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An alkyl halide C(5)H(11)Br (A) reacts w...

An alkyl halide `C_(5)H_(11)Br` (A) reacts with ethanolic `KOH` to give an alkene 'B' which reacts with `Br_(2)` to give a compound 'C' which on dehydrobromination gives an alkyne 'D'. On treatment with sodium metal in liquid ammonia one mole of 'D' give one mole of the sodium salt of 'D' and half a mole of hydrogen gas. Complete hydrogenation of 'D' yields a straight chain alkane. Identify A, B, C and D. Give the the reactions involved.

Answer

Step by step text solution for An alkyl halide C_(5)H_(11)Br (A) reacts with ethanolic KOH to give an alkene 'B' which reacts with Br_(2) to give a compound 'C' which on dehydrobromination gives an alkyne 'D'. On treatment with sodium metal in liquid ammonia one mole of 'D' give one mole of the sodium salt of 'D' and half a mole of hydrogen gas. Complete hydrogenation of 'D' yields a straight chain alkane. Identify A, B, C and D. Give the the reactions involved. by CHEMISTRY experts to help you in doubts & scoring excellent marks in Class 11 exams.

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An alkyl halide C_(5)H_(11) (A) reacts with ethanolic KOH to give an alkene 'B' which reacts with Br_(2) to give a compound 'C' which on dehydromination gives an alkyne 'D' . On treatment with sodium metal in liquid ammonia one mole of 'D' give one mole of the sodium salt of 'D' and half a mole of hydrogen gas . Complete hydrogenation of 'D' yields a straight chain alkane. Identify A, B, C and D . Give the the reactions involved.

An alkyl halide C_5H_11Br (A) reacts with ethanolic KOH to give an alkene 'B', which reacts with Br_2 to give a compound 'C', which on dehydrobromination gives an alkyne 'D'. On treatment with sodium metal in liquid ammonia , one mole of 'D' gives one mole of the sodium salt of 'D' and half a mole of hydrogen gas . complete hydrogenation of 'D' yields a straight chain alkane. Identify A,B, C and D . Give the reactions involved.

Knowledge Check

  • Benzamide is reacted with sodium metal in ethanol gives

    A
    Aniline
    B
    Benzyl amine
    C
    benzoic acid
    D
    Benzaldehyde

  • An halide C_(5)H_(11) Br on tretment with alc .KOH give 2- pentene only . The halide will be:

    A
    `CH_(3)-CH_(2)-CH_(2)-CH_(2)-CH_(2)-Br`
    B
    `CH_(3)-CH_(2)-CH_(2)-underset(Br)underset(|) CH -CH_(3)`
    C
    `CH_(3)-CH_(2)-underset(Br)underset(|) CH -CH_(2)-CH_(3)`
    D
    `CH_(3)-overset(CH_(3))overset(|)CH-underset(Br)underset(|)C-CH_(3)`

  • Hydrogenation of the above compound in the presence of sodium in liquid ammonia gives -

    A
    An optically active compound
    B
    An optically inactive compound
    C
    A racemic mixture
    D
    A diastereomeric mixture

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