Aldehydes, Ketones and Carboxylic Acids
Aldehydes, Ketones and Carboxylic Acids for JEE Main & Advanced
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:
| Structure | Common Name | IUPAC Name |
|---|---|---|
| $HCHO$ | Formaldehyde | Methanal |
| $CH_3CHO$ | Acetaldehyde | Ethanal |
| $CH_3CH_2CHO$ | Propionaldehyde | Propanal |
| $C_6H_5CHO$ | Benzaldehyde | Phenylmethanal |
| $CH_3COCH_3$ | Acetone | Propan-2-one |
| $CH_3CH_2COCH_3$ | Methyl ethyl ketone | Butan-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°)
Why are aldehydes more reactive than ketones toward nucleophilic addition?
Show solution
- Steric: Aldehydes have $H$ on C=O; ketones have 2 R groups. Less hindrance in aldehyde → easier Nu attack.
- 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.
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.
Aldehyde vs ketone in addition:
Rosenmund reduction converts:
Cyanohydrin is formed by:
2,4-DNP test confirms:
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)
Compare Tollens' and Fehling's tests for aldehydes.
Show solution
| Property | Tollens' | Fehling's |
|---|---|---|
| Reagent | $Ag(NH_3)_2^+$ in NH₃ solution | $Cu^{2+}$ + tartrate in alkali |
| Positive for | All aldehydes (aliphatic and aromatic) | Aliphatic aldehydes only |
| Result | Silver mirror | Red ppt ($Cu_2O$) |
| Ketones | Negative | Negative |
Final Answer: Both distinguish RCHO from R₂CO; only Tollens works for aromatic aldehydes.
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.
Tollens test gives:
Cannizzaro reaction requires:
Iodoform from acetone:
Aldol product of $2CH_3CHO$:
Wolff-Kishner is similar in effect to:
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
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: $CH_3COOH < HCOOH < ClCH_2COOH < CCl_3COOH$.
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.
Most acidic:
RMgX + CO₂ gives:
Acetic acid in water:
Most acidic among Cl-substituted acetic acids:
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)
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.
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 ($CH_3CH_2NH_2$).
Hell-Volhard-Zelinsky reaction:
Acid + NaHCO₃ gives:
Saponification of ester:
Hofmann bromamide gives:
Aspirin is made from:
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