Top 50 Most Important
Subjective Questions
Aldehydes, Ketones, and Carboxylic Acids. Master Nucleophilic Addition, Aldol, Cannizzaro, and Acidity trends.
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Preparation (Name Reactions)
Acid chlorides (acyl chlorides) are hydrogenated over a catalyst of Palladium deposited on Barium Sulphate to form Aldehydes.
$R-COCl + H_2 \xrightarrow{Pd/BaSO_4} R-CHO + HCl$
Role of $BaSO_4$: It acts as a catalytic poison for Palladium. It deliberately reduces the activity of Pd, stopping the further reduction of the newly formed aldehyde into a primary alcohol.
Nitriles (Cyanides) are reduced to imines using Stannous Chloride ($SnCl_2$) in the presence of Hydrochloric acid ($HCl$), which on subsequent acid hydrolysis gives the corresponding Aldehydes.
$R-CN + SnCl_2 + HCl \rightarrow R-CH=NH \xrightarrow{H_3O^+} R-CHO$
Toluene is gently oxidized by Chromyl Chloride ($CrO_2Cl_2$) in a non-polar solvent like $CS_2$. This forms a brown chromium complex. Upon acid hydrolysis, this complex breaks down to yield Benzaldehyde. Chromyl chloride prevents further oxidation to benzoic acid.
When Benzene is treated with a mixture of Carbon Monoxide ($CO$) and Hydrogen Chloride ($HCl$) in the presence of anhydrous Aluminum Chloride ($AlCl_3$) and Cuprous Chloride ($CuCl$), an aldehyde group is introduced onto the ring to form Benzaldehyde.
Acid chlorides react with dialkylcadmium ($R_2Cd$) to specifically form Ketones.
$2R'-COCl + R_2Cd \rightarrow 2R'-CO-R + CdCl_2$
(Dialkylcadmium is itself prepared by reacting a Grignard reagent with $CdCl_2$. It is less reactive than Grignard reagents, so it stops at the ketone stage without further reacting to form tertiary alcohols).
By using the Friedel-Crafts Acylation reaction.
Benzene is treated with Acetyl Chloride ($CH_3COCl$) in the presence of anhydrous $AlCl_3$. The highly electrophilic acylium ion attacks the benzene ring, yielding Acetophenone ($C_6H_5COCH_3$).
Alkynes react with water in the presence of heavy metal catalysts ($H_2SO_4$ and $HgSO_4$) at 333 K to undergo hydration. They initially form an unstable enol which immediately tautomerizes into a stable carbonyl compound.
- Ethyne forms Ethanal (an aldehyde).
- All other higher alkynes exclusively form Ketones.
Aldehydes and ketones possess polar carbonyl groups ($>C=O$) and exhibit moderately strong dipole-dipole interactions. However, they lack a hydrogen atom bonded to an electronegative oxygen, so they cannot form intermolecular hydrogen bonds like alcohols do. Hence, they vaporize at lower temperatures.
Yes. Lower members like methanal, ethanal, and propanone are miscible with water in all proportions. Although they cannot hydrogen bond with themselves, the highly electronegative oxygen atom of the carbonyl group can easily form hydrogen bonds with water molecules. Solubility decreases sharply as the bulky, hydrophobic alkyl chain grows.
This is reductive Ozonolysis. The double bond snaps completely in half.
$CH_3-C(CH_3)=CH-CH_3$ breaks at the double bond.
The left side (branched) forms Propanone (Acetone).
The right side forms Ethanal (Acetaldehyde).
Nucleophilic Addition Reactions
- Steric Factor: Aldehydes have only one bulky alkyl group, making it much easier for a nucleophile to physically approach the carbonyl carbon. Ketones have two bulky groups creating steric hindrance.
- Electronic Factor: Alkyl groups exert a +I (electron-donating) effect. Ketones have two alkyl groups that push electron density toward the carbonyl carbon, reducing its positive character (electrophilicity). Aldehydes have only one, making their carbon more electrophilic and susceptible to attack.
The addition of HCN yields a Cyanohydrin (a molecule containing both $-OH$ and $-CN$ on the same carbon).
Because pure HCN reacts very slowly, the reaction is catalyzed by a Base. The base extracts a proton from HCN to rapidly generate a high concentration of the strong nucleophile, the cyanide ion ($CN^-$).
Carbonyl compounds react with saturated aqueous $NaHSO_3$ to form a bulky, white, water-soluble crystalline bisulfite addition complex.
Utility: This reaction is highly reversible. Treating the crystalline complex with dilute mineral acid or alkali easily regenerates the original pure carbonyl compound. Therefore, it is extensively used for the separation and purification of aldehydes and ketones from non-carbonyl impurities.
When an aldehyde reacts with one equivalent of a monohydric alcohol in the presence of dry HCl gas, it forms an unstable alkoxyalcohol intermediate called a Hemiacetal.
If a second equivalent of alcohol reacts, it replaces the $-OH$ group to form a highly stable gem-dialkoxy compound known as an Acetal. (Ketones similarly form Ketals using dihydric alcohols like ethylene glycol).
Dry HCl protonates the carbonyl oxygen, dramatically increasing the electrophilicity of the carbonyl carbon, making it highly susceptible to attack by the weak alcohol nucleophile. If the acid was aqueous, the presence of water would simply drive the highly reversible equilibrium backward, hydrolyzing the acetal back into the original aldehyde.
They undergo a nucleophilic addition followed immediately by the elimination of a water molecule (condensation). The carbonyl oxygen ($=O$) and two hydrogens from the ammonia derivative ($-NH_2$) are eliminated as $H_2O$, forming compounds containing a characteristic carbon-nitrogen double bond ($>C=N-Z$).
A Schiff base is a substituted imine ($R-CH=N-R'$). It is formed when an aldehyde or ketone reacts directly with a primary amine ($R'-NH_2$) under mildly acidic conditions, eliminating water.
2,4-Dinitrophenylhydrazine (Brady's reagent) reacts with aldehydes and ketones to form highly crystalline Yellow, Orange, or Red precipitates of 2,4-dinitrophenylhydrazones.
Indication: It is the standard universal test to confirm the presence of a carbonyl group ($>C=O$).
A mildly acidic medium (pH ~3.5) is required to protonate the carbonyl oxygen, making the carbon more electrophilic.
If the pH is too low (highly acidic), the ammonia derivative itself gets protonated (forming $NH_3^+ - Z$), destroying its nucleophilic character. If too high (basic), the carbonyl oxygen is not protonated enough to trigger the reaction.
The reaction yields an Oxime.
Ethanal reacts to form Acetaldoxime.
$CH_3CHO + H_2N-OH \rightarrow CH_3CH=N-OH + H_2O$
Oxidation, Reduction & Tests
Tollens' reagent is a freshly prepared ammoniacal silver nitrate solution ($[Ag(NH_3)_2]^+$). When warmed with an Aldehyde, the aldehyde is oxidized to a carboxylate ion, and the silver ions are reduced to metallic silver, depositing as a brilliant Silver Mirror on the inner wall of the test tube.
Both aliphatic and aromatic aldehydes give this test. Ketones do not.
Fehling's reagent is a mixture of aqueous copper sulphate (Fehling A) and alkaline sodium potassium tartrate (Rochelle salt, Fehling B). On heating with an aliphatic aldehyde, the blue $Cu^{2+}$ ions are reduced to a reddish-brown precipitate of Cuprous Oxide ($Cu_2O$).
No. It is a milder oxidant than Tollens' and cannot oxidize aromatic aldehydes like Benzaldehyde. Ketones also fail this test.
Aldehydes and ketones possessing at least one methyl group attached directly to the carbonyl carbon (Methyl Ketones: $CH_3-CO-$) react with sodium hypohalite ($NaOX$, formed by $X_2 + NaOH$) to yield a haloform.
Using Iodine ($I_2$) produces a distinct yellow precipitate of Iodoform ($CHI_3$). The remainder of the molecule becomes a sodium salt of a carboxylic acid with one less carbon atom.
Use the Iodoform Test ($I_2$ + $NaOH$).
Pentan-2-one ($CH_3-CO-CH_2CH_2CH_3$) is a methyl ketone. It will react to form a yellow precipitate of iodoform ($CHI_3$).
Pentan-3-one ($CH_3CH_2-CO-CH_2CH_3$) lacks the $CH_3-CO-$ group and will not react.
The carbonyl group ($>C=O$) of aldehydes and ketones is completely stripped of oxygen and reduced directly to a methylene group ($>CH_2$) to form an Alkane.
Reagents: Zinc amalgam ($Zn-Hg$) and concentrated Hydrochloric acid ($HCl$).
Like Clemmensen, it reduces a carbonyl group ($>C=O$) to an alkane ($>CH_2$), but uses basic conditions.
Reagents: The compound is first reacted with Hydrazine ($NH_2NH_2$) to form a hydrazone, which is then heated with a strong base like KOH or NaOH in a high-boiling solvent like ethylene glycol. Nitrogen gas ($N_2$) is expelled.
The Wolff-Kishner reduction must be used.
Clemmensen reduction uses concentrated $HCl$ (strongly acidic), which would violently react with or dehydrate the acid-sensitive $-OH$ group. Wolff-Kishner operates under strongly basic conditions, leaving the acid-sensitive group perfectly intact.
Unlike aldehydes, ketones resist mild oxidation. Under vigorous conditions (strong oxidants like hot conc. $HNO_3$ or $KMnO_4$), carbon-carbon bond cleavage occurs. This complex reaction shatters the ketone into a mixture of Carboxylic Acids, each containing fewer carbon atoms than the original ketone.
During the vigorous oxidative cleavage of an unsymmetrical ketone, the carbonyl group ($>C=O$) preferentially stays attached to the smaller alkyl group.
For example, oxidizing Butan-2-one ($CH_3-CO-CH_2CH_3$) strictly yields two molecules of Ethanoic acid, rather than Propanoic + Methanoic acids.
Use Tollens' Test or Fehling's Test.
Propanal (an aldehyde) will give a silver mirror with Tollens' reagent or a red precipitate with Fehling's solution.
Propanone (a ketone) will not react with either of these mild oxidizing agents.
Aldol & Cannizzaro Reactions
The strong electron-withdrawing inductive effect (-I effect) of the highly electronegative carbonyl group pulls electron density away from the adjacent alpha-carbon, weakening the $C_\alpha - H$ bond. When a base removes this proton, the resulting carbanion (enolate ion) is highly stabilized by resonance.
Condition: The aldehyde or ketone MUST possess at least one alpha-hydrogen.
Two molecules condense in the presence of dilute alkali (like dil. NaOH) to form a $\beta$-hydroxy aldehyde (Aldol) or $\beta$-hydroxy ketone (Ketol). Upon mild heating, it rapidly loses a water molecule to yield a highly stable $\alpha,\beta$-unsaturated carbonyl compound.
Two molecules of Ethanal ($CH_3CHO$) condense:
$2CH_3CHO \xrightarrow{\text{dil. NaOH}} CH_3-CH(OH)-CH_2-CHO \text{ (3-Hydroxybutanal/Aldol)}$
Upon heating ($\Delta$), it dehydrates:
$\xrightarrow{\Delta} CH_3-CH=CH-CHO + H_2O$.
The final product is But-2-enal.
It is an aldol condensation carried out between two different aldehydes and/or ketones.
Drawback: If both different molecules possess alpha-hydrogens, they will both form enolates and attack each other and themselves. This yields a messy mixture of four distinct products, making it synthetically useless.
Condition: The aldehyde MUST lack alpha-hydrogens (e.g., Formaldehyde, Benzaldehyde).
When heated with concentrated alkali (e.g., 50% NaOH), such aldehydes undergo self-oxidation and self-reduction (disproportionation). One molecule is reduced to an alcohol, and the other is oxidized to a carboxylic acid salt.
Two molecules of methanal (formaldehyde) react in conc. KOH/NaOH:
$2HCHO + \text{conc. } KOH \xrightarrow{\Delta} CH_3OH \text{ (Methanol)} + HCOOK \text{ (Potassium formate)}$
It is a reaction between two different aldehydes that both lack alpha-hydrogens.
Example: Reaction between Benzaldehyde and Formaldehyde. Formaldehyde is exceptionally susceptible to oxidation. Thus, Formaldehyde is exclusively oxidized to sodium formate, and Benzaldehyde is exclusively reduced to Benzyl alcohol.
Benzaldehyde ($C_6H_5CHO$) strictly undergoes the Cannizzaro reaction. It does not undergo Aldol condensation.
Reason: The carbon in the benzene ring attached directly to the aldehyde group (the alpha carbon) has no hydrogen atoms bonded to it. Without an alpha-hydrogen, aldol is impossible.
The carbonyl group ($-CHO$ or $-COR$) acts as a powerful electron-withdrawing group (-R effect). It pulls pi-electrons out of the benzene ring, deactivating the ring entirely, but it drains electron density preferentially from the ortho and para positions. Thus, the meta position remains comparatively electron-rich, directing electrophiles there.
No. The $-CHO$ group is strongly deactivating. It pulls electron density away from the ring so aggressively that the weak electrophiles generated in Friedel-Crafts reactions are unable to attack. Furthermore, the Lewis acid catalyst ($AlCl_3$) strongly complexes with the carbonyl oxygen, deactivating the ring even further.
Carboxylic Acids (Acidity)
Carboxylic acids form exceptionally strong, extensive intermolecular hydrogen bonds. In the vapor phase or non-polar solvents, most carboxylic acids exist as tightly bound dimers (pairs of molecules bonded twice to each other). Breaking these dimers requires massive thermal energy.
After losing a proton, the resulting carboxylate ion is stabilized by two perfectly equivalent resonance structures where the negative charge delocalizes over two highly electronegative oxygen atoms. In phenol, the negative charge of the phenoxide ion delocalizes onto less electronegative carbon atoms, providing inferior stabilization.
EWGs (like halogens, $-NO_2$, $-CN$) exert a -I effect. They powerfully pull electron density away from the $O-H$ bond, making it easier to release $H^+$. More importantly, they disperse the negative charge on the resulting carboxylate ion, heavily stabilizing the conjugate base and thus dramatically increasing acidity.
Acetic acid < Chloroacetic acid < Trichloroacetic acid < Trifluoroacetic acid.
Acetic acid has a +I group. One Cl provides a strong -I effect. Three Cl atoms provide a massive -I effect. Three F atoms provide the strongest -I effect (because F is more electronegative than Cl), resulting in the highest acidity.
Carboxylic acids having an alpha-hydrogen are specifically halogenated at the $\alpha$-position when treated with Chlorine or Bromine in the presence of a small amount of Red Phosphorus. The product is an $\alpha$-halo carboxylic acid.
$R-CH_2-COOH \xrightarrow[\text{2. } H_2O]{1. \text{ } X_2/\text{Red P}} R-CH(X)-COOH$
Nitriles ($R-CN$) undergo complete hydrolysis when boiled with dilute mineral acids (or alkalis) to directly form Carboxylic Acids. The reaction passes through an amide intermediate.
$R-CN + 2H_2O \xrightarrow{H^+ \text{ or } OH^-, \Delta} R-COOH + NH_3$
Grignard reagents ($R-MgX$) act as strong nucleophiles and attack the electrophilic carbon of solid Carbon Dioxide (Dry Ice, $CO_2$) to form a magnesium carboxylate complex. Mild acidic hydrolysis of this complex yields the carboxylic acid containing one carbon more than the original alkyl halide.
Decarboxylation is the process of removing a carboxyl group from a molecule as $CO_2$ gas.
When the sodium salt of a carboxylic acid is strongly heated with Soda Lime (a 3:1 mixture of $NaOH$ and $CaO$), it yields an Alkane containing one carbon atom less than the original acid.
The carbonyl carbon in a carboxylic acid is attached to an $-OH$ group. The lone pair of electrons on the oxygen of the $-OH$ group is highly involved in resonance with the carbonyl group's $\pi$-bond. This delocalization severely reduces the electrophilic (positive) character of the carbonyl carbon, rendering it totally unresponsive to normal nucleophilic addition tests.
Use the Sodium Bicarbonate ($NaHCO_3$) Test.
Benzoic acid is a strong enough acid to rapidly decompose $NaHCO_3$, releasing brisk effervescence (bubbles) of $CO_2$ gas.
Phenol is a much weaker acid and cannot decompose $NaHCO_3$, so no effervescence is observed.
Chapter 8 Mastered!
You have conquered the 50 most critical subjective questions for Class 12 Chemistry, Chapter 8: Aldehydes, Ketones, and Carboxylic Acids. You are ready to ace Organic mechanisms.
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