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The dehydrohalogenation of 2-bromobutane...

The dehydrohalogenation of 2-bromobutane with alcoholic KOH gives-

A

only 2-butane

B

only 1-butene

C

2-butene as the major product

D

1-butene as the major product

Text Solution

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The correct Answer is:
C
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Dehydration of 1-butanol or 2-butanol with conc. H_2SO_4 always gives the same mixture of 2-butene (80 %) and 1-butene (20 %) but dehydrohalogenation of 1-bromobutane with alc. KOH gives 1-butene as the major product while that of 2-bromobutane gives 2-butene as the major product . Explain why ?

The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is-

Knowledge Check

  • The major product obtained in the dehydrohalogenation of neo-pentyl bromide with alcoholic KOH is

    A
    2-methylbut-1-ene
    B
    2,2-dimethylbut-1-ene
    C
    2-methylbut-2-ene
    D
    but-2-ene.

  • Chloroform with alcoholic KOH gives

    A
    potassium acetate
    B
    potassium formate
    C
    potassium chloride
    D
    potassium chlorate

  • The dehydrohalogenation of neopentyl bromide with alcoholic KOH mainly gives

    A
    2-methyl-1-butene
    B
    2-methyl-2-butene
    C
    2,2-dimethyl-1-butene
    D
    2-butene

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    The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is

    The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is

    In the dehydrohalogenation of 2-bromobutane, which conformation leads to the formation of cis-2-butene?

    When haloalkanes with beta -hydrogen atom are boiled with alcoholic solution of KOH, they undergo elimination of hyarogen halide resulting in the formation of alkenes. These reactions are called beta -elimination reactions or dehydrohalogenation reactions. These reactions follow Saytzeff's rule. Substitution and elimination reactions often compete with each other. Mostly bases behave as nucleophiles and therefore can engage in substitution or elimination reactions depending upon the alkyl halide and the reaction conditions. The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is

    When haloalkanes with Beta -hydrogen atom are boiled with alcoholic solution of KOH, they undergo elimination of hyarogen halide resulting in the formation of alkenes. These reactions are called Beta -elimination reactions or dehydrohalogenation reactions. These reactions follow Saytzeff's rule. Substitution and elimination reactions often compete with each other. Mostly bases behave as nucleophiles and therefore can engage in substitution or elimination reactions depending upon the alkyl halide and the reaction conditions. The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is

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