JEE Challenger
More from Aldehydes, Ketones and Carboxylic Acids
ChemHardAdv 2023 • Paper 1

Determine Major Products Q R and S in Reaction Scheme

Consider the following reaction scheme and choose the correct option(s) for the major products Q\mathbf{Q}, R\mathbf{R} and S\mathbf{S}.

Styrene→(ii) NaOH, H2O2, H2O(i) B2H6P→(ii) Cl2,Red phosphorus(iii) H2O(i) CrO3, H2SO4QP→(ii) NaCN(iii) H3O+,Δ(i) SOCl2R→conc. H2SO4S\begin{aligned} \text{Styrene} &\xrightarrow[\text{(ii) }\mathrm{NaOH,\, H_2O_2,\, H_2O}]{\text{(i) }\mathrm{B_2H_6}} \mathbf{P} \xrightarrow[\substack{\text{(ii) }\mathrm{Cl_2,\, \text{Red phosphorus}} \\ \text{(iii) }\mathrm{H_2O}}]{\text{(i) }\mathrm{CrO_3,\, H_2SO_4}} \mathbf{Q} \\[2em] \mathbf{P} &\xrightarrow[\substack{\text{(ii) }\mathrm{NaCN} \\ \text{(iii) }\mathrm{H_3O^+,\,\Delta}}]{\text{(i) }\mathrm{SOCl_2}} \mathbf{R} \xrightarrow{\text{conc. }\mathrm{H_2SO_4}} \mathbf{S} \end{aligned}

Options

A
Option A
B
Option B
Correct
C
Option C
D
Option D

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsCarboxylic Acids and DerivativesOxidation and Reduction (Organic)

Step-by-Step Solution

To determine the correct major products Q\mathbf{Q}, R\mathbf{R}, and S\mathbf{S}, we analyze each step of the reaction scheme systematically:

1. Formation of Compound P\mathbf{P}:

Styrene (Ph−CH=CH2\text{Ph}-\text{CH}=\text{CH}_2) undergoes hydroboration-oxidation: Ph−CH=CH2→(i) B2H6(Ph−CH2−CH2)3B→(ii) NaOH, H2O2,H2OPh−CH2−CH2OH\text{Ph}-\text{CH}=\text{CH}_2 \xrightarrow{\text{(i) } \text{B}_2\text{H}_6} \text{(Ph}-\text{CH}_2-\text{CH}_2)_3\text{B} \xrightarrow{\text{(ii) } \text{NaOH, H}_2\text{O}_2, \text{H}_2\text{O}} \text{Ph}-\text{CH}_2-\text{CH}_2\text{OH}

Hydroboration-oxidation follows anti-Markovnikov addition of water across the carbon-carbon double bond. Thus, compound P\mathbf{P} is 2-phenylethan-1-ol (Ph−CH2−CH2OH\text{Ph}-\text{CH}_2-\text{CH}_2\text{OH}).

2. Formation of Compound Q\mathbf{Q}:

Starting from compound P\mathbf{P} (Ph−CH2−CH2OH\text{Ph}-\text{CH}_2-\text{CH}_2\text{OH}):

  1. Oxidation with Jones Reagent (CrO3,H2SO4\text{CrO}_3, \text{H}_2\text{SO}_4): Primary alcohol is oxidized to 2-phenylacetic acid: Ph−CH2−CH2OH→CrO3,H2SO4Ph−CH2−COOH\text{Ph}-\text{CH}_2-\text{CH}_2\text{OH} \xrightarrow{\text{CrO}_3, \text{H}_2\text{SO}_4} \text{Ph}-\text{CH}_2-\text{COOH}

  2. Hell-Volhard-Zelinsky (HVZ) Reaction (Cl2,Red P\text{Cl}_2, \text{Red P} followed by H2O\text{H}_2\text{O}): Halogenation occurs selectively at the α\alpha-position of the carboxylic acid: Ph−CH2−COOH→(ii) Cl2,Red P,(iii) H2OPh−CH(Cl)−COOH\text{Ph}-\text{CH}_2-\text{COOH} \xrightarrow{\text{(ii) } \text{Cl}_2, \text{Red P}, \text{(iii) } \text{H}_2\text{O}} \text{Ph}-\text{CH(Cl)}-\text{COOH}

Thus, compound Q\mathbf{Q} is 2-chloro-2-phenylacetic acid (Ph−CH(Cl)−COOH\text{Ph}-\text{CH(Cl)}-\text{COOH}).

3. Formation of Compound R\mathbf{R}:

Starting from compound P\mathbf{P} (Ph−CH2−CH2OH\text{Ph}-\text{CH}_2-\text{CH}_2\text{OH}):

  1. Chlorination with SOCl2\text{SOCl}_2: Ph−CH2−CH2OH→SOCl2Ph−CH2−CH2Cl\text{Ph}-\text{CH}_2-\text{CH}_2\text{OH} \xrightarrow{\text{SOCl}_2} \text{Ph}-\text{CH}_2-\text{CH}_2\text{Cl}

  2. Nucleophilic Substitution with NaCN\text{NaCN}: Ph−CH2−CH2Cl→NaCNPh−CH2−CH2CN\text{Ph}-\text{CH}_2-\text{CH}_2\text{Cl} \xrightarrow{\text{NaCN}} \text{Ph}-\text{CH}_2-\text{CH}_2\text{CN}

  3. Acid Hydrolysis (H3O+,Δ\text{H}_3\text{O}^+, \Delta): Complete hydrolysis of the nitrile yields 3-phenylpropanoic acid: Ph−CH2−CH2CN→H3O+,ΔPh−CH2−CH2−COOH\text{Ph}-\text{CH}_2-\text{CH}_2\text{CN} \xrightarrow{\text{H}_3\text{O}^+, \Delta} \text{Ph}-\text{CH}_2-\text{CH}_2-\text{COOH}

Thus, compound R\mathbf{R} is 3-phenylpropanoic acid (Ph−CH2−CH2−COOH\text{Ph}-\text{CH}_2-\text{CH}_2-\text{COOH}).

4. Formation of Compound S\mathbf{S}:

3-phenylpropanoic acid (R\mathbf{R}) in the presence of concentrated H2SO4\text{H}_2\text{SO}_4 undergoes an intramolecular Friedel-Crafts acylation: Ph−CH2−CH2−COOH→conc. H2SO41-indanone\text{Ph}-\text{CH}_2-\text{CH}_2-\text{COOH} \xrightarrow{\text{conc. } \text{H}_2\text{SO}_4} \text{1-indanone}

  • The carboxyl hydroxyl group gets protonated and loses water to form an acylium ion (−C+=O-\text{C}^+=\text{O}).
  • Electrophilic aromatic substitution occurs intramolecularly at the ortho-position of the benzene ring, forming the five-membered fused cyclic ketone, 1-indanone.

Conclusion:

Matching the derived structures for Q\mathbf{Q}, R\mathbf{R}, and S\mathbf{S} with the given choices:

  • Q\mathbf{Q}: Ph−CH(Cl)−COOH\text{Ph}-\text{CH(Cl)}-\text{COOH}
  • R\mathbf{R}: Ph−CH2−CH2−COOH\text{Ph}-\text{CH}_2-\text{CH}_2-\text{COOH}
  • S\mathbf{S}: 1-indanone

This corresponds to Option (B).

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