401. Which reagent combination is commonly used to generate nitrous acid in situ for reactions of amines?
ⓐ. \(Br_2/NaOH\)
ⓑ. \(NaNO_2/HCl\)
ⓒ. \(CHCl_3/KOH\)
ⓓ. \(LiAlH_4\)
Correct Answer: \(NaNO_2/HCl\)
Explanation: Nitrous acid, \(HNO_2\), is unstable and is usually generated in the reaction mixture. A common method is to use sodium nitrite, \(NaNO_2\), with hydrochloric acid, \(HCl\). These reagents produce \(HNO_2\) in situ, which then reacts with amines. The behaviour of the amine toward nitrous acid depends strongly on whether the amine is primary aliphatic, primary aromatic, secondary, or tertiary.
402. What is the main organic product when a primary aliphatic amine reacts with nitrous acid?
ⓐ. A nitrile
ⓑ. An amide
ⓒ. An isocyanide
ⓓ. An alcohol
Correct Answer: An alcohol
Explanation: Primary aliphatic amines react with nitrous acid to form unstable aliphatic diazonium ions. These unstable intermediates decompose rapidly, releasing nitrogen gas. The main organic product is usually the corresponding alcohol. This reaction is different from primary aromatic amines, which can form relatively stable diazonium salts at low temperature.
403. Which equation represents the reaction of a primary aliphatic amine with nitrous acid?
ⓐ. \(RNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O\)
ⓑ. \(RNH_2 + HNO_2 \rightarrow RNC + 2H_2O\)
ⓒ. \(RNH_2 + HNO_2 \rightarrow RCONH_2 + N_2 + H_2O\)
ⓓ. \(RNH_2 + HNO_2 \rightarrow R_4N^+NO_2^-\)
Correct Answer: \(RNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O\)
Explanation: \(\textbf{Starting amine:}\)
\(RNH_2\) is a primary aliphatic amine.
\(\textbf{Reagent:}\)
Nitrous acid is represented as \(HNO_2\).
\(\textbf{Reaction behaviour:}\)
Primary aliphatic amines form unstable diazonium intermediates.
\(\textbf{Decomposition:}\)
The unstable intermediate decomposes with evolution of \(N_2\).
\(\textbf{Organic product:}\)
The corresponding alcohol, \(ROH\), is formed.
\(\textbf{Final Answer:}\)
\[\boxed{RNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O}\]
404. What is the main organic product when methylamine, \(CH_3NH_2\), reacts with nitrous acid?
ⓐ. \(CH_3CN\)
ⓑ. \(CH_3NC\)
ⓒ. \(CH_3OH\)
ⓓ. \(CH_3NO_2\)
Correct Answer: \(CH_3OH\)
Explanation: \(\textbf{Starting compound:}\)
Methylamine, \(CH_3NH_2\), is a primary aliphatic amine.
\(\textbf{Reagent:}\)
Nitrous acid, \(HNO_2\), is generated from \(NaNO_2/HCl\).
\(\textbf{Reaction type:}\)
Primary aliphatic amines react with nitrous acid to give alcohols with evolution of nitrogen gas.
\(\textbf{Apply to methylamine:}\)
The methyl group becomes attached to \(-OH\), giving methanol, \(CH_3OH\).
\(\textbf{Final Answer:}\)
Methylamine gives \(CH_3OH\).
405. What is formed when aniline reacts with \(NaNO_2/HCl\) at about \(273\,\text{K}\)?
ⓐ. Benzenediazonium sulfate
ⓑ. Benzenediazonium chloride
ⓒ. Benzylamine
ⓓ. Acetanilide
Correct Answer: Benzenediazonium chloride
Explanation: Aniline is a primary aromatic amine. When treated with \(NaNO_2/HCl\) at low temperature, it undergoes diazotisation. The product is benzenediazonium chloride, \(C_6H_5N_2^+Cl^-\). Low temperature is important because diazonium salts are more stable under cold conditions and decompose more readily at higher temperature.
406. Which equation correctly represents diazotisation of aniline?
ⓐ. \(C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5NH_3^+Cl^- + NaCl + HNO_2\)
ⓑ. \(C_6H_5NO_2 + 3H_2 \rightarrow C_6H_5NH_2 + 2H_2O\)
ⓒ. \(C_6H_5NH_2 + CH_3COCl \rightarrow C_6H_5NHCOCH_3 + HCl\)
ⓓ. \(C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O\)
Correct Answer: \(C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O\)
Explanation: \(\textbf{Starting amine:}\)
Aniline is \(C_6H_5NH_2\), a primary aromatic amine.
\(\textbf{Diazotising mixture:}\)
\(NaNO_2\) and \(HCl\) generate nitrous acid in the reaction mixture.
\(\textbf{Low-temperature product:}\)
At about \(273\,\text{K}\), aniline is converted into benzenediazonium chloride.
\(\textbf{Balanced equation:}\)
\[\boxed{C_6H_5NH_2 + NaNO_2 + 2HCl \rightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O}\]
407. Which temperature condition is most suitable for diazotisation of aniline?
ⓐ. About \(273\text{–}278\,\text{K}\)
ⓑ. About \(373\text{–}383\,\text{K}\)
ⓒ. About \(500\text{–}520\,\text{K}\)
ⓓ. About \(650\text{–}700\,\text{K}\)
Correct Answer: About \(273\text{–}278\,\text{K}\)
Explanation: Diazotisation of aniline is carried out at low temperature, usually around \(273\text{–}278\,\text{K}\). Benzenediazonium chloride is relatively stable only in cold solution. At higher temperatures, diazonium salts tend to decompose, often forming phenols in aqueous medium. Therefore, maintaining a cold reaction mixture is essential for successful diazotisation.
408. Why is diazotisation of aniline carried out at low temperature?
ⓐ. Aniline becomes a tertiary amine only below \(273\,\text{K}\).
ⓑ. \(NaNO_2\) acts as a reducing agent only above \(373\,\text{K}\).
ⓒ. The diazonium salt is more stable in cold solution.
ⓓ. The benzene ring is removed only at low temperature.
Correct Answer: The diazonium salt is more stable in cold solution.
Explanation: Benzenediazonium salts are temperature-sensitive. At low temperature, the diazonium salt can remain in solution long enough for further reactions. If the temperature is raised, it decomposes more easily, often releasing nitrogen gas and forming substitution products. This is why diazotisation is normally performed in an ice-cold medium.
409. What is commonly formed when a secondary amine reacts with nitrous acid?
ⓐ. Diazonium salt
ⓑ. \(N\)-nitrosoamine
ⓒ. Primary alcohol
ⓓ. Quaternary salt
Correct Answer: \(N\)-nitrosoamine
Explanation: Secondary amines react with nitrous acid to form \(N\)-nitrosoamines. These products contain the nitroso group attached to nitrogen. Unlike primary aliphatic amines, secondary amines do not give alcohols with evolution of nitrogen gas. Unlike primary aromatic amines, they do not form stable diazonium salts under the usual diazotisation conditions.
410. Which product is expected when dimethylamine, \((CH_3)_2NH\), reacts with nitrous acid?
ⓐ. \(CH_3OH\)
ⓑ. \(CH_3NH_2\)
ⓒ. \(CH_3CON(CH_3)_2\)
ⓓ. \((CH_3)_2NNO\)
Correct Answer: \((CH_3)_2NNO\)
Explanation: \(\textbf{Starting amine:}\)
Dimethylamine, \((CH_3)_2NH\), is a secondary amine.
\(\textbf{Reaction with nitrous acid:}\)
Secondary amines undergo nitrosation at nitrogen.
\(\textbf{Product type:}\)
The product is an \(N\)-nitrosoamine.
\(\textbf{Product formula:}\)
For dimethylamine, the product is \((CH_3)_2NNO\), commonly called \(N\)-nitrosodimethylamine.
\(\textbf{Final Answer:}\)
Dimethylamine gives \((CH_3)_2NNO\).
411. Which statement best describes the reaction of a tertiary aliphatic amine with nitrous acid?
ⓐ. Nitrous acid salt formation
ⓑ. It gives an alcohol and \(N_2\).
ⓒ. It gives a diazonium chloride.
ⓓ. It gives an isocyanide.
Correct Answer: Nitrous acid salt formation
Explanation: Tertiary aliphatic amines have no \(N-H\) bond, so they do not form \(N\)-nitrosoamines like secondary amines. They also do not form diazonium salts like primary aromatic amines. Because tertiary amines are basic, they can form salts with nitrous acid. This behaviour helps distinguish tertiary aliphatic amines from primary and secondary amines.
412. What is the usual reaction of a tertiary aromatic amine such as \(N,N\)-dimethylaniline with nitrous acid?
ⓐ. Formation of an alcohol with \(N_2\)
ⓑ. Formation of a quaternary ammonium carbonate
ⓒ. Para nitrosation of the aromatic ring
ⓓ. Reduction to an aliphatic primary amine
Correct Answer: Para nitrosation of the aromatic ring
Explanation: Tertiary aromatic amines such as \(N,N\)-dimethylaniline have a strongly activating dialkylamino group on the benzene ring. In reaction with nitrous acid, nitrosation occurs on the aromatic ring, commonly at the para position when it is available. This is different from secondary amines, which form \(N\)-nitrosoamines. It is also different from primary aromatic amines, which form diazonium salts at low temperature.
413. Which amine gives a diazonium salt with \(NaNO_2/HCl\) at \(273\,\text{K}\)?
ⓐ. \(C_6H_5NH_2\)
ⓑ. \((CH_3)_2NH\)
ⓒ. \((CH_3)_3N\)
ⓓ. \(CH_3NH_2\)
Correct Answer: \(C_6H_5NH_2\)
Explanation: \(C_6H_5NH_2\), aniline, is a primary aromatic amine. Primary aromatic amines form diazonium salts with \(NaNO_2/HCl\) at low temperature. Primary aliphatic amines form unstable diazonium intermediates that decompose to alcohols. Secondary and tertiary amines show different reactions with nitrous acid and do not give the same stable diazonium salt.
414. Which observation is associated with reaction of a primary aliphatic amine with nitrous acid?
ⓐ. Formation of a blue copper complex
ⓑ. Evolution of nitrogen gas
ⓒ. Formation of a silver mirror
ⓓ. Formation of a diazo dye directly
Correct Answer: Evolution of nitrogen gas
Explanation: Primary aliphatic amines react with nitrous acid to form unstable aliphatic diazonium ions. These intermediates decompose rapidly. Nitrogen gas, \(N_2\), is evolved during the process, and the corresponding alcohol is generally formed. The evolution of \(N_2\) is therefore an important observation for this reaction type.
415. Which comparison of nitrous-acid reactions is correct?
ⓐ. Primary aliphatic amines give stable diazonium salts, while aniline gives alcohols.
ⓑ. Secondary amines form \(N\)-nitrosoamines; primary aromatic amines form diazonium salts.
ⓒ. Tertiary aliphatic amines give isocyanides, while primary amines give amides.
ⓓ. Primary aromatic amines give quaternary salts, while secondary amines give alcohols.
Correct Answer: Secondary amines form \(N\)-nitrosoamines; primary aromatic amines form diazonium salts.
Explanation: Secondary amines react with nitrous acid to give \(N\)-nitrosoamines. Primary aromatic amines such as aniline react with \(NaNO_2/HCl\) at low temperature to form diazonium salts. Primary aliphatic amines form unstable diazonium intermediates that decompose to alcohols with evolution of \(N_2\). Tertiary aliphatic amines mainly form salts with nitrous acid.
416. Which product is obtained when ethylamine, \(C_2H_5NH_2\), reacts with nitrous acid?
ⓐ. \(C_2H_5CN\)
ⓑ. \(C_2H_5NC\)
ⓒ. \(C_2H_5N_2^+Cl^-\)
ⓓ. \(C_2H_5OH\)
Correct Answer: \(C_2H_5OH\)
Explanation: \(\textbf{Starting compound:}\)
Ethylamine, \(C_2H_5NH_2\), is a primary aliphatic amine.
\(\textbf{Reagent:}\)
Nitrous acid, \(HNO_2\), reacts with primary aliphatic amines.
\(\textbf{Intermediate behaviour:}\)
The aliphatic diazonium intermediate is unstable and decomposes.
\(\textbf{Product formation:}\)
The corresponding alcohol is formed with evolution of \(N_2\).
\(\textbf{Final Answer:}\)
Ethylamine gives ethanol, \(C_2H_5OH\).
417. Which comparison between primary aliphatic and primary aromatic amines with nitrous acid is correct?
ⓐ. Both form stable diazonium salts at room temperature.
ⓑ. Aliphatic diazonium ions decompose; aromatic diazonium salts are stable cold.
ⓒ. Aliphatic amines give azo dyes directly, while aromatic amines give only alcohols.
ⓓ. Both give \(N\)-nitrosoamines as the main products.
Correct Answer: Aliphatic diazonium ions decompose; aromatic diazonium salts are stable cold.
Explanation: Primary aliphatic amines form aliphatic diazonium ions with nitrous acid, but these ions are highly unstable. They decompose quickly, usually giving alcohols with evolution of \(N_2\). Primary aromatic amines such as aniline form arenediazonium salts that are relatively stable in cold solution. This difference is a key reason why diazonium chemistry is especially useful for aromatic amines.
418. Which reaction is most suitable for converting aniline into benzenediazonium chloride?
ⓐ. Treat with \(NaNO_2/HCl\) cold
ⓑ. Heating with \(CHCl_3/KOH\) in alcohol
ⓒ. Reducing with \(Sn/HCl\), followed by alkali
ⓓ. Reacting with \(CH_3COCl\) in base
Correct Answer: Treat with \(NaNO_2/HCl\) cold
Explanation: Aniline undergoes diazotisation with nitrous acid generated from \(NaNO_2\) and \(HCl\). The reaction is carried out at low temperature, usually around \(273\text{–}278\,\text{K}\), because benzenediazonium chloride is stable only in cold solution. Heating would favour decomposition of the diazonium salt. The other reagents correspond to carbylamine reaction, nitro reduction, and acylation respectively.
419. Which equation represents the reaction of propylamine with nitrous acid?
ⓐ. \(CH_3CH_2CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2CN + 2H_2O\)
ⓑ. \(CH_3CH_2CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2CH_2NC + 2H_2O\)
ⓒ. \(CH_3CH_2CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2CH_2OH + N_2 + H_2O\)
ⓓ. \(CH_3CH_2CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2CH_2N_2^+Cl^- + H_2O\)
Correct Answer: \(CH_3CH_2CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2CH_2OH + N_2 + H_2O\)
Explanation: \(\textbf{Starting amine:}\)
Propylamine, \(CH_3CH_2CH_2NH_2\), is a primary aliphatic amine.
\(\textbf{Reagent behaviour:}\)
Nitrous acid reacts with primary aliphatic amines through unstable diazonium intermediates.
\(\textbf{Decomposition:}\)
The intermediate decomposes with evolution of nitrogen gas, \(N_2\).
\(\textbf{Organic product:}\)
The corresponding alcohol is formed, so propylamine gives propan-1-ol.
\(\textbf{Final Answer:}\)
\[\boxed{CH_3CH_2CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2CH_2OH + N_2 + H_2O}\]
420. What is the usual product type when \(N,N\)-dimethylaniline reacts with nitrous acid under suitable conditions?
ⓐ. Benzenediazonium tetrafluoroborate
ⓑ. \(p\)-nitroso-\(N,N\)-diethylaniline
ⓒ. Phenyl isocyanide
ⓓ. \(p\)-nitroso-\(N,N\)-dimethylaniline
Correct Answer: \(p\)-nitroso-\(N,N\)-dimethylaniline
Explanation: \(N,N\)-Dimethylaniline is a tertiary aromatic amine. Its dialkylamino group strongly activates the benzene ring toward electrophilic substitution. Nitrous acid can generate a nitrosating species, and ring nitrosation occurs mainly at the para position if it is available. This differs from secondary amines, which form \(N\)-nitrosoamines, and primary aromatic amines, which form diazonium salts.
