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Aldehydes Ketones and Carboxylic Acids question

2010 · Shift 2 · Q15
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Aldehydes Ketones and Carboxylic Acids question

2010 · Shift 2 · Q15

JEE AdvancedChemistryAldehydes Ketones and Carboxylic AcidsMCQ+3 / −1
Two aliphatic aldehydes P\mathbf{P}P and Q\mathbf{Q}Q react in the presence of aqueous K2CO3\mathrm{K}_2 \mathrm{CO}_3K2​CO3​ to give compound R\mathbf{R}R, which upon treatment with HCNHCNHCN provides compound S\mathbf{S}S. On acidification and heating, S\mathbf{S}S gives the product shown below : IIT-JEE 2010 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 4 English ComprehensionThe compounds P and Q respectively are :
  1. A
    IIT-JEE 2010 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 4 English Option 1
  2. B
    IIT-JEE 2010 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 4 English Option 2
  3. C
    IIT-JEE 2010 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 4 English Option 3
  4. D
    IIT-JEE 2010 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 4 English Option 4
View written solutionFree

Correct answer: B

Step-by-Step Solution:

  1. Analyze the Final Product: The final product is 4,4-dimethylpent-2-enoic acid: CH3−C∣CH3−CH=CH−COOH≡(CH3)3C−CH=CH−COOH\text{CH}_3 - \underset{\displaystyle\text{CH}_3}{\underset{\displaystyle |}{\text{C}}} - \text{CH} = \text{CH} - \text{COOH} \quad \equiv \quad (\text{CH}_3)_3\text{C}-\text{CH}=\text{CH}-\text{COOH}CH3​−CH3​∣C​​−CH=CH−COOH≡(CH3​)3​C−CH=CH−COOH This is an α,β\alpha,\betaα,β-unsaturated carboxylic acid. The carbon skeleton consists of a tert-butyl group (CH3)3C- attached to a –CH=CH–COOH chain.

  2. Retrosynthesis - Working Backwards:

    • The overall reaction sequence starts with an aldol reaction between two aldehydes, P and Q. This suggests that the C=C double bond in the final product was formed during an aldol condensation step.
    • The aldol condensation reaction forms a new carbon-carbon bond, which typically becomes the bond between the α\alphaα and β\betaβ carbons relative to the carbonyl group. In the condensation product, this is adjacent to the C=C double bond. Let's trace back the formation of the skeleton (CH3)3C-CH=CH-C....
    • This structure is formed by the reaction between an electrophilic carbonyl component and a nucleophilic enolate component.
      • The part (CH3)3C-CH= comes from the electrophile (carbonyl acceptor). This corresponds to the aldehyde (CH3)3C-CHO (pivaldehyde or 2,2-dimethylpropanal).
      • The part =CH-C... comes from the nucleophile (enolate donor). This corresponds to the enolate of an aldehyde with the structure CH3-C.... The simplest such aldehyde is acetaldehyde, CH3CHO, which forms the enolate −CH2CHO^{-}CH2CHO−CH2CHO.

  3. Forward Reaction - Aldol Condensation of P and Q: Let's verify this by performing the forward reaction with the proposed aldehydes, P and Q. The two aldehydes are acetaldehyde (CH3CHO) and pivaldehyde ((CH3)3CCHO). The reaction is carried out in the presence of a base (aqueous K2CO3). This is a crossed aldol condensation.

    • Identify Nucleophile and Electrophile:

      • Pivaldehyde, (CH3)3CCHO, has no α\alphaα-hydrogens. Therefore, it cannot form an enolate and can only act as the electrophile (carbonyl acceptor).
      • Acetaldehyde, CH3CHO$, has acidic $\alpha$-hydrogens and will be deprotonated by the base to form the enolate ion, $^{-}CH2CHO, which acts as the nucleophile.

    • Reaction Mechanism: The enolate of acetaldehyde attacks the carbonyl carbon of pivaldehyde. (CH3)3C−C∣∣O−H+−CH2−CHO⟶(CH3)3C−C∣O−H−CH2−CHO(\text{CH}_3)_3\text{C}-\overset{\displaystyle\text{O}}{\overset{||}{\text{C}}}-\text{H} + {^-\text{CH}_2}-\text{CHO} \longrightarrow (\text{CH}_3)_3\text{C}-\underset{\displaystyle\text{O}^-}{\underset{|}{\text{C}}}H-\text{CH}_2-\text{CHO}(CH3​)3​C−C∣∣O−H+−CH2​−CHO⟶(CH3​)3​C−O−∣C​​H−CH2​−CHO Protonation of the alkoxide gives the aldol addition product, a β\betaβ-hydroxy aldehyde. (CH3)3C−C∣OHH−CH2−CHO(\text{CH}_3)_3\text{C}-\underset{\displaystyle\text{OH}}{\underset{|}{\text{C}}}H-\text{CH}_2-\text{CHO}(CH3​)3​C−OH∣C​​H−CH2​−CHO Upon heating, this intermediate undergoes dehydration (condensation) to form an α,β\alpha,\betaα,β-unsaturated aldehyde, which is compound R. (CH3)3C−CH=CH−CHO(R)(\text{CH}_3)_3\text{C}-\text{CH}=\text{CH}-\text{CHO} \quad (\mathbf{R})(CH3​)3​C−CH=CH−CHO(R)

  4. Subsequent Steps and Identification of P and Q:

    • Compound R is (CH3)3C-CH=CH-CHO. The problem states this reacts with HCN to give S, and S upon acidification and heating gives the final product, (CH3)3C-CH=CH-COOH.
    • The overall transformation from R to the final product is the oxidation of the aldehyde group (-CHO) to a carboxylic acid group (-COOH). While the specific mechanism involving HCN and H3O+/heat for this oxidation might be non-trivial or involve an uncommon pathway, the formation of the carbon skeleton of R unequivocally points to the starting materials.
    • The reactants required are acetaldehyde (CH3CHO) and pivaldehyde ((CH3)3CCHO).

  5. Evaluate the Options:

    • A: HCHO and (CH3)3CCHO. Both lack α\alphaα-hydrogens. They would undergo a Cannizzaro reaction, not aldol condensation.
    • B: CH3CHO and (CH3)3CCHO. These are the reactants identified in our analysis.
    • C: HCHO and CH3CH2CHO. The aldol product would not contain a tert-butyl group.
    • D: CH3CHO and CH3CH2CHO. The aldol products would not contain a tert-butyl group.

    Therefore, the only pair of aldehydes that can produce the required carbon skeleton is CH3CHO and (CH3)3CCHO.

    The compounds P and Q are acetaldehyde and pivaldehyde (or vice versa), which corresponds to option B.

Note on the subsequent steps: The transformation from R, (CH3)3C-CH=CH-CHO, to the final acid via HCN and H3O+/heat is not a standard textbook reaction sequence for oxidation. However, for a multiple-choice question, the identification of the aldol precursors is the key step, and only option B provides the correct precursors.

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