11th Class Chemistry Hydrocarbons

  • question_answer 87)   An alkyl halide \[{{C}_{5}}{{H}_{11}}Br(A)\]Br (A) reacts with ethanolic KOH to give an alkene ?B? which reacts with \[B{{r}_{2}}\] 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.

    Answer:

        \[\underset{\begin{smallmatrix}  (A) \\  Alkyl\,halide \end{smallmatrix}}{\mathop{{{C}_{5}}{{H}_{11}}Br}}\,\xrightarrow[\Delta ]{alc.KOH}\underset{\underset{Alkene}{\mathop{(B)}}\,}{\mathop{{{C}_{5}}{{H}_{10}}}}\,\xrightarrow[C{{S}_{2}}]{B{{r}_{2}}}\underset{\underset{Dibromo\,alkane}{\mathop{(C)}}\,}{\mathop{{{C}_{5}}{{H}_{10}}B{{r}_{2}}}}\,\]   \[\xrightarrow[\Delta ]{alc.\,KOH}\underset{\underset{Alkyne}{\mathop{(D)}}\,}{\mathop{{{C}_{5}}{{H}_{8}}}}\,\xrightarrow{{{H}_{2}}}\underset{(Straight\,chain\,alkane)}{\mathop{{{C}_{5}}{{H}_{12}}}}\,\] (D)\[\xrightarrow{Na/liqN{{H}_{3}}}{{C}_{5}}H_{7}^{+}N{{a}^{+}}+\frac{1}{2}{{H}_{2}}\]   Hydrogenation of alkyne (D) gives straight chain alkane hence all the compounds (A), (B), (C) and (D) must be straight chain compounds. Alkyne (D) forms sodium salt which proves that it is terminal alkyne. Involved reactions are given below: \[C{{H}_{3}}-\underset{1-\text{Bromopentane(A)}}{\mathop{C{{H}_{2}}CHL2CHL2}}\,-C{{H}_{2}}Br\xrightarrow[-HBr]{alc.\,KOH}\]                                 \[C{{H}_{3}}-\underset{1-Pentene(B)}{\mathop{C{{H}_{2}}C{{H}_{2}}CH}}\,=C{{H}_{2}}\xrightarrow[C{{S}_{2}}]{B{{r}_{2}}}\] \[\underset{1,2-\text{Dibromopentane}(C)}{\mathop{C{{H}_{3}}-C{{H}_{2}}C{{H}_{2}}\overset{\overset{Br}{\mathop{|}}\,}{\mathop{CH}}\,-\overset{\overset{Br}{\mathop{|}}\,}{\mathop{C{{H}_{2}}}}\,}}\,\xrightarrow[-2HBr]{alc.\,KOH}\]


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