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NEET CHEMISTRYMedium

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

A

(CH3)2CH−CH(CH3)−CH=CH2(CH_3)_2CH-CH(CH_3)-CH=CH_2

B

(CH3)3C−CH=CH2(CH_3)_3C-CH=CH_2

C

(CH3)2C=CH−CH2−CH3(CH_3)_2C=CH-CH_2-CH_3

D

(CH3)2CH−CH2−CH=CH2(CH_3)_2CH-CH_2-CH=CH_2

Step-by-Step Solution

  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 ((CH3)3C−CH=CH2(CH_3)_3C-CH=CH_2):
  • Protonation: The acid (H+H^+) adds to the double bond according to Markovnikov's rule to form a secondary carbocation: (CH3)3C−CH=CH2+H+→(CH3)3C−C+H−CH3(CH_3)_3C-CH=CH_2 + H^+ \rightarrow (CH_3)_3C-C^+H-CH_3
  • Rearrangement: The secondary carbocation undergoes a 1,2-methyl shift to form a more stable tertiary carbocation: (CH3)3C−C+H−CH3→1,2-Me shift(CH3)2C+−CH(CH3)2(CH_3)_3C-C^+H-CH_3 \xrightarrow{\text{1,2-Me shift}} (CH_3)_2C^+-CH(CH_3)_2
  • Elimination: Loss of a proton (H+H^+) from the adjacent carbons yields the most substituted alkene (Zaitsev product). Removing a proton from the adjacent secondary carbons gives tetrasubstituted alkene: (CH3)2C+−CH(CH3)2→−H+(CH3)2C=C(CH3)2(CH_3)_2C^+-CH(CH_3)_2 \xrightarrow{-H^+} (CH_3)_2C=C(CH_3)_2
  1. Product: The final product is 2,3-dimethyl-2-butene.
  2. 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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