
Complete the following reaction:
\[{{C}_{2}}{{H}_{5}}N{{H}_{2}}\text{ + HN}{{\text{O}}_{2}}\text{ }\to \]
Answer
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Hint: Try to recall between primary aliphatic amines and nitrous acid. Similarly, Alkyl amines react with nitrous acid to form an alkyl diazonium ion. But, alkyl diazonium salts are highly unstable so they are not formed as the final product. Alkyl diazonium salts dissociate to form primary alcohol.
Complete answer:
We have to find the products formed when Ethyl amine reacts with nitrous acid.
Ethyl amine has an Amino group (NH2) which combines with Nitrous acid to form an Ethyl diazonium salt.
Let us understand the reaction between Ethylamine and Nitrous acid:
${{C}_{2}}{{H}_{5}}N{{H}_{2}}\text{ + HN}{{\text{O}}_{2}}\text{ }\to \text{ }{{C}_{2}}{{H}_{5}}N_{2}^{+}$
However, the ethyl diazonium salt is very unstable and cannot be retained for practical purposes.
We know that ${{N}_{2}}$ is an excellent leaving group. So, when hydrolysed with water at 273-278 K the formed ethyl diazonium ion dissociates itself to form one new product and two by-products. It dissociates into Ethyl alcohol, Nitrogen gas is liberated and water is formed as a by-product. The solution formed is clear.
The hydrolysis reaction of Ethyl diazonium ion is as follows:
${{C}_{2}}{{H}_{5}}N_{2}^{+}\text{ + }{{\text{H}}_{2}}O\text{ }\xrightarrow{273-278K}\text{ }{{C}_{2}}{{H}_{5}}OH\text{ + }{{\text{N}}_{2}}\uparrow \text{ + }{{\text{H}}_{2}}O$
Thus, Ethyl alcohol is produced as the final product with the liberation of N2 gas.
Final answer: ${{C}_{2}}{{H}_{5}}N{{H}_{2}}\text{ + HN}{{\text{O}}_{2}}\text{ }\to \text{ }{{C}_{2}}{{H}_{5}}OH\text{ + }{{\text{N}}_{2}}\uparrow \text{ + }{{\text{H}}_{2}}O$
Note:
Alkyl Diazonium ion is produced as the intermediate product for the reaction. Being an ion it can accept any functional group to form any final product. Do remember that the aliphatic amines and aromatic amines produce the same product with the same functional group. A primary amine would yield primary alcohol, secondary amine would yield second alcohol and aniline would yield phenol.
Complete answer:
We have to find the products formed when Ethyl amine reacts with nitrous acid.
Ethyl amine has an Amino group (NH2) which combines with Nitrous acid to form an Ethyl diazonium salt.
Let us understand the reaction between Ethylamine and Nitrous acid:
${{C}_{2}}{{H}_{5}}N{{H}_{2}}\text{ + HN}{{\text{O}}_{2}}\text{ }\to \text{ }{{C}_{2}}{{H}_{5}}N_{2}^{+}$
However, the ethyl diazonium salt is very unstable and cannot be retained for practical purposes.
We know that ${{N}_{2}}$ is an excellent leaving group. So, when hydrolysed with water at 273-278 K the formed ethyl diazonium ion dissociates itself to form one new product and two by-products. It dissociates into Ethyl alcohol, Nitrogen gas is liberated and water is formed as a by-product. The solution formed is clear.
The hydrolysis reaction of Ethyl diazonium ion is as follows:
${{C}_{2}}{{H}_{5}}N_{2}^{+}\text{ + }{{\text{H}}_{2}}O\text{ }\xrightarrow{273-278K}\text{ }{{C}_{2}}{{H}_{5}}OH\text{ + }{{\text{N}}_{2}}\uparrow \text{ + }{{\text{H}}_{2}}O$
Thus, Ethyl alcohol is produced as the final product with the liberation of N2 gas.
Final answer: ${{C}_{2}}{{H}_{5}}N{{H}_{2}}\text{ + HN}{{\text{O}}_{2}}\text{ }\to \text{ }{{C}_{2}}{{H}_{5}}OH\text{ + }{{\text{N}}_{2}}\uparrow \text{ + }{{\text{H}}_{2}}O$
Note:
Alkyl Diazonium ion is produced as the intermediate product for the reaction. Being an ion it can accept any functional group to form any final product. Do remember that the aliphatic amines and aromatic amines produce the same product with the same functional group. A primary amine would yield primary alcohol, secondary amine would yield second alcohol and aniline would yield phenol.
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