In an acid/base extraction, the “right” base is chosen by comparing acid strengths (pKa values) and using the rule: a base can effectively deprotonate an acid only if the conjugate acid of the base is weaker (higher pKa) than the acid you’re trying to deprotonate.

Below is a general guide you can use to decide which base will work, even if your specific structures weren’t shown.

Key idea: pKa comparison

  • Every acid has a pKa value; lower pKa = stronger acid.
  • A proton transfer Acid1+Base2⇌Base1+Acid2\text{Acid}_1+\text{Base}_2\rightleftharpoons \text{Base}_1+\text{Acid}_2Acid1​+Base2​⇌Base1​+Acid2​ is favorable toward the weaker acid (the one with higher pKa).
  • So:
    • If pKa(your acid)<pKa(conjugate acid of base)\text{pKa}(\text{your acid})<\text{pKa}(\text{conjugate acid of base})pKa(your acid)<pKa(conjugate acid of base), deprotonation is favorable (base works).
    • If pKa(your acid)>pKa(conjugate acid of base)\text{pKa}(\text{your acid})>\text{pKa}(\text{conjugate acid of base})pKa(your acid)>pKa(conjugate acid of base), deprotonation is not favorable (base does not work).

Common organic acids and typical pKa ranges

Use this as a quick mental “pKa map” for extraction problems.

  • Strong mineral acids (HCl, HBr, H2SO4): pKa < 0
  • Carboxylic acids (e.g., benzoic acid, acetic acid): pKa ≈ 4–5
  • Protonated amines (RNH3⁺): pKa ≈ 9–11
  • Phenols (Ar–OH): pKa ≈ 10
  • Aliphatic alcohols (ROH): pKa ≈ 16–18
  • Terminal alkynes (RC≡CH): pKa ≈ 25
  • Alkanes (sp³ C–H): pKa ≈ 50

Common bases in acid/base extraction

In typical lab extractions you mainly see these aqueous bases:

  • Sodium bicarbonate, NaHCO₃
    • Base: HCO₃⁻; conjugate acid: H₂CO₃ (pKa₁ ≈ 6.3).
    • Can deprotonate carboxylic acids (pKa ≈ 4–5) but not phenols (pKa ≈ 10) or alcohols.
  • Sodium hydroxide, NaOH
    • Base: OH⁻; conjugate acid: H₂O (pKa ≈ 15.7).
    • Can deprotonate carboxylic acids and phenols , but still not most alcohols or very weak acids like alkanes.
  • Very strong non‑aqueous bases (LDA, NaH, alkoxides other than OH⁻)
    • Used for deprotonating very weak acids (e.g., terminal alkynes, certain C–H bonds), but these are usually not the ones used in standard aqueous extractions.

How this looks in the typical question

Many textbook/MCAT-style problems ask:

“Determine which base will work to deprotonate each compound in an acid/base extraction. Choose: only hydroxide, hydroxide or bicarbonate, only bicarbonate, or neither.”

The usual logic (with NaHCO₃ vs NaOH) is:

  1. Carboxylic acids (e.g., benzoic acid, acetic acid)
    • pKa ≈ 4–5, stronger than carbonic acid but much weaker than water.
 * **Both NaHCO₃ and NaOH will deprotonate them** → “Hydroxide or bicarbonate”.

 * In practice, NaHCO₃ is preferred because it’s milder and avoids side reactions.
  1. Phenols (e.g., phenol, p‑cresol)
    • pKa ≈ 10, weaker acid than carbonic acid, but stronger than water.
 * HCO₃⁻ is **not** strong enough (reaction very unfavorable).
 * OH⁻ (conjugate acid H₂O, pKa ≈ 15.7) **is** strong enough.
 * Answer → “Only hydroxide”.
  1. Neutral alcohols (ROH) like ethanol, cyclohexanol
    • pKa ≈ 16–18, weaker acids than water (pKa 15.7).
 * Neither bicarbonate nor hydroxide will significantly deprotonate them in aqueous extraction conditions.
 * Answer → “Neither hydroxide nor bicarbonate”.
  1. Extremely weak acids (alkanes, most simple C–H bonds)
    • pKa ≈ 40–50.
    • No common aqueous base (HCO₃⁻ or OH⁻) can deprotonate them.
    • Answer → “Neither hydroxide nor bicarbonate”.
  1. Very strong acids (like mineral acids or sulfonic acids) sometimes drawn in organic form
    • Much more acidic than carboxylic acids.
    • Even bicarbonate easily deprotonates them, and often you do not need NaOH in a typical extraction context.

Typical extraction strategy using these bases

In a classic lab mixture of an organic carboxylic acid , a phenol , and a neutral compound , you often see:

  • Step 1: Wash with NaHCO₃
    • Deprotonates carboxylic acid → water‑soluble carboxylate salt (moves to aqueous layer).
    • Leaves phenol and neutral compound in the organic layer.
  • Step 2: Wash organic layer with NaOH
    • Deprotonates phenol → phenoxide salt (now moves to aqueous layer).
    • Neutral compound remains in the organic layer.

This strategy uses the fact that bicarbonate is just strong enough for the stronger organic acids (carboxylic acids), while hydroxide is needed for the weaker acids (phenols).

How to apply this to your specific question

Since the diagrams of your compounds are not visible here, you can solve each one by:

  1. Identify the acidic site:
    • Carboxylic acid group: –CO₂H
    • Phenol: aromatic ring–OH
    • Alcohol: aliphatic –OH
    • Other unusual acidic protons.
  2. Estimate its pKa range using the map above.
  3. Compare to the conjugate acids:
    • H₂CO₃ (pKa ≈ 6.3) for NaHCO₃.
    • H₂O (pKa ≈ 15.7) for NaOH.
  4. Decide:
    • If pKa(acid) ≈ 4–5 → “Hydroxide or bicarbonate”.
    • If pKa(acid) ≈ 8–11 → “Only hydroxide”.
    • If pKa(acid) ≥ 15–16 → “Neither hydroxide nor bicarbonate”.

TL;DR:

  • Carboxylic acids → deprotonated by NaHCO₃ or NaOH.
  • Phenols → deprotonated only by NaOH.
  • Ordinary alcohols and most C–H bonds → deprotonated by neither in an aqueous extraction.
    Use pKa comparisons to justify each choice.

Information gathered from public forums or data available on the internet and portrayed here.