Question

Show how you would convert linoleic acid to the following fatty acid derivatives.

1. Octadecane-1-ol
2. Stearic acid
3. Octadecyl stearate
4. Nonanal
5. Nonanedioic acid
6. 9,10,12,13-tetrabromostearic acid

Short answer

Linoleic acid can be converted into the following fatty acid derivatives using certain reagents:

Step-by-step solution

Step-1. Explanation of part (a):

Linoleic acid on reaction with hydrogen and nickel catalyst undergoes hydrogenation and double bond in the structure of linoleic acid gets reduced. In next step, treatment with lithium aluminium hydride reduces the carboxylic acid group into alcoholic group and required product that is, octadecan-1-ol is obtained.

Step-2. Explanation of part (b):

Linoleic acid on hydrogenation reaction in presence of hydrogen and nickel catalyst forms the product, stearic acid. Double bond gets reduced in hydrogenation reaction. Stearic acid formed is a long-chain saturated fatty acid and known as octadecanoic acid. It is usually found in various plant and animal fats.

Step-3. Explanation of part (c):

Product formed from hydrogenation reaction of linoleic acid in part (b) and product formed from hydrogenation reaction followed by treatment with lithium aluminium hydride in part (a), reacts together to form product octadecyl stearate in the presence of acidic medium.

Step-4. Explanation of part (d):

Linoleic acid on reaction with ozone and dimethyl sulfide undergoes ozonolysis at double bond which leads to formation of products nonanal and 9-oxonanoic acid. This process is known as reductive ozonolysis and reaction stops at aldehydic stage.

Step-5. Explanation of part (e):

Linoleic acid on reaction with potassium permanganate undergoes oxidation at double bond and carboxylic acid forms at both ends of the double bond giving the required product, that is, nonanedioic acid and side product nonanoic acid.

Step-6. Explanation of part (f):

Linoleic acid on reaction with excess bromine and phosphorous tribromide followed by treatment with water, undergoes excess bromination at double bond and resultant product formed is 9,10,12,13-tetrabromostearic acid.

Formation of the product