Question

Match List I with List II

List I (Name of the Fatty Acid) & List II (Type of the fatty acid)
(A) Oleic acid & (I) $\omega$-3
(B) Petroselenic acid & (II) $\omega$-6
(C) Gamma linolenic acid & (III) $\omega$-12
(D) Eicosapentaenoic acid & (IV) $\omega$-9

• (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
• (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
• (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
• (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Hint:

To calculate the $\omega$ position of any fatty acid, subtract the highest $\Delta$ position of the double bond from the total number of carbon atoms in the chain.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Fatty acids are aliphatic carboxylic acids classified based on their carbon chain length, the number of double bonds, and the position of these double bonds.
The omega ($\omega$) notation identifies the position of the first double bond relative to the methyl ($\omega$) carbon terminal of the fatty acid chain.

Step 2: Detailed Explanation:
Let us analyze each fatty acid individually to establish the correct matches:
1. Oleic acid: This is a monounsaturated fatty acid containing 18 carbons with a single double bond at carbon 9 from the carboxyl end, denoted as 18:1 ($\Delta^9$).
Counting from the methyl terminal, the double bond is located at the 9th carbon ($18 - 9 = 9$). Thus, it is an $\omega$-9 fatty acid. (A) matches with (IV).
2. Petroselenic acid: This is an isomer of oleic acid containing 18 carbons with a single double bond at carbon 6 from the carboxyl end, denoted as 18:1 ($\Delta^6$).
Counting from the methyl terminal, the double bond is located at the 12th carbon ($18 - 6 = 12$). Thus, it is an $\omega$-12 fatty acid. (B) matches with (III).
3. Gamma linolenic acid (GLA): This is a polyunsaturated fatty acid with 18 carbons and three double bonds located at positions 6, 9, and 12 from the carboxyl end, denoted as 18:3 ($\Delta^{6,9,12}$).
The first double bond from the methyl end is at the 6th carbon ($18 - 12 = 6$). Thus, it is an $\omega$-6 fatty acid. (C) matches with (II).
4. Eicosapentaenoic acid (EPA): This is a polyunsaturated fatty acid with 20 carbons and five double bonds at positions 5, 8, 11, 14, and 17 from the carboxyl end, denoted as 20:5 ($\Delta^{5,8,11,14,17}$).
The first double bond from the methyl end is at the 3rd carbon ($20 - 17 = 3$). Thus, it is an $\omega$-3 fatty acid. (D) matches with (I).
Combining these matches, the correct sequence is (A) - (IV), (B) - (III), (C) - (II), (D) - (I).

Step 3: Final Answer:
The matching sequence is (A) - (IV), (B) - (III), (C) - (II), (D) - (I), which corresponds to option (D).