Question:

The major product formed in the following reaction is : 

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Whenever a carbocation is formed, always check for 1,2-hydride or 1,2-methyl shifts to form a more stable carbocation before finishing the reaction.
Updated On: Jan 2, 2026
  • A
  • B
  • C
  • D
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The reaction is an acid-catalyzed dehydration of an alcohol. This reaction proceeds via carbocation formation, possible rearrangement, and finally elimination to form the most stable alkene (Saytzeff rule).
Step 2: Detailed Explanation:
1. Protonation and Water Loss: The reactant is 3,3-dimethylbutan-2-ol. Protonation of the \( -OH \) group followed by the loss of water generates a secondary carbocation at C2: \( CH_3-C(CH_3)_2-\text{C}^+\text{H}-CH_3 \).
2. Rearrangement: The secondary carbocation is adjacent to a quaternary carbon. To increase stability, a 1,2-methyl shift occurs. This creates a more stable tertiary carbocation: \( CH_3-\text{C}^+(CH_3)-CH(CH_3)_2 \).
3. Elimination: A proton is lost from the adjacent carbon to form an alkene. According to Saytzeff's rule, the more substituted alkene is the major product.
4. Final Product: Loss of a proton from the C3 carbon gives 2,3-dimethylbut-2-ene (\( CH_3-C(CH_3)=C(CH_3)-CH_3 \)). This is a tetrasubstituted alkene and is highly stable.
Step 3: Final Answer:
The major product is 2,3-dimethylbut-2-ene.
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