JEE Challenger
More from Alcohols, Phenols and Ethers
ChemMedAdv 2011 • Paper 1

Structure of Intermediate Compound Q in Alkyne Reaction Sequence

Comprehension Passage

Paragraph for Question Nos. 15 and 16

An acyclic hydrocarbon P\mathbf{P}, having molecular formula C6H10\text{C}_6\text{H}_{10}, gave acetone as the only organic product through the following sequence of reactions, in which Q\mathbf{Q} is an intermediate organic compound.

Passage Diagram 0

The structure of the compound Q\mathbf{Q} is

Options

A
Option A Diagram 1
B
Option B Diagram 1
Correct
C
Option C Diagram 1
D
Option D Diagram 1

Topics & Concepts

Alcohols, Phenols and EthersAlcohols and EthersAlkanes, Alkenes, AlkynesOxidation and Reduction (Organic)

Step-by-Step Solution

To determine the structure of intermediate compound Q\mathbf{Q}, we trace the given reaction sequence step-by-step:

  1. Hydration of Compound P\mathbf{P}:
    Hydrocarbon P\mathbf{P} (C6H10\text{C}_6\text{H}_{10}) is 3,3-dimethylbut-1-yne, with the formula (CH3)3C−C≡CH(\text{CH}_3)_3\text{C}-\text{C}\equiv\text{CH}.

    Reaction of P\mathbf{P} with dil. H2SO4/HgSO4\text{dil. H}_2\text{SO}_4 / \text{HgSO}_4 results in Markovnikov addition of water across the triple bond to yield an enol intermediate:
    (CH3)3C−C≡CH→dil. H2SO4/HgSO4[(CH3)3C−C(OH)=CH2](\text{CH}_3)_3\text{C}-\text{C}\equiv\text{CH} \xrightarrow{\text{dil. H}_2\text{SO}_4/\text{HgSO}_4} \left[(\text{CH}_3)_3\text{C}-\text{C}(\text{OH})=\text{CH}_2\right]

    Tautomerization of this enol gives 3,3-dimethylbutan-2-one (pinacolone):
    [(CH3)3C−C(OH)=CH2]⟶(CH3)3C−C(=O)−CH3\left[(\text{CH}_3)_3\text{C}-\text{C}(\text{OH})=\text{CH}_2\right] \longrightarrow (\text{CH}_3)_3\text{C}-\text{C}(=\text{O})-\text{CH}_3

  2. Reduction to Form Compound Q\mathbf{Q}:
    Reduction of 3,3-dimethylbutan-2-one using NaBH4/ethanol\text{NaBH}_4/\text{ethanol} followed by workup with dilute acid yields the secondary alcohol, 3,3-dimethylbutan-2-ol:
    (CH3)3C−C(=O)−CH3→(ii) dil. acid(i) NaBH4/ethanol(CH3)3C−CH(OH)−CH3(Q)(\text{CH}_3)_3\text{C}-\text{C}(=\text{O})-\text{CH}_3 \xrightarrow[\text{(ii) dil. acid}]{\text{(i) NaBH}_4/\text{ethanol}} (\text{CH}_3)_3\text{C}-\text{CH}(\text{OH})-\text{CH}_3 \quad (\mathbf{Q})

  3. Verification via Subsequent Reactions:

    • Dehydration of Q\mathbf{Q}: Treatment of 3,3-dimethylbutan-2-ol with catalytic conc. H2SO4\text{conc. H}_2\text{SO}_4 leads to protonation and loss of a water molecule, forming a 2∘2^\circ carbocation:
      (CH3)3C−CH(OH)−CH3→H+,−H2O(CH3)3C−C+H−CH3(\text{CH}_3)_3\text{C}-\text{CH}(\text{OH})-\text{CH}_3 \xrightarrow{\text{H}^+, -\text{H}_2\text{O}} (\text{CH}_3)_3\text{C}-\overset{+}{\text{C}}\text{H}-\text{CH}_3

      A 1,2-methyl shift takes place to give a more stable 3∘3^\circ carbocation:
      (CH3)3C−C+H−CH3→1,2-methyl shift(CH3)2C+−CH(CH3)2(\text{CH}_3)_3\text{C}-\overset{+}{\text{C}}\text{H}-\text{CH}_3 \xrightarrow{1,2\text{-methyl shift}} (\text{CH}_3)_2\overset{+}{\text{C}}-\text{CH}(\text{CH}_3)_2

      Deprotonation then gives 2,3-dimethylbut-2-ene as the major alkene product:
      (CH3)2C+−CH(CH3)2→−H+(CH3)2C=C(CH3)2(\text{CH}_3)_2\overset{+}{\text{C}}-\text{CH}(\text{CH}_3)_2 \xrightarrow{-\text{H}^+} (\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2

    • Reductive Ozonolysis: Ozonolysis of 2,3-dimethylbut-2-ene (O3\text{O}_3 followed by Zn/H2O\text{Zn/H}_2\text{O}) cleaves the double bond to produce two molecules of acetone (CH3COCH3\text{CH}_3\text{COCH}_3):
      (CH3)2C=C(CH3)2→(ii) Zn/H2O(i) O32 CH3COCH3(\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2 \xrightarrow[\text{(ii) Zn/H}_2\text{O}]{\text{(i) O}_3} 2\text{ CH}_3\text{COCH}_3

Thus, the structure of compound Q\mathbf{Q} is 3,3-dimethylbutan-2-ol, (CH3)3C−CH(OH)−CH3(\text{CH}_3)_3\text{C}-\text{CH}(\text{OH})-\text{CH}_3.

This corresponds to Option (B).

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