Number of compounds which give reaction with Hinsberg's reagent is_______.
Correct Answer: 5
Solution and Explanation
Understanding Hinsberg's Test
Hinsberg's reagent (benzene sulfonyl chloride, \( C_6H_5SO_2Cl \)) is used to distinguish between primary, secondary, and tertiary amines:
Primary amines react to form sulfonamides, which are soluble in alkali.
Secondary amines react to form sulfonamides that are insoluble in alkali.
Tertiary amines do not react with Hinsberg's reagent.
Analyzing the Given Compounds
The compound containing \( \text{NH}_2 \) group (primary amine) reacts with Hinsberg's reagent.
The compound containing a secondary amine (\( -\text{NH}- \) group) also reacts.
Tertiary amines (\( -\text{N}- \)) do not react with Hinsberg's reagent.
Amides and other compounds that do not contain free primary or secondary amine groups will not react.
Counting the Reactive Compounds
After analyzing the given structures, there are \textbf{5 compounds} containing primary or secondary amines that will react with Hinsberg's reagent.
Conclusion
The number of compounds that give a reaction with Hinsberg's reagent is 5.
Top Questions on Amines
- Identify correct statements: (A) Primary amines do not give diazonium salts when treated with \({NaNO}_2\) in acidic condition.
(B) Aliphatic and aromatic primary amines on heating with \({CHCl}_3\) and ethanolic \({KOH}\) form carbylamines.
(C) Secondary and tertiary amines also give carbylamine test.
(D) Benzenesulfonyl chloride is known as Hinsberg’s reagent.
(E) Tertiary amines react with benzenesulfonyl chloride very easily.
Choose the correct answer from the options given below: Amines have a lone pair of electrons on the nitrogen atom, due to which they behave as Lewis bases. Greater the value of \( K_b \) or smaller the value of \( pK_b \), stronger is the base. Amines are more basic than alcohols, ethers, esters, etc. The basic character of aliphatic amines should increase with the increase of alkyl substitution. However, it does not occur in a regular manner, as a secondary aliphatic amine is unexpectedly more basic than a tertiary amine in aqueous solutions. Aromatic amines are weaker bases than ammonia and aliphatic amines. Electron-releasing groups such as \( -CH_3 \), \( -NH_2 \), etc., increase the basicity, while electron-withdrawing substituents such as \( -NO_2 \), \( -CN \), halogens, etc., decrease the basicity of amines. The effect of these substituents is more pronounced at the para-position than at the meta-position.
(a) Arrange the following in increasing order of their basic character. Give reason:
\[ \includegraphics[width=0.5\linewidth]{ch29i.png} \]- Amines have a lone pair of electrons on nitrogen atom due to which they behave as Lewis base. Greater the value of \( K_b \) or smaller the value of \( pK_b \), stronger is the base. Amines are more basic than alcohols, ethers, esters, etc. The basic character of aliphatic amines should increase with the increase of alkyl substitution. But it does not occur in a regular manner as a secondary aliphatic amine is unexpectedly more basic than a tertiary amine in aqueous solutions. Aromatic amines are weaker bases than ammonia and aliphatic amines. Electron releasing groups such as \(–CH_3\), \(–NH_2\), etc., increase the basicity while electron-withdrawing substituents such as \(–NO_2\), –CN, halogens, etc., decrease the basicity of amines. The effect of these substituents is more at the p₊ than at m₋ position.
- Assertion (A): Aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis. Reason (R): Aryl halides do not undergo nucleophilic substitution reaction with the anion formed by phthalimide.
- Give plausible explanation for each of the following:
Amides are less basic than amines.
Questions Asked in JEE Main exam
- Consider ‘n’ is the number of lone pair of electrons present in the equatorial position of the most stable structure of \( \text{ClF}_3 \). The ions from the following with ‘n’ number of unpaired electrons are:
A. \( \text{V}^{3+} \)
B. \( \text{Ti}^{3+} \)
C. \( \text{Cu}^{2+} \)
D. \( \text{Ni}^{2+} \)
E. \( \text{Ti}^{2+} \)
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