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Aldehydes, Ketones and Carboxylic Acids - Physical and chemical properties

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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The Carbonyl group (>C=O>C=O) is polar due to the higher electronegativity of oxygen compared to carbon, making the carbon atom electrophilic and the oxygen atom nucleophilic.

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Aldehydes and ketones have higher boiling points than non-polar hydrocarbons of similar molecular mass due to dipole-dipole interactions, but lower than alcohols because they cannot form intermolecular hydrogen bonds.

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Lower aldehydes and ketones are soluble in water because they can form hydrogen bonds with H2OH_2O molecules.

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Nucleophilic addition reactions are characteristic of carbonyl compounds. Aldehydes are generally more reactive than ketones due to steric and inductive effects.

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Oxidation: Aldehydes are easily oxidized to carboxylic acids using mild agents like Tollens' reagent ([Ag(NH3)2]+[Ag(NH_3)_2]^{+}) or Fehling's solution (Cu2+Cu^{2+} ions). Ketones require vigorous conditions.

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Reduction: Carbonyls can be reduced to hydrocarbons via Clemmensen reduction (using Znβˆ’Hg/HClZn-Hg/HCl) or Wolff-Kishner reduction (using NH2NH2/KOHNH_2NH_2/KOH).

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alpha\\alpha-Hydrogen reactions: Aldehydes/Ketones with at least one alpha\\alpha-hydrogen undergo Aldol condensation in the presence of dilute alkali.

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Cannizzaro reaction: Aldehydes with no alpha\\alpha-hydrogen (e.g., HCHOHCHO, C6H5CHOC_6H_5CHO) undergo self-oxidation and reduction (disproportionation) with concentrated alkali.

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Carboxylic acids are more acidic than alcohols and phenols because the carboxylate ion (RCOOβˆ’RCOO^-) is stabilized by two equivalent resonance structures.

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Hell-Volhard-Zelinsky (HVZ) reaction: Carboxylic acids with alpha\\alpha-hydrogen react with Cl2Cl_2 or Br2Br_2 in the presence of red phosphorus to give alpha\\alpha-halo carboxylic acids.

πŸ“Formulae

Rβˆ’CHO+HCNxrightarrowOHβˆ’Rβˆ’CH(OH)CNtext(Cyanohydrinformation)R-CHO + HCN \\xrightarrow{OH^-} R-CH(OH)CN \\text{ (Cyanohydrin formation)}

2CH3CHOxrightarrowtextdil.NaOHCH3CH(OH)CH2CHOxrightarrowDeltaCH3CH=CHCHO+H2O2CH_3CHO \\xrightarrow{\\text{dil. } NaOH} CH_3CH(OH)CH_2CHO \\xrightarrow{\\Delta} CH_3CH=CHCHO + H_2O

2HCHO+textconc.NaOHrightarrowCH3OH+HCOONa2HCHO + \\text{conc. } NaOH \\rightarrow CH_3OH + HCOONa

Rβˆ’CH2βˆ’COOHxrightarrow[ii)H2O]i)X2/textRedPRβˆ’CH(X)βˆ’COOHtext(whereX=Cl,Br)R-CH_2-COOH \\xrightarrow[ii) H_2O]{i) X_2 / \\text{Red P}} R-CH(X)-COOH \\text{ (where } X=Cl, Br)

Rβˆ’COOH+Rβ€²βˆ’OHxrightarrowH+Rβˆ’COORβ€²+H2Otext(Esterification)R-COOH + R'-OH \\xrightarrow{H^+} R-COOR' + H_2O \\text{ (Esterification)}

Ka=frac[RCOOβˆ’][H+][RCOOH]textandpKa=βˆ’logKaK_a = \\frac{[RCOO^-][H^+]}{[RCOOH]} \\text{ and } pK_a = -\\log K_a

πŸ’‘Examples

Problem 1:

Distinguish between Propanal and Propanone using a chemical test.

Solution:

Use the Tollens' Test. Propanal (CH3CH2CHOCH_3CH_2CHO) reacts with Tollens' reagent to form a silver mirror, while Propanone (CH3COCH3CH_3COCH_3) does not.

Explanation:

Aldehydes are easily oxidized by mild oxidizing agents like [Ag(NH3)2]+[Ag(NH_3)_2]^+. The reaction is: CH3CH2CHO+2[Ag(NH3)2]++3OHβˆ’rightarrowCH3CH2COOβˆ’+2Agdownarrow+4NH3+2H2OCH_3CH_2CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \\rightarrow CH_3CH_2COO^- + 2Ag \\downarrow + 4NH_3 + 2H_2O. Ketones lack the hydrogen atom attached to the carbonyl carbon, making them resistant to such mild oxidation.

Problem 2:

Why is Chloroacetic acid (ClCH2COOHClCH_2COOH) stronger than Acetic acid (CH3COOHCH_3COOH)?

Solution:

Chloroacetic acid is stronger due to the βˆ’I-I (inductive) effect of the Chlorine atom.

Explanation:

The electronegative ClCl atom withdraws electron density from the Oβˆ’HO-H bond and stabilizes the resulting carboxylate ion (ClCH2COOβˆ’ClCH_2COO^-) through the inductive effect. This makes the release of H+H^+ easier compared to CH3COOHCH_3COOH, where the βˆ’CH3-CH_3 group has a +I+I effect which destabilizes the anion.

Physical and chemical properties - Revision Notes & Key Formulas | ICSE Class 12 Chemistry