Question

Ethylamine diethylamine cannot be differentiated by
A) Hinsberg test
B) Carbylamine Test
C) Iodoform test
D) Both (A) & (B)

Answer 1

Hint: Ethylamine and diethylamine both of them are amines containing ethyl groups. Ethylamine has one ethyl group and Diethylamine has two ethyl groups. That’s why ethylamine is primary amine & Diethylamine is secondary amine. Out of given reagents, identify the reagent who reacts with primary and secondary amine in the same way.

Complete answer:
To determine by which test we cannot differentiate between Ethylamine & diethylamine. First of all, we should know about the molecular formula of Ethylamine & diethylamine.
\[
  Ethyla\min e \to C{H_3} - C{H_2} - N{H_2} \\
  Diethyla\min e \to C{H_3} - C{H_2} - NH - C{H_2} - C{H_3} \\
 \]
In ethylamine nitrogen atom is directly bonded with only one carbon atom, hence it is primary amine, In the case of diethylamine nitrogen atom is bonded with two carbon atoms, hence it is secondary amine.

In carbylamine reaction: Primary amine (aliphatic as well as aromatic) react with chloroform on heating in the presence of ethanolic solution of KOH to form isocyanides.
\[R - N{H_2} + CHCL + 3KOH\underrightarrow \vartriangle R - N\underline{\underline \to } C + 3KCL + 3{H_2}O\]
This reaction is not given by secondary amine.

Thus, we can use carbylamine to differentiate between ethylamine and diethylamine. If the sample reacts with carbylamine to form isocyanide, then the sample is ethyl amine. But if the sample does not form isocyanate with carbylamine, then the sample is diethylamine.

In Hinsberg’s Test:
When Ethylamine is shaken with benzene sulfonyl chloride and aqueous KOH solution, it gives a clear solution.
\[
  {C_6}{H_5}S{O_2}CL + C{H_3}C{H_2}N{H_2} \to {C_6}{H_5}S{O_2}NHC{H_2}C{H_3} \\
  \dfrac{{KOH}}{{ - {H_2}O}}\left[ {{C_6}{H_5}S{O_2} - N - C{H_2}C{H_3}} \right]K + {H_2}O \\
 \]
Potassium Salt (Clear Solution)

Because Diethylamine is \[{2^o}\] amine, on similar treatment forms an insoluble substance.
\[{C_6}{H_5}S{O_2}CL + {C_2}{H_3}NH{C_2}{H_5} \to {C_6}{H_5}S{O_2}N{({c_2}{H_5})_2} + HCL\]
(\[N,N - \] Diethyl benzenesulfonamide) Insoluble in \[KOH\]

If we add wer sample to benzene sulfonyl chloride, and then add potassium hydroxide solution, and if a clear solution is obtained, then wer sample is ethylamine.If we add wer sample to benzene sulfonyl chloride, and then add potassium hydroxide solution, and if an insoluble substance is obtained, then wer sample is diethylamine.

So, we can say both ethylamine & diethylamine can be differentiated by carbylamine Hinsberg's test.
In Iodoform test the mechanism involves the addition of amine to iodo carbene, no alpha hydrogen should be present to the amine group, and otherwise amine formation takes place. But ethylamine & diethylamine both have alpha hydrogen. Thus, both will give negative iodoform tests.

Hence, the correct answer is option (C).

Note: Students may confuse identifying alpha hydrogen in case of amines. In amines the hydrogen of the carbon which is directly attacked by the nitrogen are alpha hydrogen, Hydrogen attached to nitrogen are not alpha hydrogen.