How can ethanol be converted to propanone?
Answer
624k+ views
Hint: Ethanol is a two-carbon aldehyde, and propanone is a three carbon ketone. The first step is to make a Carbon- Carbon bond, and the Grignard reagent would be the best choice.
Complete step by step answer:
Reaction of Ethanol with $Zn - Hg - HCl + $$\Delta $ to give out Ethane $\left( {CH3 - CH3} \right)$[ Clemmensen Reduction]
Reaction of Ethane with $CI2$ in presence of sunlight to give $CH3 - CH2 - CI$[ Clemmensen Reduction]
Reaction of $CH3 - CH2 + CI$ with $CH3 - CI$ in presence of dry ether and $Na$ to give Propane $\left( {CH3 + CH2 - CH3} \right)$[Wurtz Reaction]
Reaction of Propane with $CI2$ in presence of sunlight to give $CH3 - CH2 - CH2 - CI$
Reaction of $CH3 - CH2 - CH2 - CI$ with $alc\;KOH$ to give $CH3 - CH = CH2$
Reaction of $H2O$ with ${H^ + }$to give $CH3 - CH\left( {OH} \right) - CH3$
Oxidizing $CH3 - CH\left( {OH} \right) - CH3$ with $K2Cr2O7$ to give Ketone as the only product and we have $CH3 - CO - CH3$(propanone) .
Note: Aldehydes and ketones can be prepared by oxidation of primary and secondary alcohols. Oxidation of alcohols involves the formation of a Carbon – Oxygen double bond with cleavage of an O-H and C-H bonds.
Complete step by step answer:
Reaction of Ethanol with $Zn - Hg - HCl + $$\Delta $ to give out Ethane $\left( {CH3 - CH3} \right)$[ Clemmensen Reduction]
Reaction of Ethane with $CI2$ in presence of sunlight to give $CH3 - CH2 - CI$[ Clemmensen Reduction]
Reaction of $CH3 - CH2 + CI$ with $CH3 - CI$ in presence of dry ether and $Na$ to give Propane $\left( {CH3 + CH2 - CH3} \right)$[Wurtz Reaction]
Reaction of Propane with $CI2$ in presence of sunlight to give $CH3 - CH2 - CH2 - CI$
Reaction of $CH3 - CH2 - CH2 - CI$ with $alc\;KOH$ to give $CH3 - CH = CH2$
Reaction of $H2O$ with ${H^ + }$to give $CH3 - CH\left( {OH} \right) - CH3$
Oxidizing $CH3 - CH\left( {OH} \right) - CH3$ with $K2Cr2O7$ to give Ketone as the only product and we have $CH3 - CO - CH3$(propanone) .
Note: Aldehydes and ketones can be prepared by oxidation of primary and secondary alcohols. Oxidation of alcohols involves the formation of a Carbon – Oxygen double bond with cleavage of an O-H and C-H bonds.
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