Question

Convert Propanoic acid into Ethanoic acid?

Answer 1

Conversion of Propanoic acid into Ethanoic acid:

Step1: Formation of Propionamide:

Propanoic acid when undergoes a reaction with an excess amount of Ammonia gives Propanamide.
$\underset{\left(\mathrm{Propanoic}\mathrm{acid}\right)}{\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}}\stackrel{\mathrm{N}{\mathrm{H}}_3/\mathrm{heat}}{\to }\underset{\left(\mathrm{Propionamide}\right)}{\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CON}{\mathrm{H}}_2}+\underset{\left(\mathrm{Water}\right)}{{\mathrm{H}}_2\mathrm{O}}$
Step 2: Formation of Ethylamine:

Propanamide on further reacting with Bromine and Potassium hydroxide (Hoffmann bromamide degradation reaction) results in Ethylamine.
$\underset{\left(\mathrm{Propionamide}\right)}{\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CON}{\mathrm{H}}_2}\stackrel{\mathrm{B}{\mathrm{r}}_2/\mathrm{KOH}}{\to }\underset{\left(\mathrm{Ethylamine}\right)}{\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2}+\underset{\left(\mathrm{Potassium}\mathrm{Bromide}\right)}{\mathrm{KBr}}$
Step 3: Formation of Ethyl alcohol:

Ethylamine converts undergo diazotization to give Ethanol.
$\underset{\left(\mathrm{Ethylamine}\right)}{\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2}\stackrel{\mathrm{HONO}}{\to }\underset{\left(\mathrm{Ethyl}\mathrm{alcohol}\right)}{\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}}+\underset{\left(\mathrm{Nitrogen}\right)}{{\mathrm{N}}_2}+\underset{\left(\mathrm{water}\right)}{{\mathrm{H}}_2\mathrm{O}}$
Step 4: Formation of Ethanal:

Ethyl alcohol when undergoes oxidation, it forms Ethanal.
$\underset{\left(\mathrm{Ethyl}\mathrm{alcohol}\right)}{\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}}\stackrel{\mathrm{oxidation}}{\to }\underset{\left(\mathrm{Ethanal}\right)}{\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}}$

Step 5: Formation of Ethanoic acid:

At last, ethanal on oxidation gives ethanoic acid as the final product.

$\underset{\left(\mathrm{Ethanal}\right)}{{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHO}}\stackrel{\mathrm{oxidation}}{\to }\underset{\left(\mathrm{Ethanoic}\mathrm{acid}\right)}{{\mathrm{CH}}_3\mathrm{COOH}}$
Hence, the conversion of Propanoic acid into Ethanoic acid has been explained above.