Question

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

• A. \(CH_3-CH(OH)-CO_2H\)
• B. \(HO-CH_2-CH_2-CO_2H\)
• C. \(HO-CH_2-CO_2H\)
• D. \(CH_3-CH_2-CO_2H\)

Hint:

Addition of \(Cl_2/H_2O\) to alkene gives chlorohydrin. \(KCN\) increases carbon chain by 1, hydrolysis converts \(CN\) to \(COOH\).

Answer 1

The Correct Option is B

Step 1: Determine molecular formula of \(A\).
Assume \(100g\):
\[ C = 85.71g \Rightarrow \frac{85.71}{12} = 7.1425 \]
\[ H = 14.29g \Rightarrow \frac{14.29}{1} = 14.29 \]
Ratio:
\[ C:H = 7.1425:14.29 = 1:2 \]
Empirical formula:
\[ CH_2 \]
Step 2: Use vapour density to find molar mass.
\[ VD = 14 \Rightarrow M = 2 \times 14 = 28 \]
Empirical mass of \(CH_2 = 14\).
So molecular formula:
\[ C_2H_4 \]
Thus \(A = C_2H_4\) (ethene).
Step 3: Reaction with \(Cl_2/H_2O\).
Ethene gives chlorohydrin:
\[ C_2H_4 \xrightarrow{Cl_2/H_2O} HO-CH_2-CH_2-Cl \]
So \(B = HO-CH_2-CH_2-Cl\).
Step 4: Reaction with \(KCN\) followed by hydrolysis.
\[ HO-CH_2-CH_2-Cl \xrightarrow{KCN} HO-CH_2-CH_2-CN \]
\[ HO-CH_2-CH_2-CN \xrightarrow{H_3O^+} HO-CH_2-CH_2-CO_2H \]
So:
\[ C = HO-CH_2-CH_2-CO_2H \]
Final Answer:
\[ \boxed{HO-CH_2-CH_2-CO_2H} \]