The percentage composition of an organic comp | vedclass

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(B) Step $1$: Determine the empirical formula of $A$.$C = 85.71 \% = \frac{85.71}{12} = 7.14$; $\frac{7.14}{7.14} = 1$$H = 14.29 \% = \frac{14.29}{1}

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$HO-CH_2-CH_2-CO_2H$

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(B) Step $1$: Determine the empirical formula of $A$.$C = 85.71 \% = \frac{85.71}{12} = 7.14$; $\frac{7.14}{7.14} = 1$$H = 14.29 \% = \frac{14.29}{1} = 14.29$; $\frac{14.29}{7.14} = 2$Empirical formula $= CH_2$.Step $2$: Determine the molecular formula of $A$.Molecular weight $= 2 	imes 	ext{vapour density} = 2 	imes 14 = 28$.$n = \frac{28}{14} = 2$.Molecular formula $= (CH_2)_2 = C_2H_4$ (Ethene).Step $3$: Reaction sequence.$A$ is $CH_2=CH_2$.$CH_2=CH_2 + Cl_2/H_2O ightarrow HO-CH_2-CH_2-Cl$ ($B$, Ethylene chlorohydrin).$HO-CH_2-CH_2-Cl + KCN ightarrow HO-CH_2-CH_2-CN$ (Nucleophilic substitution).$HO-CH_2-CH_2-CN + H_3O^+ ightarrow HO-CH_2-CH_2-COOH$ ($C$, $3$-hydroxypropanoic acid).

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 $C$.

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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 $C$.