Question

The end product in the sequence would be $ {{C}_{3}}{{H}_{7}}OH\xrightarrow[{{170}^{o}}]{{{H}_{2}}S{{O}_{4}}}A\xrightarrow{B{{r}_{2}}}B\xrightarrow[KOH]{Alc.}C $

• A. $ C{{H}_{3}}C\equiv CH $
• B. $ C{{H}_{3}}\text{ }\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}} $
• C. $ C{{H}_{2}}=C=C{{H}_{2}} $
• D. $ C{{H}_{3}}\text{ }\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\, $

Answer 1

Answer: (A)

$ C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH\xrightarrow[\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,{{170}^{o}} \\ (dehydration) \end{smallmatrix}]{{{H}_{2}}S{{O}_{4}}}\underset{\begin{smallmatrix} propene \\ \,\,\,\,'A' \end{smallmatrix}}{\mathop{C{{H}_{3}}CH=C{{H}_{2}}}}\, $ $ \xrightarrow{B{{r}_{2}}}\underset{1,\text{ }2-dibromopropane}{\mathop{\overset{\begin{smallmatrix} Br \\ | \end{smallmatrix}}{\mathop{C{{H}_{3}}CH}}\,-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,}}\,\xrightarrow[(Dehydrohalogenation)]{Alc.\,KOH} $ $ \underset{\begin{smallmatrix} propyne \\ \,\,\,\,'C' \end{smallmatrix}}{\mathop{C{{H}_{3}}C\equiv CH}}\, $