Aldehydes, Ketones & Carboxylic Acids
Complete Organic Chemistry Revision Guide for CBSE
1. The Carbonyl Group: Structure & Nature
Aldehydes, ketones, and carboxylic acids all contain the carbonyl group ($>C=O$). It is one of the most important functional groups in organic chemistry.
- Structure: The carbonyl carbon is $sp^2$ hybridized, forming three sigma ($\sigma$) bonds in a trigonal planar geometry (bond angles ~120°). The unhybridized p-orbital forms a pi ($\pi$) bond with the p-orbital of oxygen.
- Nature of the Bond: Oxygen is significantly more electronegative than carbon. Consequently, the $>C=O$ bond is highly polarized. The carbonyl carbon carries a partial positive charge ($\delta^+$), making it an electrophilic (Lewis acid) center, highly susceptible to nucleophilic attack. The oxygen carries a partial negative charge ($\delta^-$), making it a nucleophilic (Lewis base) center.
2. Master Cheat Sheet: Top 10 Name Reactions
Warning: Do not enter the exam hall without memorizing these reactions and their specific reagents!
A. Reactions Involving the $\alpha$-Hydrogen
- 1. Aldol Condensation: Aldehydes/Ketones having at least one $\alpha$-hydrogen undergo reaction in the presence of dilute alkali (dil. NaOH or Ba(OH)₂) to form $\beta$-hydroxy aldehydes (aldol) or $\beta$-hydroxy ketones (ketol). Heating removes water to give $\alpha,\beta$-unsaturated carbonyls.
$$2 CH_3CHO \xrightarrow{\text{dil. } NaOH} CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta} CH_3-CH=CH-CHO + H_2O$$
- 2. Cannizzaro Reaction: Aldehydes without an $\alpha$-hydrogen (e.g., Formaldehyde, Benzaldehyde) undergo self-oxidation and reduction (disproportionation) when heated with concentrated alkali (conc. NaOH/KOH) to form an alcohol and a salt of a carboxylic acid.
$$2 HCHO + \text{conc. } KOH \xrightarrow{\Delta} CH_3OH \text{ (Methanol)} + HCOOK \text{ (Potassium formate)}$$
B. Reduction to Alkanes
- 3. Clemmensen Reduction: Carbonyl group reduces to a $CH_2$ group on treatment with Zinc amalgam and concentrated Hydrochloric acid (Zn-Hg / conc. HCl).
$$>C=O + 4[H] \xrightarrow{Zn-Hg \text{ / conc. } HCl} >CH_2 + H_2O$$
- 4. Wolff-Kishner Reduction: Carbonyl group reduces to a $CH_2$ group on treatment with Hydrazine ($NH_2NH_2$) followed by heating with KOH in a high boiling solvent like ethylene glycol.
$$>C=O \xrightarrow{NH_2NH_2} >C=N-NH_2 \xrightarrow{KOH / \text{Ethylene glycol, } \Delta} >CH_2 + N_2$$
C. Preparation Reactions
- 5. Rosenmund Reduction: Acyl chlorides (acid chlorides) are hydrogenated over catalyst, palladium on barium sulfate ($Pd/BaSO_4$), to form aldehydes. (BaSO₄ acts as a catalytic poison to prevent further reduction to alcohol).
$$R-COCl + H_2 \xrightarrow{Pd/BaSO_4} R-CHO + HCl$$
- 6. Stephen Reaction: Nitriles are reduced to corresponding imine with stannous chloride in the presence of HCl ($SnCl_2 / HCl$), which on hydrolysis gives the corresponding aldehyde.
$$R-C\equiv N + SnCl_2 + HCl \rightarrow R-CH=NH \xrightarrow{H_3O^+} R-CHO$$
- 7. Etard Reaction: Chromyl chloride ($CrO_2Cl_2$) oxidizes the methyl group of toluene to a chromium complex, which on hydrolysis provides benzaldehyde.
$$C_6H_5CH_3 + CrO_2Cl_2 \xrightarrow{CS_2} \text{Brown Complex} \xrightarrow{H_3O^+} C_6H_5CHO$$
- 8. Gattermann-Koch Reaction: Benzene reacts with carbon monoxide and hydrogen chloride ($CO, HCl$) in the presence of anhydrous aluminium chloride ($AlCl_3 / CuCl$) to give benzaldehyde.
D. Reactions of Carboxylic Acids
- 9. Hell-Volhard-Zelinsky (HVZ) Reaction: Carboxylic acids having an $\alpha$-hydrogen are halogenated at the $\alpha$-position on treatment with chlorine or bromine in the presence of small amounts of Red Phosphorus.
$$R-CH_2-COOH \xrightarrow[\text{2. } H_2O]{\text{1. } X_2 / \text{Red P}} R-CH(X)-COOH \quad (X = Cl, Br)$$
- 10. Decarboxylation: Sodium salts of carboxylic acids lose $CO_2$ on heating with sodalime (NaOH and CaO in 3:1 ratio) to yield alkanes having one carbon less.
$$R-COONa + NaOH \xrightarrow{CaO, \Delta} R-H + Na_2CO_3$$
3. Chemical Properties & Mechanism
3.1 Nucleophilic Addition Reactions
This is the characteristic reaction of aldehydes and ketones. The nucleophile attacks the electrophilic carbonyl carbon, breaking the $\pi$ bond and pushing electrons to the oxygen. A tetrahedral intermediate is formed. Finally, protonation of the oxygen yields the addition product.
Reactivity Order: Aldehydes > Ketones
Aldehydes are more reactive than ketones towards nucleophilic addition for two major reasons:
- Steric Factor: The presence of two relatively large alkyl groups in ketones hinders the approach of the nucleophile to the carbonyl carbon compared to aldehydes (which have only one alkyl group and one small hydrogen atom).
- Electronic Factor (+I Effect): Alkyl groups are electron-releasing. Ketones have two alkyl groups that push electron density towards the carbonyl carbon, reducing its positive charge (electrophilicity) more than the single alkyl group in an aldehyde. Hence, the nucleophile is less strongly attracted to the ketone's carbon.
Important Nucleophilic Additions:
- Addition of HCN: Yields cyanohydrins. The reaction is catalyzed by a base ($OH^-$) which generates the strong nucleophile $CN^-$ from HCN.
- Addition of Grignard Reagent ($RMgX$): Yields alcohols. (Formaldehyde $\rightarrow$ 1° alcohol, other Aldehydes $\rightarrow$ 2° alcohol, Ketones $\rightarrow$ 3° alcohol).
- Addition of Ammonia Derivatives ($NH_2-Z$): Reacts in a slightly acidic medium to form products of the type $>C=N-Z$ with the loss of water.
Crucial Example: 2,4-DNP Test. Aldehydes and ketones react with 2,4-dinitrophenylhydrazine to form yellow, orange, or red precipitates of 2,4-dinitrophenylhydrazones. This is the standard test to detect the carbonyl group.
3.2 Oxidation Reactions (Distinguishing Tests)
Aldehydes are easily oxidized to carboxylic acids (they have a removable H attached to the carbonyl carbon). Ketones require vigorous conditions to oxidize, involving carbon-carbon bond cleavage. This difference is exploited in mild chemical tests.
| Test | Reagent | Observation (Aldehydes) | Observation (Ketones) |
|---|---|---|---|
| Tollens' Test | Ammoniacal Silver Nitrate $[Ag(NH_3)_2]^+$ | Silver Mirror (Ag metal deposited) | No reaction |
| Fehling's Test | Fehling A ($CuSO_4$) + Fehling B (Rochelle salt + NaOH) | Reddish-Brown Precipitate of $Cu_2O$ | No reaction |
Note: Benzaldehyde (an aromatic aldehyde) gives a positive Tollens' test but a negative Fehling's test.
The Haloform (Iodoform) Reaction
Reagent: $NaOI$ (prepared *in situ* from $I_2 + NaOH$).
Requirement: It is given ONLY by compounds containing the $CH_3-CO-$ group (methyl ketones) or compounds that can oxidize to it (like $CH_3-CH(OH)-$ such as ethanol or propan-2-ol).
Result: Formation of a brilliant yellow precipitate of Iodoform ($CHI_3$).
Usage: Excellent for distinguishing Propanone (gives test) from Propanal (no test), or Pentan-2-one (gives test) from Pentan-3-one (no test), or Acetophenone (gives test) from Benzophenone (no test).
3.3 Acidity of Carboxylic Acids
Carboxylic acids are the most acidic among common organic compounds. They turn blue litmus red. They react with electropositive metals, alkalis, and remarkably, with weaker bases like carbonates ($Na_2CO_3$) and bicarbonates ($NaHCO_3$) to evolve $CO_2$ gas with brisk effervescence (a standard test to distinguish carboxylic acids from phenols, as phenols do not react with $NaHCO_3$).
Reason for Acidity: The carboxylate ion ($R-COO^-$) formed after the loss of $H^+$ is highly stabilized by resonance. The negative charge is equally delocalized over two highly electronegative oxygen atoms. (In phenoxide, the charge is delocalized over less electronegative carbon atoms, hence carboxylic acids are stronger acids than phenols).
Effect of Substituents on Acidic Strength
- Electron Withdrawing Groups (EWG): Groups like $-CF_3$, $-NO_2$, $-CN$, $-F$, $-Cl$ stabilize the carboxylate ion by dispersing the negative charge through the -I effect. They increase the acidic strength.
Order: $CF_3COOH > CCl_3COOH > CHCl_2COOH > CH_2ClCOOH > CH_3COOH$ - Electron Donating Groups (EDG): Groups like alkyl groups ($-CH_3$) destabilize the carboxylate ion by intensifying the negative charge through the +I effect. They decrease the acidic strength.
Order: $HCOOH > CH_3COOH > CH_3CH_2COOH$
4. Most Important CBSE Board Questions
Strictly aligned with the latest CBSE sample papers, emphasizing reasoning and roadmap word problems.
Section A: 1 Mark Questions (MCQs & A-R)
Q1. Which of the following compounds will give a positive Iodoform test? 1 Mark
- Propanal
- Pentan-3-one
- Acetophenone
- Benzophenone
Explanation: The iodoform test requires the presence of a terminal methyl keto group ($CH_3-CO-$). Acetophenone is $C_6H_5-CO-CH_3$. Propanal is $CH_3CH_2CHO$. Pentan-3-one is $CH_3CH_2-CO-CH_2CH_3$. Benzophenone is $C_6H_5-CO-C_6H_5$.
Q2. The correct order of reactivity towards nucleophilic addition reaction is: 1 Mark
- Ethanal > Propanal > Propanone > Butanone
- Butanone > Propanone > Propanal > Ethanal
- Propanone > Butanone > Ethanal > Propanal
- Ethanal > Propanone > Propanal > Butanone
Explanation: Aldehydes are more reactive than ketones due to less steric hindrance and less +I effect. As the size of the alkyl group increases, steric hindrance increases and +I effect increases, decreasing reactivity.
Assertion-Reasoning
Q3. Assertion (A): Carboxylic acids do not give characteristic reactions of the carbonyl group like forming 2,4-DNP derivatives.
Reason (R): The carbonyl carbon in carboxylic acids is involved in resonance with the lone pair of the oxygen of the -OH group. 1 Mark
Explanation: Because of resonance, the $>C=O$ double bond character in carboxylic acids is significantly reduced compared to aldehydes and ketones. The electrophilic character of the carbonyl carbon is greatly diminished, preventing typical nucleophilic addition reactions.
Section B: 2 Mark Questions (Very Short Answer)
Q4. Give reasons for the following:
(i) Oxidation of propanal is easier than propanone.
(ii) Chloroacetic acid is a stronger acid than acetic acid. 2 Marks
(ii) Chlorine is an electronegative element. It exerts an electron-withdrawing inductive effect (-I effect). This effect disperses the negative charge on the carboxylate ion formed after the loss of $H^+$, stabilizing it. Acetic acid has an electron-donating methyl group (+I effect) which destabilizes the conjugate base.
Q5. How will you convert:
(i) Toluene to Benzaldehyde
(ii) Ethanal to But-2-enal 2 Marks
$$C_6H_5CH_3 \xrightarrow[\text{2. } H_3O^+]{\text{1. } CrO_2Cl_2, CS_2} C_6H_5CHO$$
(ii) Aldol Condensation: Treat ethanal with dilute NaOH, followed by heating.
$$2 CH_3CHO \xrightarrow{\text{dil. } NaOH} CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta} CH_3-CH=CH-CHO (\text{But-2-enal}) + H_2O$$
Section C: 3 Mark Questions (Short Answer)
Q6. Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Propanal and Propanone
(ii) Benzaldehyde and Acetophenone
(iii) Phenol and Benzoic acid 3 Marks
(ii) Benzaldehyde and Acetophenone: Iodoform test. Acetophenone ($C_6H_5COCH_3$) contains a methyl keto group and will give a yellow precipitate of Iodoform ($CHI_3$) when treated with $I_2$ and $NaOH$. Benzaldehyde will not give this test.
(iii) Phenol and Benzoic acid: Sodium Bicarbonate ($NaHCO_3$) test. Benzoic acid reacts with aqueous $NaHCO_3$ to produce brisk effervescence due to the evolution of $CO_2$ gas. Phenol is a weaker acid and does not react with $NaHCO_3$.
Q7. Write the mechanism of addition of hydrogen cyanide (HCN) to ethanal. 3 Marks
Step 1: Generation of Nucleophile. The base removes a proton from HCN to generate the strong nucleophile, cyanide ion ($CN^-$).
$$HCN + OH^- \rightleftharpoons CN^- + H_2O$$
Step 2: Nucleophilic Attack. The $CN^-$ ion attacks the electrophilic carbonyl carbon of ethanal. The $\pi$ electrons transfer to the electronegative oxygen atom, forming a tetrahedral alkoxide intermediate.
$$CH_3-CHO + CN^- \rightarrow [CH_3-CH(O^-)(CN)]$$
Step 3: Protonation. The intermediate extracts a proton from water (or HCN) to yield the cyanohydrin.
$$[CH_3-CH(O^-)(CN)] + H_2O \rightarrow CH_3-CH(OH)(CN) + OH^-$$
Section D: 4 Mark Case-Based Questions
Q8. Read the passage given below and answer the following questions:
(i) Why is the $\alpha$-hydrogen of aldehydes and ketones acidic in nature? (1 Mark)
(ii) Predict the product when Benzaldehyde reacts with concentrated NaOH. (1 Mark)
(iii) Differentiate between Cross-Aldol and self-Aldol condensation with suitable examples. (2 Marks)
(ii) Benzaldehyde lacks an $\alpha$-hydrogen. With conc. NaOH, it undergoes Cannizzaro reaction (disproportionation). One molecule is oxidized to Sodium benzoate ($C_6H_5COONa$) and another is reduced to Benzyl alcohol ($C_6H_5CH_2OH$).
(iii) Self-Aldol Condensation: Occurs between two molecules of the same aldehyde or ketone. E.g., two molecules of Ethanal react to give 3-Hydroxybutanal.
Cross-Aldol Condensation: Occurs between two different aldehydes or ketones. If both have $\alpha$-hydrogens, a mixture of four products is formed. E.g., reaction between Ethanal and Propanal. (If one doesn't have an $\alpha$-hydrogen, like Benzaldehyde reacting with Acetophenone, it yields a single major crossed-aldol product).
Section E: 5 Mark Long Answer Questions
Q9. An organic compound (A) with molecular formula $C_8H_8O$ forms an orange-red precipitate with 2,4-DNP reagent and gives a yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens' or Fehling's reagent, nor does it decolorize bromine water or Baeyer's reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula $C_7H_6O_2$. Identify the compounds (A) and (B) and explain the reactions involved. 5 Marks
- "Forms orange-red ppt with 2,4-DNP" $\rightarrow$ Compound (A) contains a carbonyl group (it is an aldehyde or ketone).
- "Neither reduces Tollens' nor Fehling's reagent" $\rightarrow$ Compound (A) is a ketone, not an aldehyde.
- "Gives yellow ppt with $I_2/NaOH$" $\rightarrow$ This is a positive Iodoform test. Therefore, the ketone must be a methyl ketone containing the $CH_3-CO-$ group.
- "Does not decolorize $Br_2$ water or Baeyer's reagent" $\rightarrow$ Compound (A) does not contain any aliphatic double or triple bonds ($C=C$ or $C\equiv C$). The degree of unsaturation (formula $C_8H_8O$) strongly points to an aromatic benzene ring ($C_6H_5-$).
- Putting it together: An aromatic ring ($C_6H_5$, which is $C_6H_5$) attached to a methyl keto group ($-CO-CH_3$) gives the formula $C_6H_5-CO-CH_3$ ($C_8H_8O$). Thus, Compound (A) is Acetophenone.
- "Drastic oxidation gives (B) $C_7H_6O_2$" $\rightarrow$ Oxidation of acetophenone cleaves the terminal methyl group, converting the side chain to a carboxylic acid group on the benzene ring. This forms Benzoic acid. Thus, Compound (B) is Benzoic acid.
1. Iodoform Reaction:
$$C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3 \downarrow \text{(yellow ppt)} + 3NaI + 3H_2O$$
2. Oxidation:
$$C_6H_5COCH_3 \xrightarrow{H_2CrO_4 / \Delta} C_6H_5COOH \text{ (Benzoic Acid)} + CO_2 + H_2O$$
Mastering 'Aldehydes, Ketones & Carboxylic Acids'
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The "Roadmap" Codebreaker: The 5-mark question in this chapter is almost always a roadmap. Memorize the clues: 2,4-DNP = carbonyl, Tollens = aldehyde, Iodoform = methyl ketone. With these three clues, you can solve 90% of CBSE word problems.
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Aldol vs Cannizzaro: If the molecule has an $\alpha$-hydrogen (like acetaldehyde, acetone), it undergoes Aldol with *dilute* NaOH. If it lacks an $\alpha$-hydrogen (like benzaldehyde, formaldehyde), it undergoes Cannizzaro with *concentrated* NaOH.
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Don't get tricked by Benzaldehyde: Remember that Benzaldehyde ($C_6H_5CHO$) gives a positive Tollens' test (silver mirror) but a negative Fehling's test. It is the best way to distinguish an aliphatic aldehyde from an aromatic one.
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Acidity Ordering: For carboxylic acids, look for Halogens. Fluorine pulls electrons hardest (-I effect), making the acid stronger. The closer the halogen is to the $COOH$ group, the stronger the acid.
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