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Aldehydes Ketones and Carboxylic Acids question

2019 · Shift 2 · Q1
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Aldehydes Ketones and Carboxylic Acids question

2019 · Shift 2 · Q1

JEE AdvancedChemistryAldehydes Ketones and Carboxylic AcidsMultiple correct+4 / −1
Choose the correct option(s) for the following reaction sequence JEE Advanced 2019 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 59 English Consider Q, R and S as major products.
  1. A
    JEE Advanced 2019 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 59 English Option 1
  2. B
    JEE Advanced 2019 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 59 English Option 2
  3. C
    JEE Advanced 2019 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 59 English Option 3
  4. D
    JEE Advanced 2019 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 59 English Option 4
View written solutionFree

Correct answer: C, D

The problem asks to identify the major products Q, R, and S in a given reaction sequence starting from benzene (P).

Step 1: Formation of Q from P (Benzene)

The reaction of benzene (P) with carbon monoxide (CO) and hydrogen chloride (HCl) in the presence of a catalyst system of anhydrous aluminum chloride (AlCl₃) and cuprous chloride (CuCl) is the Gattermann-Koch reaction. This reaction introduces a formyl group (-CHO) onto the benzene ring.

C6H6 (P)+CO+HCl→anhyd.AlCl3/CuClC6H5CHO (Q)+HClC_6H_6 \ (P) + CO + HCl \xrightarrow{anhyd. AlCl_3/CuCl} C_6H_5CHO \ (Q) + HClC6​H6​ (P)+CO+HClanhyd.AlCl3​/CuCl​C6​H5​CHO (Q)+HCl

The product Q is benzaldehyde.

Step 2: Formation of R from Q (Benzaldehyde)

The reaction of benzaldehyde (Q) with methylmagnesium bromide (CH₃MgBr) followed by hydrolysis (H₂O) is a Grignard reaction. The Grignard reagent acts as a nucleophile, where the methyl carbanion (CH₃⁻) attacks the electrophilic carbonyl carbon of the aldehyde.

  1. Nucleophilic Addition: C6H5CHO (Q)+CH3MgBr→C6H5CH(OMgBr)CH3C_6H_5CHO \ (Q) + CH_3MgBr \rightarrow C_6H_5CH(OMgBr)CH_3C6​H5​CHO (Q)+CH3​MgBr→C6​H5​CH(OMgBr)CH3​
  2. Hydrolysis (Workup): The intermediate alkoxide salt is protonated by water to yield a secondary alcohol. C6H5CH(OMgBr)CH3+H2O→C6H5CH(OH)CH3 (R)+Mg(OH)BrC_6H_5CH(OMgBr)CH_3 + H_2O \rightarrow C_6H_5CH(OH)CH_3 \ (R) + Mg(OH)BrC6​H5​CH(OMgBr)CH3​+H2​O→C6​H5​CH(OH)CH3​ (R)+Mg(OH)Br

The product R is 1-phenylethanol.

Step 3: Formation of S from R (1-Phenylethanol)

The reaction of 1-phenylethanol (R) with concentrated sulfuric acid (H₂SO₄) and heat (Δ) is an acid-catalyzed dehydration of an alcohol. The alcohol is protonated, water leaves as a good leaving group, forming a stable secondary benzylic carbocation. A proton is then eliminated from the adjacent carbon to form an alkene.

C6H5CH(OH)CH3 (R)→H2SO4,ΔC6H5CH=CH2 (S)+H2OC_6H_5CH(OH)CH_3 \ (R) \xrightarrow{H_2SO_4, \Delta} C_6H_5CH=CH_2 \ (S) + H_2OC6​H5​CH(OH)CH3​ (R)H2​SO4​,Δ​C6​H5​CH=CH2​ (S)+H2​O

The product S is styrene (phenylethene).

Step 4: Evaluating the Options

Let's summarize the products:

  • Q: Benzaldehyde
  • R: 1-Phenylethanol
  • S: Styrene

Now we check each option:

  • Option A: It shows Q as benzaldehyde (correct) but R as an acetal, C₆H₅CH(OCH₃)₂. This is incorrect. R is 1-phenylethanol.
  • Option B: It shows Q as benzoic acid, C₆H₅COOH. This is incorrect. Q is benzaldehyde.
  • Option C: It shows R as 1-phenylethanol and S as styrene. Both are correct as per our analysis.
  • Option D: It shows Q as benzaldehyde and R as 1-phenylethanol. Both are correct as per our analysis.

Since the question asks to choose the correct option(s), and options C and D both correctly identify some of the major products in the sequence, both are correct answers.

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