Answer
a) $H_{3}C-(CH_{2})_{11}-COO^{-}$ has a polar head ($COO^{-}$) and a singular non-polar tail ($H_{3}C-(CH_{2})_{11}$)., similar to the palmitic acid that composes the micelles. Thus, this molecule would align itself inside the micelle where the polar head is facing the water and the hydrophobic tail is facing the inside of the micelle
b) $H_{3}C-(CH_{2})_{11}-CH_{3}$ does not have a polar end. Since it is fully non-polar, it would be completely inside the hydrophobic core of the palmitic acid micelle.
Work Step by Step
Palmitic acid is a saturated fatty acid, its chemical formula is $H_{3}C-(CH_{2})_{14}-COOH$ . Fatty acids are amphipathic molecules, having both polar (COOH head) and non-polar (hydrocarbon chain) components. A micelle is a lipid molecule in aqueous solution, which has arranged itself in a spherical form in response to the polarity of the solution.
a) Water is a polar solvent. Polar ends of the fatty acids would align, so that they face the water on the periphery of the micelle. However, the nonpolar fatty acid chains would be present inside the micelle. $H_{3}C-(CH_{2})_{11}-COO^{-}$ would align itself inside the micelle, where its polar end ($COO^{-}$) would be facing the water along with the polar head of the palmitic acid micelle. The saturated fatty acid chain would be present inside the micelle unexposed to polar solvent.
b) $H_{3}C-(CH_{2})_{11}-CH_{3}$ does not have any polar end. This is completely saturated and is hydrophobic with no polar group. This would be present completely inside the micelle. This would not at all interact with water. A micelle is having a hydrophobic core, thus due to a hydrophobic moiety of the given molecule, it would be present inside the micelle.
