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NEET CHEMISTRYHydrocarbonsHard

Question

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)3CCH+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.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Hydrocarbons. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYHydrocarbonsdimethylbutenepreparedheatingfollowingcompounds

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A.Option 1 (Structure missing)
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