Answer: D) Br₂/CCl₄
Step-by-step explanation:
To distinguish between the two compounds—CH₃(CH₂)₁₀CO₂H, which is a saturated fatty acid, and CH₃(CH₂)₄CH=CH(CH₂)₄CO₂H, which contains an unsaturated carbon-carbon double bond—one needs to choose a reagent that reacts differently with saturated and unsaturated compounds.
Here's a brief analysis of the reagents listed:
A) NaOH, H2O – Sodium hydroxide in water would deprotonate the carboxylic acid group in both molecules, forming a carboxylate salt. This reaction would not help distinguish between the two.
B) Ag(NH₃)₂⁺ – This is Tollens' reagent, which is used to oxidize aldehydes to carboxylic acids. Neither of the compounds in question is an aldehyde, so this reagent would not be useful for differentiation.
C) H₂Cr₂O₇ – This is an oxidizing agent that can oxidize primary alcohols to carboxylic acids and secondary alcohols to ketones. However, neither compound has the requisite alcohol groups for this reaction, so it would not distinguish between them.
D) Br₂/CCl₄ – Bromine in carbon tetrachloride is a test for unsaturation; it will react with compounds that have carbon-carbon double bonds (alkenes) by adding across the double bond. This reaction would cause a visible decolorization of the bromine solution for the unsaturated compound, while the saturated compound would not react.
E) NH₃ – Ammonia would react with the carboxylic acid to form an ammonium salt, but this reaction would occur with both compounds and would not distinguish between them.
Based on this analysis, the best reagent to distinguish between the two compounds is:
D) Br₂/CCl₄
This reagent would react with the unsaturated fatty acid by adding bromine across the double bond, leading to a loss of the characteristic brown color of bromine. This would not occur with the saturated fatty acid. Therefore, D) Br₂/CCl₄ is the correct choice.