Suggest a reagent for conversion of ethanol to ethanoic acid.

NCERT Exemplar · Q25

Q. Suggest a reagent for conversion of ethanol to ethanoic acid.

Gujarat GsebShort· 2mImportance★★★★★

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Concept understanding — Electrophilic Addition Reactions

Electrophilic Aromatic Substitution – The First Meeting

Imagine you have a benzene ring — that perfect, flat hexagon of six carbons with alternating double bonds. It's stable, almost stubbornly so. You want to attach something new to it, say a bromine atom or a nitro group. But benzene doesn't react like an alkene. It doesn't just add across a double bond. Instead, it does something more elegant: it kicks out a hydrogen and keeps its aromatic ring intact.

That's the heart of Electrophilic Aromatic Substitution (EAS).

The Intuition: Why "Substitution" and Not "Addition"?

Benzene's stability comes from its delocalised $\pi$ electrons — a cloud above and below the ring. This cloud is electron-rich, so it attracts electrophiles (electron-loving species). But if an electrophile simply added to a double bond, the ring would break its aromaticity, losing that huge stabilisation. That would be energetically costly.

So benzene does something smarter: it lets the electrophile attack, temporarily breaks aromaticity to form a high-energy intermediate (the arenium ion), and then loses a proton to restore the aromatic ring. The net result? A hydrogen is replaced by the electrophile. The ring is back to its stable, aromatic self.

Note

The key trade-off: temporary loss of aromaticity is acceptable because the final product regains it. Addition reactions would permanently destroy aromaticity — benzene avoids that.

The Precise Statement

Electrophilic Aromatic Substitution is a reaction in which an electrophile ( $E^+$ ) replaces a hydrogen atom on an aromatic ring, proceeding through a sigma complex (arenium ion) intermediate, and restoring aromaticity after deprotonation.

The general equation:

\ce{Ar-H + E+ -> Ar-E + H+}

where Ar represents an aromatic ring.

The Mechanism in Three Steps

Generation of the electrophile – Many EAS reactions need a catalyst to create a strong enough $E^+$ . For example, bromination uses $\ce{FeBr3}$ to polarise $\ce{Br2}$ into $\ce{Br^+}$ .
Attack by the aromatic ring – The $\pi$ electrons of benzene attack the electrophile, forming a sigma complex (also called the arenium ion or Wheland intermediate). This intermediate is non-aromatic — it has four $\pi$ electrons delocalised over five carbons, and one $sp^3$ carbon bearing the electrophile and a hydrogen.

\text{Benzene} + E^+ \longrightarrow \text{Sigma complex (non-aromatic)}

Deprotonation – A base (often the counterion of the catalyst, like $\ce{FeBr4-}$ ) removes the proton from the $sp^3$ carbon. The pair of electrons from the C–H bond flows back into the ring, restoring the aromatic sextet.

\text{Sigma complex} + \text{Base} \longrightarrow \text{Product} + \ce{H+}

Watch out

The sigma complex is not aromatic. It's a high-energy intermediate. Students often mistakenly think it's still aromatic — it isn't. That's why the step is fast and the complex is short-lived.

Why This Matters for Exams

EAS is the gateway to understanding how to put groups onto benzene rings. The rate-determining step is usually the formation of the sigma complex (step 2). The regiochemistry (where the electrophile goes) depends on whether the ring already has a substituent — that's the topic of activating/deactivating groups and ortho/para vs. meta directors. …

Why this formula?

Electrophilic Addition Reactions: Why the Mechanism Works

Electrophilic addition is a cornerstone of alkene and alkyne chemistry. Instead of memorising the "arrow pushing," let's understand why the reaction proceeds the way it does — driven by electron density, stability, and charge.

1. The Core Idea: Why Alkenes React This Way

Alkenes have a π-bond — a cloud of electrons above and below the plane of the σ-bond. This π-electron cloud is:

Electron-rich (nucleophilic)
Exposed (not shielded by σ-bonds like in alkanes)

An electrophile (electron-lover) is attracted to this high electron density. The reaction is electrophilic addition because the electrophile attacks first.

Key principle: The π-bond acts as a Lewis base (electron donor). The electrophile is a Lewis acid (electron acceptor).

2. The General Mechanism (Two-Step)

Step 1: Formation of a Carbocation (or Bridged Intermediate)

The electrophile (E⁺) attacks the π-bond. The π-electrons form a new σ-bond to E⁺, leaving the other carbon with a positive charge — a carbocation.

\ce{C=C} + \ce{E+} \longrightarrow \ce{C+-C-E}

Why does this happen?

The π-bond is weaker than a σ-bond (~260 kJ/mol vs ~350 kJ/mol). Breaking the π-bond to form a σ-bond is energetically favourable because the new σ-bond is stronger. The carbocation is a high-energy intermediate, but it's stabilised by:

Hyperconjugation (alkyl groups donate electron density)
Inductive effect (alkyl groups push electrons toward the positive carbon)

Step 2: Nucleophilic Attack

A nucleophile (Nu⁻) attacks the carbocation, forming a second σ-bond.

\ce{C+-C-E} + \ce{Nu-} \longrightarrow \ce{C-Nu-C-E}

Why does this happen?

The carbocation is electron-deficient (positive charge). The nucleophile is electron-rich. Opposite charges attract — this is electrostatic and orbital overlap driven.

3. The Key "Formula" — Markovnikov's Rule

Statement: In the addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens already, and the X attaches to the carbon with fewer hydrogens.

Why does this rule hold? (The reasoning)

Consider propene: $\ce{CH3-CH=CH2}$ + HBr.

Possible carbocations:
1. Primary carbocation: $\ce{CH3-CH+-CH2Br}$ (less stable)
2. Secondary carbocation: $\ce{CH3-CHBr-CH2+}$ (more stable)

The more substituted carbocation (secondary > primary) is more stable due to:

Hyperconjugation: More alkyl groups = more C–H σ-bonds that can donate electron density into the empty p-orbital of the carbocation.
Inductive effect: Alkyl groups are electron-donating, stabilising the positive charge.

Result: The reaction proceeds via the more stable carbocation, leading to Markovnikov addition.

Markovnikov's rule is not a law — it's a consequence of carbocation stability.

4. The "Anti-Markovnikov" Exception (Why It Happens)

With HBr in the presence of peroxides (ROOR), the addition is anti-Markovnikov — Br goes to the less substituted carbon.

Why? The mechanism changes from ionic to free-radical.

Peroxide decomposes to radicals: $\ce{ROOR -> 2 RO•}$
RO• abstracts H from HBr: $\ce{RO• + HBr -> ROH + Br•}$
Br• adds to the alkene — at the less substituted carbon (because the radical formed is more stable — tertiary > secondary > primary).
The new radical abstracts H from another HBr, regenerating Br•. …

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The conversion of ethanol to ethanoic acid is an oxidation reaction. The key idea is to use a strong oxidizing agent that adds oxygen or removes hydrogen. The reagent is acidified potassium dichromate ( $K_2Cr_2O_7/H_2SO_4$ ) or acidified potassium permanganate ( $KMnO_4/H_2SO_4$ ).

This is a classic organic chemistry conversion that tests your understanding of oxidation reactions of alcohols. Ethanol is a primary alcohol, and ethanoic acid is a carboxylic acid. The journey from one to the other involves a controlled oxidation.

Why This Approach Works

The core concept is oxidation. In organic chemistry, oxidation of an alcohol means increasing the number of bonds to oxygen (or decreasing bonds to hydrogen). For a primary alcohol like ethanol ( $CH_3CH_2OH$ ), oxidation first gives an aldehyde (ethanal, $CH_3CHO$ ), and then further oxidation gives the carboxylic acid (ethanoic acid, $CH_3COOH$ ).

You cannot just "add" an oxygen atom directly. You need a reagent that can remove two hydrogen atoms from the alcohol and then add an oxygen atom to the resulting aldehyde. Strong oxidizing agents do exactly this.

Step-by-Step Reasoning

Identify the functional group change. Ethanol has an $-OH$ (alcohol) group. Ethanoic acid has a $-COOH$ (carboxylic acid) group. The carbon atom bonded to the $-OH$ in ethanol (the $C_1$ carbon) must be converted into the carboxyl carbon of the acid. This requires adding an oxygen atom and removing two hydrogen atoms from that carbon.
Recall the reagents for alcohol oxidation. For a primary alcohol, the most common and reliable oxidizing agents in your syllabus are:
- Acidified potassium dichromate ( $K_2Cr_2O_7/H_2SO_4$ ): This is the standard reagent. The orange dichromate ion ( $Cr_2O_7^{2-}$ ) is reduced to green $Cr^{3+}$ ions. The colour change is a key observation.
- Acidified potassium permanganate ( $KMnO_4/H_2SO_4$ ): This is a stronger oxidising agent. The purple permanganate ion ( $MnO_4^-$ ) is reduced to colourless $Mn^{2+}$ ions.
Understand the reaction mechanism (simplified). The reaction proceeds in two stages, but the reagent is used in excess so you don't isolate the intermediate aldehyde.
- Stage 1: Ethanol is oxidised to ethanal.

CH_3CH_2OH \xrightarrow{[O]} CH_3CHO + H_2O

    Here, $[O]$ represents the oxygen from the oxidising agent. …

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Common Mistakes: Ethanol → Ethanoic Acid (Electrophilic Addition Context)

Students often confuse this with electrophilic addition reactions (like alkene to alcohol), but this conversion is actually an oxidation reaction. Here are the most frequent errors:

✗ Mistake 1: Suggesting an electrophilic addition reagent

Wrong answer: Br₂/H₂O, H₂SO₄, H₂O/H⁺

Why it's wrong: These are reagents for electrophilic addition to alkenes/alkynes — they add atoms across a double bond. Ethanol has no π-bond; it's a saturated alcohol.

How to avoid:

First identify the functional group change: alcohol (–OH) → carboxylic acid (–COOH)
This is an oxidation (loss of H, gain of O), not an addition.
Memorise: Electrophilic addition requires a multiple bond; oxidation requires an oxidising agent.

✗ Mistake 2: Using a weak or incorrect oxidising agent

Wrong answer: KMnO₄ (without specifying conditions), H₂O₂, O₂ (alone)

Why it's wrong:

KMnO₄ in neutral/alkaline medium gives ethanal (aldehyde), not ethanoic acid.
H₂O₂ or O₂ alone are too weak to oxidise ethanol directly to acid under normal conditions.

How to avoid:

For primary alcohol → carboxylic acid, you need a strong oxidising agent under acidic conditions.
Correct choice: Acidified potassium dichromate ( $K_2Cr_2O_7/H_2SO_4$ ) or acidified potassium permanganate ( $KMnO_4/H_2SO_4$ ).

✗ Mistake 3: Forgetting the colour change / observation

Wrong: Writing only the reagent without mentioning the visual change.

Why it's important: Exam questions often ask for "reagent and observation."

How to avoid:

With $K_2Cr_2O_7/H_2SO_4$ : orange → green (Cr⁶⁺ → Cr³⁺) …

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

GUJCET 2025Set 031 markMCQ
$C_{6}H_{6} + X \xrightarrow{\text{anhydrous } AlCl_{3}} \text{Isopropylbenzene} \xrightarrow[(ii)H^{+}]{(i)O_{2}} Y + Z$
›Reveal solutionSolution

The products $Y,Z$ are fixed as phenol and acetone (cumene process); the reagent $X$ must give cumene under anhydrous $AlCl_3$ .

Cumene $\xrightarrow{(i)O_2,\ (ii)H^+}$ phenol $(C_6H_5OH)$ + acetone $(CH_3COCH_3)$ , so $Y=C_6H_5OH$ , $Z=CH_3COCH_3$ — eliminating (B) and (D). In Friedel–Crafts alkylation, $CH_3CH_2CH_2Cl$ (n-propyl chloride) rearranges via a $1^\circ\!\to\!2^\circ$ carbocation shift to the isopropyl grou …
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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. Name the product obtained in the following reaction? CH3-CH2-CH2-CH=CH2 (Pent-1-ene) --(BH3)2 / H2O2/OH^- --> ?$
(a) Pentan-3-ol
(b) Pentan-2-ol
(c) Pentan-1-ol
(d) 2-methyl butan-2-ol
›Reveal solutionSolution
Hydroboration-oxidation of pent-1-ene gives the anti-Markovnikov product pentan-1-ol.

Reagent (BH3)2 / H2O2, OH- is hydroboration-oxidation. It adds H and OH across the double bond with:

anti-Markovnikov orientation (OH goes to the less substituted, terminal carbon), …
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Q.Suggest a reagent for conversion of ethanol to ethanoic acid.

Concept understanding — Electrophilic Addition Reactions

Electrophilic Aromatic Substitution – The First Meeting

The Intuition: Why "Substitution" and Not "Addition"?

The Precise Statement

The Mechanism in Three Steps

Why This Matters for Exams

Why this formula?

Electrophilic Addition Reactions: Why the Mechanism Works

1. The Core Idea: Why Alkenes React This Way

2. The General Mechanism (Two-Step)

Step 1: Formation of a Carbocation (or Bridged Intermediate)

Step 2: Nucleophilic Attack

3. The Key "Formula" — Markovnikov's Rule

Why does this rule hold? (The reasoning)

4. The "Anti-Markovnikov" Exception (Why It Happens)

Why This Approach Works

Step-by-Step Reasoning

Method: Controlled Oxidation of Primary Alcohols

Concept First

Steps

Common Mistakes: Ethanol → Ethanoic Acid (Electrophilic Addition Context)

✗ Mistake 1: Suggesting an electrophilic addition reagent

✗ Mistake 2: Using a weak or incorrect oxidising agent

✗ Mistake 3: Forgetting the colour change / observation

Q.Suggest a reagent for conversion of ethanol to ethanoic acid.

Why This Approach Works

Step-by-Step Reasoning

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