The product (B) of the reaction is:CH(CO_2H)= | vedclass

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

Question

(C) The reaction sequence is as follows:$1$. Reaction with $2 \ mole$ of $NaOH$ converts the dicarboxylic acid into its sodium salt:$CH(CO_2H)=CH(CO_2

Vedclass · Accepted Answer

$HC \equiv CH$

Vedclass · Answer

(C) The reaction sequence is as follows:$1$. Reaction with $2 \ mole$ of $NaOH$ converts the dicarboxylic acid into its sodium salt:$CH(CO_2H)=CH(CO_2H) + 2NaOH ightarrow CH(CO_2Na)=CH(CO_2Na) + 2H_2O$Here,$(A)$ is sodium maleate (or fumarate).$2$. Kolbe's electrolytic synthesis involves the electrolysis of the aqueous solution of the sodium salt of a dicarboxylic acid.$CH(CO_2Na)=CH(CO_2Na) \xrightarrow{	ext{electrolysis}} HC \equiv CH + 2CO_2 + 2NaOH + H_2$Thus,the product $(B)$ is ethyne $(HC \equiv CH)$.

The product $(B)$ of the reaction is:
$CH(CO_2H)=CH(CO_2H)$ $\xrightarrow[2 \ mole]{NaOH} (A)$ $\xrightarrow{\text{electrolysis}} (B)$

Similar Questions

The reaction sequence is given as follows:
Product $(B)$ is:

$CaC_2 + H_2O \to A \xrightarrow{H_2SO_4/HgSO_4} B$. Identify $A$ and $B$ in the given reaction.

The reagent which is used to distinguish between propene and propyne is

One mole of $1,2$-dibromopropane on treatment with $X$ moles of $NaNH_2$ followed by treatment with ethyl bromide gave a pentyne. The value of $X$ is

What is the end product of the following sequence of operations?
$CaC_2$ $\xrightarrow{H_2O} A$ $\xrightarrow[Hg^{2+}]{dil. H_2SO_4} B$ $\xrightarrow[H_2]{Ni} C$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The product $(B)$ of the reaction is:$CH(CO_2H)=CH(CO_2H)$ $\xrightarrow[2 \ mole]{NaOH} (A)$ $\xrightarrow{\text{electrolysis}} (B)$