Reasoning Chain — Step-by-Step Mechanistic Reasoning — Notes

Identify the question: Which is the stronger acid: HCOOH or CH3COOH?
Recall the definition: Stronger acid = lower pKa = more stable conjugate base.
Write conjugate bases: HCOOH → HCOO– (formate); CH3COOH → CH3COO– (acetate).
Identify structural difference: HCOO– has H attached to carboxyl C; CH3COO– has CH3.
Apply inductive effect: H has essentially zero inductive effect. CH3 has a +I (electron-donating) effect that pushes electrons toward the carboxylate oxygens.
Assess conjugate base stability: CH3COO– has MORE electron density on the carboxylate oxygens (due to +I of CH3) → LESS stable. HCOO– has no extra electron donation → MORE stable.
Conclusion: More stable HCOO– → HCOOH gives up proton more readily → HCOOH is a stronger acid. pKa(HCOOH) = 3.75 < pKa(CH3COOH) = 4.76. Confirmed.

Compare: RCOOH (pKa ~5) vs. ROH (pKa ~16–18).
Write conjugate bases: RCOO– vs. RO–.
Analyze RCOO–: Two resonance structures where negative charge is equally shared between two oxygens. Charge delocalized over two electronegative atoms → highly stable.
Analyze RO–: Charge localized on ONE oxygen. No resonance structures (single bond from C to O). Less stable.
Apply: More stable conjugate base → acid ionizes more readily → lower pKa.
Conclusion: RCOO– >> RO– in stability → RCOOH >> ROH in acidity (by ~11–13 pKa units = 10^11–10^13 times more acidic).

SOCl2 (thionyl chloride) is an electrophile at sulfur.
The –OH oxygen of RCOOH (nucleophile) attacks S of SOCl2. One HCl is released. Product: chlorosulfite ester intermediate R–C(=O)–O–SOCl.
The intermediate is unstable. Cl– (nucleophile) attacks the acyl carbon (electrophile) in an SN2-like step.
The –O–SOCl group leaves as the anion [OSCl]– → immediately releases SO2 + Cl–.
Cl– (from step 4) protonates H+... actually the second HCl was released in step 2. Final products: RCOCl + SO2↑ + HCl↑. Both byproducts escape as gases → reaction is driven to completion.