Ethylidene chloride is a/an _______. (i) vic-dihalide (ii) gem-dihalide (iii) allylic halide (iv) vinylic hali

NCERT Exemplar · Q14

Q. Ethylidene chloride is a/an ______________.

(i) vic-dihalide

(ii) gem-dihalide

(iii) allylic halide

(iv) vinylic halide

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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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Ethylidene chloride is the common name for 1,1-dichloroethane, where both chlorine atoms are attached to the same carbon — making it a gem-dihalide. The correct option is (ii).

The first thing to understand is what the name "ethylidene chloride" actually tells you. In older nomenclature, "ethylidene" is the divalent group $\mathrm{CH_3CH\!<}$ — ethane with two hydrogens removed from the same terminal carbon (contrast "ethyl", $\mathrm{CH_3CH_2\!-}$ , formed by removing just one). Both free valencies sit on that single carbon, so "ethylidene chloride" means both are occupied by chlorine atoms — two single C–Cl bonds on the same carbon.

So the structure is $\mathrm{CH_3CHCl_2}$ . That's 1,1-dichloroethane.

Now, the classification of dihalides depends on where the two halogen atoms are located:

gem-Dihalides (geminal): both halogens on the same carbon atom. Example: $\mathrm{CH_3CHCl_2}$ .
vic-Dihalides (vicinal): halogens on adjacent carbon atoms. Example: $\mathrm{CH_2ClCH_2Cl}$ (1,2-dichloroethane).
Allylic halides: halogen attached to a carbon adjacent to a carbon-carbon double bond (allylic position). Example: $\mathrm{CH_2=CHCH_2Cl}$ .
Vinylic halides: halogen attached directly to a carbon of a carbon-carbon double bond. Example: $\mathrm{CH_2=CHCl}$ . …

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Here’s a breakdown of the common mistakes students make on this question, along with how to avoid each.

1. Confusing “Ethylidene” with “Ethylene”

The Mistake: Students see “ethyl” and immediately think of a two-carbon chain with a double bond (like ethene). This leads them to incorrectly classify the compound as a vinylic halide (option (iv)).
Why It’s Wrong: “Ethylidene” is the divalent group $\text{CH}_3\text{CH}<$ — a carbon carrying two free valencies (two H removed from the same carbon of ethane). In ethylidene chloride those two valencies hold two chlorine atoms through two single C–Cl bonds. The compound is $\text{CH}_3\text{CHCl}_2$ , which has no double bond of any kind — neither C=C nor C=Cl.
How to Avoid: Memorise the naming pattern:
- -yl = alkyl group (single bond).
- -ylidene = two hydrogens removed from the same carbon, leaving a divalent group ( $\text{CH}_3\text{CH}<$ ); its two valencies can form one double bond to a single atom or, as here, two single bonds to two separate atoms.
- -yne or -ene = carbon-carbon multiple bonds. Always draw the structure before classifying.

2. Mixing Up gem-dihalides and vic-dihalides

The Mistake: Students think any dihalide on adjacent carbons is vic, and any on the same carbon is gem. But they forget to check the carbon skeleton.
Why It’s Wrong: In ethylidene chloride ( $\text{CH}_3\text{CHCl}_2$ ), both chlorine atoms are attached to the same carbon. That makes it a gem-dihalide (option (ii)), not a vic-dihalide (which requires Cl on two adjacent carbons, e.g., $\text{ClCH}_2\text{CH}_2\text{Cl}$ ).
How to Avoid: Use the mnemonic:
- Gemini (twins) → same carbon.
- Vicinity (neighbours) → adjacent carbons. Draw the structure and number the carbons. If both halogens are on carbon #1, it’s gem.

3. Assuming “Chloride” Means a Single Chlorine Atom

The Mistake: Students see “chloride” and think only one Cl is present, leading them to classify it as an allylic halide (option (iii)) or vinylic halide.
Why It’s Wrong: There is no “di” anywhere in the common name — the two chlorines follow from the fact that ethylidene is a divalent group. “Ethylidene” ( $\text{CH}_3\text{CH}<$ ) carries two free valencies on the same carbon (exactly as Mistake 1 explains), and “chloride” tells you what occupies them — so there must be two Cl atoms, both on that carbon. The IUPAC name, 1,1-dichloroethane, makes the count explicit.
How to Avoid: Always expand the name systematically:
- Ethylidene = $\text{CH}_3\text{CH}<$ (a divalent group — two free valencies on one carbon)
- Chloride = Cl occupying those two valencies → two Cl atoms. Write the molecular formula: $\text{C}_2\text{H}_4\text{Cl}_2$ . Then draw the structure — both Cl on the same carbon: $\text{CH}_3\text{CHCl}_2$ .

4. Forgetting the Definition of Allylic and Vinylic Positions …

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

CBSE 2024Set 56/1/11 markMCQ
$PCl_5$
›Reveal solutionSolution
Acetic acid reacts with phosphorus pentachloride ( $PCl_5$ ) by replacing the hydroxyl group (-OH) with a chlorine atom, forming acetyl chloride. The correct option is (C).

Carboxylic acids are characterized by the presence of a carboxyl group ( $-COOH$ ). The hydroxyl group ( $-OH$ ) within this carboxyl group can be replaced by other atoms or groups, leading to various carboxylic acid derivatives. One common and important reaction is the conversion of a carboxylic acid into an acyl chloride. This transformation is particularly useful because acyl chlorides are highly reactive and serve as excellent intermediates for synthesizing other carboxylic acid derivatives like esters, amides, and anhydrides.

Phosphorus pentachloride ( $PCl_5$ ), along with thionyl chloride ( $SOCl_2$ ) and phosphorus trichloride ( $PCl_3$ ), are standard reagents used for this purpose. They act as chlorinating agents, effectively substituting the hydroxyl group of the carboxylic acid with a chlorine atom.

Identify the Reactants:

We are given acetic acid and $PCl_5$ .

Acetic acid has the chemical formula $CH_3COOH$ . It is a simple carboxylic acid with a methyl group attached to the carboxyl group.

$PCl_5$ is phosphorus pentachloride.

Understand the Role of $PCl_5$ :

$PCl_5$ is a powerful chlorinating agent. Its primary role in reactions with compounds containing hydroxyl groups (like alcohols or carboxylic acids) is to replace the $-OH$ group with a $-Cl$ atom. This reaction proceeds via a nucleophilic acyl substitution mechanism, where the oxygen of the hydroxyl group is protonated or activated, making it a better leaving group, and then displaced by a chloride ion.

Predict the Reaction Outcome:

In acetic acid, the hydroxyl group is part of the carboxyl group ( $CH_3C(=O)OH$ ). When acetic acid reacts with $PCl_5$ , the $-OH$ group will be replaced by a $-Cl$ atom. This results in the formation of an acyl chloride.
$PCl_5$
For acetic acid, $R = CH_3$ . Therefore, the reaction is:

$CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl$

Identify the Organic Product:

The organic product formed is $CH_3COCl$ . This compound is known as acetyl chloride. The inorganic byproducts are phosphorus oxychloride ( $POCl_3$ ) and hydrogen chloride ( $HCl$ ).

Compare with Options:

Let's examine the given options: …
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CBSE 2019Set ANNUAL1 mark
$Q. Write only the chemical equation of Hunsdiecker reaction.$
›Reveal solutionSolution

The Hunsdiecker reaction converts the silver salt of a carboxylic acid into an alkyl halide with loss of one carbon (as $CO_2$ ).

The dry silver salt of a carboxylic acid is heated with a halogen (usually bromine) in a solvent like carbon tetrachloride ( $CCl_4$ ). The silver salt loses $CO_2$ and silver halide is formed as a by-product, giving an alkyl halide containing one carbon atom less than the parent acid.

Chemical equation: …

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Q.Ethylidene chloride is a/an ______________.

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: Classification of Dihalides Based on Halogen Position

Method: Position-of-Halogens Rule

1. Confusing “Ethylidene” with “Ethylene”

2. Mixing Up gem-dihalides and vic-dihalides

3. Assuming “Chloride” Means a Single Chlorine Atom

4. Forgetting the Definition of Allylic and Vinylic Positions …

Q.Ethylidene chloride is a/an ______________.

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