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Aldehydes, Ketones and Carboxylic Acids - Alpha hydrogen reactivity in aldehydes

Grade 12ICSEChemistry

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

πŸ”‘Concepts

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The Ξ±\alpha-hydrogen atoms in aldehydes and ketones are acidic in nature. This acidity is primarily due to the strong electron-withdrawing inductive effect (βˆ’I-I effect) of the carbonyl group (>C=O>C=O) and the resonance stabilization of the resulting conjugate base, known as the enolate ion.

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Aldol Condensation: Aldehydes and ketones possessing at least one Ξ±\alpha-hydrogen undergo a reaction in the presence of dilute alkali (such as NaOHNaOH or Ba(OH)2Ba(OH)_2) to form Ξ²\beta-hydroxy aldehydes (aldols) or Ξ²\beta-hydroxy ketones (ketols).

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Dehydration of Aldols: The Ξ²\beta-hydroxy carbonyl compounds (aldols) readily lose a molecule of water upon heating to form Ξ±,Ξ²\alpha,\beta-unsaturated carbonyl compounds. The conjugation of the double bond with the carbonyl group provides extra stability.

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Cross Aldol Condensation: When aldol condensation is carried out between two different aldehydes and/or ketones, it is called cross aldol condensation. If both reactants contain Ξ±\alpha-hydrogens, a mixture of four products is typically obtained.

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Mechanism of Aldol Condensation: It involves three main steps: (1) Deprotonation of the Ξ±\alpha-carbon by a base to form an enolate ion, (2) Nucleophilic attack by the enolate ion on the carbonyl carbon of another molecule, and (3) Protonation of the alkoxide ion to form the aldol.

πŸ“Formulae

2Rβˆ’CH2CHOβ†’dil.NaOHRβˆ’CH2CH(OH)CH(R)CHO2R-CH_2CHO \xrightarrow{dil. NaOH} R-CH_2CH(OH)CH(R)CHO

Rβˆ’CH2CH(OH)CH(R)CHOβ†’Ξ”,βˆ’H2ORβˆ’CH2CH=C(R)CHOR-CH_2CH(OH)CH(R)CHO \xrightarrow{\Delta, -H_2O} R-CH_2CH=C(R)CHO

[CH2βˆ’CHO]βˆ’βŸ·[CH2=CHβˆ’O]βˆ’[CH_2-CHO]^- \longleftrightarrow [CH_2=CH-O]^-

2CH3CHO→dil.NaOHCH3CH(OH)CH2CHO→ΔCH3CH=CHCHO+H2O2CH_3CHO \xrightarrow{dil. NaOH} CH_3CH(OH)CH_2CHO \xrightarrow{\Delta} CH_3CH=CHCHO + H_2O

πŸ’‘Examples

Problem 1:

Predict the product formed when Propanal (CH3CH2CHOCH_3CH_2CHO) reacts with dilute NaOHNaOH followed by heating.

Solution:

The product is 2-methylpent-2-enal (CH3CH2CH=C(CH3)CHOCH_3CH_2CH=C(CH_3)CHO).

Explanation:

First, the base abstracts an Ξ±\alpha-hydrogen from propanal to form the enolate ion [CH3βˆ’CHβˆ’CHO]βˆ’[CH_3-CH-CHO]^-. This enolate acts as a nucleophile and attacks the carbonyl carbon of another propanal molecule to form 3-hydroxy-2-methylpentanal (the aldol). Upon heating, this aldol undergoes dehydration (loss of H2OH_2O) to form the Ξ±,Ξ²\alpha,\beta-unsaturated aldehyde, 2-methylpent-2-enal.

Problem 2:

Why does Formaldehyde (HCHOHCHO) not undergo Aldol condensation?

Solution:

Formaldehyde does not undergo Aldol condensation because it lacks Ξ±\alpha-hydrogen atoms.

Explanation:

Aldol condensation requires the formation of an enolate ion, which is generated by the removal of a hydrogen atom from the carbon atom adjacent to the carbonyl group (the Ξ±\alpha-carbon). Since Formaldehyde (HCHOHCHO) has only one carbon (the carbonyl carbon) and no Ξ±\alpha-carbon, it cannot form an enolate ion.

Problem 3:

Write the chemical equation for the Cross Aldol condensation between Benzaldehyde (C6H5CHOC_6H_5CHO) and Ethanal (CH3CHOCH_3CHO).

Solution:

C6H5CHO+CH3CHO→dil.NaOH,ΔC6H5CH=CHCHO+H2OC_6H_5CHO + CH_3CHO \xrightarrow{dil. NaOH, \Delta} C_6H_5CH=CHCHO + H_2O

Explanation:

Benzaldehyde has no Ξ±\alpha-hydrogen, while ethanal has three. Ethanal forms the enolate ion, which then attacks the more reactive carbonyl group of benzaldehyde. The resulting Ξ²\beta-hydroxy aldehyde dehydrates to form Cinnamaldehyde (3-phenylprop-2-enal).

Alpha hydrogen reactivity in aldehydes Revision - Class 12 Chemistry ICSE