Kolbe’s synthesis of sodium salt of butanoic acid gives:
a) n-hexane
b) Iso-butane
c) n-butane
d) Propane
Answer
561.3k+ views
Hint: Kolbe’s electrolysis is discovered by Herman Kolbe. It involves the electrolysis of monocarboxylic acid salts of sodium at the different electrodes. In this reaction carbon dioxide gas is evolved at the anode.
Complete step by step solution:
Kolbe’s electrolysis reaction can be termed as the sodium or potassium salts of monocarboxylic acids which on electrolysis produce alkanes.
This reaction results in the formation of symmetrical alkanes which have a higher number of carbon atoms as compared to the parent carboxylic acid salt.
Unsymmetrical alkanes cannot be synthesised by this electrolysis method as electrolysis of two different carboxylic acid salts results in the formation of mixtures of alkanes which are very difficult to separate in the chemical laboratories.
The generalised reaction of Kolbe’s electrolysis can be represented as-
$ 2RCOONa(aq) \to R - R + 2C{O_2} + NaOH + {H_2} $
Similarly by using the above template, the reaction for butanoic acid can be written as the following-
$ 2C{H_3}C{H_2}C{H_2}COONa(aq) \to C{H_3}C{H_2}C{H_2}C{H_2}C{H_2}C{H_3} + 2C{O_2} + NaOH + {H_2} $
$ Sodium Salt - bu\tan oic acid $ $ n - hexane $
During the electrolysis, carboxylate ion loses one electron and undergoes decarboxylation which produces symmetrical alkane at the anode.
The general mechanism of the reaction can be represented as-
At anode:
$ RCO{O^ - } \to RCO{O^.} + {e^ - } \to {R^.} + C{O_2} \\
2{R^.} \to R - R(alkane) \\
$
At cathode: $ N{a^ + } + {e^ - } \to Na + {H_2}O \to NaOH + \dfrac{1}{2}{H_2} $
In general we can say it is a coupling reaction of two free radicals form at the anode of the reaction.
The correct option is (a).
Note:
n-hexane is the straight chain of hydrocarbon. It appeared as a colourless liquid and odourless. It has a boiling point of $ {69^\circ }C $ . As chemical reactions need low boiling solvents to easily remove them after reaction so it is best suited as a non-polar solvent in many chemical reactions.
Complete step by step solution:
Kolbe’s electrolysis reaction can be termed as the sodium or potassium salts of monocarboxylic acids which on electrolysis produce alkanes.
This reaction results in the formation of symmetrical alkanes which have a higher number of carbon atoms as compared to the parent carboxylic acid salt.
Unsymmetrical alkanes cannot be synthesised by this electrolysis method as electrolysis of two different carboxylic acid salts results in the formation of mixtures of alkanes which are very difficult to separate in the chemical laboratories.
The generalised reaction of Kolbe’s electrolysis can be represented as-
$ 2RCOONa(aq) \to R - R + 2C{O_2} + NaOH + {H_2} $
Similarly by using the above template, the reaction for butanoic acid can be written as the following-
$ 2C{H_3}C{H_2}C{H_2}COONa(aq) \to C{H_3}C{H_2}C{H_2}C{H_2}C{H_2}C{H_3} + 2C{O_2} + NaOH + {H_2} $
$ Sodium Salt - bu\tan oic acid $ $ n - hexane $
During the electrolysis, carboxylate ion loses one electron and undergoes decarboxylation which produces symmetrical alkane at the anode.
The general mechanism of the reaction can be represented as-
At anode:
$ RCO{O^ - } \to RCO{O^.} + {e^ - } \to {R^.} + C{O_2} \\
2{R^.} \to R - R(alkane) \\
$
At cathode: $ N{a^ + } + {e^ - } \to Na + {H_2}O \to NaOH + \dfrac{1}{2}{H_2} $
In general we can say it is a coupling reaction of two free radicals form at the anode of the reaction.
The correct option is (a).
Note:
n-hexane is the straight chain of hydrocarbon. It appeared as a colourless liquid and odourless. It has a boiling point of $ {69^\circ }C $ . As chemical reactions need low boiling solvents to easily remove them after reaction so it is best suited as a non-polar solvent in many chemical reactions.
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