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The hydration of propyne in the presence of ${HgSO}_{4} / H_{2} {SO}_{4}$ produces

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{CH}_{3} CHO

{CH}_{3} {CH}_{2} CHO

{CH}_{3} {COCH}_{3}

HCHO

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detailed solution

Correct option is C

The reaction is,

${CH}_{3} C = CH \frac{{HgSO}_{4} / H_{2} {SO}_{4}}{+ H_{2} O}$ ${CH}_{3} C = CHc OH$ $\frac{{HgSO}_{4} / H_{2} {SO}_{4}}{H_{2} O} {CH}_{3} {COCH}_{3}$ $O H \overset{- H_{2} O}{\leftarrow} {CH}_{3} C_{}^{} - {CH}_{3} OH$

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

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