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Q8.9:
Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.

Solution :

1. Theoretical Foundation of Cross-Aldol Condensation

Cross-aldol condensation between two different aldehydes (propanal, $CH_3CH_2CHO$ and butanal, $CH_3CH_2CH_2CHO$) occurs when one aldehyde provides the enolate ion (nucleophile) and the other acts as the carbonyl component (electrophile). Since both aldehydes contain $\alpha$-hydrogens, four distinct crossed products are possible in addition to two self-condensation products.

2. Analysis of Potential Reactions

Each condensation requires the formation of a carbanion (enolate) at the $\alpha$-carbon of the donor molecule, followed by nucleophilic attack on the carbonyl carbon of the acceptor molecule.

Product Nucleophile (Enolate Source) Electrophile (Carbonyl Source) IUPAC Name
1 Propanal Butanal 2-methyl-3-hydroxyhexanal
2 Butanal Propanal 3-hydroxy-2-methylpentanal
3 Propanal Propanal (Self) 3-hydroxy-2-methylpentanal
4 Butanal Butanal (Self) 2-ethyl-3-hydroxyhexanal

3. Structural Formula Derivation

Product 1: Propanal as Nucleophile, Butanal as Electrophile

Reaction: $CH_3CH_2CHO + CH_3CH_2CH_2CHO \rightarrow CH_3CH_2CH_2CH(OH)CH(CH_3)CHO$

Structural Formula: $CH_3-CH_2-CH_2-CH(OH)-CH(CH_3)-CHO$

Product 2: Butanal as Nucleophile, Propanal as Electrophile

Reaction: $CH_3CH_2CH_2CHO + CH_3CH_2CHO \rightarrow CH_3CH_2CH(OH)CH(CH_2CH_3)CHO$

Structural Formula: $CH_3-CH_2-CH(OH)-CH(C_2H_5)-CHO$

4. Graphical Representation of Aldol Formation

Nucleophile: CH3-CH(-)-CHO Electrophile: R-CH=O Product: R-CH(OH)-CH(CH3)-CHO

5. Justification of Nucleophilicity and Electrophilicity

  • Nucleophile Selection: The species that loses an $\alpha$-proton to form the enolate ion, $R-CH^--CHO$, acts as the nucleophile [Per the base-catalyzed enolization mechanism].
  • Electrophile Selection: The carbonyl carbon of the non-enolate molecule acts as the electrophilic center due to the polarization of the $C=O$ bond ($C^{\delta+} = O^{\delta-}$), facilitating attack by the enolate [By Nucleophilic Addition Axioms].

Final Solution: The four products are: 1) 2-methyl-3-hydroxyhexanal (Nu: Propanal, E: Butanal), 2) 3-hydroxy-2-methylpentanal (Nu: Butanal, E: Propanal), 3) 3-hydroxy-2-methylpentanal (Self-condensation of Propanal), and 4) 2-ethyl-3-hydroxyhexanal (Self-condensation of Butanal).


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