JEE Main & Advanced

Aldehydes, Ketones and Carboxylic Acids

Aldehydes, Ketones and Carboxylic Acids for JEE Main & Advanced

1
Module 1

Aldehydes and Ketones

Nomenclature, Preparation, and Nucleophilic AdditionTopic 1

Carbonyl Group: $>C=O$. Polar; C has partial $+$ charge (δ+), O has partial $-$ (δ−).

Aldehydes: $R-CHO$ (or $H-CHO$). C=O at end of chain. Always 1° to other atoms (one H attached to carbonyl C).

Ketones: $R-CO-R'$. C=O in middle of chain.

Nomenclature:

StructureCommon NameIUPAC Name
$HCHO$FormaldehydeMethanal
$CH_3CHO$AcetaldehydeEthanal
$CH_3CH_2CHO$PropionaldehydePropanal
$C_6H_5CHO$BenzaldehydePhenylmethanal
$CH_3COCH_3$AcetonePropan-2-one
$CH_3CH_2COCH_3$Methyl ethyl ketoneButan-2-one

Preparation of Aldehydes:

1. From Alcohols (Mild Oxidation):

  • $RCH_2OH \xrightarrow{[O], mild} RCHO$ (stop at aldehyde)
  • Reagents: PCC (pyridinium chlorochromate), $MnO_2$ (for allylic alcohols), Cu/$573$K

2. From Acid Chlorides (Rosenmund Reduction): $RCOCl + H_2 \xrightarrow{Pd/BaSO_4, S} RCHO + HCl$

  • $BaSO_4$ poison prevents further reduction to alcohol

3. Stephen Reaction: $RCN + SnCl_2 + HCl \to RCH=NH \xrightarrow{H_2O} RCHO$ (nitrile → imine → aldehyde)

4. From Carboxylic Acids (DIBAL-H): $RCOOH \xrightarrow{DIBAL-H, -78°C} RCHO + Al(OH)_3$ (controlled reduction)

5. Ozonolysis of Alkenes (Reductive): $RCH=CHR' + O_3 \xrightarrow{H_2O, Zn} RCHO + R'CHO$

Preparation of Ketones:

1. From Alcohols ($2°$): $R_2CHOH \xrightarrow{[O]} R_2CO$ (various oxidants)

2. From Acid Chlorides (Friedel-Crafts Acylation): $ArH + RCOCl \xrightarrow{AlCl_3} Ar-CO-R + HCl$ (aryl ketones)

3. From Nitriles (Grignard Addition): $RCN + R'MgX \to R(R')C=N-MgX \xrightarrow{H_3O^+} RC(O)R'$

4. From Acid Chlorides + Organocopper (Gilman): $RCOCl + R'_2CuLi \to RC(O)R' + R'Cu + LiCl$

Nucleophilic Addition (typical of C=O):

C=O has $sp^2$ C with δ+; Nu attacks C; lone pair on O accepts charge.

Mechanism: $Nu^- + C=O \to Nu-C-O^- \xrightarrow{H^+} Nu-C-OH$

Reactivity: Aldehydes > Ketones in nucleophilic addition.

  • Steric: Aldehydes have H (small); ketones have 2 R groups (bulky)
  • Electronic: Two alkyl groups on ketone donate more e⁻, reducing $\delta^+$ on C

Important Nucleophilic Additions:

1. HCN Addition (gives cyanohydrins): $R_2C=O + HCN \to R_2C(OH)(CN)$ (cyanohydrin)

  • Catalyzed by base; OH⁻ generates CN⁻
  • Useful for chain extension

2. NaHSO₃ Addition (gives bisulfite adduct): $R-CHO + NaHSO_3 \to RCH(OH)(SO_3Na)$ (white crystalline ppt)

  • Test for aldehydes; reversible (acid releases CHO back)

3. Alcohol Addition (gives acetal/ketal): $RCHO + R'OH \xrightarrow{H^+} RCH(OR')_2 + H_2O$ (acetal — protective group)

  • Catalyst: H⁺; remove water to push equilibrium right

4. NH₃ and Derivatives (Imines, Schiff bases, Hydrazones):

  • $R_2C=O + R'NH_2 \to R_2C=NR' + H_2O$ (imine/Schiff base)
  • With NH₂OH (hydroxylamine): R₂C=N-OH (oxime)
  • With NH₂NH₂ (hydrazine): R₂C=N-NH₂ (hydrazone)
  • With 2,4-DNP-hydrazine: gives yellow ppt — test for carbonyl group
  • With semicarbazide ($NH_2NHCONH_2$): semicarbazone

5. Grignard Reagents (forms alcohols):

  • HCHO + RMgX → $RCH_2OH$ (1°)
  • $R'CHO + RMgX \to RR'CHOH$ (2°)
  • $R'COR'' + RMgX \to RR'R''COH$ (3°)
Worked Examples
1

Why are aldehydes more reactive than ketones toward nucleophilic addition?

Show solution
  1. Steric: Aldehydes have $H$ on C=O; ketones have 2 R groups. Less hindrance in aldehyde → easier Nu attack.
  2. Electronic: Two alkyl groups (R) on ketone donate electrons (+I); reduce δ+ on C; lessens its electrophilicity.

Order: HCHO > $RCHO > R_2CO$.

Final Answer: Less steric hindrance + less electron donation → aldehydes more reactive.

2

Predict products: (a) $CH_3CHO + HCN \to$? (b) $CH_3CHO + 2,4$-DNP $\to$?

Show solution

(a) Cyanohydrin: $CH_3CHO + HCN \to CH_3CH(OH)CN$ (acetaldehyde cyanohydrin). (b) 2,4-DNP forms yellow ppt: $CH_3CH=N-NH-C_6H_3(NO_2)_2$ (2,4-DNP hydrazone — test for aldehyde/ketone).

Final Answer: (a) Acetaldehyde cyanohydrin; (b) 2,4-DNP hydrazone of acetaldehyde.

✎ Self-Check — 5 questions0 / 5
Q1.

Aldehyde vs ketone in addition:

Q2.

Rosenmund reduction converts:

Q3.

Cyanohydrin is formed by:

Q4.

2,4-DNP test confirms:

Q5.

Stephen reaction:

Named Reactions and TestsTopic 2

Important Named Reactions:

1. Aldol Condensation: Aldehyde/ketone with $\alpha$-H + dilute NaOH/heat → $\beta$-hydroxy carbonyl (aldol) → dehydrates to $\alpha,\beta$-unsaturated carbonyl.

$2CH_3CHO \xrightarrow{NaOH, dil} CH_3CH(OH)CH_2CHO \xrightarrow{\Delta, -H_2O} CH_3CH=CHCHO$ (crotonaldehyde)

For ketones: $2CH_3COCH_3 \xrightarrow{Ba(OH)_2} CH_3C(OH)(CH_3)CH_2COCH_3 \to CH_3C(=CH_2)CH_2COCH_3$

Mechanism:

  • Base removes $\alpha$-H → enolate ion (nucleophile)
  • Enolate attacks another carbonyl C → aldol product
  • Heated → dehydrates to $\alpha,\beta$-unsaturated.

Crossed Aldol: Between two different aldehydes/ketones.

2. Cannizzaro Reaction: Aldehyde without $\alpha$-H + conc. NaOH → disproportionation:

  • One molecule oxidized (→ acid)
  • One molecule reduced (→ alcohol)

$2HCHO \xrightarrow{conc. NaOH} HCOONa + CH_3OH$ $2C_6H_5CHO \xrightarrow{conc. NaOH} C_6H_5COONa + C_6H_5CH_2OH$

Crossed Cannizzaro: With HCHO + non-$\alpha$-H aldehyde → HCHO reduced to $CH_3OH$, other one oxidized to acid.

3. Clemmensen Reduction: Reduces C=O of aldehyde/ketone to CH₂ (removes carbonyl). $R_2CO + Zn(Hg) + HCl \to R_2CH_2 + H_2O$

  • Acidic conditions; not for compounds with acid-sensitive groups.

4. Wolff-Kishner Reduction: Reduces C=O to CH₂. $R_2CO + NH_2NH_2 + KOH(alc) \to R_2CH_2 + N_2 + H_2O$

  • Basic conditions; alternative to Clemmensen.

5. Tollens' Test (Silver Mirror): For aldehydes. $RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \to RCOO^- + 2Ag \downarrow + 4NH_3 + 2H_2O$

  • Silver mirror coats the tube — positive for aldehydes; ketones don't react.

6. Fehling's Test: For aldehydes. $RCHO + 2Cu^{2+} + 5OH^- \to RCOO^- + Cu_2O \downarrow (red) + 3H_2O$

  • Red ppt of $Cu_2O$ — positive for aldehydes; aromatic aldehydes don't give Fehling.
  • Tollens works for both aliphatic and aromatic aldehydes; Fehling only aliphatic.

7. Benedict's Test: Similar to Fehling; uses $Cu^{2+}$ with sodium citrate complex.

8. Iodoform Test: For methyl ketones ($CH_3CO-$) and methylcarbinols ($CH_3CH(OH)-$): $CH_3COCH_3 + 3I_2 + 4NaOH \to CHI_3 \downarrow (yellow) + CH_3COONa + 3NaI + 3H_2O$

  • Positive: acetone, acetaldehyde, ethanol, 2-propanol, methyl ketones.
  • Negative: formaldehyde, methanol, propanone analogues without methyl.

9. Oxidation:

  • $RCHO \to RCOOH$ (easy, by many oxidants)
  • Ketones: only with strong oxidants and cleavage of C-C bonds (Baeyer-Villiger gives ester; Popoff's rule)

10. Reduction:

  • LiAlH₄ or NaBH₄ → alcohols (1° from RCHO, 2° from $R_2$CO)
  • Catalytic hydrogenation (Ni, Pt) → alcohols
  • Clemmensen/Wolff-Kishner → CH₂ (remove carbonyl)
Worked Examples
1

Compare Tollens' and Fehling's tests for aldehydes.

Show solution
PropertyTollens'Fehling's
Reagent$Ag(NH_3)_2^+$ in NH₃ solution$Cu^{2+}$ + tartrate in alkali
Positive forAll aldehydes (aliphatic and aromatic)Aliphatic aldehydes only
ResultSilver mirrorRed ppt ($Cu_2O$)
KetonesNegativeNegative

Final Answer: Both distinguish RCHO from R₂CO; only Tollens works for aromatic aldehydes.

2

Why does HCHO undergo Cannizzaro but acetaldehyde undergoes aldol with NaOH?

Show solution

Cannizzaro requires no $\alpha$-H.

  • HCHO has no $\alpha$-H (only one C); cannot form enolate; undergoes Cannizzaro disproportionation.
  • Acetaldehyde has $\alpha$-H ($CH_3$ next to CHO); base abstracts H → enolate → attacks another aldehyde → aldol.

Final Answer: Presence/absence of $\alpha$-H determines Cannizzaro vs Aldol.

✎ Self-Check — 5 questions0 / 5
Q1.

Tollens test gives:

Q2.

Cannizzaro reaction requires:

Q3.

Iodoform from acetone:

Q4.

Aldol product of $2CH_3CHO$:

Q5.

Wolff-Kishner is similar in effect to:

2
Module 2

Carboxylic Acids

Acidity, Preparation, and PropertiesTopic 1

Carboxylic Acids: Have $-COOH$ group. General formula $R-COOH$. Acidic due to delocalization of charge in $-COO^-$.

Nomenclature:

  • Common: formic acid ($HCOOH$), acetic acid ($CH_3COOH$), propionic acid ($CH_3CH_2COOH$), butyric acid, valeric acid, oxalic acid ($HOOC-COOH$)
  • IUPAC: methanoic acid, ethanoic acid, propanoic acid, butanoic acid

Aromatic: benzoic acid ($C_6H_5COOH$); naphthoic acid; etc.

Acidity of Carboxylic Acids:

$pK_a$ of acetic acid $\approx 4.74$; HCOOH $\approx 3.75$.

Stronger than alcohols and water because conjugate base ($-COO^-$) has two equivalent resonance structures — negative charge equally on both O atoms; very stable.

Effects on Acidity:

Electron Withdrawing (EW) Groups Increase Acidity:

  • F, Cl, Br: -I effect
  • NO₂, CN: strong -I and -R
  • Closer to -COOH → stronger effect

Order of acidity:

  • $CH_3COOH < CH_2ClCOOH < CHCl_2COOH < CCl_3COOH$ ($pK_a$: 4.74, 2.87, 1.35, 0.65)
  • $CH_3COOH < HCOOH$ (no $+I$ from $-CH_3$)
  • Halogen position effect: $CH_2ClCH_2COOH < CH_3CHClCOOH < ClCH_2CH_2COOH$ (closer Cl → stronger -I → more acidic, but α-substitution wins)

Wait: $CH_3CHClCOOH$ has Cl at α; $CH_2ClCH_2COOH$ has Cl at β; α more acidic.

Electron Donating (ED) Groups Decrease Acidity:

  • Alkyl groups: +I (CH₃, C₂H₅...)
  • Methoxy (resonance donation but distance matters)

For benzoic acid derivatives:

  • m-nitrobenzoic acid > p-nitrobenzoic acid (m due to inductive only; p has +R competing)
  • Actually: $p-NO_2 > m-NO_2 > o-NO_2$ for nitrobenzoic acids ($pK_a$ order). p strongest due to enhanced -I + -R.
  • $p$-methylbenzoic acid weaker than benzoic acid

Preparation:

1. From Alcohols ($1°$): $RCH_2OH \xrightarrow{[O], KMnO_4/H^+ or K_2Cr_2O_7/H^+} RCOOH$

2. From Aldehydes: $RCHO \xrightarrow{[O]} RCOOH$ (mild oxidation; e.g., Tollens, Fehling, but also $KMnO_4, K_2Cr_2O_7$)

3. From Alkenes (Cleavage): $RCH=CHR' + KMnO_4/heat \to RCOOH + R'COOH$

4. From Nitriles (Hydrolysis): $RCN + H_2O \xrightarrow{H^+ or OH^-, \Delta} RCOOH (or RCOO^- + NH_3)$

5. From Grignard Reagent + CO₂: $RMgX + CO_2 \to RCOOMgX \xrightarrow{H_3O^+} RCOOH$

  • Excellent method for synthesizing acids; extends chain by one C.

6. From Acid Chlorides, Esters, Amides (Hydrolysis): $RCOCl + H_2O \to RCOOH + HCl$ $RCOOR' + H_2O \xrightarrow{H^+ \text{ or } OH^-} RCOOH + R'OH$

Physical Properties:

  • BP: very high (higher than alcohols!) due to dimerization via H-bonds
  • $2RCOOH \rightleftharpoons (RCOOH)_2$ (cyclic dimer)
  • Soluble in water (small acids); decreases with chain length
  • Lower acids: pungent; higher: waxy solids
Worked Examples
1

Arrange in increasing acidity: $CH_3COOH$, $HCOOH$, $ClCH_2COOH$, $CCl_3COOH$.

Show solution
  • $CH_3COOH$: +I from CH₃ destabilizes anion → weakest of these
  • $HCOOH$: no +I (just H), better than $CH_3COOH$
  • $ClCH_2COOH$: 1 Cl (-I)
  • $CCl_3COOH$: 3 Cl (strongest -I)

Order: $CH_3COOH < HCOOH < ClCH_2COOH < CCl_3COOH$.

Final Answer: $CH_3COOH < HCOOH < ClCH_2COOH < CCl_3COOH$.

2

Why is the BP of $CH_3COOH$ ($118°$C) higher than $CH_3CH_2OH$ ($78°$C)?

Show solution

Both molecules form H-bonds, but carboxylic acid forms a cyclic dimer with two H-bonds between two molecules:

  • $CH_3COOH$ dimer has two strong O-H...O=C bonds — high stability requires high energy to break.
  • $CH_3CH_2OH$ forms chain-like H-bonds (one per molecule); weaker overall network.

Hence acid BP > alcohol BP.

Final Answer: Cyclic dimer in $CH_3COOH$ (2 H-bonds per pair) vs less ordered H-bonding in alcohol.

✎ Self-Check — 5 questions0 / 5
Q1.

Most acidic:

Q2.

RMgX + CO₂ gives:

Q3.

Acetic acid in water:

Q4.

Most acidic among Cl-substituted acetic acids:

Q5.

Carboxylic acid dimer formed via:

Reactions of -COOH and DerivativesTopic 2

Reactions of Carboxylic Acids:

1. Acidic Reactions:

  • With NaOH: $RCOOH + NaOH \to RCOONa + H_2O$
  • With Na/K metal: $2RCOOH + 2Na \to 2RCOONa + H_2$
  • With NaHCO₃: $RCOOH + NaHCO_3 \to RCOONa + H_2O + CO_2 \uparrow$ — distinguishes acid from phenol/alcohol (which don't release CO₂ from NaHCO₃)

2. Formation of Derivatives:

(a) Acid Chlorides: $RCOOH + SOCl_2 \to RCOCl + SO_2 + HCl$ $RCOOH + PCl_5 \to RCOCl + POCl_3 + HCl$ $RCOOH + PCl_3 \to RCOCl + H_3PO_3$

(b) Esters (Esterification, Fischer): $RCOOH + R'OH \xrightarrow{H^+, \Delta} RCOOR' + H_2O$ (reversible; use Le Chatelier — remove water or use excess alcohol)

(c) Amides: $RCOOH + NH_3 \to RCOONH_4 \xrightarrow{\Delta} RCONH_2 + H_2O$

(d) Anhydrides: $2RCOOH \xrightarrow{P_2O_5 \text{ or } \Delta} (RCO)_2O + H_2O$

3. Decarboxylation (Loss of CO₂):

  • Heat with NaOH/CaO (soda lime): $RCOONa + NaOH \to RH + Na_2CO_3$
  • Kolbe's electrolysis: $2RCOONa \xrightarrow{electrolysis} R-R + 2CO_2 + 2Na$

4. HVZ Reaction (Hell-Volhard-Zelinsky): $RCH_2COOH + Cl_2/PCl_3 \to RCHClCOOH + HCl$

  • α-Halogenation of carboxylic acids
  • $\alpha$-bromination similar with Br₂/red P

5. Reduction: $RCOOH + 4[H] \xrightarrow{LiAlH_4 \text{ or } B_2H_6} RCH_2OH + H_2O$ (1° alcohol)

Note: $NaBH_4$ does NOT reduce -COOH; only LiAlH₄ and diborane.

Important Reactions of Aromatic Acids:

Benzoic Acid ($C_6H_5COOH$):

  • $-COOH$ is m-director, deactivator
  • Nitration gives m-nitrobenzoic acid; bromination gives m-bromobenzoic acid
  • Decarboxylation: $C_6H_5COOH + NaOH/CaO \to C_6H_6 + Na_2CO_3$

Salicylic Acid (o-hydroxybenzoic acid):

  • Made by Kolbe-Schmitt
  • Reacts with acetic anhydride to give aspirin (acetylsalicylic acid):

$$C_6H_4(OH)(COOH) + (CH_3CO)_2O \to C_6H_4(OCOCH_3)(COOH) + CH_3COOH$$

Reactions of Esters (RCOOR'):

1. Hydrolysis:

  • Acidic: $RCOOR' + H_2O \xrightarrow{H^+} RCOOH + R'OH$ (reversible)
  • Basic (saponification): $RCOOR' + NaOH \to RCOO^-Na^+ + R'OH$ (irreversible — carboxylate doesn't react further)

2. Reduction: $RCOOR' + LiAlH_4 \to RCH_2OH + R'OH$

3. Ammonolysis: $RCOOR' + NH_3 \to RCONH_2 + R'OH$

4. Transesterification: $RCOOR' + R''OH \xrightarrow{H^+} RCOOR'' + R'OH$

5. Claisen Condensation: $2CH_3COOC_2H_5 + C_2H_5O^-Na^+ \to CH_3COCH_2COOC_2H_5 + C_2H_5OH$ (Acetoacetic ester from ethyl acetate; one α-H attacks the other ester's carbonyl)

Reactions of Amides (RCONH₂):

1. Hofmann Bromamide Reaction: Amide → primary amine (with one fewer C). $RCONH_2 + Br_2 + 4NaOH \to RNH_2 + 2NaBr + Na_2CO_3 + 2H_2O$ (One-carbon shorter amine, e.g., $CH_3CONH_2 \to CH_3NH_2$; ethanamide → methanamine)

2. Hydrolysis: $RCONH_2 + H_2O \to RCOOH + NH_3$ (acid or base catalyst)

3. Dehydration: $RCONH_2 \xrightarrow{P_2O_5 \text{ or } SOCl_2} R-C \equiv N + H_2O$ (forms nitrile)

Worked Examples
1

Predict product: (a) $CH_3CH_2COOH + LiAlH_4$ (b) $CH_3CH_2COCl + (CH_3CH_2)_2NH$ (c) $CH_3COOCH_3 + NaOH$ (hydrolysis)

Show solution

(a) $LiAlH_4$ reduces -COOH to $-CH_2OH$: $CH_3CH_2COOH \to CH_3CH_2CH_2OH$ (1-propanol). (b) Acid chloride + amine → amide: $CH_3CH_2COCl + (CH_3CH_2)_2NH \to CH_3CH_2CON(CH_2CH_3)_2 + HCl$. (c) Saponification (basic hydrolysis): $CH_3COOCH_3 + NaOH \to CH_3COONa + CH_3OH$.

Final Answer: (a) 1-propanol; (b) N,N-diethylpropanamide; (c) sodium acetate + methanol.

2

Predict products of Hofmann bromamide reaction on $CH_3CH_2CONH_2$.

Show solution

Hofmann bromamide degrades amide to primary amine with one fewer C: $CH_3CH_2CONH_2 + Br_2 + 4NaOH \to CH_3CH_2NH_2 + 2NaBr + Na_2CO_3 + 2H_2O$

  • Loses C from C=O (becomes carbonate)
  • 3-C propanamide → 2-C ethanamine.

Final Answer: Ethanamine ($CH_3CH_2NH_2$).

✎ Self-Check — 5 questions0 / 5
Q1.

Hell-Volhard-Zelinsky reaction:

Q2.

Acid + NaHCO₃ gives:

Q3.

Saponification of ester:

Q4.

Hofmann bromamide gives:

Q5.

Aspirin is made from:

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