A) 2-propanol
B) Propanone
C) Propanal
D) 1, 2-propane diol
Correct Answer: B
Solution :
\[\underset{propyne}{\mathop{C{{H}_{3}}-C\equiv CH}}\,+H-OH\xrightarrow[{}]{Dil.{{H}_{2}}S{{O}_{4}}}\] \[\underset{enolic\text{ }form}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}}}\,\underset{\begin{smallmatrix} keto\text{ }form \\ propanone \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} || \\ O \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{3}}}}\,\]
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