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Compounds Containing Nitrogen question

2024 · Shift 2 · Q6
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Compounds Containing Nitrogen question

2024 · Shift 2 · Q6

JEE AdvancedChemistryCompounds Containing NitrogenMultiple correct+4 / −2
The option(s) with correct sequence of reagents for the conversion of P\mathbf{P}P to Q\mathbf{Q}Q is(are) JEE Advanced 2024 Paper 2 Online Chemistry - Compounds Containing Nitrogen Question 6 English
  1. A
    i) Lindlar's catalyst, H2\mathrm{H}_2H2​; ii) SnCl2/HCl\mathrm{SnCl}_2 / \mathrm{HCl}SnCl2​/HCl; iii) NaBH4\mathrm{NaBH}_4NaBH4​; iv) H3O+\mathrm{H}_3 \mathrm{O}^{+}H3​O+
  2. B
    i) Lindlar's catalyst, H2\mathrm{H}_2H2​; ii) H3O+\mathrm{H}_3 \mathrm{O}^{+}H3​O+; iii) SnCl2/HCl\mathrm{SnCl}_2 / \mathrm{HCl}SnCl2​/HCl; iv) NaBH4\mathrm{NaBH}_4NaBH4​
  3. C
    i) NaBH4\mathrm{NaBH}_4NaBH4​; ii) SnCl2/HCl\mathrm{SnCl}_2 / \mathrm{HCl}SnCl2​/HCl; iii) H3O+\mathrm{H}_3 \mathrm{O}^{+}H3​O+; iv) Lindlar's catalyst, H2\mathrm{H}_2H2​
  4. D
    i) Lindlar's catalyst, H2\mathrm{H}_2H2​; ii) NaBH4\mathrm{NaBH}_4NaBH4​; iii) SnCl2/HCl\mathrm{SnCl}_2 / \mathrm{HCl}SnCl2​/HCl; iv) H3O+\mathrm{H}_3 \mathrm{O}^{+}H3​O+
View written solutionFree

Correct answer: C, D

Analysis of the Transformation

The conversion of starting material P to product Q requires three distinct functional group transformations:

  1. Nitro group reduction: The nitro group (-NO₂) on the benzene ring is reduced to an amino group (-NH₂).
  2. Ester reduction: The methyl ester group (-COOCH₃) is reduced to a primary alcohol (hydroxymethyl group, -CH₂OH).
  3. Alkyne reduction: The internal alkyne group (-C≡C-CH₃) is stereoselectively reduced to a cis-alkene or (Z)-alkene (-(Z)-CH=CH-CH₃).

Analysis of the Reagents

Let's evaluate the function and selectivity of the given reagents:

  • Lindlar's catalyst, H₂: This is a poisoned palladium catalyst used for the syn-hydrogenation of alkynes to cis-alkenes. It is highly selective and will not reduce nitro groups or esters under these conditions.
  • SnCl₂ / HCl: This is a classic reagent for the selective reduction of aromatic nitro groups to amino groups (anilines). It will not affect alkynes, alkenes, or alcohols. Under these strongly acidic conditions, it could potentially hydrolyze the ester, which needs to be considered when determining the sequence.
  • NaBH₄ (Sodium borohydride): This is a mild reducing agent. It reduces aldehydes and ketones to alcohols. While it typically does not reduce esters under standard conditions (for which a stronger agent like LiAlH₄ is used), in the context of this JEE question, we must assume it is intended to perform the ester reduction to a primary alcohol. It does not affect nitro groups, alkynes, or alkenes.
  • H₃O⁺: This represents an acidic workup, which can be used to protonate intermediates or to hydrolyze certain functional groups like esters.

Evaluation of the Options

We need to find the correct sequence of these reagents to achieve the desired transformation. The key is to choose an order that avoids undesirable side reactions.

Option A: i) Lindlar's catalyst, H₂; ii) SnCl₂ / HCl; iii) NaBH₄; iv) H₃O⁺

  1. P --(Lindlar's catalyst, H₂)--> The alkyne is reduced to a cis-alkene.
  2. --(SnCl₂ / HCl)--> The nitro group is reduced to an amino group. Problem: The strong acid (HCl) present can hydrolyze the ester (-COOCH₃) to a carboxylic acid (-COOH).
  3. --(NaBH₄)--> If the ester was hydrolyzed to a carboxylic acid, NaBH₄ would be ineffective as it cannot reduce carboxylic acids. This makes the sequence unreliable. Thus, Option A is incorrect.

Option B: i) Lindlar's catalyst, H₂; ii) H₃O⁺; iii) SnCl₂ / HCl; iv) NaBH₄

  1. P --(Lindlar's catalyst, H₂)--> Alkyne is reduced to a cis-alkene.
  2. --(H₃O⁺)--> The ester is deliberately hydrolyzed to a carboxylic acid.
  3. --(SnCl₂ / HCl)--> The nitro group is reduced to an amino group.
  4. --(NaBH₄)--> The final molecule contains a carboxylic acid group, which cannot be reduced by NaBH₄. Thus, Option B is incorrect.

Option C: i) NaBH₄; ii) SnCl₂ / HCl; iii) H₃O⁺; iv) Lindlar's catalyst, H₂

  1. P --(NaBH₄)--> The ester group is reduced to a primary alcohol (-CH₂OH). Nitro and alkyne groups remain unchanged. This avoids the hydrolysis issue with SnCl₂ / HCl later.
  2. --(SnCl₂ / HCl)--> The nitro group is reduced to an amino group. The alcohol and alkyne are unaffected.
  3. (H₃O⁺): This is likely a workup step, which might be oddly placed but does not invalidate the main synthetic pathway.
  4. --(Lindlar's catalyst, H₂)--> The alkyne is selectively reduced to the cis-alkene, yielding the final product Q. This sequence is chemically sound and all transformations are achieved correctly. Thus, Option C is correct.

Option D: i) Lindlar's catalyst, H₂; ii) NaBH₄; iii) SnCl₂ / HCl; iv) H₃O⁺

  1. P --(Lindlar's catalyst, H₂)--> The alkyne is reduced to a cis-alkene. The ester and nitro groups are unaffected.
  2. --(NaBH₄)--> The ester group is reduced to a primary alcohol. This step is performed before the acidic reduction, thus avoiding the hydrolysis problem.
  3. --(SnCl₂ / HCl)--> The nitro group is reduced to an amino group, yielding the final product Q.
  4. (H₃O⁺): This serves as a final workup step. This sequence is also chemically sound and achieves all the required transformations in a logical order. Thus, Option D is correct.

Both options C and D represent viable synthetic pathways. The key is to reduce the ester with NaBH₄ before subjecting the molecule to the strongly acidic conditions of the nitro group reduction.

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