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Aldehydes, Ketones and Carboxylic Acids - Physical and Chemical Properties

Grade 12CBSEChemistry

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

🔑Concepts

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Aldehydes and ketones undergo nucleophilic addition reactions because the carbonyl carbon is electrophilic due to the polar nature of the >C=O>C=O bond.

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Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric factors and the +I+I effect of two alkyl groups in ketones which reduces the electrophilicity of the carbonyl carbon.

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The boiling points of aldehydes and ketones are higher than non-polar compounds (ethers, hydrocarbons) but lower than alcohols and carboxylic acids because they cannot form intermolecular hydrogen bonds with themselves.

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Carboxylic acids have higher boiling points than alcohols of similar molecular mass because they form stable cyclic dimers through extensive intermolecular hydrogen bonding.

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Aldehydes reduce Tollens' reagent to form a silver mirror ([Ag(NH3)2]+[Ag(NH_{3})_{2}]^{+}) and Fehling's solution to form a red precipitate of Cu2OCu_{2}O, whereas ketones do not.

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Aldehydes and ketones with at least one α\alpha-hydrogen undergo Aldol condensation in the presence of dilute alkali to form β\beta-hydroxy aldehydes (aldols) or β\beta-hydroxy ketones (ketols).

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Aldehydes which do not have an α\alpha-hydrogen atom (e.g., HCHOHCHO, C6H5CHOC_{6}H_{5}CHO) undergo Cannizzaro reaction (self-oxidation and reduction) when treated with concentrated alkali.

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The acidity of carboxylic acids is enhanced by electron-withdrawing groups (EWG) like −Cl-Cl, −NO2-NO_{2}, and −CN-CN through the −I-I effect, which stabilizes the carboxylate ion (RCOO−RCOO^{-}).

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Hell-Volhard-Zelinsky (HVZ) reaction: Carboxylic acids having an α\alpha-hydrogen are halogenated at the α\alpha-position on treatment with Cl2Cl_{2} or Br2Br_{2} in the presence of small amount of red phosphorus.

📐Formulae

R−CHO+2[Ag(NH3)2]++3OH−→RCOO−+2Ag+2H2O+4NH3R-CHO + 2[Ag(NH_{3})_{2}]^{+} + 3OH^{-} \rightarrow RCOO^{-} + 2Ag + 2H_{2}O + 4NH_{3}

R−CHO+2Cu2++5OH−→RCOO−+Cu2O+3H2OR-CHO + 2Cu^{2+} + 5OH^{-} \rightarrow RCOO^{-} + Cu_{2}O + 3H_{2}O

R−CO−CH3→NaOXR−COONa+CHX3 (Iodoform Test)R-CO-CH_{3} \xrightarrow{NaOX} R-COONa + CHX_{3} \text{ (Iodoform Test)}

2HCHO→conc.NaOHCH3OH+HCOONa2HCHO \xrightarrow{conc. NaOH} CH_{3}OH + HCOONa

R−CH2−COOH→H2O(i)X2/RedPR−CH(X)−COOHR-CH_{2}-COOH \xrightarrow[H_{2}O]{(i) X_{2}/Red P} R-CH(X)-COOH

Ka=[RCOO−][H+][RCOOH]K_{a} = \frac{[RCOO^{-}][H^{+}]}{[RCOOH]}

pKa=−log⁡KapK_{a} = -\log K_{a}

💡Examples

Problem 1:

Arrange the following compounds in increasing order of their reactivity towards nucleophilic addition: CH3CHOCH_{3}CHO, CH3COCH3CH_{3}COCH_{3}, HCHOHCHO, C2H5COCH3C_{2}H_{5}COCH_{3}.

Solution:

C2H5COCH3<CH3COCH3<CH3CHO<HCHOC_{2}H_{5}COCH_{3} < CH_{3}COCH_{3} < CH_{3}CHO < HCHO

Explanation:

Reactivity decreases as the size and number of alkyl groups increase due to steric hindrance and the inductive effect (+I+I), which reduces the positive charge density on the carbonyl carbon.

Problem 2:

Identify the product formed when Propan-2-one reacts with HCNHCN.

Solution:

Acetone cyanohydrin (22-Hydroxy-22-methylpropanenitrile)

Explanation:

The cyanide ion (CN−CN^{-}) acts as a nucleophile and attacks the carbonyl carbon of (CH3)2C=O(CH_{3})_{2}C=O, followed by protonation of the oxygen to form (CH3)2C(OH)CN(CH_{3})_{2}C(OH)CN.

Problem 3:

Which is a stronger acid: Cl−CH2−COOHCl-CH_{2}-COOH or CH3−COOHCH_{3}-COOH?

Solution:

Cl−CH2−COOHCl-CH_{2}-COOH (Chloroacetic acid) is stronger.

Explanation:

The chlorine atom is an electron-withdrawing group (−I-I effect). It withdraws electron density from the carboxylate group, stabilizing the Cl−CH2−COO−Cl-CH_{2}-COO^{-} ion more effectively than the +I+I effect of the methyl group in acetic acid stabilizes CH3−COO−CH_{3}-COO^{-}.