Question

Match List I with List II
List I (Name of the fatty acid):
A. Ricinoleic acid
B. Linolenic acid
C. Palmitoleic acid
D. Erucic acid

List II (Type of the fatty acid):
I. $\omega$-3
II. $\omega$-6
III. $\omega$-7
IV. $\omega$-9
V. Hydroxylated

Choose the correct answer from the options given below:

• A. A - V, B - I, C - III, D - IV
• B. A - V, B - II, C - I, D - IV
• C. A - III, B - I, C - V, D - II
• D. A - IV, B - II, C - III, D - I

Hint:

To determine the omega ($\omega$) class of a fatty acid:
Subtract the $\Delta$ position of the final double bond from the total number of carbon atoms.
For example, for Erucic acid: $22 - 13 = 9$, which makes it an $\omega$-9 fatty acid.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Fatty acids are classified based on carbon chain length, the number and position of double bonds, and the presence of specific functional groups.
In the omega ($\omega$) nomenclature system, unsaturated fatty acids are classified by the position of the first double bond relative to the methyl (terminal) carbon of the chain.

Step 2: Detailed Explanation:
Let us match the fatty acids in List I with their corresponding structural types in List II:
1. Ricinoleic acid (A): Chemically known as 12-hydroxy-9-octadecenoic acid.
It is an 18-carbon unsaturated fatty acid that contains a hydroxyl group ($-\text{OH}$) on the twelfth carbon, making it a Hydroxylated fatty acid (V). Thus, A matches with V.
2. Linolenic acid (B): Specifically alpha-linolenic acid (ALA, 18:3 $\Delta^{9,12,15}$).
The first double bond is located at the third carbon from the terminal methyl end ($18 - 15 = 3$), classifying it as an $\omega$-3 fatty acid (I). Thus, B matches with I.
3. Palmitoleic acid (C): A 16-carbon monounsaturated fatty acid (16:1 $\Delta^9$).
The double bond is located seven carbons away from the terminal methyl end ($16 - 9 = 7$), making it an $\omega$-7 fatty acid (III). Thus, C matches with III.
4. Erucic acid (D): A 22-carbon monounsaturated fatty acid (22:1 $\Delta^{13}$).
The double bond is located nine carbons away from the terminal methyl end ($22 - 13 = 9$), classifying it as an $\omega$-9 fatty acid (IV). Thus, D matches with IV.
This matches the sequence: A - V, B - I, C - III, D - IV.

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