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Haloalkanes and Haloarenes question

2005 · Shift 0 · Q49
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Haloalkanes and Haloarenes question

2005 · Shift 0 · Q49

JEE MainChemistryHaloalkanes and HaloarenesMCQ+4 / −1
Elimination of bromine from 2-bromobutane results in the formation of-
  1. A
    equimolar mixture of 1 and 2-butene
  2. B
    predominantly 2-butene
  3. C
    predominantly 1-butene
  4. D
    predominantly 2-butyne
View written solutionFree

Correct answer: B

  1. Identify the reaction type

    Elimination of HBr from 2-bromobutane is a dehydrohalogenation reaction.

    The structure of 2-bromobutane is:

    CH3−CHBr−CH2−CH3CH_3-CHBr-CH_2-CH_3CH3​−CHBr−CH2​−CH3​

    In elimination, bromine is removed from the carbon bearing Br, and a hydrogen is removed from an adjacent carbon.

  2. Possible products

    Bromine is on carbon-2. So hydrogen can be removed from either adjacent carbon:

    • From carbon-1: CH2=CH−CH2−CH3CH_2=CH-CH_2-CH_3CH2​=CH−CH2​−CH3​ This is 1-butene.

    • From carbon-3: CH3−CH=CH−CH3CH_3-CH=CH-CH_3CH3​−CH=CH−CH3​ This is 2-butene.

  3. Apply Saytzeff (Zaitsev) rule

    In elimination reactions, the more substituted alkene is formed predominantly.

    • 1-butene is a less substituted alkene.
    • 2-butene is a more substituted alkene and is therefore more stable.

    Hence, the major product is 2-butene.

  4. Check options

    • A: equimolar mixture of 1 and 2-butene — Incorrect, because 2-butene is favored.
    • B: predominantly 2-butene — Correct.
    • C: predominantly 1-butene — Incorrect.
    • D: predominantly 2-butyne — Incorrect, since simple dehydrohalogenation of a monobromoalkane gives an alkene, not an alkyne.

  5. Final answer

    Predominantly 2-butene\boxed{\text{Predominantly 2-butene}}Predominantly 2-butene​

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