CH_3-C equiv CH xrightarrow[(ii) H_2O_2/OH^{T | vedclass

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(D) $1$. The first reaction is hydroboration-oxidation of propyne: $CH_3-C \equiv CH + BH_3-THF \rightarrow CH_3-CH=CH-BH_2$. Subsequent oxidation wit

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$(P)$ and $(Q)$ are functional isomers

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(D) $1$. The first reaction is hydroboration-oxidation of propyne: $CH_3-C \equiv CH + BH_3-THF \rightarrow CH_3-CH=CH-BH_2$. Subsequent oxidation with $H_2O_2/OH^{\Theta}$ yields the enol $CH_3-CH=CH-OH$,which tautomerizes to propanal,$CH_3-CH_2-CHO$ $(P)$.$2$. The second reaction is oxymercuration-demercuration of propyne: $CH_3-C \equiv CH + Hg(OAc)_2/H_2O \rightarrow CH_3-C(OH)=CH-HgOAc$. Reduction with $NaBH_4$ yields the enol $CH_3-C(OH)=CH_2$,which tautomerizes to acetone (propanone),$CH_3-CO-CH_3$ $(Q)$.$3$. $(P)$ is an aldehyde and $(Q)$ is a ketone. Both have the molecular formula $C_3H_6O$.$4$. Since they have the same molecular formula but different functional groups,$(P)$ and $(Q)$ are functional isomers.

$CH_3-C \equiv CH \xrightarrow[(ii) H_2O_2/OH^{\Theta}]{(i) BH_3-THF} (P)$
$CH_3-C \equiv CH \xrightarrow[(ii) NaBH_4]{(i) Hg(OAc)_2/H_2O} (Q)$
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$CH_3-C \equiv CH \xrightarrow[(ii) H_2O_2/OH^{\Theta}]{(i) BH_3-THF} (P)$$CH_3-C \equiv CH \xrightarrow[(ii) NaBH_4]{(i) Hg(OAc)_2/H_2O} (Q)$Choose the correct option