
What is the action of following on but- $2$ -ene?
(A)Dil. Alkaline $KMn{O_4}$
(B) Acidic $KMn{O_4}$
Answer
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Hint: $KMn{O_4}$ is an inorganic catalyst. The name of this catalyst is Potassium permanganate. $KMn{O_4}$ has $2$ ions ${K^ + }$ and $Mn{O_4}^ - $ ions. The IUPAC name of $KMn{O_4}$ catalyst is Potassium manganate $(VII)$ . It is widely used in various chemical industries in several laboratories. It $KMn{O_4}$ is used in a reaction, and then it mostly acts as the oxidizing agent.
Complete answer:
For the first time potassium permanganate was synthesized by, Johann Rudolf Glauber in 1659, when he reacted manganese dioxide with potassium carbonate. The conformation of the reaction was done by a change in color. Initially, the color changes to violet and then to red. The synthesis reaction of the potassium permanganate is as follows:-
$2Mn{O_2}\, + \,4KOH\, + \,{O_2} \to \,2{K_2}Mn{O_4}\, + \,2{H_2}O$
$2{K_2}Mn{O_4}\, + \,2{H_2}O\, \to \,2KMn{O_4}\, + \,2KOH\, + \,{H_2}$
For all $Mn{O_2}$ is manganese dioxide, $KOH$ is potassium hydroxide, ${K_2}Mn{O_4}$ is potassium manganate and $KMn{O_4}$ is potassium permanganate.
But- $2$ -one is an organic compound having formulas as ${C_4}{H_8}$ . It exhibits the geometrical isomerism that means it exists in cis as well as in trans form. Having molar mass equal to $56.106\,g/mol$ . With melting point equals to $ - 138.9^\circ C$ and boiling point equals $0.8$ to $3.7^\circ C$
(A)In the presence of Dil. Alkaline $KMn{O_4}$ the product formed is di-alcohol. In the case of but- $2$ -ene, Butane- $2,3$ - diol is synthesized.
(B)In the presence of acidic $KMn{O_4}$ the product formed is a carboxylic acid. In the case of but- $2$ -ene, Ethanoic acid is synthesized. Before the final product, a ring intermediate is made which leads to all differences in the use of the different types of potassium permanganate.
Note:
It is to remember that two different catalysts will never produce the same products. As the catalysts are different, so the product would be. A catalyst would never itself undergo any short of changes; it only provides an alternative path for the reaction to take place.
Complete answer:
For the first time potassium permanganate was synthesized by, Johann Rudolf Glauber in 1659, when he reacted manganese dioxide with potassium carbonate. The conformation of the reaction was done by a change in color. Initially, the color changes to violet and then to red. The synthesis reaction of the potassium permanganate is as follows:-
$2Mn{O_2}\, + \,4KOH\, + \,{O_2} \to \,2{K_2}Mn{O_4}\, + \,2{H_2}O$
$2{K_2}Mn{O_4}\, + \,2{H_2}O\, \to \,2KMn{O_4}\, + \,2KOH\, + \,{H_2}$
For all $Mn{O_2}$ is manganese dioxide, $KOH$ is potassium hydroxide, ${K_2}Mn{O_4}$ is potassium manganate and $KMn{O_4}$ is potassium permanganate.
But- $2$ -one is an organic compound having formulas as ${C_4}{H_8}$ . It exhibits the geometrical isomerism that means it exists in cis as well as in trans form. Having molar mass equal to $56.106\,g/mol$ . With melting point equals to $ - 138.9^\circ C$ and boiling point equals $0.8$ to $3.7^\circ C$
(A)In the presence of Dil. Alkaline $KMn{O_4}$ the product formed is di-alcohol. In the case of but- $2$ -ene, Butane- $2,3$ - diol is synthesized.
(B)In the presence of acidic $KMn{O_4}$ the product formed is a carboxylic acid. In the case of but- $2$ -ene, Ethanoic acid is synthesized. Before the final product, a ring intermediate is made which leads to all differences in the use of the different types of potassium permanganate.
Note:
It is to remember that two different catalysts will never produce the same products. As the catalysts are different, so the product would be. A catalyst would never itself undergo any short of changes; it only provides an alternative path for the reaction to take place.
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