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Question: Acetamide on treatment with \(LiAl{H_4}\) forms: A.ethyl amine B.ethanoic acid C.ethanimine ...

Acetamide on treatment with LiAlH4LiAl{H_4} forms:
A.ethyl amine
B.ethanoic acid
C.ethanimine
D.none of these

Explanation

Solution

Amines are the alkyl or aryl-substituted derivatives of ammonia. The number of possible amines is very large as there are three replaceable hydrogen atoms in ammonia which can be replaced by alkyl or aryl groups. Amines are classified as primary, secondary, and tertiary amines depending on whether one, two, or three alkyl or aryl groups are respectively attached to the nitrogen atom.
NH3+RHRNH2+RHR2NH+RHR3NN{H_3}\xrightarrow[{ + R}]{{ - H}}RN{H_2}\xrightarrow[{ + R}]{{ - H}}{R_2}NH\xrightarrow[{ + R}]{{ - H}}{R_3}N
If the R is alkyl then the amine is an aliphatic amine. If the r is aryl then the amine is aromatic. An amine in both alkyl and aryl groups are present is called a mixed amine. Tetra-alkyl or Tetra-aryl ammonium salts are called quaternary salts.

Complete step by step answer:
Now we will discuss the given options one by one.
-The first option is ethylamine. Acid aids on treatment Lithium aluminium hydride (LiAlH4)\left( {LiAl{H_4}} \right) dissolved in ether form primary amines which are also known as ethylamine. The reaction involved in this process is written below:

CH3CONH2LiAlH4CH3CH2NH2+H2OC{H_3}CON{H_2}\xrightarrow{{LiAl{H_4}}}C{H_3}C{H_2}N{H_2} + {H_2}O
-The second option is ethanoic acid. Aldehydes on oxidation with acidified potassium dichromate or potassium permanganate form acids containing the same number of carbon atoms. The reaction involved in this process is written below:

CH3CHOK2Cr2O7/H+CH3COOHC{H_3}CHO\xrightarrow{{{K_2}C{r_2}{O_7}/{H^ + }}}C{H_3}COOH
-The third option is ethanamine. Ethanimine is prepared when acetaldehyde reacts with ammonia. The reaction involved in this process is written below:

CH3CHO+H2NH2OCH3CHNHC{H_3}CHO + {H_2}N\xrightarrow{{ - {H_2}O}}C{H_3}CHNH
After discussing we can conclude that acetamide on treatment with LiAlH4LiAl{H_4} gives ethylamine.

Thus, the correct option is (A).

Note: Structure of amines: - We know that ammonia has a pyramidal structure and its nitrogen atom is sp3s{p^3} hybridized. Like ammonia, all amines (primary, secondary or tertiary) are pyramidal in shape. Also since lp-bp repulsions are greater than bp-bp, the bond angle between any two H- atoms or alkyl groups decreases from 10928/{109^ \circ }{28^/} to 107{107^ \circ } in primary and secondary amines. But in tertiary amines, due to steric hindrance between the three bulky alkyl groups, the bond angle increases from 107{107^ \circ } in ammonia to 108{108^ \circ } in trimethylamine.