Questions: Heteroatom Nucleophiles in Acyl Substitution

Question 1

A chemist attempts to convert an amide to an ester by treating it with excess ethanol under mild conditions. Will this proceed spontaneously?

• A. Yes — alcohols are good nucleophiles and will readily attack the amide carbonyl
• B. No — esters sit higher on the leaving-group ladder than amides, so the conversion is thermodynamically uphill without activation
• C. Yes — oxygen is more electronegative than nitrogen, making it a better leaving group
• D. No — the reaction would produce a thioester instead

Answer: (B)

The leaving-group ladder runs acid chlorides > anhydrides > thioesters > esters > amides. Converting an amide to an ester means going uphill — from a more stable derivative to a less stable one. This is thermodynamically unfavorable without activation. The reverse (ester → amide with excess amine) is spontaneous because it goes downhill. Option A tempts students who focus on nucleophile strength without thinking about thermodynamic directionality.

Question 2

In a protic solvent, which is the more reactive nucleophile toward an ester carbonyl: ethoxide (EtO⁻) or ethanethiolate (EtS⁻)?

• A. Ethoxide — oxygen is more electronegative and holds its lone pair more tightly for donation
• B. Ethanethiolate — sulfur is larger and more polarizable, lowering the activation energy for bond formation despite lower basicity
• C. Ethoxide — higher basicity always correlates with higher nucleophilicity
• D. They are equally reactive since both carry a negative charge

Answer: (B)

Nucleophilicity and basicity are not the same thing. In protic solvents, nucleophilicity tracks polarizability more than basicity: sulfur's large, diffuse electron cloud begins forming a bond at greater distance, lowering the activation energy. EtS⁻ is a weaker base than EtO⁻ (thiols have pKa ~10 vs. alcohols ~16) but a stronger nucleophile. This mirrors why iodide is a better SN2 nucleophile than fluoride despite being a weaker base.

Question 3

The tetrahedral intermediate formed during nucleophilic acyl substitution is a true reaction intermediate with a finite lifetime, not a transition state.

• T. True
• F. False

Answer: True

Unlike SN2, which proceeds through a single transition state with no intermediate, acyl substitution proceeds through a tetrahedral intermediate — a species at a local energy minimum on the reaction coordinate. It has bonds to both the incoming nucleophile and the outgoing leaving group simultaneously. Under some conditions it can even be trapped or observed. Confusing it with a transition state is a common misconception.

Question 4

Amides are poor substrates for nucleophilic acyl substitution primarily because the ammonium cation (NH₄⁺) released would be an unstable leaving group.

• T. True
• F. False

Answer: False

The poor reactivity of amides is not mainly about leaving-group stability of the released amine — it's about resistance to nucleophilic attack in the first place. Nitrogen's lone pair delocalizes into the carbonyl via resonance, giving the C–N bond partial double-bond character and reducing the carbonyl's electrophilicity. This raises the barrier to forming the tetrahedral intermediate. The leaving group argument would apply at the second step (collapse), but the rate-determining barrier is getting to the intermediate at all.

Question 5

Why does acetyl-CoA function as an activated acyl carrier in metabolism rather than a simple ester or amide? What feature of its position on the leaving-group ladder makes it suitable?

Model answer: Acetyl-CoA is a thioester, which sits in the middle of the leaving-group ladder — below acid chlorides and anhydrides (too reactive, would hydrolyze non-specifically) but above esters and amides. This intermediate reactivity makes it reactive enough to donate its acetyl group to oxygen nucleophiles (forming esters in lipid synthesis) and nitrogen nucleophiles (forming amides in protein acylation), while being stable enough not to react indiscriminately with water. The leaving-group ladder is the organizing principle: biology needs controlled reactivity, and thioesters occupy the sweet spot.

This question tests whether students can apply the leaving-group ladder beyond memorized examples to explain a biological design principle. The key insight is that 'activated' means precisely positioned on the ladder — reactive enough to be useful, stable enough to be controllable. Acid chlorides would be too reactive in the aqueous cellular environment; esters and amides are too stable to donate acyl groups efficiently without additional activation.