Chlorobenzene is formed by reaction of chlorine with benzene in the presence of $\mathrm{AlCl_3}$. Which of the following species attacks the benzene ring in this reaction? (i) $\mathrm{Cl^-}$ (ii) $\mathrm{Cl^+}$ (iii) $\mathrm{AlCl_3}$ (iv) $\mathrm{[AlCl_4]^-}$

Chlorobenzene is formed by reaction of chlorine with benzene in the presence of mathrmAlCl₃. Which of the foll

NCERT Exemplar · Q13

\mathrm{AlCl_3}

(i)

\mathrm{Cl^-}

(ii)

\mathrm{Cl^+}

(iii)

\mathrm{AlCl_3}

(iv)

\mathrm{[AlCl_4]^-}

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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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In the Friedel–Crafts chlorination of benzene, the attacking electrophile is the chlorine cation $\mathrm{Cl^+}$ , generated by the Lewis acid $\mathrm{AlCl_3}$ polarising the $\mathrm{Cl_2}$ bond. The correct option is (ii).

This is a classic electrophilic aromatic substitution (EAS) reaction — the Friedel–Crafts halogenation. Benzene’s π-electron cloud is rich and nucleophilic, but it does not react directly with neutral chlorine gas at room temperature. You need a powerful electrophile to pull electrons away from the ring and form the sigma complex. That’s where the Lewis acid comes in.

The role of $\mathrm{AlCl_3}$ is to accept a lone pair from one chlorine atom of $\mathrm{Cl_2}$ , creating a highly polarised complex. This weakens the Cl–Cl bond so much that it effectively breaks heterolytically, generating a $\mathrm{Cl^+}$ ion (or a strongly $\delta^+$ chlorine in the complex) that can attack the ring.

Let’s walk through the mechanism step by step.

Generation of the electrophile $\mathrm{AlCl_3}$ is electron-deficient (it has only six electrons in its valence shell). It coordinates to a chlorine atom of $\mathrm{Cl_2}$ , forming a complex:

\mathrm{Cl_2 + AlCl_3 \rightarrow Cl^{\delta+} \cdots Cl^{\delta-} \cdots AlCl_3}

The Al–Cl bond in the complex pulls electron density away from the $\mathrm{Cl_2}$ molecule. This makes one chlorine strongly electrophilic — essentially a $\mathrm{Cl^+}$ equivalent.

Attack on the benzene ring The π-electrons of benzene attack this electrophilic chlorine, forming a delocalised carbocation intermediate (the arenium ion or sigma complex):

\mathrm{C_6H_6 + Cl^+ \rightarrow C_6H_6Cl^+}

This step is slow and rate-determining. …

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Common Mistakes & How to Avoid Them

Mistake 1: Choosing $\mathrm{Cl^-}$ (option (i))

Why students pick it: They see chlorine is involved and assume the negative ion (chloride) is the attacking species, confusing nucleophilic with electrophilic substitution.

Why it’s wrong: Benzene is electron-rich (due to its delocalised π-system). It repels negative species. Chlorobenzene formation is an electrophilic aromatic substitution — the attacking species must be electron-deficient (an electrophile). $\mathrm{Cl^-}$ is a nucleophile, not an electrophile.

How to avoid: Always check the charge of the attacking species. If benzene is the substrate, the attacking species must be positive or at least neutral but electron-deficient. Memorise: Benzene loves positive things.

Mistake 2: Choosing $\mathrm{AlCl_3}$ (option (iii))

Why students pick it: They see $\mathrm{AlCl_3}$ is added as a catalyst and assume it directly attacks benzene.

Why it’s wrong: $\mathrm{AlCl_3}$ is a Lewis acid — it accepts electrons. Its role is to activate chlorine by forming a complex, not to attack benzene directly. The actual attacking species is generated from this interaction.

How to avoid: Distinguish between catalyst and attacking species. The catalyst helps create the electrophile but does not itself attack the ring. Ask: What does the catalyst do to the reagent? Here, $\mathrm{AlCl_3}$ polarises $\mathrm{Cl_2}$ to generate $\mathrm{Cl^+}$ .

Mistake 3: Choosing $\mathrm{[AlCl_4]^-}$ (option (iv))

Why students pick it: They see a negative charge and think it might stabilise something, or confuse it with the attacking species.

Why it’s wrong: $\mathrm{[AlCl_4]^-}$ is a byproduct formed after $\mathrm{AlCl_3}$ accepts a chloride ion. It is negatively charged and stable — it does not attack the electron-rich benzene ring.

How to avoid: Track the reaction mechanism step-by-step:

$\mathrm{AlCl_3} + \mathrm{Cl_2} \rightarrow \mathrm{Cl^+} + \mathrm{[AlCl_4]^-}$
$\mathrm{Cl^+}$ attacks benzene.
$\mathrm{[AlCl_4]^-}$ is just a spectator/counterion.

Mistake 4: Not recognising $\mathrm{Cl^+}$ as the correct electrophile (option (ii)) …

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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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←Previous questionQ12 — The position of –Br in the compound mathrmCH₃CH=CHC(Br)(CH₃)₂ can be c…→Next questionQ14 — Ethylidene chloride is a/an _______. (i) vic-dihalide (ii) gem-dihalid…

Q.Chlorobenzene is formed by reaction of chlorine with benzene in the presence of $\mathrm{AlCl_3}$ . Which of the following species attacks the benzene ring in this 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

Concept: Electrophilic Aromatic Substitution (EAS) — Role of the Catalyst

Method: Identify the Active Electrophile

Common Mistakes & How to Avoid Them

Mistake 1: Choosing $\mathrm{Cl^-}$ (option (i))

Mistake 2: Choosing $\mathrm{AlCl_3}$ (option (iii))

Mistake 3: Choosing $\mathrm{[AlCl_4]^-}$ (option (iv))

Mistake 4: Not recognising $\mathrm{Cl^+}$ as the correct electrophile (option (ii)) …

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.Chlorobenzene is formed by reaction of chlorine with benzene in the presence of AlCl3\mathrm{AlCl_3}AlCl3​. Which of the following species attacks the benzene ring in this reaction?

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Q.Chlorobenzene is formed by reaction of chlorine with benzene in the presence of $\mathrm{AlCl_3}$ . Which of the following species attacks the benzene ring in this reaction?

Unlock everything free for 14 days