The percentage composition of an organic comp | vedclass

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(B) $1$. Determine the empirical formula of $A$:$C = 85.71 \% / 12 = 7.14$; $H = 14.29 \% / 1 = 14.29$.Ratio $C:H = 7.14 : 14.29 = 1 : 2$. Empirical f

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$HO-CH_2-CH_2-CO_2H$

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(B) $1$. Determine the empirical formula of $A$:$C = 85.71 \% / 12 = 7.14$; $H = 14.29 \% / 1 = 14.29$.Ratio $C:H = 7.14 : 14.29 = 1 : 2$. Empirical formula is $CH_2$.$2$. Determine the molecular formula of $A$:Molecular weight $= 2 	imes 	ext{vapour density} = 2 	imes 14 = 28$.$n = 28 / 14 = 2$. Molecular formula is $(CH_2)_2 = C_2H_4$ (ethene).$3$. Reaction sequence:$A$ is $CH_2=CH_2$.$CH_2=CH_2 + HOCl ightarrow HO-CH_2-CH_2-Cl$ ($B$ is $2-$chloroethanol).$HO-CH_2-CH_2-Cl + KCN ightarrow HO-CH_2-CH_2-CN + KCl$.$HO-CH_2-CH_2-CN + H_3O^+ ightarrow HO-CH_2-CH_2-COOH$ ($C$ is $3-$hydroxypropanoic acid).Therefore,the correct option is $B$.

The percentage composition of an organic compound $A$ is: carbon $85.71 \%$ and hydrogen $14.29 \%$. Its vapour density is $14$. Consider the following reaction sequence:
$A$ $\xrightarrow{Cl_2 / H_2O} B$ $\xrightarrow[(ii) H_3O^+]{(i) KCN / EtOH} C$
Identify $\underline{C}$.

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The percentage composition of an organic compound $A$ is: carbon $85.71 \%$ and hydrogen $14.29 \%$. Its vapour density is $14$. Consider the following reaction sequence:$A$ $\xrightarrow{Cl_2 / H_2O} B$ $\xrightarrow[(ii) H_3O^+]{(i) KCN / EtOH} C$Identify $\underline{C}$.