CH_3-CH(CO_2K)-CH(CO_2K)-CH_3 xrightarrow{tex | vedclass

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

Question

(C) In Kolbe's electrolysis,the electrolysis of potassium salts of vicinal dicarboxylic acids results in the formation of an alkene.The reaction proce

Vedclass · Accepted Answer

trans-But$-2-$ene

Vedclass · Answer

(C) In Kolbe's electrolysis,the electrolysis of potassium salts of vicinal dicarboxylic acids results in the formation of an alkene.The reaction proceeds via a free radical mechanism:$CH_3-CH(CO_2K)-CH(CO_2K)-CH_3 \xrightarrow{	ext{electrolysis}} CH_3-CH=CH-CH_3 + 2CO_2 + H_2 + 2KOH$Between the possible geometric isomers,the trans-isomer is more stable due to reduced steric hindrance between the methyl groups.Thus,the major product is trans-But$-2-$ene.

$CH_3-CH(CO_2K)-CH(CO_2K)-CH_3 \xrightarrow{\text{electrolysis}} (A) \text{ (Major)}$
The major product $(A)$ of the above reaction is:

Similar Questions

The correct statement about $C$ and $D$ is:
$CH_3-C \equiv C-CH_3$ $\xrightarrow{H_2/Pd-BaSO_4} A$ $\xrightarrow{Br_2/CCl_4} C$
$CH_3-C \equiv C-CH_3$ $\xrightarrow{Na/Liq. NH_3} B$ $\xrightarrow{Br_2/H_2O} D$

Product $(A)$ is

In the following sequence of reactions,the alkene affords the compound $B$.
$CH_3-CH=CH-CH_3$ $\xrightarrow{O_3} A$ $\xrightarrow[Zn]{H_2O} B$
The compound $B$ is:

Explain the oxidation of alkenes in different situations with examples.

The correct stability order of alkyl substituted alkenes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$CH_3-CH(CO_2K)-CH(CO_2K)-CH_3 \xrightarrow{\text{electrolysis}} (A) \text{ (Major)}$The major product $(A)$ of the above reaction is: