The products Q and R in the following reactio | vedclass

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(C) $1$. The reaction of $CH_3-C \equiv CH$ (propyne) with $H_2$ in the presence of Lindlar's catalyst results in the formation of $CH_3-CH=CH_2$ (pro

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$Ethanal$,$Methanal$

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(C) $1$. The reaction of $CH_3-C \equiv CH$ (propyne) with $H_2$ in the presence of Lindlar's catalyst results in the formation of $CH_3-CH=CH_2$ (propene) as product $P$.$2$. The ozonolysis of propene $(CH_3-CH=CH_2)$ in the presence of $Zn/H_2O$ (reductive ozonolysis) leads to the cleavage of the double bond.$3$. The reaction is: $CH_3-CH=CH_2 + O_3 \xrightarrow{Zn/H_2O} CH_3CHO + HCHO$.$4$. Thus,the products $Q$ and $R$ are $CH_3CHO$ $(Ethanal)$ and $HCHO$ $(Methanal)$.

The products $Q$ and $R$ in the following reactions respectively are:
$H_2 + CH_3-C \equiv CH$ $\xrightarrow{\text{Lindlar's catalyst}} P$ $\xrightarrow[Zn + H_2O]{O_3} Q + R$

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The products $Q$ and $R$ in the following reactions respectively are:$H_2 + CH_3-C \equiv CH$ $\xrightarrow{\text{Lindlar's catalyst}} P$ $\xrightarrow[Zn + H_2O]{O_3} Q + R$