A) \[C{{H}_{3}}CH(N{{H}_{2}})C{{H}_{2}}N{{H}_{2}}\]
B) \[C{{H}_{3}}CHOHC{{H}_{2}}OH\]
C) \[C{{H}_{3}}\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}\]
D) \[C{{H}_{3}}C=CH\]
Correct Answer: D
Solution :
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH\xrightarrow[160{}^\circ -180{}^\circ ]{{{H}_{2}}S{{O}_{4}}}\underset{(X)}{\mathop{C{{H}_{3}}CH=C{{H}_{2}}}}\,\xrightarrow[{}]{B{{r}_{2}}}\] \[\underset{(Y)}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{CH}}\,-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,}}\,\xrightarrow[-HBr]{Alc.\,KOH}\underset{C{{H}_{3}}CH=CHBr}{\overset{C{{H}_{3}}CBr=C{{H}_{2}}}{\mathop{+}}}\,\] \[\xrightarrow[-HBr]{NaN{{H}_{2}}}\underset{(Z)}{\mathop{C{{H}_{3}}C\equiv CH}}\,\] \[NaN{{H}_{2}}\]being a stronger base than alcoholic KOH, readily brings about dehydrobromination of less reactive vinyl bromides to give propyne (Z).
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