Number of Hydroxyl Groups in Oxidation Product Q

To find the number of hydroxyl groups present in the final product $\mathbf{Q}$, we analyze the two steps of the reaction:

Step 1: Dehydration of the alcohol to form $\mathbf{P}$

• The starting material contains:
  1. A secondary alcohol group ($-\text{OH}$).
  2. An isolated carbon-carbon double bond ($\text{C}=\text{C}$).
• Heating the compound in the presence of an acid ($\text{H}^+$, heat) causes acid-catalyzed dehydration of the secondary alcohol via an $\text{E1}$ mechanism: $\text{R-OH} \xrightarrow{\text{H}^+, \Delta} \text{Diene } (\mathbf{P}) + \text{H}_2\text{O}$
• Elimination of the hydroxyl group together with an adjacent hydrogen atom introduces a new $\text{C}=\text{C}$ double bond into the molecule.
• As a result, the intermediate product $\mathbf{P}$ is a diene possessing two $\text{C}=\text{C}$ double bonds.

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Step 2: Dihydroxylation of $\mathbf{P}$ using Baeyer's reagent to form $\mathbf{Q}$

• Aqueous dilute $\text{KMnO}_4$ at cold conditions ($0^\circ\text{C}$) is Baeyer's reagent, which acts as a mild oxidizing agent.
• It selectively performs syn-dihydroxylation across alkenes, converting each carbon-carbon double bond into a vicinal cis-1,2-diol: $-\text{C}=\text{C}- \xrightarrow{\text{dil. } \text{KMnO}_4, 0^\circ\text{C}} -\text{C(OH)}-\text{C(OH)}-$
• Since compound $\mathbf{P}$ contains two $\text{C}=\text{C}$ double bonds and the reagent is used in excess, both double bonds undergo syn-dihydroxylation: $\text{Number of } -\text{OH} \text{ groups added} = 2 \times (\text{number of double bonds}) = 2 \times 2 = 4$

Thus, the final product $\mathbf{Q}$ is a tetrahydroxy compound containing 4 hydroxyl groups.