Kolbe electrolysis of potassium succinate giv | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Kolbe electrolysis of potassium succinate involves the decarboxylation of the salt of a dicarboxylic acid.Potassium succinate is $K_2C_4H_4O_4$ (o

Vedclass · Accepted Answer

$C_2H_4, \, KOH$ and $H_2$

Vedclass · Answer

(C) Kolbe electrolysis of potassium succinate involves the decarboxylation of the salt of a dicarboxylic acid.Potassium succinate is $K_2C_4H_4O_4$ (or $KOOC-CH_2-CH_2-COOK$).The reaction is: $K_2C_4H_4O_4 + 2H_2O \xrightarrow{	ext{electrolysis}} C_2H_4 + 2CO_2 + 2KOH + H_2$.Thus,the products are ethene $(C_2H_4)$,carbon dioxide $(CO_2)$,potassium hydroxide $(KOH)$,and hydrogen gas $(H_2)$.

Kolbe electrolysis of potassium succinate gives $CO_2$ and

Similar Questions

The propene reacts with $HBr$ to form

What is the number of structural isomers possible for alkene with molecular formula $C_5H_{10}$?

Out of the following,the alkene that exhibits optical isomerism is

The product obtained when $2-$methylpropan$-2-$ol is treated with alumina $(Al_2O_3)$ at $423 \ K$ is

$LDA = \text{Lithium di-isopropyl amide}$. In the given reaction,identify product $A$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module