Problem 100
Question
A positive carbylamine test is given by (1) \(\mathrm{N}, \mathrm{N}\)-dimethylaniline (2) 2,4 -dimethylaniline (3) N-methyl-o-methylaniline (4) p-methylbenzylamine (a) 2,4 (b) 2,3 (c) \(1,2,4\) (d) \(2,3,4\)
Step-by-Step Solution
Verified Answer
The answer is (a) 2,4.
1Step 1: Understand the Carbylamine Reaction
The carbylamine reaction, also known as the Hofmann isocyanide synthesis, occurs when a primary amine reacts with chloroform in the presence of alcoholic KOH to produce an isocyanide, which has a foul odor. This reaction is specific to primary amines.
2Step 2: Evaluate Each Compound for Primary Amines
We need to determine which compounds have primary amine groups (\(-\mathrm{NH}_2\)). 1. \(\mathrm{N}, \mathrm{N}\)-dimethylaniline has a secondary amine group; it's not a primary amine.2. 2,4-dimethylaniline has a primary amine group at the aniline nitrogen.3. \(\mathrm{N}\)-methyl-o-methylaniline has a secondary amine group; it's not a primary amine.4. p-methylbenzylamine has a primary amine group at the benzyl position.
3Step 3: Identify Which Options Have Primary Amines
From the evaluation, we see that compounds 2 and 4 have primary amines and can give a positive carbylamine test.
4Step 4: Compare with the Options Provided
Now, we need to match our findings with the provided options:
(a) 2,4 — Matches our findings.
(b) 2,3 — Compound 3 is not a primary amine.
(c) 1,2,4 — Compound 1 is not a primary amine.
(d) 2,3,4 — Compound 3 is not a primary amine.
Key Concepts
Hofmann isocyanide synthesisprimary amineschloroform and alcoholic KOH
Hofmann isocyanide synthesis
The Hofmann isocyanide synthesis, also known as the carbylamine reaction, is a chemical reaction that produces isocyanides or carbylamines. Isocyanides are recognizable by their distinctive, unpleasant odor. This reaction specifically involves primary amines. When a primary amine reacts with chloroform \((\text{CHCl}_3)\), in the presence of a strong base like alcoholic potassium hydroxide (KOH), it forms the isocyanide compound.
What makes this reaction special is its selectivity for primary amines. Secondary and tertiary amines do not undergo this reaction. Thus, this reaction serves as a useful test for the presence of primary amines in a compound.
What makes this reaction special is its selectivity for primary amines. Secondary and tertiary amines do not undergo this reaction. Thus, this reaction serves as a useful test for the presence of primary amines in a compound.
- Primary amine + CHCl\(_3\) + alcoholic KOH → Isocyanide (Carbylamine) with a foul smell
primary amines
Primary amines are crucial in organic chemistry due to their unique reactivity patterns. Structurally, a primary amine contains one alkyl or aryl group attached to the nitrogen atom, with the general formula R-NH\(_2\). This makes them significantly different from secondary and tertiary amines, where the nitrogen atom is bonded to two and three organic groups, respectively.
In the context of the Hofmann isocyanide synthesis or the carbylamine reaction, primary amines are the only class of amines that engage in the reaction to form isocyanides. Their ability to participate in this reaction hinges on the presence of the hydrogen atoms bonded directly to the nitrogen atom.
Primary amines are often foundational in numerous chemical syntheses and are found in many biologically active molecules and pharmaceuticals.
In the context of the Hofmann isocyanide synthesis or the carbylamine reaction, primary amines are the only class of amines that engage in the reaction to form isocyanides. Their ability to participate in this reaction hinges on the presence of the hydrogen atoms bonded directly to the nitrogen atom.
Primary amines are often foundational in numerous chemical syntheses and are found in many biologically active molecules and pharmaceuticals.
chloroform and alcoholic KOH
Chloroform and alcoholic KOH are the essential reagents in the carbylamine reaction. Chloroform \(\text{CHCl}_3\) is utilized for its role as a halogenated solvent that effectively interacts with the amine and alcoholic KOH.
The use of potassium hydroxide (KOH) in an alcoholic solution (often ethanol) turns KOH into a potent base. This strong base assists in deprotonating the amine, enabling the production of a carbanion intermediate from chloroform.
The use of potassium hydroxide (KOH) in an alcoholic solution (often ethanol) turns KOH into a potent base. This strong base assists in deprotonating the amine, enabling the production of a carbanion intermediate from chloroform.
- KOH (alcoholic form) turns into a strong base
- Facilitates the reaction to yield isocyanide
Other exercises in this chapter
Problem 97
The compound that will react most readily with \(\mathrm{NaOH}\) to form methanol is (a) \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{~N}^{+} \mathrm{I}^{-}\) (b)
View solution Problem 99
Among the following, the strongest base is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (b) \(\mathrm{p}-\mathrm{NO}_{2}-\mathrm{C}_{6} \mathrm{H}_{4}
View solution Problem 101
Among the following statements on the nitration of aromatic compounds, the false one is (a) the rate of nitration of benzene is almost the same as the hexadeute
View solution Problem 103
When nitrobenzene is treated with \(\mathrm{Br}_{2}\) in presence of \(\mathrm{FeBr}_{3}\), the major product formed is \(\mathrm{m}\)-bromonitrobenzene. The st
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