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Q7.32:

Show how would you synthesise the following alcohols from appropriate alkenes?

Solution :

Theoretical Foundation: Acid-Catalyzed Hydration of Alkenes

The synthesis of alcohols from alkenes via acid-catalyzed hydration proceeds according to Markovnikov's Rule, where the hydroxyl group ($OH^-$) attaches to the more substituted carbon atom of the double bond. This reaction follows the mechanism of electrophilic addition [Per the Mechanism of Electrophilic Addition to Alkenes].

Step 1: Analyzing the Target Molecule

The target molecule provided in the diagram is 1-methylcyclohexanol. To synthesize this via acid-catalyzed hydration, we must identify the corresponding alkene that would yield this tertiary alcohol upon the addition of $H_2O$ in the presence of $H_2SO_4$.

Step 2: Identifying the Precursor Alkene

By applying the reverse of the hydration mechanism, we remove $H_2O$ (a proton from the methyl-substituted carbon and an $OH$ group from the adjacent tertiary position) to reform the double bond. The appropriate precursor is 1-methylcyclohexene.

$CH_3$ β†’ $H_2O, H^+$ $OH$ $CH_3$

Step 3: Chemical Reaction Equation

The synthesis is performed by treating 1-methylcyclohexene with dilute sulfuric acid. The reaction proceeds via the formation of a stable tertiary carbocation [Per the Stability Order of Carbocations: $3^\circ > 2^\circ > 1^\circ$].

Reaction:
$C_7H_{12} (1\text{-methylcyclohexene}) + H_2O \xrightarrow{H^+} C_7H_{14}O (1\text{-methylcyclohexanol})$

Step 4: Step-by-Step Mechanism

  • Protonation: The double bond attacks $H^+$ from the acid to form the more stable tertiary carbocation at the $C1$ position.
  • Nucleophilic Attack: $H_2O$ acts as a nucleophile and attacks the electrophilic tertiary carbocation.
  • Deprotonation: A molecule of water removes a proton from the oxonium ion to yield the final tertiary alcohol.

Final Solution: The synthesis of 1-methylcyclohexanol is achieved by the acid-catalyzed hydration of 1-methylcyclohexene using dilute $H_2SO_4$.


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