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

Asymmetric Carbon Centers in Products of Grignard and Organometallic Reactions

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

CH3CH2CH(CH3)CH2CN→(ii) PhMgBr, then H2O(i) PhMgBr, then H3O+P\text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)CH}_2\text{CN} \xrightarrow[(ii)\text{ PhMgBr, then H}_2\text{O}]{(i)\text{ PhMgBr, then H}_3\text{O}^+} \mathbf{P}

Ph-H+CH3COCl→(ii) PhMgBr, then H2O(i) anhyd. AlCl3Q\text{Ph-H} + \text{CH}_3\text{COCl} \xrightarrow[(ii)\text{ PhMgBr, then H}_2\text{O}]{(i)\text{ anhyd. AlCl}_3} \mathbf{Q}

CH3CH2COCl→(ii) PhMgBr, then H2O(i) 12(PhCH2)2CdR\text{CH}_3\text{CH}_2\text{COCl} \xrightarrow[(ii)\text{ PhMgBr, then H}_2\text{O}]{(i)\text{ }\frac{1}{2}\text{(PhCH}_2\text{)}_2\text{Cd}} \mathbf{R}

PhCH2CHO→(iii) HCN(iv) H2SO4,Δ(i) PhMgBr, then H2O(ii) CrO3,dil. H2SO4S\text{PhCH}_2\text{CHO} \xrightarrow[\begin{array}{l} \text{(iii) HCN} \\ \text{(iv) H}_2\text{SO}_4, \Delta \end{array}]{\begin{array}{l} \text{(i) PhMgBr, then H}_2\text{O} \\ \text{(ii) CrO}_3, \text{dil. H}_2\text{SO}_4 \end{array}} \mathbf{S}

The correct statement(s) about P\mathbf{P}, Q\mathbf{Q}, R\mathbf{R}, and S\mathbf{S} is(are)

Options

A

Both P\mathbf{P} and Q\mathbf{Q} have asymmetric carbon(s).

B

Both Q\mathbf{Q} and R\mathbf{R} have asymmetric carbon(s).

C

Both P\mathbf{P} and R\mathbf{R} have asymmetric carbon(s).

Correct
D

P\mathbf{P} has asymmetric carbon(s), S\mathbf{S} does not have any asymmetric carbon.

Correct

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsAldehydes and KetonesGrignard ReagentIsomerism

Step-by-Step Solution

To determine the correct statements, let us analyze the step-by-step formation of products P\mathbf{P}, Q\mathbf{Q}, R\mathbf{R}, and S\mathbf{S}:

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

  • Reactant: CH3CH2CH(CH3)CH2CN\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_2\text{CN} (3-methylpentanenitrile)

    • Note that carbon-3 (C-3\text{C-3}) already possesses four different groups: −H-\text{H}, −CH3-\text{CH}_3, −CH2CH3-\text{CH}_2\text{CH}_3, and −CH2CN-\text{CH}_2\text{CN}.

  • Step (i): Reaction with PhMgBr\text{PhMgBr} followed by acidic hydrolysis converts the nitrile group into a ketone: CH3CH2CH(CH3)CH2CN→(i) PhMgBr, H3O+CH3CH2CH(CH3)CH2COPh\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_2\text{CN} \xrightarrow{\text{(i) PhMgBr, } \text{H}_3\text{O}^+} \text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_2\text{COPh}

  • Step (ii): Reaction of the resulting ketone with PhMgBr\text{PhMgBr} followed by H2O\text{H}_2\text{O} yields a tertiary alcohol: CH3CH2CH(CH3)CH2COPh→(ii) PhMgBr, H2OCH3CH2C∗H(CH3)CH2C(OH)(Ph)2\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_2\text{COPh} \xrightarrow{\text{(ii) PhMgBr, } \text{H}_2\text{O}} \text{CH}_3\text{CH}_2\overset{*}{\text{C}}\text{H}(\text{CH}_3)\text{CH}_2\text{C}(\text{OH})(\text{Ph})_2

  • Asymmetric Carbon Analysis:

    • The tertiary carbon −C(OH)(Ph)2-\text{C}(\text{OH})(\text{Ph})_2 is bonded to two identical phenyl groups, making it achiral.
    • However, carbon-3 (C∗\overset{*}{\text{C}}) remains attached to four distinct groups: −H-\text{H}, −CH3-\text{CH}_3, −CH2CH3-\text{CH}_2\text{CH}_3, and −CH2C(OH)(Ph)2-\text{CH}_2\text{C}(\text{OH})(\text{Ph})_2.
    • Conclusion: P\mathbf{P} has an asymmetric carbon.

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

  • Step (i): Friedel-Crafts acylation of benzene (Ph-H\text{Ph-H}) with acetyl chloride (CH3COCl\text{CH}_3\text{COCl}) using anhydrous AlCl3\text{AlCl}_3 gives acetophenone: Ph-H+CH3COCl→anhyd. AlCl3Ph-CO-CH3\text{Ph-H} + \text{CH}_3\text{COCl} \xrightarrow{\text{anhyd. AlCl}_3} \text{Ph-CO-CH}_3

  • Step (ii): Nucleophilic addition of PhMgBr\text{PhMgBr} to acetophenone yields 1,1-diphenylethanol: Ph-CO-CH3→PhMgBr, then H2OPh2C(OH)CH3\text{Ph-CO-CH}_3 \xrightarrow{\text{PhMgBr, then H}_2\text{O}} \text{Ph}_2\text{C}(\text{OH})\text{CH}_3

  • Asymmetric Carbon Analysis:

    • The central carbon atom is bonded to two identical phenyl groups (−Ph-\text{Ph}), a hydroxyl group (−OH-\text{OH}), and a methyl group (−CH3-\text{CH}_3).
    • Conclusion: Q\mathbf{Q} does NOT have any asymmetric carbon.

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

  • Step (i): Reaction of propanoyl chloride (CH3CH2COCl\text{CH}_3\text{CH}_2\text{COCl}) with organocadmium reagent 12(PhCH2)2Cd\frac{1}{2}(\text{PhCH}_2)_2\text{Cd} yields 1-phenylbutan-2-one: CH3CH2COCl→12(PhCH2)2CdCH3CH2COCH2Ph\text{CH}_3\text{CH}_2\text{COCl} \xrightarrow{\frac{1}{2}(\text{PhCH}_2)_2\text{Cd}} \text{CH}_3\text{CH}_2\text{COCH}_2\text{Ph}

  • Step (ii): Addition of PhMgBr\text{PhMgBr} followed by H2O\text{H}_2\text{O} produces a tertiary alcohol: CH3CH2COCH2Ph→PhMgBr, then H2OCH3CH2C∗(OH)(Ph)CH2Ph\text{CH}_3\text{CH}_2\text{COCH}_2\text{Ph} \xrightarrow{\text{PhMgBr, then H}_2\text{O}} \text{CH}_3\text{CH}_2\overset{*}{\text{C}}(\text{OH})(\text{Ph})\text{CH}_2\text{Ph}

  • Asymmetric Carbon Analysis:

    • The central carbon atom (C∗\overset{*}{\text{C}}) is bonded to four different groups:
      1. −OH-\text{OH}
      2. −CH2CH3-\text{CH}_2\text{CH}_3
      3. −Ph-\text{Ph}
      4. −CH2Ph-\text{CH}_2\text{Ph}
    • Conclusion: R\mathbf{R} has an asymmetric carbon.

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

  • Step (i): Reaction of PhCH2CHO\text{PhCH}_2\text{CHO} with PhMgBr\text{PhMgBr} gives 1,2-diphenylethanol: PhCH2CHO→PhMgBr, then H2OPhCH2CH(OH)Ph\text{PhCH}_2\text{CHO} \xrightarrow{\text{PhMgBr, then H}_2\text{O}} \text{PhCH}_2\text{CH}(\text{OH})\text{Ph}

  • Step (ii): Oxidation with Jones reagent (CrO3,dil. H2SO4\text{CrO}_3, \text{dil. H}_2\text{SO}_4) yields 1,2-diphenylethanone: PhCH2CH(OH)Ph→CrO3,dil. H2SO4PhCH2COPh\text{PhCH}_2\text{CH}(\text{OH})\text{Ph} \xrightarrow{\text{CrO}_3, \text{dil. H}_2\text{SO}_4} \text{PhCH}_2\text{COPh}

  • Step (iii): Cyanohydrin formation with HCN\text{HCN}: PhCH2COPh→HCNPhCH2C(OH)(CN)Ph\text{PhCH}_2\text{COPh} \xrightarrow{\text{HCN}} \text{PhCH}_2\text{C}(\text{OH})(\text{CN})\text{Ph}

  • Step (iv): Acidic hydrolysis with heating (H2SO4,Δ\text{H}_2\text{SO}_4, \Delta) hydrolyzes the −CN-\text{CN} group to −COOH-\text{COOH} and induces dehydration of the resulting α\alpha-hydroxy carboxylic acid to yield a conjugated unsaturated carboxylic acid (2,3-diphenylacrylic acid): PhCH2C(OH)(CN)Ph→H2SO4,ΔPh-CH=C(COOH)Ph\text{PhCH}_2\text{C}(\text{OH})(\text{CN})\text{Ph} \xrightarrow{\text{H}_2\text{SO}_4, \Delta} \text{Ph-CH}=\text{C}(\text{COOH})\text{Ph}

  • Asymmetric Carbon Analysis:

    • All carbon atoms in S\mathbf{S} (Ph-CH=C(COOH)Ph\text{Ph-CH}=\text{C}(\text{COOH})\text{Ph}) are sp2\text{sp}^2 hybridized.
    • Conclusion: S\mathbf{S} does NOT have any asymmetric carbon.

Summary:

  • P\mathbf{P}: Has an asymmetric carbon.
  • Q\mathbf{Q}: Does not have an asymmetric carbon.
  • R\mathbf{R}: Has an asymmetric carbon.
  • S\mathbf{S}: Does not have an asymmetric carbon.

Therefore, statements (C) and (D) are correct.

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