Ethanoic acid on heating with ammonia forms compound A which on treatment with bromine and sodium hydroxide gives compound B. Compound B on treatment with $NaN{O_2}$/dil.HCl gives compound C. The compounds A, B and C respectively are:
A. Ethanamide, methanamine, methanol
B. Propanamide, ethanamine, ethanol
C. N-ethylpropanamide, methylisonitrile, methanamine
D. Ethanamine, bromoethane, ethane diazonium chloride
Answer
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Hint: In the first reaction, the product which is formed is also known as carboxamide. In the second reaction, the name reaction used is Hofmann reaction. In the third reaction addition of water to the intermediate salt takes place.
Complete answer:The reaction of ethanoic acid with ammonia is shown below.
\[C{H_3}COOH + N{H_3}\xrightarrow{\Delta }C{H_3}CON{H_2}\]
In this reaction, ethanoic acid reacts with ammonia and on heating gives ethanamide.
The reaction of ethanamide with bromine and sodium hydroxide is shown below.
$C{H_3}CON{H_2} + B{r_2} + 3NaOH \to C{H_3}N{H_2}$
In this reaction ethanamide reacts with bromine and sodium hydroxide to form methanamine. This reaction is known as Hoffmann bromamide reaction. In this reaction a strong base is used as an alkali which attacks the amide and deprotonate it forming an anion. This reaction is generally used for converting primary amide into primary amine in which one carbon is less than the primary amide.
The reaction of methanamine with $NaN{O_2}$ and dil.HCl is shown below.
$C{H_3}N{H_2}\xrightarrow[{NaN{O_2}/dil.HCl}]{{HN{O_2}}}C{H_3}OH$
In this reaction, methanamine reacts with sodium nitrite and dil.HCl in presence of nitrous acid to form methanol. In this reaction the intermediate product formed is ethanediazonium salt which is highly unstable therefore it reacts with water to form alcohol.
Thus, compound A is Ethanamide, B is methenamine and C is methanol.
Therefore, the correct option is A.
Note:In the third reaction for the formation of alcohol from primary amine, the hydroxide ion of water is used as a nucleophile therefore the product generated is an alcohol unless chlorine is taken as the nucleophile then methyl chloride is formed.
Complete answer:The reaction of ethanoic acid with ammonia is shown below.
\[C{H_3}COOH + N{H_3}\xrightarrow{\Delta }C{H_3}CON{H_2}\]
In this reaction, ethanoic acid reacts with ammonia and on heating gives ethanamide.
The reaction of ethanamide with bromine and sodium hydroxide is shown below.
$C{H_3}CON{H_2} + B{r_2} + 3NaOH \to C{H_3}N{H_2}$
In this reaction ethanamide reacts with bromine and sodium hydroxide to form methanamine. This reaction is known as Hoffmann bromamide reaction. In this reaction a strong base is used as an alkali which attacks the amide and deprotonate it forming an anion. This reaction is generally used for converting primary amide into primary amine in which one carbon is less than the primary amide.
The reaction of methanamine with $NaN{O_2}$ and dil.HCl is shown below.
$C{H_3}N{H_2}\xrightarrow[{NaN{O_2}/dil.HCl}]{{HN{O_2}}}C{H_3}OH$
In this reaction, methanamine reacts with sodium nitrite and dil.HCl in presence of nitrous acid to form methanol. In this reaction the intermediate product formed is ethanediazonium salt which is highly unstable therefore it reacts with water to form alcohol.
Thus, compound A is Ethanamide, B is methenamine and C is methanol.
Therefore, the correct option is A.
Note:In the third reaction for the formation of alcohol from primary amine, the hydroxide ion of water is used as a nucleophile therefore the product generated is an alcohol unless chlorine is taken as the nucleophile then methyl chloride is formed.
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