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Aldehydes Ketones and Carboxylic Acids question

2008 · Shift 1 · Q16
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Aldehydes Ketones and Carboxylic Acids question

2008 · Shift 1 · Q16

JEE AdvancedChemistryAldehydes Ketones and Carboxylic AcidsMCQ+3 / −1
In the following reaction sequence, products I, J and L are formed. K represents a reagent. IIT-JEE 2008 Paper 1 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 18 English ComprehensionThe structure of compounds J and K, respectively, are:
  1. A
    IIT-JEE 2008 Paper 1 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 18 English Option 1
  2. B
    IIT-JEE 2008 Paper 1 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 18 English Option 2
  3. C
    IIT-JEE 2008 Paper 1 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 18 English Option 3
  4. D
    IIT-JEE 2008 Paper 1 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 18 English Option 4
View written solutionFree

Correct answer: A

The user wants me to solve a multi-step organic chemistry reaction sequence and identify two of the intermediate compounds/reagents.

Reaction Sequence Analysis:

  1. Starting Material: Hex-3-ynal.

    • Structure: CH3−CH2−C≡C−CH2−CHOCH_3-CH_2-C≡C-CH_2-CHOCH3​−CH2​−C≡C−CH2​−CHO
    • Functional groups: Alkyne (-C≡C-) and Aldehyde (-CHO).

  2. Step 1: Hex-3-ynal reacts with NaBH4NaBH_4NaBH4​ to form product I.

    • NaBH4NaBH_4NaBH4​ (Sodium borohydride) is a reducing agent.
    • It is a selective reducing agent, reducing aldehydes and ketones to alcohols. It does not typically reduce alkynes, alkenes, esters, or carboxylic acids under standard conditions.
    • Therefore, NaBH4NaBH_4NaBH4​ will reduce the aldehyde group (-CHO) to a primary alcohol (−CH2OH-CH_2OH−CH2​OH). The alkyne group (-C≡C-) will remain unchanged.
    • Reactant: CH3−CH2−C≡C−CH2−CHOCH_3-CH_2-C≡C-CH_2-CHOCH3​−CH2​−C≡C−CH2​−CHO
    • Product I: CH3−CH2−C≡C−CH2−CH2OHCH_3-CH_2-C≡C-CH_2-CH_2OHCH3​−CH2​−C≡C−CH2​−CH2​OH (Hex-3-yn-1-ol)

  3. Step 2: Product I reacts with PBr3PBr_3PBr3​ to form product J.

    • PBr3PBr_3PBr3​ (Phosphorus tribromide) is a reagent used to convert alcohols to alkyl bromides.
    • The hydroxyl group (-OH) of the primary alcohol in I will be replaced by a bromine atom (-Br).
    • Reactant I: CH3−CH2−C≡C−CH2−CH2OHCH_3-CH_2-C≡C-CH_2-CH_2OHCH3​−CH2​−C≡C−CH2​−CH2​OH
    • Product J: CH3−CH2−C≡C−CH2−CH2BrCH_3-CH_2-C≡C-CH_2-CH_2BrCH3​−CH2​−C≡C−CH2​−CH2​Br (1-bromohex-3-yne)

  4. Step 3: Product J reacts with Mg/ether to form product K.

    • This is the standard procedure for preparing a Grignard reagent.
    • The magnesium atom inserts into the carbon-bromine bond.
    • Reactant J: CH3−CH2−C≡C−CH2−CH2BrCH_3-CH_2-C≡C-CH_2-CH_2BrCH3​−CH2​−C≡C−CH2​−CH2​Br
    • Product K (Grignard reagent): CH3−CH2−C≡C−CH2−CH2MgBrCH_3-CH_2-C≡C-CH_2-CH_2MgBrCH3​−CH2​−C≡C−CH2​−CH2​MgBr

  5. Step 4: Product K reacts with CO2CO_2CO2​ followed by H3O+H_3O^+H3​O+ to form product L.

    • This is the carboxylation of a Grignard reagent to form a carboxylic acid.
    • The Grignard reagent K acts as a nucleophile (carbanion R−R^-R−) and attacks the electrophilic carbon of carbon dioxide (O=C=O).
    • Nucleophilic attack: CH3−CH2−C≡C−CH2−CH2−CH_3-CH_2-C≡C-CH_2-CH_2^-CH3​−CH2​−C≡C−CH2​−CH2−​ attacks CO2CO_2CO2​.
    • Intermediate: A carboxylate salt R−COO−MgBr+R-COO^-MgBr^+R−COO−MgBr+.
    • Reactant K: CH3−CH2−C≡C−CH2−CH2MgBrCH_3-CH_2-C≡C-CH_2-CH_2MgBrCH3​−CH2​−C≡C−CH2​−CH2​MgBr
    • Reaction with CO2CO_2CO2​: CH3−CH2−C≡C−CH2−CH2−COO−MgBr+CH_3-CH_2-C≡C-CH_2-CH_2-COO^-MgBr^+CH3​−CH2​−C≡C−CH2​−CH2​−COO−MgBr+
    • Acid workup with H3O+H_3O^+H3​O+: The carboxylate salt is protonated to form the carboxylic acid.
    • Product L: CH3−CH2−C≡C−CH2−CH2−COOHCH_3-CH_2-C≡C-CH_2-CH_2-COOHCH3​−CH2​−C≡C−CH2​−CH2​−COOH (Hept-4-ynoic acid)

Question: The structure of compounds J and K, respectively, are?

  • Structure of J: CH3−CH2−C≡C−CH2−CH2BrCH_3-CH_2-C≡C-CH_2-CH_2BrCH3​−CH2​−C≡C−CH2​−CH2​Br
  • Structure of K: CH3−CH2−C≡C−CH2−CH2MgBrCH_3-CH_2-C≡C-CH_2-CH_2MgBrCH3​−CH2​−C≡C−CH2​−CH2​MgBr

Let's check the options:

The problem provides images for the options. I need to interpret the structures from the image.

  • Option A:

    • J: CH3CH2−C≡C−CH2CH2BrCH_3CH_2-C≡C-CH_2CH_2BrCH3​CH2​−C≡C−CH2​CH2​Br (This matches my derived structure for J)
    • K: CH3CH2−C≡C−CH2CH2MgBrCH_3CH_2-C≡C-CH_2CH_2MgBrCH3​CH2​−C≡C−CH2​CH2​MgBr (This matches my derived structure for K)

  • Option B:

    • J: CH3CH2−C≡C−CH2CH2OHCH_3CH_2-C≡C-CH_2CH_2OHCH3​CH2​−C≡C−CH2​CH2​OH (This is structure I, not J)
    • K: CH3CH2−C≡C−CH2CH2MgBrCH_3CH_2-C≡C-CH_2CH_2MgBrCH3​CH2​−C≡C−CH2​CH2​MgBr (This is structure K)
    • So, Option B is incorrect because the structure of J is wrong.

  • Option C:

    • J: CH3CH2−C≡C−CH2CH2BrCH_3CH_2-C≡C-CH_2CH_2BrCH3​CH2​−C≡C−CH2​CH2​Br (This matches my derived structure for J)
    • K: CH3CH2−CH=CH−CH2CH2MgBrCH_3CH_2-CH=CH-CH_2CH_2MgBrCH3​CH2​−CH=CH−CH2​CH2​MgBr (This is incorrect. The alkyne has been reduced to an alkene, which is not what happens. Mg/ether does not reduce alkynes.)
    • So, Option C is incorrect.

  • Option D:

    • J: CH3CH2−CH=CH−CH2CH2BrCH_3CH_2-CH=CH-CH_2CH_2BrCH3​CH2​−CH=CH−CH2​CH2​Br (This is incorrect. The alkyne has been reduced. PBr3PBr_3PBr3​ does not reduce alkynes.)
    • K: CH3CH2−C≡C−CH2CH2MgBrCH_3CH_2-C≡C-CH_2CH_2MgBrCH3​CH2​−C≡C−CH2​CH2​MgBr (This is structure K)
    • So, Option D is incorrect because the structure of J is wrong.

Conclusion:

Based on the step-by-step analysis, Option A correctly identifies both compound J and compound K.

  • J is 1-bromohex-3-yne.
  • K is the Grignard reagent, (hex-3-yn-1-yl)magnesium bromide.

The stored correct answer is A. My derived answer matches the stored answer.

Detailed breakdown for the explanation section:

  • Step 1: Reaction of Hex-3-ynal with NaBH₄

    • Start with the structure of Hex-3-ynal: CH₃-CH₂-C≡C-CH₂-CHO.
    • Identify the reagent: NaBH₄ (Sodium borohydride).
    • Explain the function of NaBH₄: It's a selective reducing agent that reduces aldehydes to primary alcohols. It does not affect the carbon-carbon triple bond.
    • Show the reaction: CH₃-CH₂-C≡C-CH₂-CHO + NaBH₄ → CH₃-CH₂-C≡C-CH₂-CH₂OH.
    • Name the product I: Hex-3-yn-1-ol.

  • Step 2: Reaction of I with PBr₃

    • Start with the structure of I: CH₃-CH₂-C≡C-CH₂-CH₂OH.
    • Identify the reagent: PBr₃ (Phosphorus tribromide).
    • Explain the function of PBr₃: It converts primary and secondary alcohols into alkyl bromides by substituting the -OH group with a -Br atom.
    • Show the reaction: CH₃-CH₂-C≡C-CH₂-CH₂OH + PBr₃ → CH₃-CH₂-C≡C-CH₂-CH₂Br.
    • Name the product J: 1-bromohex-3-yne. This is the first part of the answer.

  • Step 3: Reaction of J with Mg/ether

    • Start with the structure of J: CH₃-CH₂-C≡C-CH₂-CH₂Br.
    • Identify the reagents: Mg (Magnesium) in dry ether.
    • Explain the reaction: This is the formation of a Grignard reagent. The magnesium atom inserts itself between the carbon and the bromine atom.
    • Show the reaction: CH₃-CH₂-C≡C-CH₂-CH₂Br + Mg --(ether)--> CH₃-CH₂-C≡C-CH₂-CH₂MgBr.
    • Identify this product as reagent K. This is the second part of the answer.

  • Step 4: (Optional but good for completeness) Reaction of K with CO₂ and H₃O⁺

    • Start with the structure of K: CH₃-CH₂-C≡C-CH₂-CH₂MgBr.
    • Explain the reaction: The Grignard reagent acts as a nucleophile, attacking the electrophilic carbon of CO₂. Subsequent acidic workup (H₃O⁺) protonates the carboxylate intermediate to yield a carboxylic acid.
    • Show the reaction:
      1. CH₃-CH₂-C≡C-CH₂-CH₂MgBr + CO₂ → CH₃-CH₂-C≡C-CH₂-CH₂-COOMgBr
      2. CH₃-CH₂-C≡C-CH₂-CH₂-COOMgBr + H₃O⁺ → CH₃-CH₂-C≡C-CH₂-CH₂-COOH + Mg(OH)Br
    • Name the final product L: Hept-4-ynoic acid.

  • Final Answer Derivation:

    • From the analysis, J is CH₃-CH₂-C≡C-CH₂-CH₂Br.
    • From the analysis, K is CH₃-CH₂-C≡C-CH₂-CH₂MgBr.
    • Compare these structures with the given options.
    • Option A shows these exact structures for J and K.
    • Options B, C, and D have incorrect structures for either J or K or both.
    • Therefore, Option A is the correct answer.
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