JEE Challenger
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ChemMedAdv 2023 • Paper 2

Identification and Properties of Major Products in Grignard Reactions

In the following reactions, P\mathbf{P}, Q\mathbf{Q}, R\mathbf{R}, and S\mathbf{S} are the major products.

The correct statement about P\mathbf{P}, Q\mathbf{Q}, R\mathbf{R}, and S\mathbf{S} is

Question Diagram 1

Options

A

P\mathbf{P} is a primary alcohol with four carbons.

B

Q\mathbf{Q} undergoes Kolbe's electrolysis to give an eight-carbon product.

Correct
C

R\mathbf{R} has six carbons and it undergoes Cannizzaro reaction.

D

S\mathbf{S} is a primary amine with six carbons.

Topics & Concepts

Haloalkanes and HaloarenesGrignard ReagentHaloalkanes and HaloarenesOxidation and Reduction (Organic)

Step-by-Step Solution

To determine the correct statement, we analyze the reaction steps to identify products P\mathbf{P}, Q\mathbf{Q}, R\mathbf{R}, and S\mathbf{S}.

The starting compound in all the reaction sequences is isobutyl chloride (1-chloro-2-methylpropane), (CH3)2CHCH2Cl(CH_3)_2CHCH_2Cl.

1. Formation and Analysis of Product P\mathbf{P}:

  1. Grignard Reagent Formation: (CH3)2CHCH2Cl+Mg→dry ether(CH3)2CHCH2MgCl(CH_3)_2CHCH_2Cl + \text{Mg} \xrightarrow{\text{dry ether}} (CH_3)_2CHCH_2MgCl
  2. Protonation (Hydrolysis): (CH3)2CHCH2MgCl+H2O→(CH3)2CHCH3+Mg(OH)Cl(CH_3)_2CHCH_2MgCl + \text{H}_2\text{O} \rightarrow (CH_3)_2CHCH_3 + \text{Mg(OH)Cl}
  • Product P\mathbf{P} is isobutane (2-methylpropane), which is an alkane containing 4 carbons, not an alcohol.
  • Thus, Option A is incorrect.

2. Formation and Analysis of Product Q\mathbf{Q}:

  1. Grignard Addition to CO2\text{CO}_2: (CH3)2CHCH2MgCl+CO2→dry ether(CH3)2CHCH2COOMgCl(CH_3)_2CHCH_2MgCl + \text{CO}_2 \xrightarrow{\text{dry ether}} (CH_3)_2CHCH_2COOMgCl

  2. Acid Hydrolysis: (CH3)2CHCH2COOMgCl→H3O+(CH3)2CHCH2COOH(3-methylbutanoic acid)(CH_3)_2CHCH_2COOMgCl \xrightarrow{\text{H}_3\text{O}^+} (CH_3)_2CHCH_2COOH \quad \text{(3-methylbutanoic acid)}

  3. Neutralization: (CH3)2CHCH2COOH+NaOH→(CH3)2CHCH2COONa+H2O(CH_3)_2CHCH_2COOH + \text{NaOH} \rightarrow (CH_3)_2CHCH_2COONa + \text{H}_2\text{O} Therefore, product Q\mathbf{Q} is sodium 3-methylbutanoate.

  4. Kolbe's Electrolysis of Q\mathbf{Q}: During Kolbe's electrolysis, carboxylate ions undergo oxidative decarboxylation at the anode to generate alkyl radicals, which then dimerize: 2(CH3)2CHCH2COONa→electrolysis(CH3)2CHCH2−CH2CH(CH3)2+2CO2+2e−+2Na+2(CH_3)_2CHCH_2COONa \xrightarrow{\text{electrolysis}} (CH_3)_2CHCH_2-CH_2CH(CH_3)_2 + 2\text{CO}_2 + 2 e^- + 2\text{Na}^+ The resulting hydrocarbon is 2,5-dimethylhexane, an eight-carbon alkane (C8H18\text{C}_8\text{H}_{18}).

  • Thus, Option B is correct.

3. Formation and Analysis of Product R\mathbf{R}:

  1. Nucleophilic Addition to Acetaldehyde: (CH3)2CHCH2MgCl+CH3CHO→then H2O(CH3)2CHCH2−CH(OH)−CH3(4-methylpentan-2-ol)(CH_3)_2CHCH_2MgCl + \text{CH}_3\text{CHO} \xrightarrow{\text{then }\text{H}_2\text{O}} (CH_3)_2CHCH_2-CH(OH)-CH_3 \quad \text{(4-methylpentan-2-ol)}
  2. Oxidation: (CH3)2CHCH2−CH(OH)−CH3→CrO3(CH3)2CHCH2−CO−CH3(4-methylpentan-2-one)(CH_3)_2CHCH_2-CH(OH)-CH_3 \xrightarrow{\text{CrO}_3} (CH_3)_2CHCH_2-CO-CH_3 \quad \text{(4-methylpentan-2-one)}
  • Product R\mathbf{R} is a ketone containing 6 carbon atoms and α\alpha-hydrogens. Because it is a ketone and contains α\alpha-hydrogens, it cannot undergo the Cannizzaro reaction (which is given by non-enolizable aldehydes).
  • Thus, Option C is incorrect.

4. Formation and Analysis of Product S\mathbf{S}:

  1. Nitrile Synthesis: (CH3)2CHCH2Cl+NaCN→ethanol(CH3)2CHCH2CN+NaCl(CH_3)_2CHCH_2Cl + \text{NaCN} \xrightarrow{\text{ethanol}} (CH_3)_2CHCH_2CN + \text{NaCl}
  2. Reduction to Primary Amine: (CH3)2CHCH2CN→H2/Ni(CH3)2CHCH2CH2NH2(3-methylbutan-1-amine)(CH_3)_2CHCH_2CN \xrightarrow{\text{H}_2/\text{Ni}} (CH_3)_2CHCH_2CH_2NH_2 \quad \text{(3-methylbutan-1-amine)}
  3. Carbylamine Reaction: (CH3)2CHCH2CH2NH2+CHCl3+3KOH→Δ(CH3)2CHCH2CH2NC+3KCl+3H2O(CH_3)_2CHCH_2CH_2NH_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} (CH_3)_2CHCH_2CH_2NC + 3\text{KCl} + 3\text{H}_2\text{O}
  4. Reduction of Isocyanide: (CH3)2CHCH2CH2NC→(1)LiAlH4,(2)H2O(CH3)2CHCH2CH2NHCH3(CH_3)_2CHCH_2CH_2NC \xrightarrow{(1) \text{LiAlH}_4, (2) \text{H}_2\text{O}} (CH_3)_2CHCH_2CH_2NHCH_3
  • Product S\mathbf{S} is N,3N,3-dimethylbutan-1-amine, which is a secondary amine containing 6 carbon atoms (not a primary amine).
  • Thus, Option D is incorrect.

Conclusion:

The correct statement is B.

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