Answer
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Hint: Propanoic acid has the formula \[C{{H}_{3}}C{{H}_{2}}COOH\]. It is used in producing polymers and chemicals. It is also used in industrial processes as well as in biological applications. Ethanoic acid is also known as acetic acid. It has the chemical formula of \[C{{H}_{3}}COOH\]. It has an acrid and sharp smell. The common name of acetic acid is vinegar.
Complete step-by-step answer:
Here we need to decrease one carbon atom and two hydrogen atoms from propanoic acid to get ethanoic acid. So to convert propanoic acid into the ethanoic acid we need to heat the propanoic acid with ammonia which in reaction produces an ammonium salt which is known as propionamide. Then we will react with the ammonium salt called propionamide with bromine and alcoholic or aqueous solution of potassium hydroxide or KOH to give the product ethylamine. So now one carbon atom has been reduced from the compound. This reaction is named as Hoffmann bromamide degradation reaction. We know that if we react the nitrous acid with the amine the production of alcohol takes place. So now we will treat the ethylamine with the nitrous acid. After all of these reactions then by oxidising the ethyl alcohol we will get the product ethanol which again undergoes oxidation reaction and gives the desired product that is ethanoic acid. Look at the following reactions :
We react propanoic acid with ammonia to give propionamide:
\[C{{H}_{3}}C{{H}_{2}}COOH\xrightarrow{N{{H}_{3}}/heat}C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}+{{H}_{2}}O\]
Now propionamide reacts with bromine and potassium hydroxide to give ethylamine:
\[C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}\xrightarrow{B{{r}_{2}}/KOH}C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}+KBr\]
Then the ethylamine reacts with nitrous acid to give ethyl alcohol:
\[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\xrightarrow{HONO}C{{H}_{3}}C{{H}_{2}}OH+{{N}_{2}}+{{H}_{2}}O\]
Now the ethyl alcohol undergoes oxidation to form ethanal:
\[C{{H}_{3}}C{{H}_{2}}OH\xrightarrow{oxidation}C{{H}_{3}}C{{H}_{2}}CHO\]
Now the ethanol undergoes oxidation to form ethanoic acid:
\[C{{H}_{3}}C{{H}_{2}}CHO\xrightarrow{oxidation}C{{H}_{3}}COOH\]\[\]
Note: If we need to convert the ethanoic acid into propanoic acid then we need to treat the acid with the \[LiAl{{H}_{4}}\]to get ethyl alcohol. Then treat the ethyl alcohol with phosphorus trichloride to get ethyl iodide. Then react it with potassium cyanide to get ethyl nitrite which undergoes the hydrolysis process and then we get the propanoic acid.
Complete step-by-step answer:
Here we need to decrease one carbon atom and two hydrogen atoms from propanoic acid to get ethanoic acid. So to convert propanoic acid into the ethanoic acid we need to heat the propanoic acid with ammonia which in reaction produces an ammonium salt which is known as propionamide. Then we will react with the ammonium salt called propionamide with bromine and alcoholic or aqueous solution of potassium hydroxide or KOH to give the product ethylamine. So now one carbon atom has been reduced from the compound. This reaction is named as Hoffmann bromamide degradation reaction. We know that if we react the nitrous acid with the amine the production of alcohol takes place. So now we will treat the ethylamine with the nitrous acid. After all of these reactions then by oxidising the ethyl alcohol we will get the product ethanol which again undergoes oxidation reaction and gives the desired product that is ethanoic acid. Look at the following reactions :
We react propanoic acid with ammonia to give propionamide:
\[C{{H}_{3}}C{{H}_{2}}COOH\xrightarrow{N{{H}_{3}}/heat}C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}+{{H}_{2}}O\]
Now propionamide reacts with bromine and potassium hydroxide to give ethylamine:
\[C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}\xrightarrow{B{{r}_{2}}/KOH}C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}+KBr\]
Then the ethylamine reacts with nitrous acid to give ethyl alcohol:
\[C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}\xrightarrow{HONO}C{{H}_{3}}C{{H}_{2}}OH+{{N}_{2}}+{{H}_{2}}O\]
Now the ethyl alcohol undergoes oxidation to form ethanal:
\[C{{H}_{3}}C{{H}_{2}}OH\xrightarrow{oxidation}C{{H}_{3}}C{{H}_{2}}CHO\]
Now the ethanol undergoes oxidation to form ethanoic acid:
\[C{{H}_{3}}C{{H}_{2}}CHO\xrightarrow{oxidation}C{{H}_{3}}COOH\]\[\]
Note: If we need to convert the ethanoic acid into propanoic acid then we need to treat the acid with the \[LiAl{{H}_{4}}\]to get ethyl alcohol. Then treat the ethyl alcohol with phosphorus trichloride to get ethyl iodide. Then react it with potassium cyanide to get ethyl nitrite which undergoes the hydrolysis process and then we get the propanoic acid.
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