The hydration of propyne in the presence of HgSO4/H2SO4 produces

# The hydration of propyne in the presence of ${\mathrm{HgSO}}_{4}/{\mathrm{H}}_{2}{\mathrm{SO}}_{4}$ produces

1. A

${\mathrm{CH}}_{3}\mathrm{CHO}$

2. B

${\mathrm{CH}}_{3}{\mathrm{CH}}_{2}\mathrm{CHO}$

3. C

${\mathrm{CH}}_{3}{\mathrm{COCH}}_{3}$

4. D

$\mathrm{HCHO}$

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### Solution:

The reaction is,

${\mathrm{CH}}_{3}\mathrm{C}=\mathrm{CH}\frac{{\mathrm{HgSO}}_{4}/{\mathrm{H}}_{2}{\mathrm{SO}}_{4}}{+{\mathrm{H}}_{2}\mathrm{O}}$$\frac{{\mathrm{HgSO}}_{4}/{\mathrm{H}}_{2}{\mathrm{SO}}_{4}}{{\mathrm{H}}_{2}\mathrm{O}}{\mathrm{CH}}_{3}{\mathrm{COCH}}_{3}$

Thus, The hydration of propyne in the presence of ${\mathrm{HgSO}}_{4}/{\mathrm{H}}_{2}{\mathrm{SO}}_{4}$ produces ${\mathrm{CH}}_{3}{\mathrm{COCH}}_{3}$.

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