DAT Organic Chemistry: Choosing and Using Bases

DAT High-Yield Concept

Base Selection Rule: To achieve complete deprotonation, the conjugate acid of the base must have a higher $\text{p}K_{\text{a}}$ (weaker acid) than the target acidic proton.
Terminal Alkyne Deprotonation: Hydroxide ( $\ce{OH^-}$ ) and alkoxides ( $\ce{RO^-}$ ) are not strong enough to deprotonate terminal alkynes ( $\text{p}K_{\text{a}} \approx 25$ ); stronger bases like amide ( $\ce{NH2^-}$ ) or hydride ( $\ce{H^-}$ ) are required.
Selective Deprotonation: In polyfunctional molecules, a weaker base selectively removes the most acidic proton (lowest $\text{p}K_{\text{a}}$ ) without deprotonating less acidic functional groups.

Fundamentals of Base Selection

In organic synthesis, selecting an appropriate base depends directly on the relative acidities of the target proton and the conjugate acid formed upon deprotonation. Brønsted-Lowry acid-base equilibrium favors the side with the weaker acid (higher $\text{p}K_{\text{a}}$ ) and weaker base (equilibrium position calculations are covered in depth in their own lesson). To fully deprotonate a functional group, the base's conjugate acid should typically have a $\text{p}K_{\text{a}}$ at least 4–5 units higher than that of the substrate.

PKa of Alcohols Values are taken from Evans pKa Table http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf (Source: Mfomich · CC0 · Wikimedia Commons)

For example, acetic acid ( $\text{p}K_{\text{a}} \approx 4.75$ ) reacts with hydroxide ( $\ce{OH^-}$ ) to form acetate and water ( $\text{p}K_{\text{a}} = 15.7$ ). Because water is a much weaker acid than acetic acid, the reaction proceeds quantitatively forward.

Deprotonation of acetic acid by a hydroxide ion (Source: loltim · CC BY-SA 4.0 · Wikimedia Commons)

Conversely, attempting to deprotonate ethyne ( $\ce{HC \equiv CH}$ , $\text{p}K_{\text{a}} \approx 25$ ) with hydroxide yields water ( $\text{p}K_{\text{a}} = 15.7$ ). Because water is a stronger acid than ethyne, hydroxide cannot effectively deprotonate a terminal alkyne. Instead, a stronger base like sodium amide ( $\ce{NaNH2}$ ), whose conjugate acid ammonia ( $\ce{NH3}$ ) has a $\text{p}K_{\text{a}} \approx 38$ , must be employed.

Hydroxide is not strong enough; amide is — Picking a base is a two-line calculation, not a feel. Write down the pKa of the proton you want to remove, then the pKa of your base's conjugate acid, and take the base only if its conjugate acid is the weaker one. Hydroxide fails on a terminal alkyne because water at 15.7 is nine pKa units more acidic than the alkyne, so the equilibrium sits on the wrong side by about a billion to one. Sodium amide works because ammonia at 38 is far weaker, which is exactly why NaNH2 is the standard reagent for making acetylides.

Base Selection Strategy and $\text{p}K_{\text{a}}$ Thresholds

Base	Conjugate Acid	Conjugate Acid $\text{p}K_{\text{a}}$ `	Suitable Target Acids ( $\text{p}K_{\text{a}}$ Threshold)
Carboxylate ( $\ce{RCOO^-}$ )	$\ce{RCOOH}$	~4.7	Strong mineral acids ( $\text{p}K_{\text{a}} < 0$ )
Bicarbonate ( $\ce{HCO3^-}$ )	$\ce{H2CO3}$	~6.4	Carboxylic acids ( $\text{p}K_{\text{a}} \approx 4-5$ )
Hydroxide / Alkoxide ( $\ce{OH^-}$ / $\ce{RO^-}$ )	$\ce{H2O}$ / $\ce{ROH}$	~15.7 – 16	Phenols ( $\text{p}K_{\text{a}} \approx 10$ ), $\beta$ -dicarbonyls ( $\text{p}K_{\text{a}} \approx 9-11$ )
Amide / Hydride ( $\ce{NH2^-}$ / $\ce{H^-}$ )	$\ce{NH3}$ / $\ce{H2}$	~35 – 38	Terminal alkynes ( $\text{p}K_{\text{a}} \approx 25$ ), esters/ketones ( $\alpha$ -protons, $\text{p}K_{\text{a}} \approx 19-25$ )

DAT Mnemonic: "SWAAP" — Strongest Weak Acid Always Protonates first! The most acidic proton is always removed first by any added base.

Worked Example 1: Selecting a Base for Terminal Alkynes

Problem: Determine whether sodium hydroxide ( $\ce{NaOH}$ ) or sodium amide ( $\ce{NaNH2}$ ) is suitable for converting ethyne into an acetylide ion for subsequent nucleophilic substitution.

Reasoning: Ethyne loses an $\text{sp}$ -hybridized proton ( $\text{p}K_{\text{a}} \approx 25$ ). Reaction with hydroxide gives: $\ce{HC \equiv CH + OH^- <=> HC \equiv C^- + H2O}$ Water ( $\text{p}K_{\text{a}} = 15.7$ ) is more acidic than ethyne ( $\text{p}K_{\text{a}} = 25$ ), so equilibrium strongly favors reactants. Reaction with amide gives: $\ce{HC \equiv CH + NH2^- -> HC \equiv C^- + NH3}$ Ammonia ( $\text{p}K_{\text{a}} \approx 38$ ) is a much weaker acid than ethyne, driving complete deprotonation.

Conclusion: $\ce{NaNH2}$ must be used.

Exam Trap: DAT questions frequently offer $\ce{NaOH}$ as a distractor for alkyne deprotonation. Remember that alkoxides and hydroxide cannot deprotonate terminal alkynes.

Selective Deprotonation in Polyfunctional Molecules

When a single molecule contains multiple acidic functional groups, chemoselectivity can be achieved by carefully choosing base strength. A weak base will selectively deprotonate only the most acidic site (lowest $\text{p}K_{\text{a}}$ ), leaving less acidic sites intact.

Chemoselectivity in Acid-Base Chemistry: To deprotonate group A (lower

\text{p}K_{\text{a}}

) without deprotonating group B (higher

\text{p}K_{\text{a}}

), choose a base whose conjugate acid has a

\text{p}K_{\text{a}}

between that of group A and group B.

Worked Example 2: Differentiating Carboxylic Acids and Phenols

Problem: 4-Hydroxybenzoic acid contains both a carboxylic acid group ( $\text{p}K_{\text{a}} \approx 4.5$ ) and a phenolic hydroxyl group ( $\text{p}K_{\text{a}} \approx 10$ ). Design conditions to selectively deprotonate the carboxylic acid.

Reasoning: To deprotonate only the carboxylic acid ( $\text{p}K_{\text{a}} \approx 4.5$ ) and not the phenol ( $\text{p}K_{\text{a}} \approx 10$ ), select a base with a conjugate acid $\text{p}K_{\text{a}}$ between 4.5 and 10, such as sodium bicarbonate ( $\ce{NaHCO3}$ ), whose conjugate acid carbonic acid has $\text{p}K_{\text{a}} \approx 6.4$ : $\ce{HO-C6H4-COOH + HCO3^- -> HO-C6H4-COO^- + H2O + CO2}$ Bicarbonate is strong enough to deprotonate the carboxylic acid but too weak to deprotonate the phenol ( $\ce{HO-C6H4-COO^- + HCO3^- <=-> ^-O-C6H4-COO^- + H2CO3}$ is unfavorable). Using a stronger base like $\ce{NaOH}$ would deprotonate both groups, forming a dicationic/dianionic species.

Conclusion: Use 1 equivalent of $\ce{NaHCO3}$ for selective carboxylate formation.

Critical Thinking Questions

Why is sodium hydride ( $\ce{NaH}$ ) commonly chosen over sodium hydroxide ( $\ce{NaOH}$ ) when preparing alkoxides from alcohols in high yield without equilibrium reversibility?
Suppose a synthetic target contains both a primary alcohol ( $\text{p}K_{\text{a}} \approx 16$ ) and a terminal alkyne ( $\text{p}K_{\text{a}} \approx 25$ ). Which functional group is deprotonated first upon addition of 1 equivalent of $\ce{NaNH2}$ ?
Explain why sodium bicarbonate ( $\ce{NaHCO3}$ ) solution can be used in an extraction liquid-liquid separator to separate a mixture of benzoic acid and phenol.

References

Sources returned by Google Search grounding during lesson verification.

How to Choose an Acid or a Base to Protonate or Deprotonate a Given Compound - Chemistry Steps — chemistrysteps.com
Acid-Base Reactions: Introducing Ka and pKa - Master Organic Chemistry — masterorganicchemistry.com

Choosing and Using Bases

Summary

Practice Questions

AI Tutor