Question

Acid D formed in above reaction is :

• A. Gluconic acid
• B. Succinic acid
• C. Oxalic acid
• D. Malonic acid

Answer 1

Answer: (B)

Succinic acid

Answer 2

Step-by-step Reaction Analysis:
1.Reaction of C$_2$H$_5$Br with alcoholic KOH:
C$_2$H$_5$Br (ethyl bromide) undergoes dehydrohalogenation with alcoholic KOH to form ethene (CH$_2$ = CH$_2$).
This step is represented as:
$\text{C}_2\text{H}_5\text{Br} \xrightarrow{\text{alc. KOH}} \text{CH}_2 = \text{CH}_2 \; (\text{A})$
2. Addition of Br$_2$ in CCl$_4$:
Ethene reacts with bromine (Br$_2$) in the presence of carbon tetrachloride (CCl$_4$) to form 1,2-dibromoethane (BrCH$_2$CH$_2$Br).
This step is represented as:
$\text{CH}_2 = \text{CH}_2 \xrightarrow{\text{Br}_2/\text{CCl}_4} \text{BrCH}_2\text{CH}_2\text{Br} \; (\text{B})$
3. Reaction of 1,2-dibromoethane with excess KCN:
The compound BrCH$_2$CH$_2$Br reacts with excess potassium cyanide (KCN) to form 1,2-dicyanoethane (NCCH$_2$CH$_2$CN).
This step is represented as:
$\text{BrCH}_2\text{CH}_2\text{Br} \xrightarrow{\text{Excess KCN}} \text{NCCH}_2\text{CH}_2\text{CN} \; (\text{C})$
4. Hydrolysis of 1,2-dicyanoethane:
Hydrolysis of 1,2-dicyanoethane (NCCH$_2$CH$_2$CN) in the presence of excess H$_3$O$^+$ (acidic medium) yields succinic acid (HOOCCH$_2$CH$_2$COOH).
This step is represented as:
$\text{NCCH}_2\text{CH}_2\text{CN} \xrightarrow{\text{H}_3\text{O}^+\text{ (Excess)}} \text{HOOCCH}_2\text{CH}_2\text{COOH} \; (\text{D})$
Conclusion: The acid D formed is succinic acid.