Identify Z in the following reaction sequence {CH₃CH₂CH₂OH} {{Conc.}\,{H₂SO₄} X → Y [(ii)\,{NaNH}₂]{(i)\,{Alc.KOH}} Z

Correct Answer is : CH3{CH₃CH3 - C ≡ CH

{CH₃ - {CH(NH₂)} - {CN₂NH₂

Correct Answer is : CH3{CH₃CH3 - C ≡ CH

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VITEEE 2015 Paper

Section: Chemistry

Question : 79 of 120

Marks: +1, -0

\mathrm{CH_3CH_2CH_2OH}

\xrightarrow{\text{Conc.}\,\mathrm{H_2SO_4}}

\xrightarrow[(ii)\,\mathrm{NaNH}_2]{(i)\,\mathrm{Alc.KOH}}

Solution:

\mathrm{CH_3CH_2CH_2OH}

\xrightarrow{\text{Conc.}\,\mathrm{H_2SO_4}}

\mathrm{CH_3CH=CH_2}

\xrightarrow{\mathrm{Br}_2}

\xrightarrow[(ii)\,\mathrm{NaNH}_2]{(i)\,\mathrm{Alc.KOH}}

\underset{\mathrm{Y}}{\mathrm{CH_3CHBr-CH_2Br}}

[

\underset{\mathrm{B}}{\underset{\mathrm{A}}{\mathrm{CH_3C(BR)}}}

\mathrm{CH_2}

\mathrm{CH_3CH}

= CHBr]

\xrightarrow[-\mathrm{HBr}]{\mathrm{NaNH}_2}

\underset{\mathrm{Z}}{\mathrm{CH_3C}}

≡ CH

Alcoholic KOH brings about dehydrobromination of Y and give a mixture of vinyl bromide (A and B) while

\mathrm{NaNH_2}

being a strong base than ale. KOH readily brings about dehydrobromination of less reactive vinyl bromide to give propyne

\mathrm{CH_3\,C}

= CH i.e. (Z).

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