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NEET CHEMISTRYHard

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

A

(Option 1 Structure Missing)

B

(CH3)3C-CH=CH2

C

(CH3)2C=CH-CH2-CH3

D

(CH3)2CH-CH2-CH=CH2

Step-by-Step Solution

  1. Reaction Type: Heating an alkene or alcohol with a strong acid (like H2SO4H_2SO_4) typically leads to the formation of a more substituted alkene via the formation of a carbocation intermediate and subsequent rearrangement.
  2. Analysis of Option 2 ((CH3)3C-CH=CH2): This compound is 3,3-dimethylbut-1-ene.
  • Protonation: The acid protonates the double bond to form a secondary carbocation: (CH3)3C−CH+−CH3(CH_3)_3C-CH^+-CH_3.
  • Rearrangement: A 1,2-methyl shift occurs to form a more stable tertiary carbocation: (CH3)2C+−CH(CH3)2(CH_3)_2C^+-CH(CH_3)_2.
  • Elimination: Elimination of a proton from the adjacent carbon yields the most substituted, stable alkene (Saytzeff product): 2,3-dimethylbut-2-ene ((CH3)2C=C(CH3)2)((CH_3)_2C=C(CH_3)_2).
  1. Conclusion: The rearrangement of the carbon skeleton in 3,3-dimethylbut-1-ene under acidic conditions leads directly to the formation of 2,3-dimethyl-2-butene.
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