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Multiple Choice
Which of the following fatty acids would be the most unstable and most prone to rancidity?
A
Stearic acid (18:0)
B
Palmitic acid (16:0)
C
Linolenic acid (18:3, \(\omega\)-3)
D
Oleic acid (18:1, \(\omega\)-9)
Verified step by step guidance
1
Understand the concept of fatty acid stability: Fatty acid stability is influenced by the degree of unsaturation (number of double bonds). Saturated fatty acids (no double bonds) are more stable, while unsaturated fatty acids (with double bonds) are less stable due to their susceptibility to oxidation.
Analyze the fatty acids provided: Stearic acid (18:0) and Palmitic acid (16:0) are saturated fatty acids, meaning they have no double bonds. Oleic acid (18:1, ω-9) has one double bond, making it monounsaturated. Linolenic acid (18:3, ω-3) has three double bonds, making it polyunsaturated.
Explain the relationship between unsaturation and rancidity: The more double bonds a fatty acid has, the more prone it is to oxidation. Oxidation leads to the breakdown of fatty acids, causing rancidity. Polyunsaturated fatty acids are the most prone to rancidity due to their high number of double bonds.
Compare the fatty acids: Linolenic acid (18:3, ω-3) has the highest degree of unsaturation (three double bonds), making it the most unstable and prone to rancidity. Oleic acid (18:1, ω-9) is less prone to rancidity due to having only one double bond. Stearic acid (18:0) and Palmitic acid (16:0) are saturated and the most stable.
Conclude the reasoning: Based on the degree of unsaturation, Linolenic acid (18:3, ω-3) is the most unstable and prone to rancidity among the fatty acids listed.