JEE Challenger
More from Aldehydes, Ketones and Carboxylic Acids
ChemMedAdv 2019 • Paper 2

Matching Reactions with Intermediate and Final Products for Substrates Three and Four

Comprehension Passage

Answer Q.17 and Q.18 by appropriately matching the lists based on the information given in the paragraph.

List-I includes starting materials and reagents of selected chemical reactions. List-II gives structures of compounds that may be formed as intermediate products and/or final products from the reactions of List-I.

Which of the following options represents the correct combination considering List-I and List-II?

Question Diagram 1

Options

A

(III), (S), (R)

B

(IV), (Q), (U)

C

(III), (T), (U)

D

(IV), (Q), (R)

Correct

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsOxidation and Reduction (Organic)Aldehydes and KetonesCarboxylic Acids and DerivativesAlcohols and Ethers

Step-by-Step Solution

To determine the correct combination for Question 18, let us analyze the reactions given in List-I for substrates (III) and (IV):

Analysis of Reaction (III):

  1. Starting Material: Methyl 2-(chloromethyl)benzoate, C6H4(CH2Cl)(COOCH3)\text{C}_6\text{H}_4(\text{CH}_2\text{Cl})(\text{COOCH}_3).
  2. Step (i) & (ii):
    • Reaction with KCN\text{KCN} substitutes the chlorine atom via an SN2\text{S}_\text{N}2 mechanism to yield the nitrile: −CH2CN-\text{CH}_2\text{CN}.
    • Acidic hydrolysis (H3O+,Δ\text{H}_3\text{O}^+, \Delta) hydrolyzes the −CH2CN-\text{CH}_2\text{CN} group to −CH2COOH-\text{CH}_2\text{COOH} and the ester group −COOCH3-\text{COOCH}_3 to −COOH-\text{COOH}.
    • This yields 2-(carboxymethyl)benzoic acid (T).
  3. Step (iii):
    • Reduction with LiAlH4\text{LiAlH}_4 reduces both carboxylic acid groups: C6H4(CH2COOH)(COOH)→LiAlH4C6H4(CH2CH2OH)(CH2OH)(Q)\text{C}_6\text{H}_4(\text{CH}_2\text{COOH})(\text{COOH}) \xrightarrow{\text{LiAlH}_4} \text{C}_6\text{H}_4(\text{CH}_2\text{CH}_2\text{OH})(\text{CH}_2\text{OH}) \quad \mathbf{(Q)}
  4. Step (iv):
    • Intramolecular acid-catalyzed dehydration of the diol with concentrated H2SO4\text{H}_2\text{SO}_4 forms the 6-membered cyclic ether isochroman (R).

Intermediates and final products for (III): (T)\mathbf{(T)}, (Q)\mathbf{(Q)}, (R)\mathbf{(R)}.

Analysis of Reaction (IV):

  1. Starting Material: Methyl 2-(2-methoxy-2-oxoethyl)benzoate, C6H4(CH2COOMe)(COOMe)\text{C}_6\text{H}_4(\text{CH}_2\text{COOMe})(\text{COOMe}).
  2. Step (i):
    • Reduction with LiAlH4\text{LiAlH}_4 converts both ester groups into alcohols:
      • −COOMe→LiAlH4−CH2OH-\text{COOMe} \xrightarrow{\text{LiAlH}_4} -\text{CH}_2\text{OH}
      • −CH2COOMe→LiAlH4−CH2CH2OH-\text{CH}_2\text{COOMe} \xrightarrow{\text{LiAlH}_4} -\text{CH}_2\text{CH}_2\text{OH}
    • This forms the diol intermediate 2-(2-hydroxyethyl)benzyl alcohol (Q).
  3. Step (ii):
    • Treatment of diol (Q) with concentrated H2SO4\text{H}_2\text{SO}_4 causes intramolecular dehydration to yield the cyclic ether isochroman (R) as the final product.

Intermediates and final products for (IV): (Q)\mathbf{(Q)} and (R)\mathbf{(R)}.

Matching with the Options:

  • (A) (III), (S), (R) →\rightarrow Incorrect
  • (B) (IV), (Q), (U) →\rightarrow Incorrect
  • (C) (III), (T), (U) →\rightarrow Incorrect
  • (D) (IV), (Q), (R) →\rightarrow Correct

Correct Answer: D

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