
Complete the following reaction equation:
${C_2}{H_5}N{H_2} + {C_6}{H_5}S{O_2}Cl \to $
Answer
438k+ views
Hint: The chemical molecule ethylamine, commonly known as ethanamine, has the formula ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}_{\text{2}}}$. The odour of this colourless gas is similar to that of ammonia. It condenses to a liquid miscible with almost all solvents at slightly below room temperature. As is typical of amines, it is a nucleophilic base. In the chemical industry and organic synthesis, ethylamine is frequently used.
Complete answer:
The Hinsberg reaction is a test for primary, secondary, and tertiary amine detection. The amine is thoroughly shaken with Hinsberg reagent in the presence of aqueous alkali in this test (either KOH or NaOH). A substrate is treated with a reagent containing an aqueous sodium hydroxide solution and benzenesulfonyl chloride. A soluble sulfonamide salt is formed by a primary amine. The main amine's sulfonamide is precipitated after acidification of this salt. In the same process, a secondary amine forms an insoluble sulfonamide. A tertiary amine will stay insoluble if it does not react with the initial reagent (benzene sulfonyl chloride). Benzene sulphonyl chloride is Hinsberg's reagent. N-ethyl benzene sulphonamide is produced when ethylamine interacts with benzene sulphonyl chloride and is soluble in NaOH. Amines act as nucleophiles, engaging the electrophile sulfonyl chloride and displacing chloride. Sulfonamides formed by primary and secondary amines are weakly soluble and precipitate out of solution as solids.
\[PhS{O_2}Cl{\text{ }} + {\text{ }}2{\text{ }}RR'NH{\text{ }} \to {\text{ }}PhS{O_2}NRR'{\text{ }} + {\text{ }}\left[ {RR'N{H_2}^ + } \right]C{l^ - }\]
When benzene sulfonyl chloride reacts with primary amines, a sulfonamide product is formed that is alkali soluble.
${C_2}{H_5}N{H_2} + {C_6}{H_5}S{O_2}Cl \to {C_6}{H_5}S{O_2}NH{C_2}{H_5} + HCl$
Note:
When benzene sulfonyl chloride reacts with secondary amines, a sulfonamide product is formed that is not alkali soluble. Between a tertiary amine and the benzene sulfonyl chloride reagent, no such reaction occurs. Sulfonyl chloride hydrolysis is aided by tertiary amines. This reaction produces salts that are water soluble. As a result, the Hinsberg reagent reacts differentially with primary, secondary, and tertiary amines. These changes may be seen in the sulfonamide product's alkali solubility.
Complete answer:
The Hinsberg reaction is a test for primary, secondary, and tertiary amine detection. The amine is thoroughly shaken with Hinsberg reagent in the presence of aqueous alkali in this test (either KOH or NaOH). A substrate is treated with a reagent containing an aqueous sodium hydroxide solution and benzenesulfonyl chloride. A soluble sulfonamide salt is formed by a primary amine. The main amine's sulfonamide is precipitated after acidification of this salt. In the same process, a secondary amine forms an insoluble sulfonamide. A tertiary amine will stay insoluble if it does not react with the initial reagent (benzene sulfonyl chloride). Benzene sulphonyl chloride is Hinsberg's reagent. N-ethyl benzene sulphonamide is produced when ethylamine interacts with benzene sulphonyl chloride and is soluble in NaOH. Amines act as nucleophiles, engaging the electrophile sulfonyl chloride and displacing chloride. Sulfonamides formed by primary and secondary amines are weakly soluble and precipitate out of solution as solids.
\[PhS{O_2}Cl{\text{ }} + {\text{ }}2{\text{ }}RR'NH{\text{ }} \to {\text{ }}PhS{O_2}NRR'{\text{ }} + {\text{ }}\left[ {RR'N{H_2}^ + } \right]C{l^ - }\]
When benzene sulfonyl chloride reacts with primary amines, a sulfonamide product is formed that is alkali soluble.
${C_2}{H_5}N{H_2} + {C_6}{H_5}S{O_2}Cl \to {C_6}{H_5}S{O_2}NH{C_2}{H_5} + HCl$
Note:
When benzene sulfonyl chloride reacts with secondary amines, a sulfonamide product is formed that is not alkali soluble. Between a tertiary amine and the benzene sulfonyl chloride reagent, no such reaction occurs. Sulfonyl chloride hydrolysis is aided by tertiary amines. This reaction produces salts that are water soluble. As a result, the Hinsberg reagent reacts differentially with primary, secondary, and tertiary amines. These changes may be seen in the sulfonamide product's alkali solubility.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
