Question:
Which of the following reactions will not give a primary amine?
Which of the following reactions will not give a primary amine?
Show Hint
When reducing amides, nitriles, or using the Hofmann Bromamide reaction, primary amines are obtained. However, the reduction of isocyanides produces a secondary amine instead.
Updated On: Jan 22, 2025
- \( \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \)
- \( \text{CH}_3\text{CN} \xrightarrow{\text{LiAlH}_4} \)
- \( \text{CH}_3\text{NC} \xrightarrow{\text{LiAlH}_4} \)
- \( \text{CH}_3\text{CONH}_2 \xrightarrow{\text{LiAlH}_4} \)
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
The reactions and their respective products are as follows:
(a) Hofmann Bromamide Reaction: \[ \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{CH}_3\text{NH}_2. \] This reaction involves the conversion of an amide (\( \text{CONH}_2 \)) to a primary amine (\( \text{CH}_3\text{NH}_2 \)) with the loss of one carbon atom.
(b) Reduction of Nitriles: \[ \text{CH}_3\text{CN} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{NH}_2. \] Nitriles (\( \text{CN} \)) are reduced to primary amines (\( \text{CH}_3\text{CH}_2\text{NH}_2 \)) using lithium aluminium hydride (\( \text{LiAlH}_4 \)).
(c) Reduction of Isocyanides (Methyl Isocyanide): \[ \text{CH}_3\text{NC} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{NHCH}_3. \] Isocyanides (\( \text{NC} \)) are reduced to secondary amines (\( \text{CH}_3\text{NHCH}_3 \)) using lithium aluminium hydride. Hence, this reaction does not produce a primary amine.
(d) Reduction of Amides: \[ \text{CH}_3\text{CONH}_2 \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{NH}_2. \] Amides (\( \text{CONH}_2 \)) are reduced to primary amines (\( \text{CH}_3\text{CH}_2\text{NH}_2 \)) using lithium aluminium hydride. Conclusion: Among the given reactions, option (3) does not produce a primary amine; it produces a secondary amine instead. Final Answer: \[ \boxed{\text{CH}_3\text{NC} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{NHCH}_3 \text{ (Secondary Amine)}}. \]
(a) Hofmann Bromamide Reaction: \[ \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2/\text{KOH}} \text{CH}_3\text{NH}_2. \] This reaction involves the conversion of an amide (\( \text{CONH}_2 \)) to a primary amine (\( \text{CH}_3\text{NH}_2 \)) with the loss of one carbon atom.
(b) Reduction of Nitriles: \[ \text{CH}_3\text{CN} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{NH}_2. \] Nitriles (\( \text{CN} \)) are reduced to primary amines (\( \text{CH}_3\text{CH}_2\text{NH}_2 \)) using lithium aluminium hydride (\( \text{LiAlH}_4 \)).
(c) Reduction of Isocyanides (Methyl Isocyanide): \[ \text{CH}_3\text{NC} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{NHCH}_3. \] Isocyanides (\( \text{NC} \)) are reduced to secondary amines (\( \text{CH}_3\text{NHCH}_3 \)) using lithium aluminium hydride. Hence, this reaction does not produce a primary amine.
(d) Reduction of Amides: \[ \text{CH}_3\text{CONH}_2 \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{NH}_2. \] Amides (\( \text{CONH}_2 \)) are reduced to primary amines (\( \text{CH}_3\text{CH}_2\text{NH}_2 \)) using lithium aluminium hydride. Conclusion: Among the given reactions, option (3) does not produce a primary amine; it produces a secondary amine instead. Final Answer: \[ \boxed{\text{CH}_3\text{NC} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{NHCH}_3 \text{ (Secondary Amine)}}. \]
Was this answer helpful?
0
0
Top Questions on Amines
- The most reactive amine towards dilute hydrochloric acid is:
- Number of compounds which give reaction with Hinsberg's reagent is_______.
- Number of amine compounds from the following giving solids which are soluble in NaOH upon reaction with Hinsberg's reagent is _____.
- An amine $(\text{X})$ is prepared by ammonolysis of benzyl chloride. On adding $p$-toluenesulphonyl chloride to it, the solution remains clear.Molar mass of the amine $(\text{X})$ formed is _____ $\text{g mol}^{-1}$.(Given molar mass in $\text{g mol}^{-1}$: $\text{C} = 12, \, \text{H} = 1, \, \text{O} = 16, \, \text{N} = 14$)
- 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.
View More Questions
Questions Asked in MHT CET exam
- The area of the region \( \{(x, y): 0 \leq y \leq x^2 + 1, \, 0 \leq y \leq x + 1, \, 0 \leq x \leq 2\ \) is:}
- The solution of the differential equation \( \sqrt{1 - y^2} \, dx + x \, dy - \sin^{-1} y \, dy = 0 \) is:
- MHT CET - 2024
- Geometric Progression
- If the vector equation of the line \[ \frac{x - 2}{2} = \frac{2y - 5}{-3} = z + 1, \] is given by: \[ \vec{r} = \left(2\hat{i} + \frac{5}{2}\hat{j} - \hat{k}\right) + \lambda\left(2\hat{i} - \frac{3}{2}\hat{j} + p\hat{k}\right), \] then \( p \) is equal to:
- MHT CET - 2024
- Random Variables
- Let \( X \) be the discrete random variable representing the number (\( x \)) appeared on the face of a biased die when it is rolled. The probability distribution of \( X \) is as follows: \[ X = x: \quad 1, \, 2, \, 3, \, 4, \, 5, \, 6 \] \[ P(X = x): \quad 0.1, \, 0.15, \, 0.3, \, 0.25, \, k, \, k \] The variance of \( X \) is:
- MHT CET - 2024
- Differential equations
- Which of the following is correct?
- MHT CET - 2024
- Vector Algebra
View More Questions