CH_3-C equiv CH xrightarrow[dil. H_2SO_4]{HgS | vedclass

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(B) In reaction $1$,the hydration of propyne with $HgSO_4/dil. H_2SO_4$ follows Markovnikov's rule to form acetone $(CH_3-CO-CH_3)$. In reaction $2$,h

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$NaOI$

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(B) In reaction $1$,the hydration of propyne with $HgSO_4/dil. H_2SO_4$ follows Markovnikov's rule to form acetone $(CH_3-CO-CH_3)$. In reaction $2$,hydroboration-oxidation follows anti-Markovnikov's rule to form propanal $(CH_3-CH_2-CHO)$. Acetone,being a methyl ketone,gives a positive iodoform test with $NaOI$ (forming a yellow precipitate of $CHI_3$),while propanal does not. Therefore,they can be differentiated using $NaOI$.

$CH_3-C \equiv CH \xrightarrow[dil. H_2SO_4]{HgSO_4} (A)$
$CH_3-C \equiv CH \xrightarrow[(1) BH_3.THF]{(2) H_2O_2/HO^{-}} (B)$
Products $(A)$ and $(B)$ can be differentiated by:

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$CH_3-C \equiv CH \xrightarrow[dil. H_2SO_4]{HgSO_4} (A)$ $CH_3-C \equiv CH \xrightarrow[(1) BH_3.THF]{(2) H_2O_2/HO^{-}} (B)$ Products $(A)$ and $(B)$ can be differentiated by: