Identify the products A and B formed in the following reaction: (a) mathrmCH₃-CH₂-CH=CH-CH₃ + HCl arrow A + B

NCERT Exemplar · Q57

Q. Identify the products A and B formed in the following reaction:

(a)

\mathrm{CH_3-CH_2-CH{=}CH-CH_3 + HCl \rightarrow A + B}

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Concept understanding — Carboxylic Acid Halogenation

Carboxylic Acid Halogenation: The Hell–Volhard–Zelinsky Reaction

Imagine you have a carboxylic acid — say, propanoic acid ( $\mathrm{CH_3CH_2COOH}$ ). You want to replace one of the hydrogen atoms on the carbon chain with a halogen (like bromine or chlorine). But here's the catch: the carboxylic acid group ( $-\mathrm{COOH}$ ) is already quite reactive. If you just add bromine directly, nothing useful happens — the $\alpha$ -carbon (the carbon right next to the $-\mathrm{COOH}$ group) is not reactive enough to attack bromine on its own.

The trick is to activate the $\alpha$ -carbon first. This is exactly what the Hell–Volhard–Zelinsky (HVZ) reaction does.

The Intuition

The $-\mathrm{COOH}$ group is electron-withdrawing. That makes the $\alpha$ -carbon slightly positive (electrophilic), but not enough to react with a halogen directly. To make it work, we convert the acid into an acyl halide (like $\mathrm{RCOBr}$ ) using $\mathrm{PBr_3}$ or $\mathrm{PCl_3}$ . The acyl halide is even more electron-withdrawing, which makes the $\alpha$ -hydrogen more acidic — it can be removed by a base (like a catalytic amount of $\mathrm{PBr_3}$ or $\mathrm{Br_2}$ itself) to form an enol or enolate intermediate. This enol then attacks a halogen molecule, giving an $\alpha$ -haloacyl halide. Finally, water hydrolyses it back to the $\alpha$ -halo carboxylic acid.

In short: activate → enolize → halogenate → hydrolyse.

The Precise Statement

\mathrm{RCH_2COOH \xrightarrow[1.\ \mathrm{Br_2,\ PBr_3}]{2.\ \mathrm{H_2O}} RCHBrCOOH}

The reaction is regioselective: halogenation occurs exclusively at the $\alpha$ -carbon (the carbon adjacent to the $-\mathrm{COOH}$ group). No other position on the chain is halogenated.

Step-by-Step Mechanism

Formation of acyl bromide The carboxylic acid reacts with $\mathrm{PBr_3}$ (or $\mathrm{PCl_3}$ ) to form an acyl bromide:

\mathrm{RCH_2COOH + PBr_3 \rightarrow RCH_2COBr + H_3PO_3}

Enolization A catalytic amount of $\mathrm{PBr_3}$ or $\mathrm{Br_2}$ acts as a Lewis acid, making the $\alpha$ -hydrogen more acidic. A base (often $\mathrm{Br}^-$ from the reaction) abstracts this hydrogen, forming an enol:

\mathrm{RCH_2COBr \rightleftharpoons RCH=C(OH)Br}

Halogenation The enol attacks a $\mathrm{Br_2}$ molecule, giving the $\alpha$ -bromoacyl bromide:

\mathrm{RCH=C(OH)Br + Br_2 \rightarrow RCHBrCOBr + HBr}

Hydrolysis Water hydrolyses the acyl bromide back to the carboxylic acid:

\mathrm{RCHBrCOBr + H_2O \rightarrow RCHBrCOOH + HBr}

Note

The $\mathrm{PBr_3}$ is catalytic — it is regenerated in the hydrolysis step. Only a small amount is needed.

Why This Matters

The $\alpha$ -halo carboxylic acid is a versatile intermediate. You can:

Substitute the halogen with $\mathrm{OH}$ to get $\alpha$ -hydroxy acids (like lactic acid).
Substitute with $\mathrm{NH_3}$ to get $\alpha$ -amino acids (the building blocks of proteins). …

Why this formula?

Carboxylic Acid Halogenation — The Hell-Volhard-Zelinsky (HVZ) Reaction

Let's start with the core reaction and then unpack why it works the way it does.

The Reaction in a Nutshell

Carboxylic acids undergo α-halogenation (replacement of an α-hydrogen with a halogen) only under specific conditions:

\ce{R-CH2-COOH + Br2 ->[PBr3 (cat.)] R-CHBr-COOH + HBr}

The key reagents: Br₂ (or Cl₂) + a catalytic amount of PBr₃ (or PCl₃). The product is an α-halo carboxylic acid.

Why Does This Happen? The Step-by-Step Reasoning

1. The Problem: Carboxylic Acids Are Not Enolizable Directly

A carboxylic acid has a carbonyl group ( $\ce{C=O}$ ), but the α-hydrogen is not acidic enough to be removed by a base like $\ce{OH-}$ .
Why? The conjugate base (carboxylate ion, $\ce{RCOO-}$ ) is more stable than an enolate. So enolate formation is disfavoured.

Key insight: We need to activate the carbonyl first.

2. The Solution: Convert to an Acyl Halide (More Electrophilic)

PBr₃ reacts with the carboxylic acid to form an acyl bromide:

\ce{3 R-COOH + PBr3 -> 3 R-COBr + H3PO3}

The acyl bromide has a better leaving group (Br⁻ vs OH⁻) and a more electrophilic carbonyl carbon. This makes enolization easier.

3. Enolization of the Acyl Halide

A small amount of HBr (from the reaction) or Br₂ itself can act as a Lewis acid to polarize the carbonyl.
The α-hydrogen is now removable by a weak base (like Br⁻ or the enol itself), forming an enol:

\ce{R-CH2-COBr <=> R-CH=C(OH)Br}

This enol is nucleophilic at the α-carbon.

4. Halogenation of the Enol

The enol attacks Br₂ (or Cl₂) at the α-position:

\ce{R-CH=C(OH)Br + Br2 -> R-CHBr-C(OH)Br2 -> R-CHBr-COBr + HBr}

The product is an α-bromo acyl bromide.

5. Regeneration of the Acid

The α-bromo acyl bromide reacts with water (or with another molecule of carboxylic acid) to give the α-bromo carboxylic acid:

\ce{R-CHBr-COBr + H2O -> R-CHBr-COOH + HBr}

The HBr produced can re-enter the cycle, making the process catalytic in PBr₃.

The Key Formula(e) — Why They Hold

Overall Stoichiometry

\ce{R-CH2-COOH + Br2 ->[PBr3 (cat.)] R-CHBr-COOH + HBr}

Why this holds:

One Br₂ molecule provides one Br atom for substitution and one for HBr.
The catalyst (PBr₃) is not consumed — it is regenerated in the cycle. …

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$CH_3CH_2CH{=}CHCH_3$ is pent-2-ene: BOTH alkene carbons carry exactly one alkyl substituent each (not one primary and one secondary), so protonating either carbon gives a secondary carbocation of comparable stability to the other. There is no primary-carbocation pathway here at all, so the reaction gives 2-chloropentane and 3-chloropentane in roughly comparable amounts, not as a clean major/minor pair.

This is electrophilic addition of HCl to an alkene, but the usual sharp Markovnikov major/minor split only appears when the two alkene carbons are substituted to different DEGREES (e.g. one carbon bearing two alkyl groups, the other bearing none or one). Here that is not the case.

Numbering the chain

Number so the double bond gets the lowest locant (this makes it pent-2-ene, matching how the compound would actually be named): $C1(CH_3)-C2(CH)=C3(CH)-C4(CH_2)-C5(CH_3)$ , double bond between C2 and C3.

C2 is bonded to: C1 (one alkyl group, a methyl), one H, and the double bond.
C3 is bonded to: C4 (one alkyl group, the start of an ethyl chain), one H, and the double bond.

Both alkene carbons carry exactly one alkyl substituent each. Neither is a terminal $=CH_2$ , so there is no way to generate a primary carbocation from this alkene at all.

Both carbocations are secondary

Protonating C2 places the positive charge on C3, which is then bonded to C2 and C4 — a secondary carbocation.
Protonating C3 places the positive charge on C2, which is then bonded to C1 and C3 — also a secondary carbocation.

Since both possible carbocations are secondary and structurally very similar (one flanked by a methyl and the chain, the other by an ethyl-chain carbon and the chain), neither is meaningfully more stable than the other. Chloride ion then attacks whichever cation formed, giving: …

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Here are the common mistakes students make when solving this reaction, along with how to avoid each.

Mistake 1: Forcing Markovnikov's Rule to Give a Single "Major" Product

The Mistake: Students assume Markovnikov's rule must always identify one clearly major and one clearly minor product, so they label one of 2-chloropentane or 3-chloropentane as "major" without real justification.

Why it's wrong: Markovnikov's rule (in its full form) says H⁺ adds to the carbon that leads to the MORE STABLE carbocation. Here, both possible additions to $\mathrm{CH_3-CH_2-CH=CH-CH_3}$ (pent-2-ene) lead to a carbocation with exactly the same DEGREE of substitution:

Protonating C3 gives a cation at C4, flanked by a methyl group (C5) and a propyl-length chain (via C3-C2-C1) — a secondary carbocation.
Protonating C4 gives a cation at C3, flanked by an ethyl group (via C2-C1) and another ethyl-length fragment (via C4-C5) — also a secondary carbocation.

Neither is meaningfully more stabilized than the other (both are simple secondary carbocations with ordinary alkyl substituents, no branching right at the cationic centre to favour one side).

How to Avoid: When protonating an unsymmetrical alkene, always draw BOTH possible carbocations and compare their DEGREE (primary/secondary/tertiary) and any special stabilization (allylic, benzylic, branching at the cationic carbon). If both carbocations come out the same degree with no other stabilizing difference, expect a genuine mixture in comparable amounts — don't force a "major" answer that isn't chemically justified.

Mistake 2: Miscounting Hydrogens and Concluding the Alkene Must Be Symmetrical

The Mistake: Since both alkene carbons (C3, C4) have exactly 1 H each, students sometimes conclude the molecule is symmetrical and expect only one product.

Why it's wrong: Equal H-count on the two alkene carbons does NOT mean the molecule is symmetrical. C3 is attached to an ethyl group (C2-C1) while C4 is attached to a methyl group (C5) — these are different substituents, so the alkene is genuinely unsymmetrical. It just happens that the resulting carbocations, despite this asymmetry, end up comparably stable.

How to Avoid: Check the FULL substituent (not just the H count) on each alkene carbon before concluding anything about symmetry.

Mistake 3: Writing Only One Product (Ignoring the Mixture)

The Mistake: The question asks for A and B, but students write only one product, assuming one is definitively correct.

How to Avoid: When two carbocation pathways are comparably stable, both products form and both must be reported:

A: $\mathrm{CH_3-CH_2-CHCl-CH_2-CH_3}$ (3-chloropentane)
B: $\mathrm{CH_3-CH_2-CH_2-CHCl-CH_3}$ (2-chloropentane)

Mistake 4: Miscounting the Chain When Naming the Products …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$\mathrm{H_3CCH_2COOH} \xrightarrow{X} \mathrm{CH_3CHBrCOOH}$
(i) $\mathrm{Br_2}$ /P red,
(ii) $\mathrm{H_2O}$ (B)
(i) $\mathrm{Br_2}$ / $\mathrm{CCl_4}$ ,
(ii) $\mathrm{H_2O}$ (C) $\mathrm{Br_2}$ / $\mathrm{OH^-}$ (D) $\mathrm{PBr_3}$
›Reveal solutionSolution
This tests recognition of the Hell–Volhard–Zelinsky reaction, which converts a carboxylic acid with $\alpha$ -hydrogens into its $\alpha$ -bromo (or $\alpha$ -chloro) derivative using $\mathrm{Br_2}$ /red phosphorus followed by hydrolysis.

Concept and Intuition

Carboxylic acids do not directly react with $\mathrm{Br_2}$ at their $\alpha$ -carbon the way ketones/aldehydes do via their enol, because carboxylic acids exist overwhelmingly in the keto (acid) form and enolize very poorly. The Hell–Volhard–Zelinsky (HVZ) reaction solves this: red phosphorus reacts with $\mathrm{Br_2}$ to generate a small amount of $\mathrm{PBr_3}$ , which converts the carboxylic acid into its far more easily enolizable acid bromide ( $\mathrm{RCOBr}$ ). This acid bromide's enol form reacts readily with $\mathrm{Br_2}$ to substitute a hydrogen at the $\alpha$ -carbon, giving the $\alpha$ -bromo acid bromide. Finally, this reactive intermediate is hydrolysed with water to yield the stable $\alpha$ -bromo carboxylic acid.

Step-by-Step Solution

Identify the required transformation: $\mathrm{CH_3CH_2COOH}$ (propanoic acid) → $\mathrm{CH_3CHBrCOOH}$ (2-bromopropanoic acid) — bromine is introduced specifically at the $\alpha$ -carbon (the carbon next to $-\mathrm{COOH}$ ).

Recognize this pattern as $\alpha$ -halogenation of a carboxylic acid, which is precisely what the HVZ reaction accomplishes.

The HVZ reagents are (i) $\mathrm{Br_2}$ with catalytic red phosphorus, which in situ forms $\mathrm{PBr_3}$ and converts the acid to the acid bromide, enabling $\alpha$ -bromination via the enol form; then (ii) hydrolysis with $\mathrm{H_2O}$ converts the resulting $\alpha$ -bromo acid bromide back to the free $\alpha$ -bromo carboxylic acid.

This matches option (A) exactly: (i) $\mathrm{Br_2}$ /P red, (ii) $\mathrm{H_2O}$ . …
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AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$\mathrm{CH_3CH_2COOH} \rightarrow \mathrm{CH_3CH_2COCl}$
›Reveal solutionSolution
This tests the preferred laboratory reagent for converting a carboxylic acid to its acyl chloride; thionyl chloride ( $\mathrm{SOCl_2}$ ) is preferred because its by-products are gases that leave the product clean.

Concept and Intuition

Several reagents can convert a carboxylic acid $-\mathrm{COOH}$ into an acyl chloride $-\mathrm{COCl}$ (e.g. $\mathrm{PCl_3}$ , $\mathrm{PCl_5}$ , $\mathrm{SOCl_2}$ ), but thionyl chloride is generally the preferred choice in practice. The reaction is:
$\mathrm{RCOOH + SOCl_2 \rightarrow RCOCl + SO_2\uparrow + HCl\uparrow}$
Both by-products, sulphur dioxide and hydrogen chloride, are gases that simply escape the reaction mixture, so the desired acyl chloride is obtained in a purer form without needing extensive work-up to remove solid phosphorus-containing by-products (as would be needed with $\mathrm{PCl_3}$ / $\mathrm{PCl_5}$ , which leave behind phosphorous acid or phosphorus oxychloride residues).

Step-by-Step Solution

Identify the required transformation: $\mathrm{CH_3CH_2COOH}$ (propanoic acid) → $\mathrm{CH_3CH_2COCl}$ (propanoyl chloride), i.e., replacing $-\mathrm{OH}$ with $-\mathrm{Cl}$ at the carbonyl carbon.

Recognize this as acyl chloride formation from a carboxylic acid.

Among the given options, only $\mathrm{SOCl_2}$ (option C) is a reagent used for this transformation; HCl, HOCl, and NaOCl (options A, B, D) are not reagents that convert carboxylic acids to acid chlorides. …
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AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
$CH_2CH_2OH$
›Reveal solutionSolution
Oxidation of the primary alcohol to a carboxylic acid, followed by Hell-Volhard-Zelinsky alpha-chlorination, gives an alpha-chloro carboxylic acid as the final product.

Concept and Intuition

This sequence chains together two classic named reactions: (1) $CrO_3/H_2SO_4$ (Jones-type oxidation) fully oxidizes a primary alcohol through the aldehyde stage to a carboxylic acid; (2) the Hell-Volhard-Zelinsky (HVZ) reaction uses $Cl_2$ (or $Br_2$ ) with red phosphorus to selectively halogenate a carboxylic acid specifically at its alpha carbon (via an enol-like acyl halide intermediate), which is a hallmark reaction for functionalizing the position next to $-COOH$ .

Step-by-Step Solution

Start: 2-cyclohexylethanol, $C_6H_{11}-CH_2-CH_2-OH$ (a cyclohexane ring bearing a $-CH_2CH_2OH$ side chain).

Step (i), $CrO_3$ / $H_2SO_4$ : oxidizes the terminal $-CH_2OH$ all the way to $-COOH$ (strong oxidant, primary alcohol → carboxylic acid), giving cyclohexylacetic acid, $C_6H_{11}-CH_2-COOH$ .

Step (ii), $Cl_2$ /red P (HVZ reaction): red phosphorus first converts a small amount of the acid to the acyl chloride, which enolizes and is chlorinated specifically at the alpha carbon (the carbon adjacent to the carbonyl), giving $C_6H_{11}-CHCl-COOH$ (after the acyl chloride hydrolyzes back to the acid). …
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AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
$Q. In the following reaction sequence, identify product 'Q' and reagent 'R': [FIGURE] (Me-CH2-C\equivC-CH2-Br is treated with$
(i) Mg/Ether
(ii) $CO_2$
(iii) $H_3O^+$ to give product Q; Q is then treated with reagent R to give Me-C≡C-CH2-C(=O)Cl) (A) [FIGURE] (Me-CH2-C≡C-CH2-COOH) and $SOCl_2$ (B) [FIGURE] (Me-CH=CH-CH2-CH2-COOH, trans double bond) and $SO_2Cl_2$ (C) [FIGURE] (Me-CH=CH-CH2-COOH, cis double bond) and $SOCl_2$ (D) [FIGURE] (Me-CH2-C≡C-CH2-COOH) and $CH_3SO_2Cl_2$
›Reveal solutionSolution
Grignard formation + $CO_2$ + $H_3O^+$ is a standard carboxylation that never touches the triple

bond; converting the resulting acid to an acid chloride is the textbook job of $SOCl_2$ .

Concept and Intuition

Forming a Grignard reagent from an alkyl (here propargylic) halide is a simple oxidative-insertion

of Mg into the C–Br bond — it does not isomerise or rearrange the carbon skeleton or move the

triple bond under these conditions (no acid/base or transition-metal catalyst is present to trigger

an allenic/propargylic shift). Bubbling $CO_2$ through a Grignard reagent, then quenching with

aqueous acid, is the standard way to extend a carbon chain by one carbon and install a $-COOH$

group at the position that held the $-MgBr$ .

Once a carboxylic acid needs to become an acid chloride, the reagent of choice (retaining the

carbon skeleton, releasing only gaseous by-products $SO_2$ and HCl so the product is easy to

purify) is thionyl chloride, $SOCl_2$ .

Step-by-Step Solution

$Me{-}CH_2{-}C{\equiv}C{-}CH_2{-}Br \xrightarrow{Mg/Ether} Me{-}CH_2{-}C{\equiv}C{-}CH_2{-}MgBr$ — simple insertion, skeleton and triple-bond position unchanged.

$\xrightarrow{(ii)\ CO_2}$ the Grignard carbon attacks $CO_2$ to give a magnesium carboxylate: $Me{-}CH_2{-}C{\equiv}C{-}CH_2{-}COOMgBr$ .

$\xrightarrow{(iii)\ H_3O^+}$ protonation gives the free acid, Q $= Me{-}CH_2{-}C{\equiv}C{-}CH_2{-}COOH$ (a hex-3-ynoic acid).

Converting Q's $-COOH$ to $-COCl$ with retention of the carbon skeleton requires $R = SOCl_2$ : $R{-}COOH + SOCl_2 \to R{-}COCl + SO_2\uparrow + HCl\uparrow$ . …
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Q.Identify the products A and B formed in the following reaction:

Concept understanding — Carboxylic Acid Halogenation

Carboxylic Acid Halogenation: The Hell–Volhard–Zelinsky Reaction

The Intuition

The Precise Statement

Step-by-Step Mechanism

Why This Matters

Why this formula?

Carboxylic Acid Halogenation — The Hell-Volhard-Zelinsky (HVZ) Reaction

The Reaction in a Nutshell

Why Does This Happen? The Step-by-Step Reasoning

1. The Problem: Carboxylic Acids Are Not Enolizable Directly

2. The Solution: Convert to an Acyl Halide (More Electrophilic)

3. Enolization of the Acyl Halide

4. Halogenation of the Enol

5. Regeneration of the Acid

The Key Formula(e) — Why They Hold

Overall Stoichiometry

Numbering the chain

Both carbocations are secondary

Concept: Electrophilic Addition to Unsymmetrical Alkenes (Markovnikov’s Rule)

Method: Markovnikov Addition

Step 1 – Identify the double bond and the two alkene carbons

Step 2 – Count hydrogens on each alkene carbon

Step 3 – Add H⁺ and Cl⁻ in both possible ways

Mistake 1: Forcing Markovnikov's Rule to Give a Single "Major" Product

Mistake 2: Miscounting Hydrogens and Concluding the Alkene Must Be Symmetrical

Mistake 3: Writing Only One Product (Ignoring the Mixture)

Mistake 4: Miscounting the Chain When Naming the Products …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Identify the products A and B formed in the following reaction:

Numbering the chain

Both carbocations are secondary

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