Question
Explain Wolf$-$Kishner reduction reaction. Write preparation of propanone by using ethanoyl chloride and dimethyl cadmium.
✓
Answer
The Wolf$-$Kishner reduction is a reaction that reduces the carbonyl group of aldehydes and ketones to a methylene group $(CH_2)$ on treatment with zinc$–$amalgam and hydrazine followed by heating with sodium or potassium hydroxide in high boiling solvent like ethylene glycol $($Wolf$-$Kishner reduction$).$
Wolf$-$Kishner reduction is used to synthesize straight chain alkyl substituted benzenes which is not possible by FriedelCrafts alkylation reaction.
The preparation of propanone $($acetone$)$ from ethanoyl chloride and dimethyl cadmium is a process that involves the reaction of ethanoyl chloride with dimethyl cadmium. In this reaction, two molecules of ethanoyl chloride react with one molecule of dimethyl cadmium to yield propanone and cadmium chloride as products. The reaction proceeds as follows:
$\underset{\text { (Ethanoyl chloride) }}{2 CH _3- COCl }+\underset{\text { (Dimethyl cadmium) }}{\left( CH _3\right)_2 Cd } \longrightarrow 2 CH _3-\underset{\text { Propanone(Acetone) }}{ CO - CH _3+ CdCl _2}$
This process allows for the formation of propanone, a ketone, through the acylation of the methyl group from dimethyl cadmium with ethanoyl chloride, followed by the elimination of cadmium chloride.
Wolf$-$Kishner reduction is used to synthesize straight chain alkyl substituted benzenes which is not possible by FriedelCrafts alkylation reaction.
The preparation of propanone $($acetone$)$ from ethanoyl chloride and dimethyl cadmium is a process that involves the reaction of ethanoyl chloride with dimethyl cadmium. In this reaction, two molecules of ethanoyl chloride react with one molecule of dimethyl cadmium to yield propanone and cadmium chloride as products. The reaction proceeds as follows:
$\underset{\text { (Ethanoyl chloride) }}{2 CH _3- COCl }+\underset{\text { (Dimethyl cadmium) }}{\left( CH _3\right)_2 Cd } \longrightarrow 2 CH _3-\underset{\text { Propanone(Acetone) }}{ CO - CH _3+ CdCl _2}$
This process allows for the formation of propanone, a ketone, through the acylation of the methyl group from dimethyl cadmium with ethanoyl chloride, followed by the elimination of cadmium chloride.
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