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

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(C) The reaction of propyne with $Hg^{+2}/dil. H_2SO_4$ (oxymercuration-demercuration) yields acetone $(CH_3COCH_3)$,which is $(P)$.The hydroboration-

Vedclass · Accepted Answer

$NaOI$

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(C) The reaction of propyne with $Hg^{+2}/dil. H_2SO_4$ (oxymercuration-demercuration) yields acetone $(CH_3COCH_3)$,which is $(P)$.The hydroboration-oxidation of propyne yields propanal $(CH_3CH_2CHO)$,which is $(Q)$.Acetone $(P)$ is a methyl ketone and gives a positive iodoform test with $NaOI$ (forming yellow precipitate of $CHI_3$).Propanal $(Q)$ is an aldehyde and does not contain the $CH_3CO-$ group,so it gives a negative iodoform test.Both $P$ and $Q$ are carbonyl compounds and would react with $2,4-DNP$ and $NaHSO_3$,so they cannot be differentiated by these reagents.Therefore,they can be differentiated by $NaOI$.

$CH_3-C \equiv CH \xrightarrow[dil. H_2SO_4]{Hg^{+2}} (P)$,$CH_3-C \equiv CH \xrightarrow[(2) H_2O_2]{(1) BH_3, THF} (Q)$. $P$ and $Q$ can be differentiated by

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