X of enthanamine was subjected to reaction with \(NaNO_2/HCl\) followed by hydrolysis to liberate \(N_2\) and HCl. The HCl generated was completely neutralised by 0.2 moles of NaOH. X is ____ g.
Correct Answer: 9
Solution and Explanation
The reaction sequence is as follows:
Step 1: Reaction of ethanamine with \(\text{NaNO}_2\) and \(\text{HCl}\)
Ethanamine (\(\text{CH}_3\text{CH}_2\text{NH}_2\)) reacts with \(\text{NaNO}_2\) and \(\text{HCl}\) to form an unstable diazonium salt (\(\text{CH}_3\text{CH}_2\text{N}_2^+\text{Cl}^-\)):
\[\text{CH}_3\text{CH}_2\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{N}_2^+\text{Cl}^- + \text{NaCl} + \text{H}_2\text{O}.\]
Step 2: Hydrolysis of the diazonium salt
The diazonium salt decomposes upon hydrolysis, liberating \(\text{N}_2\), \(\text{HCl}\), and ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)):
\[\text{CH}_3\text{CH}_2\text{N}_2^+\text{Cl}^- + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{N}_2 + \text{HCl}.\]
Step 3: Calculation of the mass of ethanamine
The reaction generates \(0.2\) moles of \(\text{HCl}\), which neutralizes \(0.2\) moles of \(\text{NaOH}\). From the stoichiometry of the reaction, 1 mole of ethanamine produces 1 mole of \(\text{HCl}\). Therefore, the moles of ethanamine (\(\text{CH}_3\text{CH}_2\text{NH}_2\)) required are \(0.2\) moles.
The molar mass of ethanamine is:
\[\text{Molar mass of } \text{CH}_3\text{CH}_2\text{NH}_2 = 12 + 3 + 3 + 12 + 2 + 14 + 1 = 44 \, \text{g/mol}.\]
The mass of ethanamine is:
\[\text{Mass} = \text{Moles} \times \text{Molar mass} = 0.2 \times 44 = 8.8 \, \text{g} \, (\text{Nearest Integer to } 9).\]
Final Answer: \(X = 9\).
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
- A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is \( -3 \), then the magnitude of the radius of curvature of the mirror is:
- JEE Main - 2025
- Spherical Mirrors
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to:
- JEE Main - 2025
- Differential Equations
- The number of natural numbers, between 212 and 999, such that the sum of their digits is 15, is _____.
- JEE Main - 2025
- Number Systems
- Let \( P_n = \alpha^n + \beta^n \), \( P_{10} = 123 \), \( P_9 = 76 \), \( P_8 = 47 \), and \( P_1 = 1 \). The quadratic equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
- JEE Main - 2025
- Sequences and Series
- If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:
- JEE Main - 2025
- Matrices and Determinants