Specific Reactions

Carboxylic acids undergo a rich variety of reactions at both the carboxyl group (–COOH) and at the alpha-carbon adjacent to it. Two reactions of particular importance at the senior level are the Hell–Volhard–Zelinsky (HVZ) halogenation and decarboxylation using soda-lime. These reactions illustrate how the carbonyl group activates nearby C–H bonds and how the carboxylate unit can be removed entirely to shorten a carbon chain. The HVZ reaction selectively places a halogen atom (usually Br or Cl) at the alpha-carbon of a carboxylic acid in the presence of a phosphorus trihalide catalyst. This alpha-haloacid is a versatile synthetic intermediate: it can be converted into alpha-amino acids, alpha-hydroxy acids, or unsaturated acids by simple substitution or elimination. Understanding this reaction requires knowledge of enols, acyl halides, and electrophilic halogenation. Decarboxylation with soda-lime (a mixture of NaOH and CaO) is the simplest way to remove a –COOH group from an organic molecule, producing an alkane with one fewer carbon. When a sodium or calcium carboxylate salt is heated with soda-lime, carbon dioxide is expelled and the remaining carbanion picks up a proton to give the alkane. This reaction is classically used to determine the structure of unknown acids and to synthesise lower-molecular-mass hydrocarbons.

Key takeaways
  • HVZ Reaction – Role of Phosphorus Trihalide
  • HVZ Reaction – Enolisation and Bromination
  • HVZ Reaction – Selectivity
  • Decarboxylation – Preparation and Mechanism
  • Decarboxylation – Bond Cleavage
  • Decarboxylation – Practical Observations

HVZ Reaction – Role of Phosphorus Trihalide

A carboxylic acid R–COOH does not form an enol directly under normal acidic conditions in the same way a ketone does. However, when treated with PBr₃ (or PCl₃), the acid is first converted in situ to an acyl bromide (R–COBr) via nucleophilic substitution at phosphorus. Acyl halides enolise far more readily than the free acid because the C=O of the acyl halide is more electrophilic, lowering the pKa of the alpha-C–H bond.

HVZ Reaction – Enolisation and Bromination

The acyl bromide tautomerises to its enol form: R–C(OH)=CH– (if R contains alpha-H). The electron-rich C=C double bond of the enol attacks Br₂ (or Cl₂) in an electrophilic addition, with the halogen acting as the electrophile. This places Br at the alpha-carbon. The intermediate alpha-halo acyl bromide is then hydrolysed by the excess carboxylic acid in the reaction mixture, regenerating an acyl bromide bond and producing the final alpha-bromo carboxylic acid (R–CHBr–COOH).

HVZ Reaction – Selectivity

The reaction is selective for the alpha-position because only the alpha-C–H bonds are activated by the adjacent carbonyl. Unlike direct free-radical halogenation, HVZ gives predominantly the alpha-monohalogenated product when one equivalent of Br₂ is used. With excess Br₂ and prolonged reaction, di- and tri-halogenated products can form. The product alpha-bromo acid is typically isolated after aqueous workup.

Decarboxylation – Preparation and Mechanism

The sodium salt of the carboxylic acid (RCOONa, prepared by neutralising RCOOH with NaOH) is mixed intimately with solid soda-lime (NaOH/CaO) and heated strongly in a test tube fitted with a delivery tube. CaO serves to absorb water and prevent hydrolysis of the product, while also hardening the NaOH to keep it free-flowing at high temperatures.

Decarboxylation – Bond Cleavage

Heat causes heterolytic cleavage of the C–C bond between the carbonyl carbon and the alpha-carbon. The carboxylate anion –COO⁻ loses CO₂ (expelled as a gas), generating a carbanion R⁻. This carbanion immediately abstracts a proton from the NaOH lattice, giving the alkane R–H. The overall transformation is: RCOONa + NaOH → R–H + Na₂CO₃. The product alkane has one fewer carbon atom than the original acid.

Decarboxylation – Practical Observations

The gaseous product (alkane) is collected over water or by downward displacement of air. Effervescence is observed as CO₂ is also produced (absorbed by excess NaOH to give Na₂CO₃). The residue is sodium carbonate. For example, sodium ethanoate gives methane: CH₃COONa + NaOH → CH₄ + Na₂CO₃. This is a classic laboratory preparation of methane.

The Key Equation

\ce{RCH_2COOH + Br_2 ->[$\text{(i) PBr}_3$][$\text{(ii) H}_2\text{O}$] RCHBrCOOH + HBr}
  • \ce{RCH_2COOH} — Carboxylic acid (with alpha-H) (–): The starting material; R is any alkyl or aryl group. The two alpha-hydrogen atoms on the CH₂ group are the ones susceptible to replacement.
  • \ce{Br_2} — Bromine (–): The halogenating agent. Acts as an electrophile towards the enol intermediate formed from the acyl bromide.
  • \ce{PBr_3} — Phosphorus tribromide (catalyst/reagent) (–): Converts the carboxylic acid to an acyl bromide in situ, greatly facilitating enolisation at the alpha-carbon. Regenerated at the end of the cycle.
  • \ce{RCHBrCOOH} — Alpha-bromo carboxylic acid (–): The product; bromine has been selectively placed at the carbon directly adjacent (alpha) to the carboxyl group.

The Key Equation

\ce{CH_3COONa + NaOH ->[$\Delta$, CaO] CH_4 ^ + Na_2CO_3}
  • \ce{CH_3COONa} — Sodium ethanoate (–): The sodium salt of acetic acid; the reactant that undergoes decarboxylation.
  • \ce{NaOH} — Sodium hydroxide (in soda-lime) (–): Provides the base and the proton source for the carbanion intermediate. Mixed with CaO to form soda-lime.
  • \ce{CH_4} — Methane (–): Gaseous product; one carbon fewer than the starting acid. Collected by gas displacement methods.
  • \ce{Na_2CO_3} — Sodium carbonate (–): Solid residue formed as CO₂ is absorbed by excess NaOH. Confirms that CO₂ was generated during the reaction.

Worked Example

Problem. Propanoic acid (CH₃CH₂COOH) is treated with one equivalent of Br₂ in the presence of a catalytic amount of PBr₃, followed by aqueous workup. (a) Identify all possible monobromination products and state which is formed. (b) Write the balanced equation for the HVZ product. (c) If instead the sodium salt of propanoic acid is heated with soda-lime, what organic product forms and how many carbons does it contain?

\ce{CH_3CH_2COOH + Br_2 ->[$\text{PBr}_3$] CH_3CHBrCOOH + HBr} \quad \text{and} \quad \ce{CH_3CH_2COONa + NaOH ->[$\Delta$] CH_3CH_3 + Na_2CO_3}

Answer. HVZ product: 2-bromopropanoic acid. Decarboxylation product: ethane (2 carbons).

Figure
Diagram
HVZ Mechanism and Decarboxylation SummaryHVZ Reaction Pathway① RCOOH + PBr₃→ RCOBr (acyl bromide)② RCOBr ⇌ RC(OH)=CH–(enolisation – fast)③ Enol + Br₂ → α-Br product→ RCHBrCOOHDecarboxylation Pathway① RCOONa + NaOH → heat / CaO (soda-lime)② RCOO⁻ → R⁻ + CO₂ ↑③ R⁻ + H⁺(NaOH) → R–H→ Alkane (n–1 carbons)Example: Sodium ethanoate → MethaneCH₃–COO⁻Na⁺+ NaOHΔ, CaOCH₄ ↑+ Na₂CO₃3 carbons → 1 carbon (–CO₂)Example: Propanoic acid → 2-Bromopropanoic acid (HVZ)CH₃CH₂COOH+ Br₂PBr₃ cat.CH₃CHBrCOOH+ HBrAlpha-carbon (C-2) selectively brominatedα
Origins · 1 minJacob Volhard and how this was discovered
JV
Jacob Volhard
1834–1910 · German

Jacob Volhard was a German organic chemist who, together with Georg Hell and Nikolai Zelinsky, developed the Hell–Volhard–Zelinsky (HVZ) reaction in the 1880s—a landmark method for selectively introducing a halogen atom at the alpha-carbon of a carboxylic acid. Volhard also made fundamental contributions to analytical chemistry, including the argentometric titration method for determining halide ions that still bears his name (Volhard's method). He was a professor at the University of Halle and later Munich, where he supervised numerous influential students. Volhard was a student of Justus von Liebig, one of the founding fathers of organic chemistry, making him part of a remarkable intellectual lineage in 19th-century German chemistry.

  • Introduction to Carboxylic Acids — Background knowledge on the structure, naming, and physical properties of carboxylic acids is essential before studying their specific reactions such as HVZ and decarboxylation.
  • Nucleophilic Acyl Substitution — The first step in the HVZ reaction converts the carboxylic acid to an acyl halide via nucleophilic substitution at the carbonyl carbon; understanding this mechanism clarifies why PBr₃ is necessary.
  • Keto–Enol Tautomerism — Enolisation of the acyl bromide intermediate is the critical step in HVZ; a solid understanding of keto–enol equilibria and the reactivity of enols as nucleophiles is directly required.
  • Synthesis of Alpha-Amino Acids — The alpha-bromo acid produced by HVZ is a direct precursor to alpha-amino acids via nucleophilic substitution with ammonia (Gabriel synthesis or direct aminolysis), linking HVZ to biochemistry.
  • Alkanes: Properties and Reactions — Decarboxylation with soda-lime is a classic synthesis of alkanes; understanding alkane structure helps predict the product (R–H) obtained when a carboxylate salt loses CO₂.