Question

Find the end product in the given reaction.
\[C{H_3}CON{H_2}\xrightarrow{{B{r_2} + KOH,\Delta }}X\xrightarrow{{CHC{l_3}/KOH}}Y\xrightarrow{{LiAl{H_4}}}Z\]
(A) \[C{H_3}C{H_2}N{H_2}\]
(B) \[C{H_3}N{H_2}\]
(C) \[{C_2}{H_5}{N^ + } \equiv {C^ - }\]
(D) \[{(C{H_3})_2}NH\]

Answer 1

Hint: In order to find the end product of the given reaction, we must be aware of the basic chemical reactions that are taking place. We should also be aware why the reagents are being used in that given chemical reaction.

Complete step by step answer:
- Let us move to the given question. Acetamide will react to Bromine and Potassium hydroxide to give methylamine. The by-products formed in this reaction are water, potassium bromide and potassium carbonate. Methylamine which is formed is the primary amine.
\[C{H_3}CON{H_2}\xrightarrow{{B{r_2} + KOH}}C{H_3}N{H_2} + {H_2}O + KBr + {K_2}C{O_3}\]
- This methylamine on heating with chloroform and potassium hydroxide to give methyl isocyanate. The by-products formed in this reaction are potassium chloride and water molecules. The reaction is given below for this reaction.
\[C{H_3}N{H_2}\xrightarrow{{CHC{l_3} + KOH}}C{H_3}N \equiv C + KCl + {H_2}O\]
- Methyl isocyanate reacts with Lithium aluminium hydride to give Dimethylamine. The lithium aluminium hydride is the reducing agent.
\[C{H_3}N \equiv C\xrightarrow{{LiAl{H_4}}}{(C{H_3})_2}NH\]
- Therefore, the end-product of this reaction is Dimethylamine. This dimethylamine that is formed is the secondary amine. The correct option is option “D” .

Note: While writing a chemical reaction, we should never forget to write the reagents. The reaction will be complete only when the reagents are written above the arrow. We can also mention conditions like temperature, pressure and heat in the arrow.