Compound I undergoes hydroboration-oxidation reaction with $(\text{BH}_3)_2$ followed by treatment with $\text{H}_2\text{O}_2$ and aqueous $\text{NaOH}$ to produce another compound II, which upon oxidation with $\text{CrO}_3$ gives 2,3-dimethyl-cyclohexanone as the product. What is the structure of I?
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Hydroboration-oxidation is an anti-Markovnikov hydration.
To get a ketone at C1 with a methyl at C2, the double bond must have been between C1 and C2, with C2 being the more substituted carbon (bearing the methyl) so that $-\text{OH}$ adds selectively to C1.
Step 1 : Understanding the Question:
This question asks us to find the structure of alkene I which, upon hydroboration-oxidation to form alcohol II and subsequent oxidation with chromium trioxide ($\text{CrO}_3$), yields 2,3-dimethylcyclohexanone as the final product. Step 2 : Key Formulas and Approach:
We will work backward from the final product to determine the intermediate and reactant:
• Oxidation: Oxidation of secondary alcohol II with $\text{CrO}_3$ yields the ketone 2,3-dimethylcyclohexanone. This means II must be 2,3-dimethylcyclohexan-1-ol.
• Hydroboration-Oxidation: Hydroboration-oxidation adds $-\text{H}$ and $-\text{OH}$ across a double bond in a syn, anti-Markovnikov (less-substituted carbon) fashion. Step 3 : Detailed Explanation:
Let us analyze the structures:
• The Final Product: 2,3-dimethylcyclohexanone
The structure has a carbonyl group at position 1, and methyl groups at positions 2 and 3.
To obtain this ketone upon oxidation, alcohol II must have its hydroxyl group ($-\text{OH}$) at position 1:
\[ \text{Compound II} = \text{2,3-dimethylcyclohexan-1-ol} \]
• Evaluating Alkene I Options: • Option (a): 1,2-dimethylcyclohexene
The double bond is between C1 and C2, both of which are tertiary carbons (each is bonded to a methyl group).
Hydroboration-oxidation of this symmetric alkene would place the $-\text{OH}$ group on either C1 or C2, yielding 1,2-dimethylcyclohexan-1-ol (a tertiary alcohol, which cannot be oxidized to a ketone). So Option (a) is incorrect.
• Option (b): 2,3-dimethylcyclohexene
The double bond is between C1 and C2. C2 has a methyl group, while C1 is unsubstituted. C3 has the second methyl group.
During hydroboration-oxidation, the boron atom adds to the less-substituted carbon of the double bond (C1), which upon oxidation yields the alcohol:
\[ \text{2,3-dimethylcyclohexan-1-ol} \]
Oxidation of this secondary alcohol with $\text{CrO}_3$ converts the secondary alcohol at C1 into a carbonyl group, yielding:
\[ \text{2,3-dimethylcyclohexanone} \]
This matches the target product. Step 4 : Final Answer:
The starting alkene I must be 2,3-dimethylcyclohexene, represented by Structure (b).
This corresponds to Option (B).
