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    The Alkene Formed As A Major Product in The Above

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    The major product of the Hofmann elimination reaction described is the least substituted alkene (C2H4=C2H4). [The Hofmann elimination reaction involves treating a quaternary ammonium salt with excess methyl iodide to produce a tertiary amine and an alkene. The major product is the least substituted alkene according to Hofmann's rule.] Option A, butylene, is the least substituted alkene and therefore the major product of this reaction.

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    The Alkene Formed As A Major Product in The Above

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    shineyrbl
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    The major product of the Hofmann elimination reaction described is the least substituted alkene (C2H4=C2H4). [The Hofmann elimination reaction involves treating a quaternary ammonium salt with excess methyl iodide to produce a tertiary amine and an alkene. The major product is the least substituted alkene according to Hofmann's rule.] Option A, butylene, is the least substituted alkene and therefore the major product of this reaction.

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    The major product of the Hofmann elimination reaction described is the least substituted alkene (C2H4=C2H4). [The Hofmann elimination reaction involves treating a quaternary ammonium salt with excess methyl iodide to produce a tertiary amine and an alkene. The major product is the least substituted alkene according to Hofmann's rule.] Option A, butylene, is the least substituted alkene and therefore the major product of this reaction.

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    The Alkene Formed As A Major Product in The Above

    Uploaded by

    shineyrbl
    The major product of the Hofmann elimination reaction described is the least substituted alkene (C2H4=C2H4). [The Hofmann elimination reaction involves treating a quaternary ammonium salt with excess methyl iodide to produce a tertiary amine and an alkene. The major product is the least substituted alkene according to Hofmann's rule.] Option A, butylene, is the least substituted alkene and therefore the major product of this reaction.

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    The alkene formed as a major product in the


    above elimination reactions is:

    A.

    B.

    C.

    D.

    CBSE Chemistry Grade 11 Alkenes


    Last updated date: 22nd Dec 2023 • Total views: 298.8k • View

    Answer NCERT Solutions CBSE

    Answer Veri!ed

    298.8k+ views

    Hint: The given reaction can be explained on


    the basis of Hofmann elimination which can
    be de!ned as the process of creating
    tertiary amines and alkenes i.e. amines and
    alkenes having three groups attached with
    itself from the treatment of quaternary
    ammonium with excess methyl iodide.

    Complete answer:
    This process can also be referred to as
    exhaustive methylation. The Hofmann
    elimination process is named after the
    scientist who discovered this process and
    this is discovered by German chemist
    August Wilhelm Von Hofmann.
    The Hofmann rule states that when there is
    an asymmetrical amine present in the
    molecule then the major product of that
    alkene is least substituted in nature and we
    can also say that as a result it also behaves
    as a least stable product. The given
    compound is quaternary ammonium salt in

    which four β hydrogen atoms and on

    further heating these quaternary


    ammonium salts it gives Hofmann
    elimination reactions in which it abstract
    most acidic hydrogen and major product
    formed is least substituted alkene i.e.

    C H2 = C H2 C H2 = C H2 , from this we
    can easily conclude that option A is correct.

    Note:
    If there are no beta hydrogen atoms present
    in the molecule then Ho"mann elimination
    reaction is not possible to occur. The most
    important example of the Hofmann
    elimination process or the exhaustive
    methylation process is the synthesis of
    trans-cyclooctene.

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