Aldehydes Ketones Carboxylic Acids JEE Notes 2026 | Reactions, Mechanisms, PYQs & Trends

1

Structure of Carbonyl & Carboxyl Group

Aldehydes, ketones, and carboxylic acids are among the most important classes of organic compounds containing the carbonyl group (C=O). This chapter is one of the highest-weightage chapters in JEE Organic Chemistry, with 2-3 questions appearing in almost every session. Understanding the structure and reactivity of the carbonyl group is the foundation for mastering all subsequent reactions.

1.1 Structure of the Carbonyl Group (C=O)

Key Structural Features of C=O

The carbonyl group consists of a carbon atom double-bonded to an oxygen atom (C=O).

Hybridization: Both C and O are sp² hybridized

Bond angle: Approximately 120° (trigonal planar geometry)

Bond length: C=O bond length = 1.23 Å (shorter than C–O single bond = 1.43 Å)

Electronegativity difference: Oxygen (3.5) is more electronegative than carbon (2.5), creating a polar bond

\stackrel{\delta+}{C} = \stackrel{\delta-}{O}

The carbonyl carbon is electrophilic (partially positive) and the oxygen is nucleophilic (partially negative). This polarity is the driving force for nucleophilic addition reactions.

Important for JEE — Resonance in Carbonyl Group

The carbonyl group shows two resonance structures:

R-\overset{+}{C}(=O)-R' \longleftrightarrow R-\overset{-}{C}(-\overset{+}{O})-R'

The second resonance structure (with charges) is a minor contributor but explains the electrophilic character of carbonyl carbon and nucleophilic character of oxygen.

1.2 Aldehydes vs Ketones — Structural Comparison

Property	Aldehydes (RCHO)	Ketones (RCOR')
General Formula	R–CHO (R = H or alkyl/aryl)	R–CO–R' (R, R' = alkyl/aryl)
Carbonyl C bonded to	At least one H	Two carbon groups (no H on C=O)
Functional group	–CHO (aldehydic)	–CO– (ketonic)
IUPAC suffix	-al	-one
Reactivity (N.A.)	More reactive	Less reactive
Oxidation	Easily oxidized to acids	Resist oxidation (need strong oxidants)
Tollens' test	Positive (silver mirror)	Negative
Fehling's test	Positive (red ppt) — except aromatic	Negative

1.3 Structure of the Carboxyl Group (–COOH)

Carboxyl Group Structure

The carboxyl group (–COOH) is a combination of:

A carbonyl group (C=O)
A hydroxyl group (–OH)

R-C\underset{\displaystyle OH}{\overset{\displaystyle O}{=}}

Hybridization: Carbon is sp² hybridized. The –OH oxygen is also sp² hybridized.

Resonance: The lone pair on –OH oxygen participates in resonance with C=O, creating partial double bond character for C–OH bond:

R-C(=O)-OH \longleftrightarrow R-C(-O^{-})=\overset{+}{O}H

Due to this resonance, the C–OH bond in carboxylic acids has partial double bond character, making it stronger and less reactive than a typical C–OH single bond. This is why –OH is not easily replaced in carboxylic acids (unlike alcohols).

Key Insight — Why Carboxylic Acids Are Acidic

The acidic nature of carboxylic acids is due to:

Resonance stabilization of the carboxylate anion (RCOO⁻)
The negative charge is delocalized over two equivalent oxygen atoms
Both C–O bonds in RCOO⁻ have equal bond length (1.27 Å) — between single (1.43 Å) and double (1.23 Å)

R-COO^- \longleftrightarrow R-C(=O)-O^- \longleftrightarrow R-C(-O^-)=O

PYQ — JEE Main 2024

Question: The C–O bond length in the carboxylate ion (RCOO⁻) is:

(a) Same as C=O bond length (1.23 Å)

(b) Same as C–O single bond length (1.43 Å)

(c) Intermediate between C=O and C–O bond lengths (~1.27 Å)

(d) Longer than C–O single bond

Solution:

In the carboxylate ion (RCOO⁻), the negative charge is delocalized over both oxygen atoms through resonance.

Due to resonance, both C–O bonds have partial double bond character.

This makes both bonds equivalent with a bond length that is:

- Greater than C=O (1.23 Å) — because it's not a full double bond

- Less than C–O (1.43 Å) — because it has partial double bond character

The measured bond length is 1.27 Å, which is intermediate between the two.

\text{C–O bond in RCOO}^- \approx 1.27 \text{ Å (intermediate)}

Answer: (c) Intermediate between C=O and C–O bond lengths (~1.27 Å)

2

Preparation of Aldehydes & Ketones

There are numerous methods for preparing aldehydes and ketones, ranging from oxidation of alcohols to specialized name reactions. JEE frequently tests these preparation methods, especially the name reactions like Rosenmund reduction, Stephen reaction, Etard reaction, and Gattermann-Koch reaction.

2.1 Preparation of Aldehydes

Important — Most Repeated PYQ Topic

Rosenmund Reduction: Reduction of acid chlorides to aldehydes

R-COCl + H_2 \xrightarrow{Pd/BaSO_4} R-CHO + HCl

Key Points:

Catalyst: Pd on BaSO₄ (BaSO₄ acts as poison to prevent over-reduction to alcohol)
Also called Rosenmund-von Braun reaction
Formaldehyde (HCHO) cannot be prepared by this method (no HCOCl exists)
Can be used for both aliphatic and aromatic aldehydes

Stephen Reaction (Reduction of Nitriles)

R-C\equiv N + SnCl_2 + HCl \rightarrow R-CH=NH \xrightarrow{H_3O^+} R-CHO

Key Points:

Reduces nitriles (cyanides) to aldehydes
Intermediate: imine (R-CH=NH), which is hydrolyzed to aldehyde
Uses SnCl₂/HCl (mild reducing agent)
Useful for preparing both aliphatic and aromatic aldehydes

Etard Reaction (Aromatic Aldehydes Only)

\text{C}_6\text{H}_5\text{CH}_3 + \text{CrO}_2\text{Cl}_2 \rightarrow \text{C}_6\text{H}_5\text{CHO} \text{ (via chromium complex)}

Key Points:

Toluene is oxidized to benzaldehyde
Reagent: Chromyl chloride (CrO₂Cl₂) in CS₂
Intermediate: chromium complex, hydrolyzed to give aldehyde
Only methyl group attached to benzene ring is oxidized
Side-chain oxidation stops at aldehyde stage (not over-oxidized to acid)

High PYQ Frequency — Gattermann-Koch Reaction

Gattermann-Koch Reaction: Formylation of benzene

\text{C}_6\text{H}_6 + CO + HCl \xrightarrow{AlCl_3/CuCl} \text{C}_6\text{H}_5CHO

Key Points:

Benzene is treated with CO and HCl in presence of anhydrous AlCl₃ and CuCl
Produces benzaldehyde
Equivalent to introducing a –CHO group on aromatic ring
CuCl acts as co-catalyst
Note: This is NOT Gattermann reaction (which uses HCN/HCl and is for phenols/ethers)

DIBAL-H Reduction (From Esters/Nitriles)

R-C\equiv N \xrightarrow{DIBAL-H} R-CHO\] \[R-COOR' \xrightarrow{DIBAL-H} R-CHO

DIBAL-H (Diisobutylaluminium hydride) reduces nitriles and esters to aldehydes at low temperature (–78°C).

2.2 Preparation of Ketones

From Acid Chlorides + Dialkylcadmium

2R-COCl + R_2''Cd \rightarrow 2R-CO-R'' + CdCl_2

Key Points:

Acid chlorides react with dialkylcadmium (R₂Cd) to give ketones
Grignard reagent cannot be used (it over-reacts to give 3° alcohol)
Dialkylcadmium is less reactive than Grignard — stops at ketone stage
Useful for preparing unsymmetrical ketones

From Nitriles (with Grignard Reagent)

R-C\equiv N + R'MgX \rightarrow R-C(=NMgX)-R' \xrightarrow{H_3O^+} R-CO-R'

Grignard reagent adds to nitrile; hydrolysis gives ketone.

Friedel-Crafts Acylation (Aromatic Ketones)

\text{C}_6\text{H}_6 + CH_3COCl \xrightarrow{AlCl_3} \text{C}_6\text{H}_5COCH_3 + HCl

Key Points:

Aromatic ketones prepared by Friedel-Crafts acylation
Acyl chloride (RCOCl) + benzene + AlCl₃ → aryl ketone
No poly-substitution (–COR is deactivating)
Product is always a phenyl ketone (Ar–CO–R)

From Alcohols (Oxidation)

R-CH_2OH \xrightarrow{PCC} R-CHO \text{ (aldehyde)}\] \[R_2CHOH \xrightarrow{K_2Cr_2O_7/H^+} R-CO-R \text{ (ketone)}

Key Points:

PCC (Pyridinium chlorochromate) oxidizes 1° alcohol to aldehyde (stops at aldehyde)
Collins reagent (CrO₃·2C₅H₅N) also stops at aldehyde
K₂Cr₂O₇/H⁺ or KMnO₄/H⁺ over-oxidize 1° alcohol to carboxylic acid
2° alcohols are oxidized to ketones (no over-oxidation)
Oppenauer oxidation: Uses Al[OCH(CH₃)₂]₃ — selectively oxidizes 1° alcohol to aldehyde

2.3 From Hydrocarbons

Ozonolysis of Alkenes

R-CH=CH-R' + O_3 \rightarrow \text{ozonide} \xrightarrow{Zn/H_2O} R-CHO + R'-CHO\] \[R_2C=CR_2 + O_3 \rightarrow R_2CO + R_2CO \text{ (ketones)}

Ozonolysis followed by reductive workup (Zn/H₂O) gives aldehydes and/or ketones depending on the substitution pattern of the alkene. Use of H₂O₂ (oxidative workup) would over-oxidize aldehydes to acids.

Hydration of Alkynes

HC\equiv CH + H_2O \xrightarrow{HgSO_4/H_2SO_4} CH_3CHO \text{ (acetaldehyde)}\] \[RC\equiv CH + H_2O \xrightarrow{HgSO_4/H_2SO_4} R-CO-CH_3 \text{ (ketone)}

Key Points:

Acetylene gives acetaldehyde (only case giving aldehyde)
All other terminal alkynes give methyl ketones (Markovnikov addition)
Internal alkynes give mixtures of ketones
Follows Markovnikov's rule: H goes to carbon with more H's

PYQ — JEE Main 2024 (Jan 30 Shift 2)

Question: In Rosenmund reduction, the role of BaSO₄ in Pd/BaSO₄ catalyst is to:

(a) Increase the reactivity of the catalyst

(b) Act as a poison to prevent over-reduction to alcohol

(c) Provide acidic medium

(d) Reduce the acid chloride to aldehyde

Solution:

Rosenmund Reduction: RCOCl + H₂ → RCHO + HCl

Catalyst: Pd supported on BaSO₄

Without BaSO₄, Pd would catalyze over-reduction:

RCOCl + 2H₂ → RCH₂OH + HCl (unwanted alcohol formation)

BaSO₄ acts as a poison/catalyst modifier that:

- Reduces the activity of Pd catalyst

- Prevents further reduction of aldehyde to alcohol

- Stops the reaction at the aldehyde stage

Similarly, thioquinanthrene or quinoline/S (Lindlar-type poison) can also be used.

\text{BaSO}_4 \text{ acts as poison to prevent over-reduction}

Answer: (b) Act as a poison to prevent over-reduction to alcohol

Common Confusion — Gattermann vs Gattermann-Koch

Gattermann-Koch: Benzene + CO + HCl + AlCl₃/CuCl → Benzaldehyde (only for aromatic hydrocarbons)
Gattermann: Phenol/ether + HCN + HCl + AlCl₃ → forms imine → hydrolysis → aldehyde (for activated aromatic rings)
Don't confuse the two — Gattermann-Koch is specifically for benzene, Gattermann is for phenols/ethers

3

Nucleophilic Addition Reactions

Nucleophilic addition is the most characteristic reaction of aldehydes and ketones. The electrophilic carbonyl carbon (δ+) is attacked by nucleophiles, and the resulting alkoxide is protonated. Understanding the mechanism and reactivity order is crucial for JEE.

3.1 General Mechanism of Nucleophilic Addition

Most Important — Mechanism (High PYQ Frequency)

\text{Step 1: } Nu^- \text{ attacks } C^{\delta+} \rightarrow \text{tetrahedral alkoxide}\] \[\text{Step 2: } \text{Protonation of } O^- \rightarrow \text{addition product}

Detailed Mechanism:

The nucleophile (:Nu⁻) attacks the electrophilic carbonyl carbon (Cδ+)
The π bond breaks; electrons move to oxygen, forming an alkoxide ion
The carbon changes from sp² (trigonal planar) to sp³ (tetrahedral)
The alkoxide is protonated (by water/solvent) to give the addition product

R_2C=O + :Nu^- \rightarrow R_2C(Nu)-O^- \xrightarrow{H^+} R_2C(Nu)-OH

3.2 Reactivity Order — Aldehydes vs Ketones

Most Repeated PYQ — Reactivity Order

\text{Reactivity: } HCHO > CH_3CHO > CH_3CH_2CHO > CH_3COCH_3 > CH_3COR > R_2CO

Two factors govern reactivity:

1. Electronic (Inductive) Effect

Alkyl groups are +I groups (electron-donating). They increase electron density on carbonyl carbon, making it less electrophilic → less reactive.

Aldehydes (1 alkyl group) > Ketones (2 alkyl groups)

2. Steric Effect

Bulkier groups around C=O create steric hindrance, making it harder for nucleophile to attack.

Ketones have 2 groups → more hindered → less reactive

JEE Shortcut — Aromatic vs Aliphatic

Aromatic aldehydes (like C₆H₅CHO) are less reactive than aliphatic aldehydes because:

The phenyl group donates electrons by +R (resonance) effect to the carbonyl carbon
This reduces the electrophilic character of C=O
Also, the phenyl group is bulky → steric hindrance

Extended order: HCHO > CH₃CHO > C₆H₅CHO > CH₃COCH₃ > C₆H₅COCH₃ > C₆H₅COC₆H₅

3.3 Addition of HCN (Cyanohydrin Formation)

R-CHO + HCN \rightarrow R-CH(OH)-CN \text{ (cyanohydrin)}\] \[R-CO-R' + HCN \rightarrow R-C(OH)(CN)-R' \text{ (cyanohydrin)}

Key Points:

Product is a cyanohydrin (α-hydroxy nitrile)
HCN is a weak nucleophile; reaction is base-catalyzed
The CN⁻ ion is the actual nucleophile (generated by base)
Reactivity order: Aldehydes > Ketones (same as general order)
Ketones are less reactive due to steric and electronic reasons
Benzaldehyde does NOT form cyanohydrin easily (aromatic ring deactivates C=O)
Product has one more C atom — useful for chain extension

3.4 Addition of NaHSO₃ (Bisulfite Addition)

R-CHO + NaHSO_3 \rightarrow R-CH(OH)-SO_3Na \text{ (bisulfite adduct)}

Key Points:

NaHSO₃ is a strong nucleophile (HSO₃⁻ ion)
Product is a crystalline solid (used for purification/separation)
Adduct decomposes with acid or base to regenerate aldehyde/ketone
Reactivity: Aldehydes > Methyl ketones > Other ketones
Cannot add to: acetophenone (C₆H₅COCH₃), benzophenone (too bulky)

3.5 Addition of Ammonia Derivatives

Most Repeated PYQ — Ammonia Derivatives Table

Ammonia derivatives (derivatives of NH₃ where one or more H are replaced) react with aldehydes/ketones to form imines/Schiff bases (C=N–). The general reaction involves loss of water (condensation):

R_2C=O + H_2N-Z \rightarrow R_2C=N-Z + H_2O

Reagent (H₂N–Z)	Name	Product Name	Product Structure
NH₃	Ammonia	Imine (unstable)	R₂C=NH
H₂N–NH₂	Hydrazine	Hydrazone	R₂C=NNH₂
H₂N–NH-C₆H₅	Phenylhydrazine	Phenylhydrazone	R₂C=NNHC₆H₅
H₂N–OH	Hydroxylamine	Oxime	R₂C=NOH
H₂N–NHCONH₂	Semicarbazide	Semicarbazone	R₂C=NNHCONH₂
H₂N–NH-C₆H₃(NO₂)₂	2,4-DNP	2,4-Dinitrophenylhydrazone	R₂C=NNHC₆H₃(NO₂)₂

JEE Pro Tip — pH Dependence

The addition of ammonia derivatives is pH-dependent:

Optimal pH ≈ 5 (weakly acidic)
Too acidic: The amine (H₂N–Z) gets protonated → loses nucleophilicity
Too basic: The carbonyl is not activated → no reaction
At pH 5: carbonyl O is protonated (increasing electrophilicity of C), but amine N is still free to act as nucleophile

3.6 Addition of Grignard Reagent

R-CHO + R'MgX \rightarrow R-CH(OMgX)-R' \xrightarrow{H_3O^+} R-CH(OH)-R'\] \[R-CO-R' + R''MgX \rightarrow R-C(OMgX)(R')-R'' \xrightarrow{H_3O^+} R-C(OH)(R')-R''

Key Points:

Aldehyde + Grignard → 1° alcohol (from HCHO) or 2° alcohol (from RCHO)
Ketone + Grignard → 3° alcohol
HCHO + RMgX → RCH₂OH (1° alcohol with one extra C)
Reaction is irreversible (Grignard is a very strong nucleophile)

PYQ — JEE Main 2023

Question: The correct order of reactivity towards nucleophilic addition reaction is:

(a) HCHO < CH₃CHO < CH₃COCH₃ < C₆H₅COCH₃

(b) HCHO > CH₃CHO > CH₃COCH₃ > C₆H₅COCH₃

(c) C₆H₅COCH₃ > CH₃COCH₃ > CH₃CHO > HCHO

(d) CH₃CHO > HCHO > CH₃COCH₃ > C₆H₅COCH₃

Solution:

Factor 1: Electronic Effect

- HCHO: No alkyl group → most electrophilic C=O → most reactive

- CH₃CHO: One +I group (CH₃) → less electrophilic → less reactive

- CH₃COCH₃: Two +I groups → even less electrophilic → less reactive

- C₆H₅COCH₃: Phenyl group (+R effect) + CH₃ → least reactive

Factor 2: Steric Effect

- HCHO (smallest) → least hindered → most reactive

- C₆H₅COCH₃ (bulkiest) → most hindered → least reactive

Both factors give the same order:

\[HCHO > CH_3CHO > CH_3COCH_3 > C_6H_5COCH_3\]

Answer: (b) HCHO > CH₃CHO > CH₃COCH₃ > C₆H₅COCH₃

4

Reduction & Oxidation of Aldehydes and Ketones

Aldehydes and ketones undergo reduction to alcohols and (in the case of aldehydes) oxidation to carboxylic acids. The choice of reducing agent determines the product — and JEE tests this knowledge extensively.

4.1 Reduction to Alcohols

Catalytic Reduction (H₂/Ni, Pt, or Pd)

R-CHO + H_2 \xrightarrow{Ni} R-CH_2OH \text{ (1° alcohol)}\] \[R-CO-R' + H_2 \xrightarrow{Ni} R-CH(OH)-R' \text{ (2° alcohol)}

NaBH₄ (Sodium Borohydride) — Mild Reducing Agent

R-CHO + NaBH_4 \rightarrow R-CH_2OH \text{ (1° alcohol)}\] \[R-CO-R' + NaBH_4 \rightarrow R-CH(OH)-R' \text{ (2° alcohol)}

Key Points:

Reduces only aldehydes and ketones (selective)
Does NOT reduce esters, carboxylic acids, or amides
Milder than LiAlH₄
Source of H⁻ (hydride ion)

LiAlH₄ (Lithium Aluminium Hydride) — Strong Reducing Agent

R-CHO \xrightarrow{LiAlH_4} R-CH_2OH\] \[R-CO-R' \xrightarrow{LiAlH_4} R-CH(OH)-R'\] \[R-COOH \xrightarrow{LiAlH_4} R-CH_2OH\] \[R-COOR' \xrightarrow{LiAlH_4} R-CH_2OH + R'OH

Key Points:

Reduces aldehydes, ketones, carboxylic acids, esters, amides, nitriles
Does NOT reduce isolated C=C (double bond)
More powerful than NaBH₄
Reaction must be in dry ether (LiAlH₄ reacts violently with water)

4.2 Reduction to Hydrocarbons (C=O → CH₂)

High PYQ Frequency — Clemmensen vs Wolff-Kishner

Clemmensen Reduction

R-CO-R' \xrightarrow{Zn-Hg/HCl} R-CH_2-R'

Medium: Acidic (conc. HCl)
Reagent: Zn-Hg amalgam
NOT suitable for: Acid-sensitive compounds

Wolff-Kishner Reduction

R-CO-R' \xrightarrow{NH_2NH_2/KOH} R-CH_2-R'

Medium: Basic (KOH, ethylene glycol)
Reagent: Hydrazine (NH₂NH₂) + KOH
NOT suitable for: Base-sensitive compounds

Both reduce C=O to CH₂ — choice depends on compatibility with other functional groups in the molecule.

4.3 Oxidation of Aldehydes

Important — Aldehydes vs Ketones Oxidation

Aldehydes are easily oxidized to carboxylic acids by mild oxidizing agents:

R-CHO + [O] \rightarrow R-COOH

Common oxidizing agents:

Tollens' reagent [Ag(NH₃)₂]⁺ — gives silver mirror (Ag↓)
Fehling's solution — gives red-brown precipitate (Cu₂O)
Benedict's solution — similar to Fehling's
K₂Cr₂O₇/H⁺, KMnO₄/H⁺ — strong oxidants

Critical — Aromatic Aldehydes Exception

Benzaldehyde (C₆H₅CHO) does NOT give Fehling's test!

Fehling's test is positive for aliphatic aldehydes only
Aromatic aldehydes give Tollens' test but NOT Fehling's test
Reason: Fehling's oxidation is slower; the C₆H₅ group stabilizes C=O by resonance
Ketones are NOT oxidized by mild oxidants (need strong oxidants + heat → cleavage occurs)

4.4 Oxidation of Ketones (Baeyer-Villiger Oxidation)

Baeyer-Villiger Oxidation

R-CO-R' + mCPBA \rightarrow R-CO-O-R' \text{ (ester)}

Key Points:

Ketone is converted to an ester
Oxidant: Peracid (mCPBA, CF₃CO₃H, or peracetic acid)
Oxygen is inserted between C=O and one of the alkyl groups
The migrating group is the one with higher migratory aptitude: H > 3° alkyl > 2° > 1° > methyl

PYQ — JEE Main 2022

Question: Which of the following reagents can be used to convert acetaldehyde (CH₃CHO) to ethyl alcohol (CH₃CH₂OH)?

(a) Zn-Hg/HCl

(b) NH₂NH₂/KOH

(c) NaBH₄

(d) AlCl₃

Solution:

Analysis of each option:

(a) Zn-Hg/HCl is Clemmensen reduction — reduces C=O to CH₂ (gives ethane, not ethanol)

CH_3CHO \xrightarrow{Zn-Hg/HCl} CH_3CH_3 \text{ (ethane)}

(b) NH₂NH₂/KOH is Wolff-Kishner reduction — also reduces C=O to CH₂ (gives ethane)

CH_3CHO \xrightarrow{NH_2NH_2/KOH} CH_3CH_3

(c) NaBH₄ is a mild reducing agent — reduces C=O to CH-OH (gives ethanol) ✓

CH_3CHO \xrightarrow{NaBH_4} CH_3CH_2OH \text{ (ethanol)}

(d) AlCl₃ is a Lewis acid — used in Friedel-Crafts, not for reduction

\text{NaBH}_4 \text{ reduces aldehyde to 1° alcohol}

Answer: (c) NaBH₄

5

Aldol Condensation & Cannizzaro Reaction

These two reactions are the most frequently tested name reactions from this chapter in JEE. The key distinction: Aldol requires α-hydrogen, while Cannizzaro requires NO α-hydrogen. Mastering this distinction alone can fetch you marks in multiple questions.

5.1 Aldol Condensation

Most Repeated PYQ — Aldol Condensation

Definition: Two molecules of an aldehyde or ketone containing at least one α-hydrogen undergo condensation in the presence of a dilute base (like dil. NaOH, Na₂CO₃) or dilute acid to form a β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol).

2CH_3CHO \xrightarrow{dil.\ NaOH} CH_3-CH(OH)-CH_2-CHO \text{ (aldol)}

On heating, the aldol undergoes dehydration (loss of H₂O) to give an α,β-unsaturated aldehyde:

CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta} CH_3-CH=CH-CHO + H_2O

The complete process (aldol + dehydration) is called Aldol Condensation.

Mechanism of Aldol Condensation

Step 1 (Enolate formation): OH⁻ removes α-H from aldehyde, forming a resonance-stabilized enolate ion

CH_3CHO + OH^- \rightarrow ^-CH_2CHO + H_2O

Step 2 (Nucleophilic addition): Enolate ion (nucleophile) attacks the carbonyl carbon of another aldehyde molecule
Step 3 (Protonation): The alkoxide ion is protonated by water to give β-hydroxy aldehyde (aldol)
Step 4 (Dehydration): On heating, aldol loses water to form α,β-unsaturated compound

JEE Shortcut — Which Compounds Give Aldol?

Gives Aldol: Compounds with at least one α-H atom

✓ CH₃CHO (acetaldehyde) — 3 α-H
✓ CH₃COCH₃ (acetone) — 6 α-H
✓ CH₃CH₂CHO (propanal) — 2 α-H
✓ C₆H₅CH₂CHO (phenylacetaldehyde) — 2 α-H
✓ C₆H₅COCH₃ (acetophenone) — 3 α-H

Does NOT give Aldol: Compounds without α-H

✗ HCHO (formaldehyde)
✗ C₆H₅CHO (benzaldehyde)
✗ (CH₃)₃C-CHO (pivaldehyde/trimethylacetaldehyde)
✗ C₆H₅COC₆H₅ (benzophenone)

5.2 Crossed Aldol Condensation

When two different aldehydes/ketones undergo aldol condensation, it is called crossed aldol. The product is a mixture of 4 products (unless one has no α-H).

Best case: One reactant has no α-H (like HCHO or C₆H₅CHO) — it can only act as electrophile (not nucleophile), giving cleaner product.

C_6H_5CHO + CH_3CHO \xrightarrow{dil.\ NaOH} C_6H_5CH=CH-CHO \text{ (cinnamaldehyde)}

This is a crossed aldol between benzaldehyde (no α-H) and acetaldehyde (has α-H).

5.3 Cannizzaro Reaction

Most Repeated PYQ — Cannizzaro Reaction

Definition: Aldehydes that do NOT have α-hydrogen atoms undergo self-oxidation-reduction (disproportionation) when treated with concentrated alkali (NaOH/KOH). One molecule is reduced to alcohol and another is oxidized to carboxylic acid salt.

2HCHO + NaOH \rightarrow CH_3OH + HCOONa\] \[2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa

Compounds that undergo Cannizzaro Reaction

HCHO (formaldehyde) — no α-carbon, so no α-H
C₆H₅CHO (benzaldehyde) — α-carbon is aromatic C, has no H attached
(CH₃)₃C–CHO (pivaldehyde) — α-carbon is quaternary C, has no H
2,4,6-trinitrobenzaldehyde — electron-withdrawing groups prevent enolization
Furfural (furan-2-carbaldehyde)

JEE Shortcut — Aldol vs Cannizzaro (Decision Tree)

Step 1: Check if the compound has α-H
Step 2: If YES (α-H present) + dilute base → Aldol Condensation
Step 3: If NO (α-H absent) + concentrated base → Cannizzaro Reaction

Crossed Cannizzaro: When formaldehyde (HCHO) is mixed with another aldehyde without α-H, HCHO is preferentially oxidized (since it's more reactive) and the other is reduced.

PYQ — JEE Main 2024 (Jan 29 Shift 1)

Question: Which of the following compounds will undergo Cannizzaro reaction?

(a) CH₃CHO

(b) CH₃COCH₃

(c) C₆H₅CHO

(d) CH₃CH₂CHO

Solution:

Condition for Cannizzaro: The aldehyde must NOT have α-hydrogen

Check each option:

(a) CH₃CHO → α-carbon is CH₃ → has 3 α-H → gives ALDOL, not Cannizzaro

(b) CH₃COCH₃ → α-carbon is CH₃ → has 6 α-H → gives ALDOL (with acetone)

(c) C₆H₅CHO → α-carbon is the aromatic C of phenyl → NO α-H → gives CANNIZZARO ✓

(d) CH₃CH₂CHO → α-carbon is CH₂ → has 2 α-H → gives ALDOL

2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa

Answer: (c) C₆H₅CHO

Common Mistake — Aldol vs Cannizzaro

Mistake: Assuming aromatic aldehydes (like C₆H₅CHO) always give Cannizzaro
Fact: C₆H₅CH₂CHO (phenylacetaldehyde) HAS α-H (on CH₂) → gives ALDOL
Mistake: Confusing α-carbon with the carbon attached to functional group
Fact: α-carbon is the carbon directly adjacent to C=O. Check for H on THIS carbon.
Key distinction: C₆H₅CHO → Cannizzaro (no α-H), but C₆H₅CH₂CHO → Aldol (has α-H on CH₂)

6

Haloform Reaction & Other Important Reactions

The haloform reaction is an identification test as well as a synthetic method. Along with other reactions like Kolbe-Schmitt, Reimer-Tiemann, and electrophilic substitutions, these constitute the bulk of JEE name reaction questions.

6.1 Haloform Reaction

Most Repeated PYQ — Iodoform Test

Definition: Aldehydes/ketones with CH₃CO– group (methyl ketones) and alcohols that can be oxidized to CH₃CO– group react with NaOH + X₂ (or NaOX) to give a haloform (CHX₃) and a carboxylate salt.

CH_3COR + 3NaOI \rightarrow CI_3COR \xrightarrow{NaOH} CHI_3\downarrow + RCOONa\] \[\text{(yellow ppt of iodoform)}

Compounds giving Haloform/Iodoform test:

Methyl ketones (CH₃COR): CH₃COCH₃, CH₃COC₂H₅, C₆H₅COCH₃ (acetophenone)
Acetaldehyde (CH₃CHO) — the only aldehyde that gives iodoform test
Secondary alcohols with CH₃CH(OH)– group: CH₃CH(OH)CH₃ (isopropanol), CH₃CH(OH)C₂H₅
Ethanol (CH₃CH₂OH) — gets oxidized to CH₃CHO which gives iodoform

JEE Favorite Trap — Which does NOT give Iodoform?

Common wrong answers in JEE:

✗ HCHO — no CH₃ group → no iodoform
✗ C₆H₅CHO (benzaldehyde) — no CH₃CO group → no iodoform
✗ Diethyl ketone (CH₃CH₂COCH₂CH₃) — no CH₃CO group → no iodoform
✗ 1-Propanol (CH₃CH₂CH₂OH) — 1° alcohol without CH₃CH(OH) → no iodoform
✗ 3-Pentanol (CH₃CH₂CH(OH)CH₂CH₃) — no CH₃CH(OH) → no iodoform

6.2 Electrophilic Substitution Reactions of Aromatic Aldehydes

Kolbe-Schmitt Reaction

\text{C}_6\text{H}_5OH + CO_2 \xrightarrow{NaOH, 125°C, 4-7 atm} \text{C}_6\text{H}_4(OH)COONa \xrightarrow{H^+} \text{C}_6\text{H}_4(OH)COOH

This reaction produces salicylic acid (2-hydroxybenzoic acid) from sodium phenoxide and CO₂. The carboxyl group enters ortho position.

Reimer-Tiemann Reaction

\text{C}_6\text{H}_5OH + CHCl_3 + NaOH \rightarrow \text{C}_6\text{H}_4(OH)CHO \text{ (salicylaldehyde)}

This reaction introduces a –CHO group at the ortho position of phenol, producing salicylaldehyde. The reagent is chloroform (CHCl₃) and NaOH (dichlorocarbene intermediate).

6.3 Polymerization Reactions

Polymerization of Formaldehyde and Acetaldehyde

Paraformaldehyde: nHCHO → (CH₂O)ₙ on evaporation of formaldehyde solution
Trioxane: 3HCHO → cyclic trimer (1,3,5-trioxane) with H₂SO₄
Paraformaldehyde: (HCHO)ₙ where n = 6 to 100, formed on concentrating HCHO solution
Paraldehyde: 4CH₃CHO → cyclic tetramer with dil. H₂SO₄ at room temperature
Metaldehyde: (CH₃CHO)ₙ with HCl at low temperature (used as fuel tablet)

PYQ — JEE Main 2023

Question: Which of the following compounds will give a yellow precipitate with iodine and NaOH?

(a) CH₃CH₂OH (ethanol)

(b) C₆H₅CHO (benzaldehyde)

(c) CH₃COCH₂CH₃ (butanone)

(d) CH₃CH₂CH₂OH (1-propanol)

Solution:

Condition for iodoform test: Compound must have CH₃CO– group OR must be oxidizable to CH₃CO– group.

Analysis:

(a) CH₃CH₂OH (ethanol) → oxidized to CH₃CHO → has CH₃CO– group → gives iodoform ✓

CH_3CH_2OH \xrightarrow{[O]} CH_3CHO \xrightarrow{NaOH/I_2} CHI_3\downarrow + HCOONa

(b) C₆H₅CHO (benzaldehyde) → no CH₃CO– group → no iodoform

(c) CH₃COCH₂CH₃ (butanone) → has CH₃CO– group → gives iodoform ✓

CH_3COCH_2CH_3 \xrightarrow{NaOH/I_2} CHI_3\downarrow + CH_3CH_2COONa

(d) CH₃CH₂CH₂OH (1-propanol) → oxidized to CH₃CH₂CHO → no CH₃CO– → no iodoform

Both (a) and (c) give iodoform test.

\text{Both (a) and (c) give yellow precipitate of CHI}_3

Answer: Both (a) and (c)

7

Tests for Aldehydes & Ketones

Identification tests form a significant portion of JEE questions. The key is knowing which tests distinguish aldehydes from ketones, and which tests work for specific types of aldehydes (aliphatic vs aromatic).

7.1 Summary of All Tests

Test	Reagent	Aliphatic Aldehyde	Aromatic Aldehyde	Ketone
Tollens' Test	[Ag(NH₃)₂]⁺	✓ Silver mirror	✓ Silver mirror	✗ No reaction
Fehling's Test	Cu²⁺/tartrate/OH⁻	✓ Red ppt (Cu₂O)	✗ No reaction	✗ No reaction
Benedict's Test	Cu²⁺/citrate/OH⁻	✓ Red ppt	✗ No reaction	✗ No reaction
Schiff's Test	Schiff's reagent	✓ Pink color	✓ Pink color (slow)	✗ No color
2,4-DNP Test	2,4-Dinitrophenylhydrazine	✓ Yellow/orange ppt	✓ Yellow/orange ppt	✓ Yellow/orange ppt
NaHSO₃ Test	Sodium bisulfite	✓ White crystalline ppt	✓ White crystalline ppt	✓ (only methyl ketones)
Iodoform Test	NaOH/I₂	✓ (only CH₃CHO)	✗ No reaction	✓ (only CH₃COR)

Important — Tollens' Test (Silver Mirror Test)

Reagent: Ammoniacal silver nitrate [Ag(NH₃)₂]⁺OH⁻

R-CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow R-COO^- + 2Ag\downarrow + 4NH_3 + 2H_2O

Observation: Silver mirror on the walls of the test tube.

Given by: All aldehydes (aliphatic and aromatic). NOT given by ketones.

Important — Fehling's Test

Reagent: Fehling's A (CuSO₄) + Fehling's B (NaOH + sodium potassium tartrate/Rochelle salt)

R-CHO + 2Cu^{2+} + 5OH^- \rightarrow R-COO^- + Cu_2O\downarrow + 3H_2O

Observation: Red-brown precipitate of Cu₂O.

Given by: Only aliphatic aldehydes. NOT given by aromatic aldehydes or ketones.

PYQ — JEE Main 2024 (Apr 4 Shift 1)

Question: A compound gives silver mirror with Tollens' reagent but does NOT give red precipitate with Fehling's solution. The compound can be:

(a) CH₃CHO (acetaldehyde)

(b) C₆H₅CHO (benzaldehyde)

(c) CH₃COCH₃ (acetone)

(d) C₂H₅OH (ethanol)

Solution:

Analysis of given conditions:

Condition 1: Gives silver mirror with Tollens' → must be an aldehyde (aliphatic or aromatic)

→ Eliminates (c) acetone (ketone) and (d) ethanol (alcohol)

Condition 2: Does NOT give red ppt with Fehling's → must be an aromatic aldehyde or ketone

→ Eliminates (a) CH₃CHO (aliphatic aldehyde — DOES give Fehling's test)

Remaining candidate: (b) C₆H₅CHO (benzaldehyde)

- Benzaldehyde gives Tollens' test (silver mirror) ✓

- Benzaldehyde does NOT give Fehling's test (aromatic) ✓

\text{C}_6\text{H}_5\text{CHO — aromatic aldehyde}

Answer: (b) C₆H₅CHO (benzaldehyde)

8

Preparation of Carboxylic Acids

Carboxylic acids can be prepared by oxidation of primary alcohols, aldehydes, and alkyl benzenes, as well as by hydrolysis of nitriles, amides, esters, and Grignard reagents with CO₂.

8.1 From Primary Alcohols, Aldehydes, and Alkenes

R-CH_2OH \xrightarrow{KMnO_4/H^+ \text{ or } K_2Cr_2O_7/H^+} R-COOH\] \[R-CHO \xrightarrow{[O]} R-COOH

Oxidizing agents: KMnO₄/H⁺, K₂Cr₂O₇/H⁺, CrO₃/H⁺ (Jones reagent)

8.2 From Alkyl Benzenes (Side-chain Oxidation)

Important — Side-chain Oxidation (JEE Favorite)

\text{C}_6\text{H}_5CH_3 \xrightarrow{KMnO_4/KOH, \Delta} \text{C}_6\text{H}_5COOK \xrightarrow{H^+} \text{C}_6H_5COOH

Key Points:

Any alkyl side chain on benzene ring (–CH₃, –C₂H₅, –C₃H₇, etc.) is oxidized to –COOH
The entire side chain is oxidized to –COOH regardless of length
Requires hot KMnO₄/KOH (strongly oxidizing, basic medium)
Toluene → benzoic acid; Ethylbenzene → benzoic acid; Isopropylbenzene → benzoic acid
Exception: If the carbon attached to ring has NO H (like in tert-butylbenzene C₆H₅C(CH₃)₃), oxidation does NOT occur

8.3 From Nitriles and Amides

R-CN + 2H_2O \xrightarrow{H^+ \text{ or } OH^-} R-COOH + NH_3\] \[R-CONH_2 + H_2O \xrightarrow{H^+ \text{ or } OH^-} R-COOH + NH_3

Nitriles and amides are hydrolyzed to carboxylic acids under acidic or basic conditions.

8.4 From Grignard Reagents

R-MgX + CO_2 \rightarrow R-COOMgX \xrightarrow{H_3O^+} R-COOH

Key Points:

Grignard reagent attacks CO₂ (electrophile)
Adds one carbon atom (chain extension)
Produces a carboxylic acid with one more C than the original Grignard
Must be done in dry ether (Grignard reacts with water)

8.5 From Acyl Halides, Anhydrides, and Esters

R-COCl + H_2O \rightarrow R-COOH + HCl\] \[(RCO)_2O + H_2O \rightarrow 2R-COOH\] \[R-COOR' + H_2O \xrightarrow{H^+ \text{ or } OH^-} R-COOH + R'OH

Acyl halides and anhydrides are easily hydrolyzed. Esters require acid or base catalyst (base-catalyzed is saponification, irreversible).

PYQ — JEE Main 2022

Question: Oxidation of toluene to benzoic acid using KMnO₄/KOH requires:

(a) Cold KMnO₄/acidic medium

(b) Hot KMnO₄/alkaline medium

(c) Cold KMnO₄/neutral medium

(d) Hot KMnO₄/acidic medium

Solution:

The side-chain oxidation of alkyl benzenes requires strong conditions:

- Hot KMnO₄ (heating required for complete oxidation)

- Alkaline (KOH) medium

The reaction proceeds as:

\text{C}_6H_5CH_3 \xrightarrow{KMnO_4/KOH, \Delta} \text{C}_6H_5COOK \xrightarrow{H^+} \text{C}_6H_5COOH

First, potassium benzoate (salt) is formed, which on acidification gives benzoic acid.

Cold/neutral KMnO₄ would give cis-dihydroxylation of the aromatic ring (not side-chain oxidation).

\text{Hot KMnO}_4/\text{alkaline medium (KOH)}

Answer: (b) Hot KMnO₄/alkaline medium

9

Chemical Reactions of Carboxylic Acids

Carboxylic acids undergo reactions involving: (a) the acidic O–H bond, (b) the C–OH bond, and (c) the α-hydrogen (HVZ reaction). Understanding these three categories is essential for JEE.

9.1 Reactions Involving O–H Bond Cleavage (Acidic Character)

Reactions with Metals, Bases, and Carbonates

With active metals (Na, K):

2R-COOH + 2Na \rightarrow 2R-COONa + H_2\uparrow

With NaOH/Na₂CO₃/NaHCO₃:

R-COOH + NaOH \rightarrow R-COONa + H_2O

R-COOH + NaHCO_3 \rightarrow R-COONa + H_2O + CO_2\uparrow

Key Point: The reaction with NaHCO₃ (evolution of CO₂ gas) is a test for carboxylic acids — phenols do NOT react with NaHCO₃ (weaker acids), so this distinguishes carboxylic acids from phenols.

9.2 Reactions Involving C–OH Bond Cleavage

Most Repeated PYQ — Esterification & Nucleophilic Acyl Substitution

Esterification (Fischer Esterification):

R-COOH + R'OH \xrightarrow{H^+} R-COOR' + H_2O

Key Points:

Acid-catalyzed (usually conc. H₂SO₄)
Reversible reaction (ester hydrolysis is the reverse)
Mechanism: Nucleophilic acyl substitution (addition-elimination)
The –OH of acid is replaced by –OR' of alcohol (NOT the –OH of alcohol)

Conversion to Acid Chlorides, Anhydrides, Amides

With PCl₅, PCl₃, or SOCl₂:

R-COOH + PCl_5 \rightarrow R-COCl + POCl_3 + HCl

R-COOH + SOCl_2 \rightarrow R-COCl + SO_2 + HCl

SOCl₂ is preferred — gaseous by-products (SO₂, HCl) escape, giving pure acid chloride.

With NH₃ (forms amide):

R-COOH + NH_3 \rightarrow R-COONH_4 \xrightarrow{\Delta} R-CONH_2 + H_2O

9.3 Reduction of Carboxylic Acids

R-COOH \xrightarrow{LiAlH_4} R-CH_2OH \text{ (1° alcohol)}\] \[R-COOH \xrightarrow{B_2H_6/THF} R-CH_2OH

Key Points:

LiAlH₄ reduces carboxylic acids to 1° alcohols
NaBH₄ does NOT reduce carboxylic acids (too mild)
B₂H₆ (diborane) in THF also reduces COOH to CH₂OH — selective for COOH in presence of other groups
Catalytic reduction (H₂/Ni) does NOT reduce COOH easily

9.4 HVZ Reaction (α-Halogenation)

High PYQ Frequency — HVZ Reaction

Hell-Volhard-Zelinsky (HVZ) Reaction: Carboxylic acids with α-hydrogen atoms react with Cl₂ or Br₂ in the presence of red phosphorus to give α-halo carboxylic acids.

R-CH_2-COOH + Br_2 \xrightarrow{P \text{ (red)}} R-CHBr-COOH + HBr

Key Points:

Specifically halogenates the α-carbon (carbon adjacent to –COOH)
Requires red phosphorus as catalyst
Works with Cl₂ or Br₂ (not I₂ — too slow)
Only ONE α-H is replaced per HVZ step (but with excess X₂, all α-H can be replaced)
Acetic acid → mono-, di-, tri-chloroacetic acid
The reaction proceeds via acyl halide intermediate (RCOX), which enolizes more readily
Test for α-H: If a carboxylic acid gives HVZ reaction, it has α-H

9.5 Decarboxylation

R-COOH \xrightarrow{NaOH, CaO, \Delta} R-H + CO_2

Key Points:

Sodium salt of carboxylic acid + soda lime (NaOH + CaO) → alkane
CaO prevents NaOH from attacking glass
Kolbe's electrolysis: Electrolysis of sodium/potassium salt of carboxylic acid → alkane (at anode)

2R-COONa + 2H_2O \xrightarrow{\text{electrolysis}} R-R + 2CO_2 + 2NaOH + H_2

PYQ — JEE Main 2021

Question: The reaction of acetic acid with Cl₂ in presence of red phosphorus gives:

(a) Chloroacetic acid (ClCH₂COOH)

(b) Dichloroacetic acid (Cl₂CHCOOH)

(c) Trichloroacetic acid (Cl₃CCOOH)

(d) Acetyl chloride (CH₃COCl)

Solution:

This is the HVZ reaction:

CH_3COOH + Cl_2 \xrightarrow{P \text{ (red)}} ClCH_2COOH + HCl

The reaction specifically replaces ONE α-hydrogen with halogen (per HVZ step).

Acetic acid (CH₃COOH) has 3 α-H on the CH₃ group.

With limited Cl₂ (1 equivalent), only ONE H is replaced → monochloroacetic acid (ClCH₂COOH)

With excess Cl₂, all three α-H can be replaced → trichloroacetic acid (Cl₃CCOOH)

But the standard HVZ product (with stoichiometric Cl₂) is mono-chloroacetic acid.

\text{Product: ClCH}_2\text{COOH (monochloroacetic acid)}

Answer: (a) Chloroacetic acid (ClCH₂COOH)

10

Acidic Strength of Carboxylic Acids

Comparing the acidic strength of carboxylic acids is one of the most frequently tested topics in JEE. The key is understanding how inductive effect, resonance, and hybridization of substituents affect the stability of the carboxylate anion.

10.1 Factors Affecting Acidic Strength

Most Repeated PYQ — Acidic Strength Comparison

Rule: Any factor that stabilizes the carboxylate anion (RCOO⁻) increases acid strength, and any factor that destabilizes it decreases acid strength.

Acid-Strengthening Factors

Electron-withdrawing groups (–I effect)
Electronegative substituents
sp > sp² > sp³ (hybridization)
Closer to –COOH group
Greater number of EWG

Acid-Weakening Factors

Electron-donating groups (+I effect)
Alkyl groups
sp³ < sp² < sp
Farther from –COOH group
Resonance donation (+R)

10.2 Inductive Effect Order

Common Substituent Effects

–I groups (acid-strengthening):

NO_2 > CN > F > Cl > Br > I > CF_3 > OH > OCH_3 > C\equiv CH > Ph

+I groups (acid-weakeninging):

\[(CH_3)_3C- > (CH_3)_2CH- > CH_3CH_2- > CH_3-\]

10.3 Important Acidic Strength Comparisons

Acid	pKa	Reason
FCH₂COOH	2.66	Strong –I effect of F
ClCH₂COOH	2.86	–I effect of Cl (weaker than F)
BrCH₂COOH	2.90	–I effect of Br (weaker than Cl)
ICH₂COOH	3.12	–I effect of I (weakest halogen)
HCOOH (formic acid)	3.75	No alkyl group (no +I effect)
C₆H₅COOH (benzoic acid)	4.20	sp² C has –I effect (+R is weak)
CH₃COOH (acetic acid)	4.76	+I effect of CH₃ destabilizes anion
CH₃CH₂COOH	4.87	Greater +I effect
(CH₃)₃CCOOH	5.03	Strongest +I effect

Key Order — Must Memorize

\[FCH_2COOH > ClCH_2COOH > BrCH_2COOH > ICH_2COOH > HCOOH > C_6H_5COOH > CH_3COOH > CH_3CH_2COOH\]

Acid strength decreases from left to right (pKa increases).

10.4 Effect of Number and Distance of Substituents

Number of EWG: More EWG → stronger acid

Cl_3CCOOH > Cl_2CHCOOH > ClCH_2COOH > CH_3COOH\] \[(pKa: 0.65) \quad (1.29) \quad (2.86) \quad (4.76)

Distance from –COOH: Closer EWG → stronger acid (inductive effect decreases with distance)

ClCH_2COOH > ClCH_2CH_2COOH > ClCH_2CH_2CH_2COOH\] \[(\alpha) \quad\quad\quad\quad (\beta) \quad\quad\quad\quad\quad\quad (\gamma)

10.5 Substituted Benzoic Acids

Effect of Substituents on Benzoic Acid

Acid-Strengthening (at any position)

–NO₂ (–I and –R)
–CN (–I and –R)
–X (halogens: –I > +R)
–CF₃ (strong –I)
–COOH, –CHO, –COR (–R)

Ortho effect: o-substituted benzoic acids are generally stronger than p- and m- isomers (due to steric inhibition of resonance)

Acid-Weakening (at ortho/para)

–OH, –OR (+R > –I)
–NH₂, –NHR, –NR₂ (+R > –I)
–CH₃, –C₂H₅ (+I and +H)
–O⁻ (strong +R)

At meta position, only inductive effect matters (resonance doesn't reach meta)

JEE Shortcut — Halogens on Benzoic Acid

For halogen-substituted benzoic acids:

Halogens are –I (acid-strengthening) AND +R (acid-weakening)
At meta: Only –I operates → strongest effect
At para: –I + +R → net weaker than meta
At ortho: Ortho effect (steric) → strongest among the three
Order: o-chloro > m-chloro > p-chloro > benzoic acid

PYQ — JEE Main 2024 (Apr 6 Shift 1)

Question: The correct decreasing order of acidic strength is:

(a) FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH

(b) CH₃COOH > BrCH₂COOH > ClCH₂COOH > FCH₂COOH

(c) ClCH₂COOH > FCH₂COOH > BrCH₂COOH > CH₃COOH

(d) FCH₂COOH > BrCH₂COOH > ClCH₂COOH > CH₃COOH

Solution:

Key principle: Electron-withdrawing groups (–I) stabilize the carboxylate anion → increase acid strength.

Electronegativity order of halogens: F > Cl > Br > I

→ –I effect order: F > Cl > Br > I

→ Acid-strengthening order: FCH₂COOH > ClCH₂COOH > BrCH₂COOH > ICH₂COOH

CH₃COOH has an electron-donating group (+I) → weakest acid

\[FCH_2COOH > ClCH_2COOH > BrCH_2COOH > CH_3COOH\]

pKa values: FCH₂COOH (2.66) < ClCH₂COOH (2.86) < BrCH₂COOH (2.90) < CH₃COOH (4.76)

Answer: (a) FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH

11

Name Reactions Summary & Formula Sheet

This section provides a comprehensive summary of all name reactions from this chapter. This is your one-stop reference for quick revision before the exam.

11.1 Complete Name Reactions Table

Name Reaction	Starting Material	Reagent	Product
Rosenmund	RCOCl	H₂/Pd-BaSO₄	RCHO
Stephen	RCN	SnCl₂/HCl, then H₃O⁺	RCHO
Etard	C₆H₅CH₃	CrO₂Cl₂/CS₂	C₆H₅CHO
Gattermann-Koch	C₆H₆	CO + HCl/AlCl₃/CuCl	C₆H₅CHO
Clemmensen	R₂C=O	Zn-Hg/HCl	R₂CH₂
Wolff-Kishner	R₂C=O	NH₂NH₂/KOH/Δ	R₂CH₂
Aldol Condensation	RCH₂CHO	dil. NaOH	β-hydroxy aldehyde → α,β-unsaturated
Cannizzaro	RCHO (no α-H)	conc. NaOH	RCH₂OH + RCOONa
Haloform	CH₃COR/CH₃CHO	NaOH/X₂	CHX₃ + RCOONa
HVZ	RCH₂COOH	X₂/P (red)	RCHXCOOH
Kolbe-Schmitt	C₆H₅ONa	CO₂/125°C/4-7 atm	Salicylic acid
Reimer-Tiemann	C₆H₅OH	CHCl₃/NaOH	Salicylaldehyde
Friedel-Crafts Acylation	C₆H₆	RCOCl/AlCl₃	C₆H₅COR
Baeyer-Villiger	RCOR'	mCPBA	Ester (RCOOR' or R'COOR)
Tollens Test	RCHO	[Ag(NH₃)₂]⁺	RCOO⁻ + Ag↓
Fehling Test	RCHO (aliphatic)	Cu²⁺/tartrate	RCOO⁻ + Cu₂O↓

11.2 Quick Decision Flowchart

Identify the Reaction Type

Has α-H? → YES → Aldol (dil. base) / NO → Cannizzaro (conc. base)
Has CH₃CO– or CH₃CH(OH)– group? → YES → Iodoform test
Aldehyde (aromatic) + Ag(NH₃)₂⁺? → Tollens' positive but Fehling's negative
Need to reduce C=O to CH₂? → Clemmensen (acid) or Wolff-Kishner (base)
Need to reduce C=O to CH-OH? → NaBH₄ or LiAlH₄ or H₂/Ni
Need α-halo acid? → HVZ reaction
Acid chloride → aldehyde? → Rosenmund reduction
Nitrile → aldehyde? → Stephen reaction or DIBAL-H
Toluene → benzaldehyde? → Etard reaction or Gattermann-Koch (from benzene)
Phenol + CO₂ → salicylic acid? → Kolbe-Schmitt

PYQ — JEE Main 2023

Question: Match the following reactions with their names:

Column I: (P) CH₃COCl + H₂/Pd-BaSO₄ → CH₃CHO

Column I: (Q) CH₃CHO + dil. NaOH → CH₃CH(OH)CH₂CHO

Column I: (R) 2C₆H₅CHO + conc. NaOH → C₆H₅CH₂OH + C₆H₅COONa

Column I: (S) CH₃COOH + Cl₂/P → ClCH₂COOH

Column II: (1) Aldol (2) Cannizzaro (3) Rosenmund (4) HVZ

Solution:

(P) CH₃COCl + H₂/Pd-BaSO₄ → CH₃CHO

Acid chloride reduced to aldehyde with Pd/BaSO₄ → Rosenmund reduction (3)

(Q) CH₃CHO + dil. NaOH → Aldol product

Two aldehyde molecules with α-H condense with dilute base → Aldol condensation (1)

(R) 2C₆H₅CHO + conc. NaOH → alcohol + acid salt

Aldehyde without α-H, disproportionation with conc. base → Cannizzaro reaction (2)

(S) CH₃COOH + Cl₂/P → ClCH₂COOH

α-halogenation of carboxylic acid → HVZ reaction (4)

P-3, \ Q-1, \ R-2, \ S-4

Answer: P-3, Q-1, R-2, S-4

Common Mistakes Summary

Mistake: Using NaBH₄ to reduce esters → Fact: NaBH₄ only reduces aldehydes/ketones
Mistake: Using Clemmensen for base-sensitive compounds → Use Wolff-Kishner instead
Mistake: Confusing Cannizzaro with Aldol → Cannizzaro needs NO α-H, Aldol needs α-H
Mistake: Thinking all aldehydes give Fehling's test → Aromatic aldehydes don't
Mistake: Not recognizing ethanol gives iodoform → It gets oxidized to CH₃CHO first
Mistake: Using H₂/Pd to reduce acid chloride → This gives alcohol (over-reduction); use Rosenmund instead
Mistake: Confusing Gattermann with Gattermann-Koch → Different reagents and substrates
Mistake: Forgetting that phenol does NOT react with NaHCO₃ → Only carboxylic acids do

Previous Year Questions Analysis

JEE Main (Last 10 Years)

✓ Aldol Condensation: 18%
✓ Cannizzaro Reaction: 15%
✓ Iodoform/Haloform Test: 14%
✓ Acidic Strength Comparison: 12%
✓ Nucleophilic Addition Reactivity: 10%
✓ Identification Tests (Tollens, Fehling): 10%
✓ Name Reactions (Rosenmund, Stephen, HVZ): 12%
✓ Reduction (Clemmensen, Wolff-Kishner): 9%

JEE Advanced (Last 10 Years)

✓ Crossed Aldol/Cannizzaro: 20%
✓ Multi-step Synthesis: 18%
✓ Reaction Mechanisms: 15%
✓ Acidic Strength (substituted benzoic acids): 12%
✓ Identification of Unknown Compounds: 15%
✓ Baeyer-Villiger Oxidation: 10%
✓ Kolbe-Schmitt/Reimer-Tiemann: 10%

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)

JEE Advanced: 8-16 marks (2-4 questions)

Difficulty Level: Easy to Moderate

Time Required: 4-5 hours practice

Download Complete PYQ PDF (2015-2024)

Practice Problem Set

Level 1: Basic (JEE Main Standard)

Which of the following gives aldol condensation: HCHO, CH₃CHO, C₆H₅CHO, (CH₃)₃CCHO?
Identify the product: CH₃COCl + H₂/Pd-BaSO₄ → ?
What is the product of Cannizzaro reaction of formaldehyde?
Which test distinguishes acetaldehyde from acetone?
Write the product of HVZ reaction on propanoic acid.
Arrange in order of acidic strength: CH₃COOH, ClCH₂COOH, FCH₂COOH, HCOOH

Level 2: Intermediate (JEE Main/Advanced)

Predict the product of crossed aldol between C₆H₅CHO and CH₃CHO.
Which compounds give iodoform test: ethanol, 2-propanol, 1-propanol, benzaldehyde, acetophenone?
A compound C₃H₆O gives silver mirror with Tollens' but negative iodoform test. Identify it.
Arrange: o-nitrobenzoic acid, m-nitrobenzoic acid, p-nitrobenzoic acid, benzoic acid (acidic strength)
Write the mechanism of nucleophilic addition of HCN to acetaldehyde.
Convert acetic acid to acetone in 3 steps.

Level 3: Advanced (JEE Advanced/Olympiad)

An organic compound A (C₄H₈O) gives positive iodoform test and negative Tollens' test. On NaBH₄ reduction, it gives B (C₄H₁₀O). Identify A and B.
Predict the product of Baeyer-Villiger oxidation of acetophenone.
A compound gives positive Tollens' test but negative Fehling's test. On oxidation gives C₇H₆O₂. Identify the compound.
Arrange in order of acidic strength: (COOH)₂, CH₂(COOH)₂, succinic acid, glutaric acid. Explain.
Complete the reaction: C₆H₅OH + CHCl₃ + NaOH → ? Name the reaction.
Compound A (C₄H₈O₂) on treatment with NaHCO₃ gives CO₂. On HVZ reaction gives B. On reduction with LiAlH₄ gives C (C₄H₁₀O). Identify A, B, C.

Get Solutions PDF with Step-by-Step Working

> FAQ

Frequently Asked Questions

What is nucleophilic addition reaction in aldehydes and ketones?

Nucleophilic addition reaction is the characteristic reaction of aldehydes and ketones where a nucleophile attacks the electrophilic carbonyl carbon. The carbonyl carbon is partially positive due to the electronegativity of oxygen. The nucleophile adds to the carbonyl carbon, and a proton adds to the oxygen. Examples include addition of HCN, NaHSO3, Grignard reagents, and ammonia derivatives.

What is aldol condensation?

Aldol condensation is a reaction in which two molecules of an aldehyde or ketone containing at least one alpha-hydrogen undergo condensation in the presence of a dilute base (like NaOH) or dilute acid to form a beta-hydroxy aldehyde (aldol) or beta-hydroxy ketone (ketol). On heating, the aldol undergoes dehydration to give an alpha,beta-unsaturated compound. Example: 2CH3CHO → CH3CH(OH)CH2CHO (aldol) → CH3CH=CHCHO + H2O.

What is Cannizzaro reaction?

Cannizzaro reaction is a disproportionation reaction of aldehydes that do NOT have alpha-hydrogen atoms (like HCHO, C6H5CHO, (CH3)3C-CHO). When treated with concentrated alkali (NaOH/KOH), one molecule is reduced to alcohol and another is oxidized to carboxylic acid salt. Example: 2HCHO + NaOH → CH3OH + HCOONa.

Which compounds give iodoform test?

Compounds that give iodoform test are: (1) Aldehydes/ketones with CH3CO- group (like CH3CHO, CH3COCH3, CH3COR), (2) Alcohols that can be oxidized to CH3CO- group (like CH3CH2OH, CH3CH(OH)R), (3) Methyl ketones. The test involves treatment with NaOH and I2 (or NaOI), producing a yellow precipitate of CHI3 (iodoform).

What is the order of reactivity of aldehydes and ketones towards nucleophilic addition?

The order of reactivity is: HCHO > CH3CHO > CH3CH2CHO > CH3COCH3 > CH3COR > R2CO. This is because: (1) Aldehydes are more reactive than ketones (less steric hindrance, more electrophilic carbon), (2) Electron-donating alkyl groups decrease electrophilicity of carbonyl carbon, (3) Steric hindrance increases from aldehydes to ketones.

What is HVZ reaction?

Hell-Volhard-Zelinsky (HVZ) reaction is the reaction of a carboxylic acid with chlorine or bromine in the presence of red phosphorus to give alpha-halo carboxylic acids. The reaction selectively halogenates the alpha-carbon. Example: CH3COOH + Br2/P → CH2BrCOOH (alpha-bromoacetic acid). The reaction proceeds via the acyl halide intermediate.

What is the difference between Clemmensen and Wolff-Kishner reduction?

Both reduce carbonyl group (C=O) to CH2. Clemmensen reduction uses Zn-Hg/HCl (acidic medium), while Wolff-Kishner reduction uses NH2NH2/KOH/ethylene glycol (basic medium). Clemmensen is not suitable for acid-sensitive compounds, while Wolff-Kishner is not suitable for base-sensitive compounds. Both convert RCHO → RCH3 and RCOR' → RCH2R'.

Why is benzoic acid stronger acid than acetic acid?

Benzoic acid (C6H5COOH, pKa = 4.2) is stronger than acetic acid (CH3COOH, pKa = 4.76) because the phenyl group (C6H5-) is sp2 hybridized and exerts a -I (electron-withdrawing inductive) effect, which stabilizes the carboxylate anion more than the sp3-hybridized methyl group. However, phenyl also shows +R (resonance) effect which slightly reduces its acid-strengthening ability, making benzoic acid weaker than formic acid.

How many questions come from aldehydes ketones carboxylic acids in JEE Mains?

On average 2-3 questions (8-12 marks) appear from this chapter in every JEE Main session. It is one of the highest weightage chapters in organic chemistry. Most questions are from aldol condensation, Cannizzaro reaction, iodoform test, acidic strength comparison, and identification of compounds using chemical tests.

What is Rosenmund reduction?

Rosenmund reduction is the reduction of an acyl chloride (acid chloride) to an aldehyde using hydrogen gas in the presence of palladium catalyst supported on barium sulfate (Pd/BaSO4). The BaSO4 acts as a poison to prevent over-reduction to alcohol. Example: CH3COCl + H2/Pd-BaSO4 → CH3CHO + HCl.

Aldehydes, Ketones & Carboxylic Acids notes, reactions & PYQs for JEE 2026

📑 Table of Contents

Structure of Carbonyl & Carboxyl Group

1.1 Structure of the Carbonyl Group (C=O)

Key Structural Features of C=O

⭐ Important for JEE — Resonance in Carbonyl Group

1.2 Aldehydes vs Ketones — Structural Comparison

1.3 Structure of the Carboxyl Group (–COOH)

Carboxyl Group Structure

💡 Key Insight — Why Carboxylic Acids Are Acidic

📝 PYQ — JEE Main 2024

Preparation of Aldehydes & Ketones

2.1 Preparation of Aldehydes

⭐ Important — Most Repeated PYQ Topic

Stephen Reaction (Reduction of Nitriles)

Etard Reaction (Aromatic Aldehydes Only)

🔥 High PYQ Frequency — Gattermann-Koch Reaction

DIBAL-H Reduction (From Esters/Nitriles)

2.2 Preparation of Ketones

From Acid Chlorides + Dialkylcadmium

From Nitriles (with Grignard Reagent)

Friedel-Crafts Acylation (Aromatic Ketones)

From Alcohols (Oxidation)

2.3 From Hydrocarbons

Ozonolysis of Alkenes

Hydration of Alkynes

📝 PYQ — JEE Main 2024 (Jan 30 Shift 2)

⚠️ Common Confusion — Gattermann vs Gattermann-Koch

Nucleophilic Addition Reactions

3.1 General Mechanism of Nucleophilic Addition

⭐ Most Important — Mechanism (High PYQ Frequency)

3.2 Reactivity Order — Aldehydes vs Ketones

🔥 Most Repeated PYQ — Reactivity Order

💡 JEE Shortcut — Aromatic vs Aliphatic

3.3 Addition of HCN (Cyanohydrin Formation)

3.4 Addition of NaHSO₃ (Bisulfite Addition)

3.5 Addition of Ammonia Derivatives

⭐ Most Repeated PYQ — Ammonia Derivatives Table

💡 JEE Pro Tip — pH Dependence

3.6 Addition of Grignard Reagent

📝 PYQ — JEE Main 2023

Reduction & Oxidation of Aldehydes and Ketones

4.1 Reduction to Alcohols

Catalytic Reduction (H₂/Ni, Pt, or Pd)

NaBH₄ (Sodium Borohydride) — Mild Reducing Agent

LiAlH₄ (Lithium Aluminium Hydride) — Strong Reducing Agent

4.2 Reduction to Hydrocarbons (C=O → CH₂)

🔥 High PYQ Frequency — Clemmensen vs Wolff-Kishner

4.3 Oxidation of Aldehydes

⭐ Important — Aldehydes vs Ketones Oxidation

⚠️ Critical — Aromatic Aldehydes Exception

4.4 Oxidation of Ketones (Baeyer-Villiger Oxidation)

Baeyer-Villiger Oxidation

📝 PYQ — JEE Main 2022

Aldol Condensation & Cannizzaro Reaction

5.1 Aldol Condensation

🔥 Most Repeated PYQ — Aldol Condensation

Mechanism of Aldol Condensation

💡 JEE Shortcut — Which Compounds Give Aldol?

5.2 Crossed Aldol Condensation

5.3 Cannizzaro Reaction

🔥 Most Repeated PYQ — Cannizzaro Reaction

Compounds that undergo Cannizzaro Reaction

🎯 JEE Shortcut — Aldol vs Cannizzaro (Decision Tree)

📝 PYQ — JEE Main 2024 (Jan 29 Shift 1)

⚠️ Common Mistake — Aldol vs Cannizzaro

Haloform Reaction & Other Important Reactions

6.1 Haloform Reaction

🔥 Most Repeated PYQ — Iodoform Test

⚠️ JEE Favorite Trap — Which does NOT give Iodoform?

6.2 Electrophilic Substitution Reactions of Aromatic Aldehydes

Kolbe-Schmitt Reaction

Reimer-Tiemann Reaction

6.3 Polymerization Reactions

Polymerization of Formaldehyde and Acetaldehyde

📝 PYQ — JEE Main 2023

Tests for Aldehydes & Ketones

7.1 Summary of All Tests

⭐ Important — Tollens' Test (Silver Mirror Test)

⭐ Important — Fehling's Test

Table of Contents

Important for JEE — Resonance in Carbonyl Group

Key Insight — Why Carboxylic Acids Are Acidic

PYQ — JEE Main 2024

Important — Most Repeated PYQ Topic

High PYQ Frequency — Gattermann-Koch Reaction

PYQ — JEE Main 2024 (Jan 30 Shift 2)

Common Confusion — Gattermann vs Gattermann-Koch

Most Important — Mechanism (High PYQ Frequency)

Most Repeated PYQ — Reactivity Order

JEE Shortcut — Aromatic vs Aliphatic

Most Repeated PYQ — Ammonia Derivatives Table

JEE Pro Tip — pH Dependence

PYQ — JEE Main 2023

High PYQ Frequency — Clemmensen vs Wolff-Kishner

Important — Aldehydes vs Ketones Oxidation

Critical — Aromatic Aldehydes Exception

PYQ — JEE Main 2022

Most Repeated PYQ — Aldol Condensation

JEE Shortcut — Which Compounds Give Aldol?

Most Repeated PYQ — Cannizzaro Reaction

JEE Shortcut — Aldol vs Cannizzaro (Decision Tree)

PYQ — JEE Main 2024 (Jan 29 Shift 1)

Common Mistake — Aldol vs Cannizzaro

Most Repeated PYQ — Iodoform Test

JEE Favorite Trap — Which does NOT give Iodoform?

PYQ — JEE Main 2023

Important — Tollens' Test (Silver Mirror Test)

Important — Fehling's Test

PYQ — JEE Main 2024 (Apr 4 Shift 1)

Important — Side-chain Oxidation (JEE Favorite)

PYQ — JEE Main 2022

Most Repeated PYQ — Esterification & Nucleophilic Acyl Substitution

High PYQ Frequency — HVZ Reaction

PYQ — JEE Main 2021

Most Repeated PYQ — Acidic Strength Comparison

Key Order — Must Memorize

JEE Shortcut — Halogens on Benzoic Acid

PYQ — JEE Main 2024 (Apr 6 Shift 1)

PYQ — JEE Main 2023

Common Mistakes Summary

Previous Year Questions Analysis

Practice Problem Set

Why This Chapter is Important

Must-Remember Formulas & Values

5-Step Exam Strategy

7-Day Mastery Roadmap