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.

Showing the 12 most recent of 14 on this concept.

COMEDK 2026Set 2026-M1 markMCQ
$A \rightarrow B$
›Reveal solutionSolution
The reaction sequence starts with hydrolysis of acetonitrile to ethanoic acid, followed by α‑chlorination via the Hell‑Volhard‑Zelinsky reaction, giving 2‑chloroethanoic acid. The correct option is (C).

Concept & Intuition

The problem asks you to identify two things: the name of the reaction that converts intermediate A into final product B, and the structure/name of B. The first step is a simple acid‑catalysed hydrolysis of a nitrile (CH₃CN) to a carboxylic acid (ethanoic acid, A). The second step uses Cl₂ in the presence of red phosphorus and water under acidic conditions — this is the classic setup for the Hell‑Volhard‑Zelinsky (HVZ) reaction, which selectively halogenates the α‑carbon of a carboxylic acid. The product is therefore an α‑chloro acid.

Step‑by‑step reasoning

First step: hydrolysis of acetonitrile

$\mathrm{CH_3CN} \xrightarrow[\mathrm{H_2O}]{\mathrm{H^+}} \mathrm{CH_3COOH}$
Nitriles are hydrolysed by hot aqueous acid to give carboxylic acids. Here, acetonitrile (CH₃CN) becomes ethanoic acid (acetic acid). So compound A is ethanoic acid.

Second step: the reaction conditions

The reagents are $\mathrm{Cl_2 / \text{Red Phosphorus}}$ in the presence of $\mathrm{H^+/H_2O}$ . Red phosphorus reacts with chlorine to form PCl₃ in situ, which converts the carboxylic acid into an acyl chloride. The acyl chloride then enolises and is chlorinated at the α‑position. Water present hydrolyses the acyl chloride back to the carboxylic acid. This is the Hell‑Volhard‑Zelinsky (HVZ) reaction.

Applying HVZ to ethanoic acid

Ethanoic acid has the structure $\mathrm{CH_3COOH}$ . The α‑carbon is the carbon adjacent to the carboxyl group — that is, the methyl carbon. Chlorination replaces one hydrogen on that carbon:

$\mathrm{CH_3COOH} \xrightarrow{\mathrm{Cl_2 / P, H^+/H_2O}} \mathrm{ClCH_2COOH}$
The product is 2‑chloroethanoic acid (also called chloroacetic acid). The “2‑” indicates the chlorine is on the carbon next to the carboxyl group.

Matching with the options …
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KCET 2025Set D-41 markMCQ
$_2$
›Reveal solutionSolution

Ethanol + conc. H $_2$ SO $_4$ is temperature-controlled: 413 K gives bimolecular dehydration (ether); 443 K gives intramolecular dehydration (alkene). The question specifies 413 K.

Step 1 — The competing pathways.

Concentrated H $_2$ SO $_4$ protonates the –OH of ethanol, converting it into $\mathrm{-\overset{+}{O}H_2}$ — an excellent leaving group (water). What happens next depends on whether a nucleophile or elimination takes over, and temperature is the switch:
$\mathrm{2C_2H_5OH \xrightarrow[413\ \text{K}]{\text{conc. H}_2\text{SO}_4} C_2H_5\!-\!O\!-\!C_2H_5 + H_2O} \quad \text{(ether — bimolecular)}$ $\mathrm{C_2H_5OH \xrightarrow[443\ \text{K}]{\text{conc. H}_2\text{SO}_4} CH_2{=}CH_2 + H_2O} \quad \text{(alkene — intramolecular)}$
Step 2 — Why 413 K gives the ether.

At $413\ \text{K}$ ( $=413-273 = 140^\circ\text{C}$ ), a second, un-protonated ethanol molecule — still abundant in the mixture — acts as the nucleophile and attacks the carbon of the protonated ethanol via $\mathrm{S_N}2$ , displacing water:
$\mathrm{C_2H_5\!-\!\overset{+}{O}H_2 + HO\!-\!C_2H_5 \longrightarrow C_2H_5\!-\!\overset{+}{O}(H)\!-\!C_2H_5 + H_2O}$
Loss of H $^+$ from that oxonium ion then gives diethyl ether. This is called bimolecular dehydration because two alcohol molecules combine, losing one molecule of water between them. This is in fact the industrial route to diethyl ether.

Step 3 — Why not ethene (option C). …
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KCET 2024Set B-21 markMCQ

_4

›Reveal solution

Hydrolysis needs the central atom to expand its octet using vacant $d$ -orbitals; carbon has none, so $CCl_4$ alone resists water.

Step 1 — The mechanism of tetrahalide hydrolysis

For $MCl_4$ ( $M$ = Si, Ge, Sn), water attacks the central atom:

SiCl_4 + 4H_2O \longrightarrow Si(OH)_4 + 4HCl

The first step is nucleophilic attack by the oxygen lone pair of $H_2O$ on $M$ , giving a five- (or six-) coordinate intermediate/transition state. To accept that extra bond pair the central atom must expand its octet — which requires energetically accessible vacant $d$ -orbitals in its valence shell.

Step 2 — Which elements have those $d$ -orbitals?

Halide	Central atom	Period	Valence shell	Vacant $d$ available?	Hydrolyses?
$CCl_4$	C	2	$2s\,2p$	No — there is no $2d$ subshell	No
$SiCl_4$	Si	3	$3s\,3p\,3d$	Yes ( $3d$ )	Yes
$GeCl_4$	Ge	4	$4s\,4p\,4d$	Yes ( $4d$ )	Yes
$SnCl_4$	Sn	5	$5s\,5p\,5d$	Yes ( $5d$ )	Yes

Carbon's valence shell ( $n = 2$ ) simply has no $d$ subshell, so its maximum covalency is 4. It cannot form the five-coordinate transition state, and the alternative route — spontaneous ionisation of $C-Cl$ — is far too endothermic.

Step 3 — The supporting steric argument …

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KCET 2024Set B-21 markMCQ
$_2$
›Reveal solutionSolution
DDT is the historically first chlorinated organic insecticide; the other options are either a later one or not insecticides.

Step 1 — Recall the history

DDT — p,p'-dichlorodiphenyltrichloroethane — was first synthesised in 1873, but its insecticidal property was discovered by Paul Hermann Müller in 1939 (for which he received the 1948 Nobel Prize in Physiology or Medicine). From the 1940s it was used worldwide against malarial mosquitoes and agricultural pests. It is universally cited as the first chlorinated organic insecticide.

Structure: two chlorobenzene rings joined to a $-CH-CCl_3$ bridge, i.e. a heavily chlorinated hydrocarbon.

Step 2 — Screen the other options

(A) Gammexane — this is the $\gamma$ -isomer of benzene hexachloride ( $C_6H_6Cl_6$ , BHC/lindane). It is a chlorinated organic insecticide, but it came into use after DDT, so it is not the first. This is the intended distractor. ✗

(B) Chloroform ( $CHCl_3$ ) — a chlorinated organic compound, but it is a solvent/anaesthetic, not an insecticide. ✗ …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{C}_3 \mathrm{H}_5 \mathrm{N}$
›Reveal solutionSolution

Propanenitrile hydrolyses to propanoic acid, then HVZ gives $\alpha$ -chloro acid, and ammonolysis gives $\text{2-aminopropanoic acid}$ (alanine). Official key: (A).

Working

The formula $\mathrm{C_3H_5N}$ is that of propanenitrile, $\text{CH}_3\text{CH}_2\text{C}\equiv\text{N}$ .

Step 1 — Hydrolysis of the nitrile gives the carboxylic acid $\text{X}$ :
$\text{CH}_3\text{CH}_2\text{CN} \xrightarrow{\mathrm{H_3O^+}} \underset{\text{X: propanoic acid}}{\text{CH}_3\text{CH}_2\text{COOH}}$
Step 2 — Hell–Volhard–Zelinsky ( $\mathrm{Cl_2/P}$ ) chlorinates the $\alpha$ -carbon, giving $\text{Y}$ :

$\text{CH}_3\text{CH}_2\text{COOH} \xrightarrow{\mathrm{Cl_2/P}} \underset{\text{Y: 2-chloropropanoic acid}}{\text{CH}_3\text{CHCl-COOH}}$ …
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KCET 2023Set D-21 markMCQ
$_2$
›Reveal solutionSolution

Carbon monoxide co-ordinates to the haem iron far more strongly than dioxygen does, converting haemoglobin into carboxyhaemoglobin (HbCO).

1. How haemoglobin normally works

Haemoglobin (Hb) carries a haem group with an $\mathrm{Fe^{2+}}$ centre. Molecular oxygen binds reversibly to this iron:
$\mathrm{Hb} + \mathrm{O_2} \rightleftharpoons \mathrm{HbO_2} \quad (\text{oxyhaemoglobin})$
The reversibility is essential — $\mathrm{O_2}$ must be released in the tissues.

2. What CO does

CO is a strong $\sigma$ -donor / $\pi$ -acceptor ligand. Its back-bonding with $\mathrm{Fe^{2+}}$ makes the Fe–CO bond far stronger than the Fe– $\mathrm{O_2}$ bond (affinity roughly 200–300× greater):
$\mathrm{Hb} + \mathrm{CO} \longrightarrow \mathrm{HbCO} \quad (\text{carboxyhaemoglobin})$
Because the binding is essentially irreversible on a physiological timescale, those haem sites are permanently blocked. The blood's oxygen-carrying capacity collapses, causing hypoxia — the classic symptom of CO poisoning from incomplete combustion.

3. Why the others fail …
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KCET 2023Set D-21 markMCQ
$_2$
›Reveal solutionSolution
In the electrolysis of brine (concentrated NaCl solution) with inert electrodes, the key is to compare the standard reduction potentials of the competing ions. Chloride ions are preferentially oxidized over water at the anode, producing chlorine gas, while water is reduced at the cathode to give hydrogen gas. The correct answer is (D).

The electrolysis of brine is a classic industrial process (the chlor-alkali process). The confusion usually arises because students remember that in the electrolysis of dilute NaCl or of water itself, oxygen is produced at the anode. But brine is concentrated sodium chloride. The concentration of chloride ions is high, and this shifts the competition in their favor.

At the anode, two oxidation reactions are possible:

$2Cl^- \rightarrow Cl_2 + 2e^-$ (standard potential $E^\circ = -1.36$ V)

$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ (standard potential $E^\circ = -1.23$ V)

The reaction with the less negative (or more positive) potential is thermodynamically easier. On paper, water oxidation ( $-1.23$ V) looks easier than chloride oxidation ( $-1.36$ V). However, the oxidation of water has a large overpotential at inert electrodes like platinum or graphite — it needs extra voltage to get going. In concentrated chloride solution, the actual potential required to evolve oxygen is higher than that for chlorine. So chlorine is liberated at the anode.

At the cathode, the competition is between:

$Na^+ + e^- \rightarrow Na$ ( $E^\circ = -2.71$ V)

$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ ( $E^\circ = -0.83$ V)

Reduction of water to hydrogen is far easier (much less negative potential) than depositing sodium metal. Sodium would immediately react with water anyway. So hydrogen gas is liberated at the cathode, and the solution becomes alkaline (NaOH forms).

Let's go through the options step by step.

Option (A): O $_2$ liberates at anode

This is true only for very dilute NaCl or for electrolysis of pure water. In concentrated brine, the overpotential for oxygen evolution makes chlorine the preferred product. So (A) is false for brine.

Option (B): H $_2$ liberates at anode

Hydrogen is produced at the cathode, not the anode. Anodes are where oxidation happens. This is completely wrong.

Option (C): Na deposits on cathode …
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KCET 2022Set B-31 markMCQ
$\alpha$
›Reveal solutionSolution
HVZ halogenation proceeds through an enol of the acyl halide, which can only form if the acid carries an $\alpha$ -hydrogen — ethanoic acid has three, methanoic acid has none.

1. What the HVZ reaction actually does

Carboxylic acids with an $\alpha$ -H react with $Cl_2/Br_2$ in the presence of a small amount of red phosphorus to give the $\alpha$ -halo acid:
$R\text{--}CH_2\text{--}COOH \xrightarrow[\;(ii)\;H_2O\;]{(i)\;X_2/\text{red P}} R\text{--}CHX\text{--}COOH$
2. Why an $\alpha$ -H is mandatory

The mechanism is:

Red P + $X_2$ generates $PX_3$ , which converts the acid into the acyl halide $R\text{--}CH_2\text{--}COX$ .

The acyl halide tautomerises to its enol: $R\text{--}CH=C(OH)X$ . This step consumes an $\alpha$ -hydrogen.

The nucleophilic enol double bond attacks $X_2$ , putting the halogen on the $\alpha$ -carbon.

Hydrolysis / exchange regenerates the $\alpha$ -halo acid.

Without an $\alpha$ -H there is no enolisable position, step 2 fails, and the whole sequence stops.

3. Apply it to the two acids

Acid Structure $\alpha$ -carbon $\alpha$ -H present?
Ethanoic acid $CH_3\text{--}COOH$ the $CH_3$ carbon Yes — 3 of them
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KCET 2020Set A-11 markMCQ
Q.Identify 'X' in the following reaction
- 6Cl $_2$ (excess) $\xrightarrow[\text{dark, cold}]{\text{Anhydrous AlCl}_3}$ X + 6HCl
(A) (B) (C) (D)
›Reveal solutionSolution
$AlCl_3$ + dark/cold $\Rightarrow$ electrophilic substitution; six $Cl_2$ and six $HCl$ liberated $\Rightarrow$ all six ring H's replaced $\Rightarrow$ hexachlorobenzene.

Step 1 — Read the reagent and the conditions.

Benzene reacts with chlorine along two completely different routes:

Cl $_2$ / anhydrous $AlCl_3$ (or $FeCl_3$ ), dark, cold $\rightarrow$ electrophilic substitution. The Lewis acid polarises $Cl-Cl$ to generate $Cl^{+}$ , which attacks the $\pi$ cloud; the ring's aromaticity is restored by loss of $H^{+}$ , which appears as $HCl$ .

Cl $_2$ / UV light, sunlight, no Lewis acid $\rightarrow$ free-radical addition of three $Cl_2$ across the three $\pi$ bonds giving benzene hexachloride (BHC / gammexane), $C_6H_6Cl_6$ , with a saturated ring and no HCl formed.

Here the conditions are anhydrous $AlCl_3$ , dark and cold $\Rightarrow$ substitution.

Step 2 — Use the stoichiometry as the clincher.
$\mathrm{C_6H_6} + 6\,\mathrm{Cl_2} \xrightarrow{\text{anhyd. }AlCl_3,\ \text{dark, cold}} \mathrm{C_6Cl_6} + 6\,\mathrm{HCl}$
Six molecules of $HCl$ are released, i.e. six ring hydrogens have been replaced. An addition reaction produces no $HCl$ at all. So $X$ must have all six H's swapped for Cl while the aromatic ring survives.

Step 3 — Pick the structure. …
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KCET 2020Set A-11 markMCQ
$\text{CH}_3COCl$
›Reveal solutionSolution
Hinsberg's reagent is benzenesulfonyl chloride ( $\text{C}_6\text{5SO}_2Cl$ ), used to distinguish primary, secondary, and tertiary amines via sulfonamide formation.

The question tests a classic name reaction in organic chemistry — the Hinsberg test. This test is a practical way to tell apart the three classes of amines based on their reactivity with a sulfonyl chloride. The key idea is that the reagent must have a reactive sulfonyl chloride group ( $-\text{SO}_2Cl$ ) that can attack the nitrogen of an amine. Let's see why only one option fits.

What Hinsberg's reagent does

It reacts with amines to form sulfonamides. A primary amine gives a sulfonamide that is soluble in alkali (because it has an acidic N–H hydrogen). A secondary amine gives a sulfonamide that is insoluble in alkali (no acidic hydrogen). A tertiary amine does not react at all under the test conditions — it may form a salt but no stable sulfonamide. The reagent must therefore be a sulfonyl chloride, not an acyl chloride or an amide.

Check each option

(A) $\text{CH}_3COCl$ / pyridine — This is acetyl chloride in pyridine. It's an acylating agent, used for acetylation (like in the Schotten–Baumann reaction), not for the Hinsberg test. It would give amides, not sulfonamides.

(B) $(\text{CH}_3\text{CO})_2O$ / pyridine — Acetic anhydride, again an acylating agent. Same reason: not a sulfonyl chloride.

(C) $\text{C}_6\text{H}_5\text{SO}_2Cl$ — Benzenesulfonyl chloride. This is exactly the compound: it has the $-\text{SO}_2Cl$ group attached to a benzene ring. This is the classic Hinsberg reagent.

(D) $\text{C}_6\text{H}_5\text{SO}_2\text{NH}_2$ — Benzenesulfonamide. This is already a sulfonamide — it has no reactive chloride to attack an amine. It cannot act as a reagent for the test.

Why the others are wrong …
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KCET 2019Set A-11 markMCQ
$Q. Propanoic acid undergoes HVZ reaction to give chloropropanoic acid. The product obtained is (A) stronger acid than propanoic acid (B) weaker acid than propanoic acid (C) as stronger as propanoic acid (D) stronger than dichloropropanoic acid$
›Reveal solutionSolution
The HVZ reaction replaces an $\alpha$ -hydrogen with chlorine, and the electron-withdrawing inductive effect of chlorine makes the resulting $\alpha$ -chloropropanoic acid a stronger acid than propanoic acid. The correct option is (A).

The key here is understanding how substituents affect acid strength in carboxylic acids. The acidity of a carboxylic acid depends on the stability of its conjugate base — the carboxylate anion. Any group that pulls electron density away from the carboxylate ion (an electron-withdrawing group) stabilizes the negative charge and makes the acid stronger. Conversely, electron-donating groups destabilize the anion and weaken the acid.

In the HVZ (Hell–Volhard–Zelinsky) reaction, propanoic acid is treated with bromine (or chlorine) in the presence of a catalytic amount of $PBr_3$ or $PCl_3$ . This selectively replaces one $\alpha$ -hydrogen (the hydrogen on the carbon adjacent to the carboxyl group) with a halogen atom. So propanoic acid ( $CH_3CH_2COOH$ ) becomes 2-chloropropanoic acid ( $CH_3CHClCOOH$ ).

Now, chlorine is highly electronegative. It pulls electron density toward itself through the sigma bonds — this is the inductive effect. In 2-chloropropanoic acid, the chlorine is attached to the $\alpha$ -carbon, which is directly next to the carboxyl group. This means the electron-withdrawing effect is transmitted very effectively to the carboxylate group, stabilizing the conjugate base.

Let’s walk through the reasoning step by step.

Identify the change caused by the HVZ reaction.

Propanoic acid: $CH_3CH_2COOH$

After HVZ with chlorine: $CH_3CHClCOOH$ (2-chloropropanoic acid)

One $\alpha$ -hydrogen is replaced by a chlorine atom.

Recall the inductive effect on acidity.

The carboxylate anion ( $RCOO^-$ ) is stabilized by any group that pulls electron density away from the negative charge. Chlorine is strongly electron-withdrawing ( $-I$ effect). The closer the chlorine is to the carboxyl group, the stronger its effect. Here it is on the $\alpha$ -carbon — the closest possible position without being directly on the carboxyl carbon.

Compare the acid strengths.

Propanoic acid has no electron-withdrawing substituent on the $\alpha$ -carbon. Its $pK_a$ is about 4.87.

2-Chloropropanoic acid has a chlorine on the $\alpha$ -carbon. Its $pK_a$ is about 2.86.

A lower $pK_a$ means a stronger acid. So 2-chloropropanoic acid is significantly stronger than propanoic acid. …
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KCET 2018Set A-11 markMCQ
$\mathrm{CH_3COCH_3}$
›Reveal solutionSolution

Haloform requires a methyl ketone ( $\mathrm{CH_3\!-\!CO\!-\!R}$ ) or a methyl carbinol; only acetone qualifies.

Step 1 — The structural requirement.

The haloform reaction proceeds when a compound carries the group
$\mathrm{CH_3\!-\!\underset{\displaystyle \|}{\mathrm{C}}\!-\!}\;\;(\text{a methyl ketone, } \mathrm{CH_3COR})$
or a $\mathrm{CH_3\!-\!CH(OH)\!-\!}$ group (a methyl carbinol, which the reagent first oxidises to the methyl ketone). Ethanol and acetaldehyde also work for the same reason.

Step 2 — Why that group.

The three hydrogens on the methyl adjacent to $\mathrm{C{=}O}$ are acidic. In base they are removed one by one and replaced by halogen, giving $\mathrm{CX_3CO}\!-\!$ . The strongly electron-withdrawing $\mathrm{CX_3}$ makes $\mathrm{CX_3^-}$ a good leaving group, so $\mathrm{OH^-}$ attacks the carbonyl and expels $\mathrm{CHX_3}$ (the haloform):
$\mathrm{CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3\downarrow + CH_3COONa + 3NaI + 3H_2O}$
Step 3 — Screen the options. …
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Acid	Structure	$\alpha$ -carbon	$\alpha$ -H present?
Ethanoic acid	$CH_3\text{--}COOH$	the $CH_3$ carbon	Yes — 3 of them

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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 …

Step 1 — The mechanism of tetrahalide hydrolysis

Step 2 — Which elements have those $d$ -orbitals?

Step 3 — The supporting steric argument …

Step 1 — Recall the history

Step 2 — Screen the other options

1. How haemoglobin normally works

2. What CO does

3. Why the others fail …

1. What the HVZ reaction actually does

2. Why an $\alpha$ -H is mandatory

3. Apply it to the two acids

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

Numbering the chain

Both carbocations are secondary

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