Question

In the reaction sequence:
\[ \text{C}_2\text{H}_5\text{I} \xrightarrow{\text{Alcoholic KOH}} X \xrightarrow{\text{Br}_2/\text{CCl}_4} Y \xrightarrow{\text{KCN}} Z \xrightarrow{H_3O^+} A \] The product \(A\) is

• A. succinic acid
• B. malonic acid
• C. oxalic acid
• D. maleic acid

Hint:

Alkyl halide \(\xrightarrow{\text{alc. KOH}}\) alkene, then \(\text{Br}_2\) adds, then KCN replaces halogens with CN, and hydrolysis of CN gives dicarboxylic acid.

Answer 1

The Correct Option is A

Step 1: Identify \(X\).
\(\text{C}_2\text{H}_5\text{I}\) with alcoholic KOH undergoes dehydrohalogenation (elimination) to form ethene:
\[ \text{C}_2\text{H}_5\text{I} \xrightarrow{\text{alc. KOH}} \text{CH}_2=\text{CH}_2 \ (X) \] Step 2: Identify \(Y\).
Ethene adds bromine in presence of \(CCl_4\) to form 1,2-dibromoethane:
\[ \text{CH}_2=\text{CH}_2 \xrightarrow{\text{Br}_2/CCl_4} \text{BrCH}_2-\text{CH}_2\text{Br} \ (Y) \] Step 3: Identify \(Z\).
1,2-dibromoethane reacts with KCN and both bromines are replaced by CN groups, giving dicyanoethane:
\[ \text{BrCH}_2-\text{CH}_2\text{Br} \xrightarrow{\text{KCN}} \text{NCCH}_2-\text{CH}_2\text{CN} \ (Z) \] Step 4: Final hydrolysis to acid \(A\).
On acidic hydrolysis, both \(-CN\) groups convert to \(-COOH\):
\[ \text{NCCH}_2-\text{CH}_2\text{CN} \xrightarrow{H_3O^+} \text{HOOC-CH}_2-\text{CH}_2-\text{COOH} \] This is succinic acid.
Final Answer: \[ \boxed{\text{succinic acid}} \]