Problem 9
Question
Which of the following reacts with \(\mathrm{NaNO}_{2}+\mathrm{HCl}\) at \(273-278 \mathrm{~K}\) to give alcohol/phenol? (a) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NHCH}_{3}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\)
Step-by-Step Solution
Verified Answer
(b) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \) aniline reacts to give phenol.
1Step 1: Understanding the Reaction Conditions
The reaction involves sodium nitrite (\( \mathrm{NaNO}_2 \)) and hydrochloric acid (\( \mathrm{HCl} \)) at a temperature of 273-278 K. This is a typical setup for a diazotization reaction, where primary aromatic amines are converted into diazonium salts.
2Step 2: Identify the Type of Amines
We need to identify which of the given compounds can undergo a diazotization reaction:- (a) \( \mathrm{CH}_{3} \mathrm{NH}_{2} \) is a primary aliphatic amine.- (b) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \) is a primary aromatic amine, also known as aniline.- (c) \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NHCH}_{3} \) is a secondary amine.- (d) \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} \) is a secondary amine.
3Step 3: Determine Eligibility for Diazotization
For diazotization, the compound must be a primary amine and aromatic to form stable diazonium salts. Only anilines (primary aromatic amines) undergo this reaction smoothly and can further undergo hydrolysis to form phenols.
4Step 4: Identify the Correct Compound for Reaction
Only compound (b), \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \) (aniline), is a primary aromatic amine. Aniline reacts with \( \mathrm{NaNO}_2 \) and \( \mathrm{HCl} \) to form a diazonium salt, which upon hydrolysis yields phenol.
5Step 5: Conclusion
The given reaction conditions specifically facilitate the diazotization of aniline to form phenol, as no other type of amine in the given options fits this reaction scheme.
Key Concepts
Primary Aromatic AminesDiazonium SaltsPhenol Formation
Primary Aromatic Amines
Primary aromatic amines are an essential class of organic compounds that have a fundamental role in diazotization reactions. These compounds are characterized by having an amino group (\(\text{NH}_2\)) directly attached to an aromatic ring, like benzene. A common example of a primary aromatic amine is aniline, which is represented as \(\text{C}_6\text{H}_5\text{NH}_2\).
The key aspect of these amines is their ability to participate in reactions that are uniquely suited to the chemical behavior of aromatic rings. They are distinct from primary aliphatic amines, where the amino group is attached to an aliphatic (non-aromatic) carbon chain.
The key aspect of these amines is their ability to participate in reactions that are uniquely suited to the chemical behavior of aromatic rings. They are distinct from primary aliphatic amines, where the amino group is attached to an aliphatic (non-aromatic) carbon chain.
- The aromatic ring in primary aromatic amines can stabilize certain intermediates through resonance.
- This stabilization makes them the ideal candidates for specific reactions, like diazotization.
Diazonium Salts
Diazonium salts are pivotal intermediates in many organic synthesis processes, particularly in transforming primary aromatic amines into other functional groups.
These salts are created when a primary aromatic amine reacts with \(\text{NaNO}_2\) and \(\text{HCl}\). The reaction proceeds through a diazotization mechanism, forming a diazonium ion, which can be represented generically as \(\text{ArN}_2^+\text{Cl}^-\).
The formation of diazonium salts is temperature-sensitive, and typically carried out at cool temperatures (273-278 K) to ensure the stability of the salt.
These salts are created when a primary aromatic amine reacts with \(\text{NaNO}_2\) and \(\text{HCl}\). The reaction proceeds through a diazotization mechanism, forming a diazonium ion, which can be represented generically as \(\text{ArN}_2^+\text{Cl}^-\).
The formation of diazonium salts is temperature-sensitive, and typically carried out at cool temperatures (273-278 K) to ensure the stability of the salt.
- Diazonium salts are known for their versatility and are used to generate a wide range of aromatic compounds.
- One of the most common uses is in converting diazonium salts to phenols via hydrolysis.
Phenol Formation
Phenol formation from diazonium salts is a critical step in organic chemistry transformations, primarily involving the hydrolysis of diazonium salts. When a diazonium salt is formed, it can be converted into a phenol by the addition of water (a process known as hydrolysis).
In this specific context, the phenol formation from aniline goes through the intermediate stage of forming a diazonium salt. Upon warming or in the presence of water, the diazonium group (\(\text{N}_2^+\)) leaves, and a hydroxyl group (\(\text{OH}\)) takes its place on the aromatic ring, resulting in phenol.
In this specific context, the phenol formation from aniline goes through the intermediate stage of forming a diazonium salt. Upon warming or in the presence of water, the diazonium group (\(\text{N}_2^+\)) leaves, and a hydroxyl group (\(\text{OH}\)) takes its place on the aromatic ring, resulting in phenol.
- The reaction is represented as \(\text{ArN}_2^+\text{Cl}^- + \text{H}_2\text{O} \rightarrow \text{ArOH}\).
- Phenols have unique chemical properties, mainly due to the hydroxyl group's electron-donating nature.
Other exercises in this chapter
Problem 6
Carbylamine test is performed in alcoholic KOH by heating a mixture of (a) trihalogenated methane and a primary amine (b) an alkyl halide and a primary amine (c
View solution Problem 7
Acetamide is treated separately with the following reagents. Which one of these would give methylamine? (a) sodalime (b) \(\mathrm{PCl}_{5}\) (c) \(\mathrm{NaOH
View solution Problem 10
Hinsberg reagent is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{Cl}\) (c) \(\mathrm{C}_{6} \ma
View solution Problem 11
When acetamide is treated with \(\mathrm{NaOBr}\), the product formed is (a) \(\mathrm{CH}_{3} \mathrm{OH}\) (b) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (c) \(\math
View solution