(iii) in the following conversions : CH3CH2CH2-OH is converted to CH3-CH2-COOH using reagent
(i) ? followed by
(ii) H3O+; CH3-CH2-COOH is further converted, using reagent
(iii) ? followed by
(iv) H2O, to CH3-CH(Cl)-COOH (α-Chloropropanoic acid). Which of the products, propanoic acid or α-Chloropropanoic acid has more acidic character ? State the reason.
Goa GbshseGBSHSE Class 12 Board Exam 2024Subjective· 2mImportance★★★★★
The Intuition: Why Carboxylic Acids Are Hard to Halogenate
Imagine you want to attach a halogen atom (like bromine) to a carbon chain. With an alkane, you can just use Br₂ and light — the reaction is straightforward. But with a carboxylic acid, things get tricky. The carboxyl group (−COOH) is strongly electron-withdrawing. It pulls electron density away from the carbon next to it (the α-carbon), making that C–H bond less reactive toward radical halogenation. Direct bromination of a carboxylic acid with Br₂ alone is painfully slow and gives messy mixtures.
So how do chemists get around this? They use a clever trick: convert the acid into a more reactive intermediate first.
The HVZ Reaction: What It Does
The Hell-Volhard-Zelinsky (HVZ) reaction is a method to introduce a halogen atom specifically at the α-position of a carboxylic acid. The reagents are:
A carboxylic acid (with at least one α-hydrogen)
Cl₂ or Br₂ (chlorine or bromine)
A catalytic amount of red phosphorus (P4)
The product is an α-halo carboxylic acid.
R−CH2−COOHBr2red PR−CHBr−COOH+HBr
The reaction works for any carboxylic acid that has at least one hydrogen on the carbon adjacent to the carboxyl group. If the α-carbon is fully substituted (no α-H), the reaction does not occur.
The Mechanism: Step by Step
The magic happens in three stages. Let's trace it with propanoic acid (CH3CH2COOH) and bromine.
Step 1: Formation of the acyl bromide. Red phosphorus reacts with Br₂ to form PBr₃ (phosphorus tribromide). This PBr₃ then converts some of the carboxylic acid into an acyl bromide:
3R−COOH+PBr3→3R−COBr+H3PO3
The acyl bromide is the key intermediate. It is much more reactive than the original acid.
Step 2: Enolization and bromination. The acyl bromide has a slightly acidic α-hydrogen (more acidic than in the acid itself). A trace of HBr (from step 1) catalyzes the formation of an enol:
R−CH2−COBrHBrR−CH=C(OH)Br
This enol attacks Br₂, giving the α-bromo acyl bromide:
R−CH=C(OH)Br+Br2→R−CHBr−COBr+HBr
The HBr released continues the catalytic cycle.
Step 3: Hydrolysis. Finally, water (added at the end of the reaction) hydrolyzes the α-bromo acyl bromide back to the α-bromo carboxylic acid:
R−CHBr−COBr+H2O→R−CHBr−COOH+HBr
Note
The red phosphorus is catalytic — it is consumed to make PBr₃, but PBr₃ is regenerated in the cycle. Only a small amount is needed.
A primary alcohol is oxidised to a carboxylic acid by a strong oxidiser, and that acid is then chlorinated at its alpha-carbon by chlorine with red phosphorus; the chlorinated acid is stronger because the electron-withdrawing chlorine stabilises its conjugate base. Reasoning from these steps is what identifies the reagents and the more acidic product below. …
Propan-1-ol is oxidised to propanoic acid by KMnO4/K2Cr2O7; propanoic acid is α-chlorinated by Cl2/red P (HVZ reaction); the chlorinated acid is the stronger acid due to the –I effect of chlorine.
Reagent (i) — CH3CH2CH2OH → CH3CH2COOH: A primary alcohol is oxidised all the way to a carboxylic acid (with the same number of carbons) by a strong oxidising agent such as alkaline KMnO4 or acidified K2Cr2O7; (ii) H3O+ is simply the acidic work-up step that liberates the free carboxylic acid from its salt after oxidation. So (i) = alkaline KMnO4 (or acidified K2Cr2O7), and the reaction is: CH3CH2CH2OH(i) alk. KMnO4(ii) H3O+CH3CH2COOH.
Reagent (iii) — CH3CH2COOH → CH3-CH(Cl)-COOH: Carboxylic acids having an α-hydrogen react with chlorine (or bromine) in the presence of a small amount of red phosphorus to substitute the α-hydrogen with the halogen; this is the Hell–Volhard–Zelinsky (HVZ) reaction, going through an acid halide intermediate that is hydrolysed back to the acid by (iv) H2O. So (iii) = Cl2/red phosphorus, giving: CH3CH2COOH(iii) Cl2,red P(iv) H2OCH3−CH(Cl)−COOH (α-chloropropanoic acid).
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2025Set ANNUAL1 mark
Q.Acetic acid reacts with chlorine in presence of red phosphorus to give chloroacetic acid. Name the reaction.
›Reveal solutionSolution
Halogenation of a carboxylic acid's α-carbon using Cl₂/Br₂ with red phosphorus is the classic HVZ reaction.
Carboxylic acids possessing an α-hydrogen undergo selective halogenation at the α-carbon when treated with chlorine or bromine in the presence of a small amount of red phosphorus (which generates the reactive phosphorus trihalide catalytically in situ). This substitution of an α-hydrogen by a halogen is called the Hell–Volhard–Zelinsky (HVZ) reaction:
Red phosphorus converts a bit of the halogen into a reactive phosphorus trihalide, which activates the carboxylic acid's alpha-hydrogen for substitution by a halogen.
The Hell-Volhard-Zelinsky (HVZ) reaction is used to introduce a halogen atom at the alpha-carbon (the carbon adjacent to -COOH) of a carboxylic acid that has at least one α-hydrogen. The carboxylic acid is treated with chlorine or bromine (X2) in the presence of a small amount of red phosphorus (which reacts with the halogen to generate phosphorus trihalide, PX3, in situ — this converts the acid to its more reactive acid halide intermediate, allowing enolisation and halogenation at the alpha carbon):
Red phosphorus converts a small amount of the acid to its acyl halide, which enolises and gets halogenated at the α-carbon; the halogen is then relayed to fresh acid molecules, α-halogenating the whole batch.
The Hell–Volhard–Zelinsky (HVZ) reaction is used to introduce a halogen atom at the α-carbon (the carbon next to the −COOH group) of a carboxylic acid that has at least one α-hydrogen.
Procedure: The carboxylic acid is treated with chlorine or bromine (Cl2 or Br2) in the presence of a small (catalytic) amount of red phosphorus.