
R is:
A. sulphanilamide
B. p-bromo sulphanilamide
C. o-bromo sulphanilic acid
D. sulphanilic acid
Answer
152.7k+ views
Hint: The reaction in the question involves the conversion of a carboxylic acid into an amide. Following this, Hoffmann bromamide reaction takes place in which the amide converts into an amine.
Complete step by step answer: Let us now look at what is happening in the given reaction step by step.
Firstly, benzoic acid is heated in the presence of ammonia. As a result, it produces the product P. When benzoic acid is reacted with ammonia, it reacts to give ammonium benzoate. Ammonium benzoate is a salt of ammonia with benzoic acid.
On heating ammonium benzoate again, it loses a molecule of water. As a result, it forms benzamide. Therefore, the product P is benzamide.
After this, the benzamide is treated with NaOBr. Here, the reaction that will take place is Hoffmann Bromamide reaction. In this, the primary amide converts into a primary amine which has one less carbon atom than the corresponding amide. This takes place when the amide is heated with in the existence of NaOH which is NaOBr. Thus, benzamide reacts with NaOBr to give product aniline as the primary amine. Hence, the product Q is aniline.
After this, aniline is treated with conc. to form a salt of aniline and sulphide. This process is known as sulphonation. Here, aniline reacts with conc. to produce aniline hydrogen sulphide.
This aniline hydrogen sulphide formed is heated at 460K to form sulphanilic acid. Therefore, the product R formed is sulphanilic acid.
Hence, the correct answer is option (D).
Note: After the sulphonation of aniline, the product formed, that is, aniline hydrogen sulphide, the sulphonate group is present at the para position. This is due to the steric hindrance which is caused when the sulphonate group approaches the ortho position.
Complete step by step answer: Let us now look at what is happening in the given reaction step by step.
Firstly, benzoic acid is heated in the presence of ammonia. As a result, it produces the product P. When benzoic acid is reacted with ammonia, it reacts to give ammonium benzoate. Ammonium benzoate is a salt of ammonia with benzoic acid.
On heating ammonium benzoate again, it loses a molecule of water. As a result, it forms benzamide. Therefore, the product P is benzamide.
After this, the benzamide is treated with NaOBr. Here, the reaction that will take place is Hoffmann Bromamide reaction. In this, the primary amide converts into a primary amine which has one less carbon atom than the corresponding amide. This takes place when the amide is heated with
After this, aniline is treated with conc.
This aniline hydrogen sulphide formed is heated at 460K to form sulphanilic acid. Therefore, the product R formed is sulphanilic acid.
Hence, the correct answer is option (D).
Note: After the sulphonation of aniline, the product formed, that is, aniline hydrogen sulphide, the sulphonate group is present at the para position. This is due to the steric hindrance which is caused when the sulphonate group approaches the ortho position.
Latest Vedantu courses for you
Grade 9 | CBSE | SCHOOL | English
Vedantu 9 CBSE Pro Course - (2025-26)
School Full course for CBSE students
₹37,300 per year
Recently Updated Pages
JEE Main 2022 (June 29th Shift 2) Maths Question Paper with Answer Key

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 29th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 26th Shift 2) Chemistry Question Paper with Answer Key

JEE Main 2022 (June 26th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (June 29th Shift 1) Physics Question Paper with Answer Key

Trending doubts
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electrical Field of Charged Spherical Shell - JEE

Ideal and Non-Ideal Solutions Raoult's Law - JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Classification of Drugs

Compressibility Factor Z | Plot of Compressibility Factor Z Vs Pressure for JEE

Other Pages
Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

JEE Main Chemistry Online Mock Test for Class 12

NCERT Solutions for Class 12 Chemistry In Hindi Chapter 10 Haloalkanes and Haloarenes In Hindi Mediem

Collision - Important Concepts and Tips for JEE

Brief Information on Alpha, Beta and Gamma Decay - JEE Important Topic

NCERT Solutions for Class 12 Chemistry Chapter 9 Amines
