A) acetone
B) propene
C) propanol
D) propanal
E) 2-propanol
Correct Answer: A
Solution :
When propyne is treated with dilute\[{{H}_{2}}S{{O}_{4}}\]in presence of mercury (II) sulphate then acetone is formed. \[C{{H}_{3}}C\equiv CH+{{H}_{2}}O\xrightarrow[HgS{{O}_{4}}]{dill.\,{{H}_{2}}S{{O}_{4}}}\] \[\underset{enolic\text{ }form}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{2}}}}\,\underset{acetone}{\mathop{C{{H}_{3}}-\overset{\begin{smallmatrix} O \\ |\,| \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}}}\,\]
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