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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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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