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Fats

>> Parts of this equation/concept include:

A -	Identifying Fatty Acids
B -	Esterification of Carboxylic Acids

Fatty acids have a long (more than six) chain of carbons ending with a carboxylic acid group. They can be classified based on the number of carbon–carbon double bonds as saturated (none), monounsaturated (one), or polyunsaturated (more than one).

>> Example 1

Identify the following as saturated, monounsaturated, or polyunsaturated fatty acids. ("None of the above" is also a possible answer.)

1. 
   
   
   
2. 
   
   
   
3. CH₃COOH
   
   
   
4. 

Solution:

1. This is a polyunsaturated fatty acid. It is a fatty acid, since it has the carboxylic acid group on the left end and it contains a long (10 carbon) carbon chain. It is polyunsaturated since it contains three double bonds between carbons.
2. This is a saturated fatty acid. It ends with a carboxylic acid group (on the left) and contains eight carbons beyond the COOH group. The carbons are not explicitly written in this structure but are implied where two line segments meet. If anything other than a CH₂ group (CH₃ at the end) was at that point, it would have been noted.
3. This is not a fatty acid. Although it has a carboxylic acid group, one carbon is not sufficient to give it hydrophobic (water-fearing) properties. Although there is no firm number that indicates a fatty acid, at least six carbons is usual.
4. This is a monounsaturated fatty acid. There is one (mono) carbon–carbon double bond. It has a long carbon chain ending (on the right) with a carboxylic acid group.

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Carboxylic acids react in a condensation reaction with alcohols to create an ester. The hydroxyl groups form water and an ether link with the carbonyl carbon and the carbon bonded to the alcohol group. An ester is the combination of the carbonyl and ether on the same carbon.

>> Example 2

Identify the ester group in the following molecules.

1. 
   
   
   
2. 
   
   
   
3. 

Solution:

1. This is not an ester. The ether carbon is not bonded to the carbonyl carbon. It is, however, a peptide bond, with the secondary amine bonded to the carbonyl carbon.

2. This is an ester. The ester group is circled below.

   
   
   
   

3. This is an ester. The ester group is circled below.

   

>> Example 3

What is the product (other than water) of the reaction between CH₃(CH₂)₁₀COOH and 1-propanol?

Solution:

1-propanol has the formula CH₃CH₂CH₂OH. Prop for three carbons, ol for the OH group, and 1 for its position on the end. The water can be formed by removing the OH of the carboxylic acid and the hydrogen of the alcohol. Thus the resulting ester has 11 carbons, then the carbonyl group, then the ether, then three more carbons. The structure is

In lipids the fatty acids react with glycerol.

Therefore three fatty acids and one glycerol are needed to create a lipid.

>> Example 4

What is the product of the reaction between glycerol and three dodecanoic acid molecules?

Solution:

Each dodecanoic acid forms an ester with an alcohol group of the glycerol. The dodec root means that the acid has a total of 12 carbons. The last carbon is a COOH group (oic acid). The an means that this fatty acid is saturated, with no double bonds. (Otherwise it would have been en.) Therefore the resulting structure is

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