Question

How can I convert formaldehyde to acetone?

Answer 1

First of all here we have to know that complete conversion of acetone takes place from a series of steps. So, we need to familiarize and separately write each step in order to convert from formaldehyde followed by a series of reactions to finally get acetone. Each step includes treatment of acetaldehyde with methyl magnesium bromide in presence of dry ether to produce magnesium complex. This magnesium complex on further hydrolysis with acid results in isopropyl alcohol which on acidification results to form acetone.

Complete solution:
We just shortly described the complete process in a hint. Now let us briefly describe each step of the whole process in a detailed manner.
Actually before starting the conversion of the first product formaldehyde as given in question , we have to prepare methyl magnesium bromide because in presence of this methyl magnesium bromide, formaldehyde is proceeding its first reaction. So, let us describe each step starting from this.
Step1: Preparing methyl magnesium bromide
First of all we have to prepare methyl magnesium bromide as mentioned above in presence of dry ether and the preparation is as follows,
$C{H_3}Br + Mg\mathop \to \limits^{dryether} C{H_3}MgBr$
Step2: Addition of Grignard to formaldehyde.
In this step it is preceded by addition of Grignard reagent or methyl magnesium bromide which is prepared above to formaldehyde and finally to form an alcohol in presence of ${H^ + }$ ion.
$C{H_2}( = O) + C{H_3}MgBr\mathop \to \limits^{{H^ + }} C{H_3}C{H_2}OH$
Step3: Ethanol is oxidized
Ethanol which is produced as a second step is then oxidized by an oxidation agent which is $PCC$(pyridinium chlorochromate) to form aldehyde. Another major point we have to know about this
oxidation agent $PCC$ is a milder version of chromic acid which oxidizes primary alcohols to aldehydes and does not oxidize aldehydes to carboxylic acids. The oxidation is as follows,
$C{H_3}C{H_2}OH\mathop \to \limits^{PCC} C{H_3}CH( = O)$
Step4: Addition of another mole of $C{H_3}MgBr$
In this step addition of Grignard reagent or methyl magnesium bromide is again done as in second step but here it is added to acetaldehyde produced in third step to form another alcohol in presence of ${H^ + }$ion. Addition is as follows,
$C{H_3}CH( = O) + C{H_3}MgBr\mathop \to \limits^{{H^ + }} C{H_3}CHOHC{H_3}$
Step5: propan- $2$ -ol is oxidized.
This is the final step to get the last product which is acetone as given in question. Here propan- $2$ -ol is oxidized to form acetone which is the final product in presence of potassium dichromate and ${H^ + }$ ion. Final oxidation is as follows,
$C{H_3}CHOHC{H_3}\mathop \to \limits^{{K_2}C{r_2}{O_{7,}}{H^ + }} C{H_3}C( = O)C{H_3}$
In this way we can convert formaldehyde to acetone.

Note: To deal with this question , actually we need to familiarize well with the above question to convert it easily. The main thing we need to remember here is we can convert any aldehyde rather than formaldehyde to acetone by properly following the above steps. In some steps the product is oxidized also, therefore we should have a clear idea about various oxidation agents to write the reaction in a correct manner because some products will be obtained only if the particular oxidation agent is added. This should be followed not only in the practical session but strictly on while writing reaction or conversion also.