2,3-dimethyl-2-butene can be prepared by heating which of the following compounds with a strong acid?

Question

Accepted Answer

1. **Reaction Type:** The reaction involves the treatment of an alkene with a strong acid, leading to the formation of a carbocation intermediate, followed by rearrangement and elimination to form a more stable alkene (isomerisation).
2. **Mechanism for Option B ($(CH_3)_3C-CH=CH_2$):**
   - **Protonation:** The acid ($H^+$) adds to the double bond according to Markovnikov's rule to form a secondary carbocation:
     $$ (CH_3)_3C-CH=CH_2 + H^+ ightarrow (CH_3)_3C-C^+H-CH_3 $$
   - **Rearrangement:** The secondary carbocation undergoes a **1,2-methyl shift** to form a more stable tertiary carbocation:
     $$ (CH_3)_3C-C^+H-CH_3 \xrightarrow{	ext{1,2-Me shift}} (CH_3)_2C^+-CH(CH_3)_2 $$
   - **Elimination:** Loss of a proton ($H^+$) from the adjacent carbons yields the most substituted alkene (Zaitsev product). Removing a proton from the adjacent secondary carbons gives tetrasubstituted alkene:
     $$ (CH_3)_2C^+-CH(CH_3)_2 \xrightarrow{-H^+} (CH_3)_2C=C(CH_3)_2 $$
3. **Product:** The final product is 2,3-dimethyl-2-butene.
4. **Other Options:** The other options do not yield the specific carbon skeleton of 2,3-dimethyl-2-butene through simple acid-catalyzed rearrangement.

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