH=CH-CH}}_{\text{2}}\text{-CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{COOH}$ fatty acid should be drawn.

Concept Introduction:

Fatty acids are long-chain carboxylic acid, which may or may not have unsaturation in the molecule. They react with glycerol to form triglycerides. The reaction is called esterification as a carboxylic acid group reacts with the alcoholic group to form an ester group.

Unsaturated fatty acids can be defined as the long-chain fatty acids which have a long hydrocarbon chain with −COOH group. In unsaturated fatty acids, the carbon chain must have at least one double bond.

Answer to Problem 33P

The stereoisomer of cis-fatty acid must be trans-isomer as given below;

  

Explanation of Solution

Fatty acids are long-chain carboxylic acids, which consist of two parts; long hydrocarbon chains and polar −COOH group. The double bond can arrange in two ways, i.e., cis and trans in the hydrocarbon chain of the fatty acid molecule. Hence, the stereoisomer of cis-fatty acid ${\text{CH}}_{\text{3}}\text{-CH}\text{-}{\text{CH=CH-CH}}_{\text{2}}\text{-CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{COOH}$ must be trans-fatty acid.

  

Interpretation Introduction

(d)

Interpretation:

The structure of wax formed by the reaction of ${\text{CH}}_{\text{3}}\text{-CH}\text{-}{\text{CH=CH-CH}}_{\text{2}}\text{-CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{COOH}$ and ${\text{CH}}_{\text{3}}\text{-(CH}\text{)}{}_{\text{10}}\text{-OH}$ should be drawn.

Concept Introduction:

Fatty acids are long-chain carboxylic acid, which may or may not have unsaturation in the molecule. They react with glycerol to form triglycerides. The reaction is called esterification as a carboxylic acid group reacts with the alcoholic group to form an ester group.

Unsaturated fatty acids can be defined as the long-chain fatty acids which have a long hydrocarbon chain with −COOH group. In unsaturated fatty acids, the carbon chain must have at least one double bond.

Answer to Problem 33P

The skeleton formula of wax formed by the reaction of ${\text{CH}}_{\text{3}}\text{-CH}\text{-}{\text{CH=CH-CH}}_{\text{2}}\text{-CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{COOH}$ and ${\text{CH}}_{\text{3}}\text{-(CH}\text{)}{}_{\text{10}}\text{-OH}$ must be ${\text{CH}}_{\text{3}}\text{-CH}\text{-}{\text{CH=CH-(CH}}_{\text{2}}{\text{)}}_{\text{5}}\text{-COO-(CH}\text{)}{}_{\text{10}}{\text{-CH}}_{\text{3}}$.

Explanation of Solution

Fatty acids are long-chain carboxylic acids, which consist of two parts; long hydrocarbon chains and polar −COOH group. Waxes are a good example of hydrolyzable lipids, which are composed of fatty acid and higher alcohols. They have an ester functional group which is formed between $\text{–COOH and –OH}$ functional group of the fatty acid and alcohols respectively. The general reaction of wax formation of fatty acid and higher alcohols can be written as:

  ${\text{R-CH}}_{\text{2}}{\text{-COOH + HO-CH}}_{\text{2}}\text{-R }\stackrel{}{\to }{\text{R-CH}}_{\text{2}}{\text{-COO-CH}}_{\text{2}}{\text{-R + H}}_{\text{2}}\text{O}$

Hence, the reaction of ${\text{CH}}_{\text{3}}\text{-CH}\text{-}{\text{CH=CH-CH}}_{\text{2}}\text{-CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{COOH}$ and ${\text{CH}}_{\text{3}}\text{-(CH}\text{)}{}_{\text{10}}\text{-OH}$ must be written as:

  $\begin{array}{l}{\text{CH}}_{\text{3}}\text{-CH}\text{-}{\text{CH=CH-CH}}_{\text{2}}\text{-CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}\text{CH}\text{-}{\text{COOH + CH}}_{\text{3}}\text{-(CH}\text{)}{}_{\text{10}}\text{-OH }\\ \downarrow \\ {\text{CH}}_{\text{3}}\text{-CH}\text{-}{\text{CH=CH-(CH}}_{\text{2}}{\text{)}}_{\text{5}}\text{-COO-(CH}\text{)}{}_{\text{10}}{\text{-CH}}_{\text{3}}{\text{ + H}}_{\text{2}}\text{O}\end{array}$