Question

Acid D formed in above reaction is :

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

Answer 1

The Correct Option is B

To solve the given problem and identify the acid D formed in the reaction, we will analyze each step of the reaction:

1. First Step:  
   The reaction starts with \(C_2H_5Br\) reacting with alcoholic KOH. This is a dehydrohalogenation reaction, which leads to the formation of an alkene: 
   \(C_2H_5Br \xrightarrow[]{alc.\,KOH} C_2H_4\) (Ethylene)
2. Second Step: 
   Ethylene reacts with \(Br_2\) in carbon tetrachloride \((CCl_4)\), resulting in vicinal dibromide: 
   \(C_2H_4 + Br_2 \xrightarrow[]{CCl_4} BrCH_2CH_2Br\) (1,2-Dibromoethane)
3. Third Step: 
   1,2-Dibromoethane reacts with excess KCN which leads to substitution reaction replacing bromine with cyanide group: 
   \(BrCH_2CH_2Br + 2KCN \rightarrow NCCH_2CH_2CN\) (Ethylene cyanide)
4. Fourth Step: 
   Hydrolysis of Ethylene cyanide with excess water in the presence of an acid (H₃O⁺) converts the nitrile groups to carboxylic acid: 
   \(NCCH_2CH_2CN \xrightarrow[]{H_3O^+} HOOCCH_2CH_2COOH\) (Succinic acid)

Thus, the acid D formed in the above reaction is Succinic acid.

The correct answer is 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.