Name the reagents used in the following reactions: (i) Oxidation of a primary alcohol to carboxylic acid. (ii)

Exercises · 7.21

Q. Name the reagents used in the following reactions:

(i) Oxidation of a primary alcohol to carboxylic acid.

(ii) Oxidation of a primary alcohol to aldehyde.

(iii) Bromination of phenol to 2,4,6-tribromophenol.

(iv) Benzyl alcohol to benzoic acid.

(v) Dehydration of propan-2-ol to propene.

(vi) Butan-2-one to butan-2-ol.

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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 key idea is to match each organic transformation with the specific reagent that achieves it cleanly. The answers are: (i) Acidified $\ce{K2Cr2O7}$ or $\ce{KMnO4}$ ,

(ii) PCC (pyridinium chlorochromate),

(iii) Bromine water,

(iv) Acidified $\ce{KMnO4}$ ,

(v) Concentrated $\ce{H2SO4}$ at 443 K,

(vi) $\ce{NaBH4}$ or $\ce{LiAlH4}$ .

Let’s go through each one. The logic here is simple: you need to know which reagent does what, and why. For oxidation, the strength of the oxidising agent and the reaction conditions decide whether you stop at an aldehyde or go all the way to a carboxylic acid. For the others, it’s about recognising characteristic reactions — like how phenol is so reactive that bromine water gives a tribromo product instantly, or how ketones are reduced to secondary alcohols.

1. Oxidation of a primary alcohol to carboxylic acid.

A primary alcohol ( $\ce{R-CH2OH}$ ) can be oxidised in two stages: first to an aldehyde ( $\ce{R-CHO}$ ), then to a carboxylic acid ( $\ce{R-COOH}$ ). To get the acid directly, you need a strong oxidising agent that doesn’t stop at the aldehyde. The classic choices are acidified potassium dichromate ( $\ce{K2Cr2O7/H2SO4}$ ) or acidified potassium permanganate ( $\ce{KMnO4/H2SO4}$ ). Both are strong enough to push the oxidation all the way.

Tip

In the lab, $\ce{K2Cr2O7/H2SO4}$ is more common because it gives a clear colour change from orange to green as it gets reduced to $\ce{Cr^{3+}}$ .

2. Oxidation of a primary alcohol to aldehyde.

Here you want to stop at the aldehyde stage. If you use a strong oxidising agent, it will over-oxidise. So you need a milder, specific reagent. The go-to is PCC (pyridinium chlorochromate, $\ce{C5H5NH+CrO3Cl-}$ ) in anhydrous conditions (e.g., in dichloromethane). It oxidises primary alcohols to aldehydes without touching them further. Another option is the Swern oxidation (DMSO + oxalyl chloride), but PCC is the standard for exams.

Watch out

Never use $\ce{K2Cr2O7}$ or $\ce{KMnO4}$ here — they will give the carboxylic acid, not the aldehyde.

3. Bromination of phenol to 2,4,6-tribromophenol.

Phenol is highly activated towards electrophilic substitution because the –OH group donates electrons into the ring. Bromine water ( $\ce{Br2/H2O}$ ) is a mild source of electrophilic bromine. The reaction is so fast that all three ortho and para positions get substituted instantly, giving a white precipitate of 2,4,6-tribromophenol. No catalyst is needed.

\ce{C6H5OH + 3Br2 -> 2,4,6-Br3C6H2OH + 3HBr}

4. Benzyl alcohol to benzoic acid.

Benzyl alcohol ( $\ce{C6H5CH2OH}$ ) is a primary alcohol attached to a benzene ring. To oxidise it to benzoic acid ( $\ce{C6H5COOH}$ ), you need a strong oxidising agent. Acidified potassium permanganate ( $\ce{KMnO4/H2SO4}$ ) works perfectly — it oxidises the – $\ce{CH2OH}$ group all the way to – $\ce{COOH}$ . Note that the benzene ring is unaffected under these conditions. …

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Here is the clear solution method for the given reactions, following a Concept-First approach.

Method: Reagent Mapping for Functional Group Transformations

This method relies on identifying the starting functional group and the target functional group, then selecting the specific reagent that achieves that transformation without over-oxidizing or under-reacting.

Steps for Each Reaction

(i) Oxidation of a primary alcohol to carboxylic acid.

Concept: Primary alcohols ( $RCH_2OH$ ) can be oxidized to carboxylic acids ( $RCOOH$ ) using a strong oxidizing agent that does not stop at the aldehyde stage.
Reagent: Acidified potassium dichromate ( $K_2Cr_2O_7/H_2SO_4$ ) or Acidified potassium permanganate ( $KMnO_4/H_2SO_4$ ).
Why: The dichromate ion ( $Cr_2O_7^{2-}$ ) in acidic medium provides a strong oxidizing environment that pushes the reaction past the aldehyde to the acid.

(ii) Oxidation of a primary alcohol to aldehyde.

Concept: To stop at the aldehyde ( $RCHO$ ), you must use a milder oxidizing agent that cannot further oxidize the aldehyde to the acid.
Reagent: PCC (Pyridinium Chlorochromate) in anhydrous conditions (e.g., $CH_2Cl_2$ ).
Why: PCC is a specific, mild oxidant that converts primary alcohols to aldehydes without over-oxidation. (Note: $K_2Cr_2O_7$ would give the acid here).

(iii) Bromination of phenol to 2,4,6-tribromophenol.

Concept: Phenol ( $C_6H_5OH$ ) is highly activated towards electrophilic substitution due to the -OH group. Bromine water is sufficient; no catalyst (like $FeBr_3$ ) is needed.
Reagent: Bromine water ( $Br_2/H_2O$ ).
Why: The reaction is so fast that a dilute solution of bromine water instantly gives a white precipitate of 2,4,6-tribromophenol.

(iv) Benzyl alcohol to benzoic acid.

Concept: Benzyl alcohol ( $C_6H_5CH_2OH$ ) is a primary alcohol attached to a benzene ring. Strong oxidation cleaves the side chain to the carboxylic acid.
Reagent: Alkaline potassium permanganate ( $KMnO_4/KOH$ ) followed by acidification, or Acidified $K_2Cr_2O_7$ .
Why: The strong oxidizing agent ( $KMnO_4$ ) oxidizes the $CH_2OH$ group directly to $COOH$ .

(v) Dehydration of propan-2-ol to propene.

Concept: Dehydration of an alcohol ( $ROH$ ) to an alkene ( $RCH=CH_2$ ) requires an acid catalyst and heat to remove a water molecule.
Reagent: Concentrated sulphuric acid ( $H_2SO_4$ ) at 443 K (170°C). …

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Here are the common mistakes students make on this question, along with the concept-first reasoning to avoid each.

General Mistake: Confusing Reagent Names with Reaction Conditions

Students often write the chemical formula but forget the specific reagent name or medium (e.g., acid/base). In exams, the name must be exact.

(i) Oxidation of a primary alcohol to carboxylic acid.

Common Mistake: Writing only KMnO4 or K2Cr2O7 without specifying acidic medium.

Why it’s wrong: In neutral/alkaline medium, KMnO4 can stop at aldehyde or give incomplete oxidation.
Correct Reagent: Acidified potassium dichromate ( $K_2Cr_2O_7/H_2SO_4$ ) or acidified potassium permanganate ( $KMnO_4/H_2SO_4$ ).
How to avoid: Remember: Carboxylic acid from primary alcohol needs strong oxidizing agent in acidic medium. Always write “acidified” or “ $H_2SO_4$ ”.

(ii) Oxidation of a primary alcohol to aldehyde.

Common Mistake: Using the same strong oxidant as in (i) (e.g., acidified $K_2Cr_2O_7$ ).

Why it’s wrong: Strong oxidants over-oxidize the aldehyde to carboxylic acid.
Correct Reagent: PCC (Pyridinium chlorochromate) or Collins reagent ( $CrO_3 \cdot 2Py$ ).
How to avoid: Link aldehyde with mild, anhydrous oxidants. PCC is the exam favourite. Do not write $K_2Cr_2O_7$ here.

(iii) Bromination of phenol to 2,4,6-tribromophenol.

Common Mistake: Writing $Br_2$ alone, or $Br_2/FeBr_3$ (as in benzene bromination).

Why it’s wrong: Phenol is highly activated — it does not need a Lewis acid catalyst. $Br_2$ alone in water works instantly.
Correct Reagent: Bromine water ( $Br_2/H_2O$ ).
How to avoid: Remember: Phenol + Br2 water → white precipitate of tribromophenol. No catalyst needed.

(iv) Benzyl alcohol to benzoic acid.

Common Mistake: Writing $K_2Cr_2O_7/H_2SO_4$ (which works, but is not the specific reagent often expected).

Why it’s wrong: While acidified dichromate works, the question expects a strong oxidizing agent that is commonly used for this specific conversion.
Correct Reagent: Alkaline potassium permanganate ( $KMnO_4/KOH$ ) followed by acidification, or acidified $KMnO_4$ .
How to avoid: For benzyl alcohol → benzoic acid, think of $KMnO_4$ (hot, conc.) as the standard. Write “alkaline $KMnO_4$ ” or “acidified $KMnO_4$ ”.

(v) Dehydration of propan-2-ol to propene.

Common Mistake: Writing only “heat” or “ $H_2SO_4$ ” without specifying concentrated.

Why it’s wrong: Dilute acid does not cause dehydration; it may just dissolve the alcohol.
Correct Reagent: Concentrated sulphuric acid ( $Conc. H_2SO_4$ ) at 170°C (or Alumina $Al_2O_3$ at 350°C).
How to avoid: Always write “conc. $H_2SO_4$ ” and mention heat (170°C). For exam, “conc. $H_2SO_4$ , heat” is sufficient.

(vi) Butan-2-one to butan-2-ol. …

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Q.Name the reagents used in the following reactions:

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