Hinsberg Test
In Hinsberg test, secondary amine gives product that is:
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More nitrogen organics Practice Questions
Hinsberg reagent (benzenesulfonyl chloride) is used to distinguish:
Hinsberg reagent (benzenesulfonyl chloride) is used to distinguish:
Gabriel phthalimide synthesis is used to prepare:
Gabriel phthalimide synthesis is used to prepare:
Reduction of nitrobenzene with Sn/HCl gives initially:
Reduction of nitrobenzene with Sn/HCl gives initially:
Diazonium salts couple with phenols in alkaline solution to form:
Diazonium salts couple with phenols in alkaline solution to form:
Diazotization of aniline performed at low temperature (~0–5°C) because:
Diazotization of aniline performed at low temperature (~0–5°C) because:
Replacement of diazonium with –OH occurs by heating aqueous solution yielding:
Replacement of diazonium with –OH occurs by heating aqueous solution yielding:
Carbylamine test requires chloroform and alcoholic KOH; characteristic product is:
Carbylamine test requires chloroform and alcoholic KOH; characteristic product is:
Hofmann bromamide reaction converts an amide to:
Hofmann bromamide reaction converts an amide to:
Aniline is less basic than methylamine due to:
Aniline is less basic than methylamine due to:
Amides are far less basic than amines because:
Amides are far less basic than amines because:
Gas phase basicity order (strongest base first) is typically:
Gas phase basicity order (strongest base first) is typically:
Treatment of benzenediazonium chloride with CuCN (Sandmeyer) yields:
Treatment of benzenediazonium chloride with CuCN (Sandmeyer) yields:
About This Question
- Subject
- chemistry
- Chapter
- nitrogen organics
- Topic
- hinsberg test
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Insoluble sulfonamide (does not dissolve)
- The Reagent: Benzenesulfonyl chloride (C6H5SO2Cl).
- Primary Amine (1∘): Forms a sulfonamide with a remaining H on the Nitrogen (R−NH−SO2Ph). This Hydrogen is acidic and reacts with NaOH to form a salt, making it soluble.
- Secondary Amine (2∘): Forms a sulfonamide with NO remaining H atoms on the Nitrogen (R2N−SO2Ph). Since there is no acidic proton, it cannot react with NaOH.
- Observation: The product is formed but remains insoluble in the alkaline medium.
- Tertiary Amine (3∘): Does not react at all with the reagent.
This medium difficulty chemistry question is from the chapter nitrogen organics, covering the topic of hinsberg test. It appeared in the 2025 exam.
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