JEE Challenger
More from Aldehydes, Ketones and Carboxylic Acids
ChemMedAdv 2010 • Paper 2

Identification of Organic Compound R

Comprehension Passage

Paragraph for Questions 12 to 14.
Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3\text{K}_2\text{CO}_3 to give compound R, which upon treatment with HCN provides compound S. On acidification and heating, S gives the product shown below :

Passage Diagram 0

The compound R is

Options

A
Option A Diagram 1
Correct
B
Option B Diagram 1
C
Option C Diagram 1
D
Option D Diagram 1

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsAldehydes and KetonesAlcohols and Ethers

Step-by-Step Solution

The reaction involves a crossed-aldol condensation between 2-methylpropanal (isobutyraldehyde) and formaldehyde (HCHO\text{HCHO}) in the presence of a dilute base.

  1. Enolate Formation:
    2-Methylpropanal, (CH3)2CHCHO\text{(CH}_3)_2\text{CHCHO}, possesses one reactive α\alpha-hydrogen atom. Deprotonation by a base (OH−\text{OH}^-) yields the enolate nucleophile:
    (CH3)2CHCHO+OH−⇌−C(CH3)2CHO+H2O\text{(CH}_3)_2\text{CHCHO} + \text{OH}^- \rightleftharpoons {}^-\text{C}(\text{CH}_3)_2\text{CHO} + \text{H}_2\text{O}

  2. Nucleophilic Addition to Formaldehyde:
    Formaldehyde (HCHO\text{HCHO}) lacks α\alpha-hydrogens and has a highly electrophilic carbonyl carbon. The enolate ion attacks formaldehyde selectively:
    HCHO+−C(CH3)2CHO⟶O−−CH2−C(CH3)2−CHO\text{HCHO} + {}^-\text{C}(\text{CH}_3)_2\text{CHO} \longrightarrow \text{O}^--\text{CH}_2-\text{C}(\text{CH}_3)_2-\text{CHO}

  3. Protonation:
    Protonation of the alkoxide intermediate yields the aldol product, compound R:
    O−−CH2−C(CH3)2−CHO+H2O⟶HO−CH2−C(CH3)2−CHO+OH−\text{O}^--\text{CH}_2-\text{C}(\text{CH}_3)_2-\text{CHO} + \text{H}_2\text{O} \longrightarrow \text{HO}-\text{CH}_2-\text{C}(\text{CH}_3)_2-\text{CHO} + \text{OH}^-

Compound R is 3-hydroxy-2,2-dimethylpropanal, which matches the chemical structure shown in option A.

Correct Answer: A

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