Q29-3P - Organic Chemistry (Mcmurry)

Question

PROBLEM 29-3 How many molecules of acetyl $\mathbf{CoA}$ are produced by catabolism of the following fatty acids, and how many passages of the $\mathbf{\beta }\mathbf{-}$oxidation pathway are needed?

(a) Palmitic acid, ${\mathbf{CH}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{2}}\mathbf{\right)}}_{\mathbf{14}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}$.

(b) Arachidic acid, ${\mathbf{CH}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{2}}\mathbf{\right)}}_{\mathbf{18}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}$.

Step-by-Step Solution

Answer

The number of passages of the oxidation pathway was needed,

1. For Palmitic acid, ${CH}_3{\left({\mathrm{CH}}_2\right)}_{14}{\mathrm{CO}}_2\mathrm{H}$ ,7 passages and 8 acetyl COA;
2. For Arachidic acid, ${CH}_3{\left({\mathrm{CH}}_2\right)}_{18}{\mathrm{CO}}_2\mathrm{H}$ ,9 passages and 10 acetyl COA;

Introduction to the Concept

As a result of catabolic breakdown, $\mathbf{\beta }\mathbf{-}$oxidation of fatty acids damages the carbon chain. Because most fatty acids have an even number of carbon atoms, following $\mathbf{\beta }\mathbf{-}$oxidation, none are left over. The number of acetyl COA molecules generated as the principal end product is n/2 and the number of breakdown passages is n/2-1 for the given reaction when mere is an n-carbon fatty acid chain.

Solution Explanation

1. Palmitic acid is a fatty acid with 16 carbons. The number of passages involved is $8-1=7$, while the number of acetyl COA generated is $\frac{16}{2}=8$.
2. Similarly, the total number of acetyl COA generated by the 20 carbon fatty acid Arachidic acid is $\frac{20}{2}=10$.

The total number of passages is $10-1=9$.

The quantity of acetyl A molecules generated and the breakdown sequences involved are determined by the length of the fatty acid chain undergoing catabolic breakdown by $\mathbf{\beta }\mathbf{-}$oxidation.

The number of acetyl COA molecules in an n-carbon chain is n/2, while the number of passages is n/2-1.