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Aldehydes Ketones and Carboxylic Acids question

2008 · Shift 1 · Q17
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Aldehydes Ketones and Carboxylic Acids question

2008 · Shift 1 · Q17

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 14 English ComprehensionThe structure of product L is :
  1. A
    IIT-JEE 2008 Paper 1 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 14 English Option 1
  2. B
    IIT-JEE 2008 Paper 1 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 14 English Option 2
  3. C
    IIT-JEE 2008 Paper 1 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 14 English Option 3
  4. D
    IIT-JEE 2008 Paper 1 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 14 English Option 4
View written solutionFree

Correct answer: C

The user wants me to determine the structure of the final product L in a given multi-step reaction sequence.

Step 1: Analyze the starting material and the first reaction.

  • The starting material is Hex-3-ynal. Its structure is CH3−CH2−C≡C−CH2−CHOCH_3-CH_2-C≡C-CH_2-CHOCH3​−CH2​−C≡C−CH2​−CHO.
  • The first reagent is Sodium Borohydride (NaBH4NaBH_4NaBH4​). NaBH4NaBH_4NaBH4​ is a selective reducing agent that reduces aldehydes and ketones to their corresponding alcohols. It does not affect the carbon-carbon triple bond.
  • The aldehyde group (-CHO) is reduced to a primary alcohol group (−CH2OH-CH_2OH−CH2​OH).
  • So, product I is Hex-3-yn-1-ol: CH3−CH2−C≡C−CH2−CH2OHCH_3-CH_2-C≡C-CH_2-CH_2OHCH3​−CH2​−C≡C−CH2​−CH2​OH.

Hex-3-ynal→NaBH4Hex-3-yn-1-ol (I)\text{Hex-3-ynal} \xrightarrow{NaBH_4} \text{Hex-3-yn-1-ol (I)}Hex-3-ynalNaBH4​​Hex-3-yn-1-ol (I) CH3−CH2−C≡C−CH2−CHO→NaBH4CH3−CH2−C≡C−CH2−CH2OHCH_3-CH_2-C≡C-CH_2-CHO \xrightarrow{NaBH_4} CH_3-CH_2-C≡C-CH_2-CH_2OHCH3​−CH2​−C≡C−CH2​−CHONaBH4​​CH3​−CH2​−C≡C−CH2​−CH2​OH

Step 2: Analyze the second reaction.

  • Product I (Hex-3-yn-1-ol) is reacted with Phosphorus Tribromide (PBr3PBr_3PBr3​).
  • PBr3PBr_3PBr3​ is a reagent used to convert primary and secondary alcohols into alkyl bromides. The hydroxyl group (-OH) is replaced by a bromine atom (-Br).
  • So, product J is 1-bromohex-3-yne: CH3−CH2−C≡C−CH2−CH2BrCH_3-CH_2-C≡C-CH_2-CH_2BrCH3​−CH2​−C≡C−CH2​−CH2​Br.

Hex-3-yn-1-ol (I)→PBr31-bromohex-3-yne (J)\text{Hex-3-yn-1-ol (I)} \xrightarrow{PBr_3} \text{1-bromohex-3-yne (J)}Hex-3-yn-1-ol (I)PBr3​​1-bromohex-3-yne (J) CH3−CH2−C≡C−CH2−CH2OH→PBr3CH3−CH2−C≡C−CH2−CH2BrCH_3-CH_2-C≡C-CH_2-CH_2OH \xrightarrow{PBr_3} CH_3-CH_2-C≡C-CH_2-CH_2BrCH3​−CH2​−C≡C−CH2​−CH2​OHPBr3​​CH3​−CH2​−C≡C−CH2​−CH2​Br

Step 3: Analyze the third reaction (formation of Grignard reagent).

  • Product J (1-bromohex-3-yne) is reacted with Magnesium in ether (Mg/ether). This reagent is represented by K.
  • This reaction forms a Grignard reagent. The magnesium atom inserts itself between the carbon and the bromine atom.
  • The intermediate Grignard reagent is hex-3-ynylmagnesium bromide: CH3−CH2−C≡C−CH2−CH2MgBrCH_3-CH_2-C≡C-CH_2-CH_2MgBrCH3​−CH2​−C≡C−CH2​−CH2​MgBr.

1-bromohex-3-yne (J)→K=Mg/etherHex-3-ynylmagnesium bromide\text{1-bromohex-3-yne (J)} \xrightarrow{K = Mg/ether} \text{Hex-3-ynylmagnesium bromide}1-bromohex-3-yne (J)K=Mg/ether​Hex-3-ynylmagnesium bromide CH3−CH2−C≡C−CH2−CH2Br→Mg/etherCH3−CH2−C≡C−CH2−CH2MgBrCH_3-CH_2-C≡C-CH_2-CH_2Br \xrightarrow{Mg/ether} CH_3-CH_2-C≡C-CH_2-CH_2MgBrCH3​−CH2​−C≡C−CH2​−CH2​BrMg/ether​CH3​−CH2​−C≡C−CH2​−CH2​MgBr

Step 4: Analyze the final reaction.

  • The Grignard reagent is reacted with carbon dioxide (CO2CO_2CO2​) followed by an acidic workup (H3O+H_3O^+H3​O+).
  • The Grignard reagent acts as a nucleophile (carbanion), attacking the electrophilic carbon of CO2CO_2CO2​.
  • This forms a magnesium carboxylate salt, which is then protonated by the acid to yield a carboxylic acid.
  • This process adds a -COOH group to the carbon chain, increasing the chain length by one carbon.
  • The final product L is Hept-4-ynoic acid: CH3−CH2−C≡C−CH2−CH2−COOHCH_3-CH_2-C≡C-CH_2-CH_2-COOHCH3​−CH2​−C≡C−CH2​−CH2​−COOH.

Hex-3-ynylmagnesium bromide→1.CO22.H3O+Hept-4-ynoic acid (L)\text{Hex-3-ynylmagnesium bromide} \xrightarrow{1. CO_2 \quad 2. H_3O^+} \text{Hept-4-ynoic acid (L)}Hex-3-ynylmagnesium bromide1.CO2​2.H3​O+​Hept-4-ynoic acid (L) CH3−CH2−C≡C−CH2−CH2MgBr→1.CO22.H3O+CH3−CH2−C≡C−CH2−CH2−COOHCH_3-CH_2-C≡C-CH_2-CH_2MgBr \xrightarrow{1. CO_2 \quad 2. H_3O^+} CH_3-CH_2-C≡C-CH_2-CH_2-COOHCH3​−CH2​−C≡C−CH2​−CH2​MgBr1.CO2​2.H3​O+​CH3​−CH2​−C≡C−CH2​−CH2​−COOH

Step 5: Compare the structure of L with the given options.

  • The derived structure for L is CH3−CH2−C≡C−CH2−CH2−COOHCH_3-CH_2-C≡C-CH_2-CH_2-COOHCH3​−CH2​−C≡C−CH2​−CH2​−COOH.
  • Option A: CH3−CH2−CH=CH−CH(Br)−CH2OHCH_3-CH_2-CH=CH-CH(Br)-CH_2OHCH3​−CH2​−CH=CH−CH(Br)−CH2​OH - Incorrect. Contains a double bond, a bromine, and a hydroxyl group.
  • Option B: CH3−CH2−CH(OH)−C≡C−CH2−COOHCH_3-CH_2-CH(OH)-C≡C-CH_2-COOHCH3​−CH2​−CH(OH)−C≡C−CH2​−COOH - Incorrect. Contains an extra hydroxyl group.
  • Option C: CH3−CH2−C≡C−CH2−CH2−COOHCH_3-CH_2-C≡C-CH_2-CH_2-COOHCH3​−CH2​−C≡C−CH2​−CH2​−COOH - Correct. This structure matches our derived product L.
  • Option D: CH3−CH2−CH=CH−CH2−CH2−COOHCH_3-CH_2-CH=CH-CH_2-CH_2-COOHCH3​−CH2​−CH=CH−CH2​−CH2​−COOH - Incorrect. Contains a double bond instead of a triple bond.

Therefore, the structure of product L is represented by option C.

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