MCAT® ExamChemical and Physical Foundations of Biological SystemsHard

A student is working with a solution containing a mixture of two organic compounds: benzoic acid (pKa ≈ 4.2) and aniline (pKb ≈ 9.4). They want to separate these two compounds using an acid-base extraction procedure. Which of the following steps would be most effective in separating benzoic acid from aniline when starting with an organic solvent layer?

  1. AAdd an aqueous solution of NaOH to convert both compounds into their conjugate bases.
  2. BAdd an aqueous solution of NaHCO₃ (sodium bicarbonate) to selectively deprotonate benzoic acid.
  3. CAdd an aqueous solution of HCl to convert both compounds into their conjugate acids.
  4. DAdd an aqueous solution of a strong base like KOH to selectively deprotonate aniline.
Show answer & explanation

Correct answer: B. Add an aqueous solution of NaHCO₃ (sodium bicarbonate) to selectively deprotonate benzoic acid.

Acid-base extraction relies on selectively converting one component of a mixture into its charged (ionic) form, which is soluble in the aqueous layer, while the other component remains uncharged and soluble in the organic layer. Benzoic acid is a carboxylic acid (weak acid, pKa 4.2). Aniline is a weak base (pKb 9.4, so its conjugate acid has pKa 14-9.4 = 4.6). To separate them, we need to selectively ionize one. Sodium bicarbonate (NaHCO₃) is a weak base (conjugate acid H₂CO₃ has pKa1 6.4, so NaHCO₃ acts as a base with pKb ≈ 7.6). Its conjugate acid (H₂CO₃) has a pKa (6.4) that is greater than benzoic acid's pKa (4.2) but less than aniline's conjugate acid pKa (4.6). Therefore, NaHCO₃ is strong enough to deprotonate benzoic acid (making it water-soluble as benzoate) but not strong enough to protonate aniline significantly to make it water-soluble. This allows for selective separation. Options A and B would likely convert both or neither effectively for selective separation. Option D would not work as aniline is a base and would need an acid to be protonated.

Why the other options are wrong

  • A. NaOH is a strong base and would deprotonate both benzoic acid and potentially deprotonate aniline's conjugate acid if aniline was already protonated, making the separation non-selective.
  • C. HCl is a strong acid and would protonate both aniline (making it water-soluble) and keep benzoic acid uncharged, but this doesn't help separate them if both become aqueous or if the goal is to target benzoic acid specifically with a weaker base.
  • D. KOH is a strong base. It would deprotonate benzoic acid, but it would not selectively deprotonate aniline (aniline is already a base, not an acid to be deprotonated by KOH).

Acid-Base Extraction Principles

Acid-base extraction is a technique used to separate organic compounds based on their differential solubility in aqueous solutions at different pH values. By converting a compound into its charged (ionic) form, it becomes water-soluble and can be extracted from an organic layer.

  • Acids are extracted with bases (e.g., NaHCO₃, NaOH).
  • Bases are extracted with acids (e.g., HCl).
  • The choice of acid/base depends on the pKa/pKb values.
  • Weak acids/bases allow for selective extraction.

Memory trick: Ionize One, Leave One Organic.

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