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NEET CHEMISTRYHydrocarbonsMedium

Question

Which of the following statement(s) is/are correct about the elimination reaction of 2-Bromopentane to form pent-2-ene? (a) β\beta-Elimination reaction (b) Follows Zaitsev rule (c) Dehydrohalogenation reaction (d) Dehydration reaction

A

(a), (c), (d)

B

(b), (c), (d)

C

(a), (b), (d)

D

(a), (b), (c)

Step-by-Step Solution

  1. β\beta-Elimination: The reaction involves the removal of a hydrogen atom from the β\beta-carbon (the carbon adjacent to the one holding the halogen) and the halogen from the α\alpha-carbon. Hence, it is a β\beta-elimination reaction .
  2. Zaitsev Rule: In dehydrohalogenation, if more than one alkene can be formed, the major product is the more substituted alkene (Zaitsev product). For 2-bromopentane, pent-2-ene (more substituted) is the major product compared to pent-1-ene .
  3. Dehydrohalogenation: The reaction entails the elimination of a hydrogen halide (HBr), so it is termed dehydrohalogenation .
  4. Dehydration: This term specifically refers to the removal of a water molecule (H2OH_2O), typically from alcohols (e.g., ethanol to ethene) . Since 2-bromopentane is a haloalkane, not an alcohol, this is not a dehydration reaction.

Therefore, statements (a), (b), and (c) are correct.

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.

CHEMISTRYHydrocarbonsfollowingstatementscorrecteliminationreaction

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