
$A + C{S_2} + HgC{l_2}$ gives ${C_2}{H_5} - N = C = S$ . Thus, compound A is:
A) ${C_2}{H_5}N{H_2}$
B) ${C_2}{H_5}NH{C_2}{H_5}$
C) $C{H_3} - CH = NOH$
D) $C{H_3}C{H_2}N{O_2}$
Answer
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Hint: The reaction taking place here is Hofmann mustard oil reaction. Since, the given product is alkyl isothiocyanate, carefully analyze the reaction with the given options to find out which compound gives the desired product.
Complete step by step answer:
Let us first understand Hofmann mustard oil reaction;
This test is used to differentiate between primary, secondary and tertiary amines with the help of $C{S_2}$ and $HgC{l_2}$ .
Now, let us understand which amines give which type of product.
Primary amines react with $C{S_2}$ and $HgC{l_2}$ , they form a product which has a mustard oil like odour and the product is an alkyl isothiocyanate. For example,
$\,{C_2}{H_5}N{H_2} + C{S_2} + HgC{l_2} \to {C_2}{H_5} - N = C = S\,$
Secondary amines react with $C{S_2}$ and form dithiocarbamic acid, further on treatment with mercuric chloride they do not react.
${C_2}{H_5}NH{C_2}{H_5} + C{S_2} \to S = C(SH)N{({C_2}{H_5})_{ 2}}\xrightarrow{{HgC{l_2}}}$ no product.
Tertiary amines do not give this reaction as they do not have free hydrogen atoms to react with.
Thus, option A is the correct answer which is known as ethyl amine.
There is an alternative name test which is also used to distinguish between the primary, secondary and tertiary amines. The name of the test is Hinsberg’s test. In this test, the respective amines react with benzene sulphonyl chloride to give respective products.
When an amine is treated with benzene sulfonyl chloride in excess of potassium hydroxide, the following observations are seen:
A primary amine gives a clear solution which on acidification gives an insoluble compound.
A secondary amine gives an insoluble compound which does not dissolve in acid.
A tertiary amine gives an insoluble compound which dissolves in acid.
So, option A is the correct answer for this question.
Note:
Note that only primary amines give Hofmann mustard oil reaction. In this reaction, there is a black precipitate which is formed at the end of reaction which indicates the amine as primary amine. And this black precipitate is due to mercuric sulphate $\,HgS\,$.
Complete step by step answer:
Let us first understand Hofmann mustard oil reaction;
This test is used to differentiate between primary, secondary and tertiary amines with the help of $C{S_2}$ and $HgC{l_2}$ .
Now, let us understand which amines give which type of product.
Primary amines react with $C{S_2}$ and $HgC{l_2}$ , they form a product which has a mustard oil like odour and the product is an alkyl isothiocyanate. For example,
$\,{C_2}{H_5}N{H_2} + C{S_2} + HgC{l_2} \to {C_2}{H_5} - N = C = S\,$
Secondary amines react with $C{S_2}$ and form dithiocarbamic acid, further on treatment with mercuric chloride they do not react.
${C_2}{H_5}NH{C_2}{H_5} + C{S_2} \to S = C(SH)N{({C_2}{H_5})_{ 2}}\xrightarrow{{HgC{l_2}}}$ no product.
Tertiary amines do not give this reaction as they do not have free hydrogen atoms to react with.
Thus, option A is the correct answer which is known as ethyl amine.
There is an alternative name test which is also used to distinguish between the primary, secondary and tertiary amines. The name of the test is Hinsberg’s test. In this test, the respective amines react with benzene sulphonyl chloride to give respective products.
When an amine is treated with benzene sulfonyl chloride in excess of potassium hydroxide, the following observations are seen:
A primary amine gives a clear solution which on acidification gives an insoluble compound.
A secondary amine gives an insoluble compound which does not dissolve in acid.
A tertiary amine gives an insoluble compound which dissolves in acid.
So, option A is the correct answer for this question.
Note:
Note that only primary amines give Hofmann mustard oil reaction. In this reaction, there is a black precipitate which is formed at the end of reaction which indicates the amine as primary amine. And this black precipitate is due to mercuric sulphate $\,HgS\,$.
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