301. An unknown amine forms no sulphonamide with Hinsberg reagent, remains unreacted in excess alkali, and dissolves when dilute acid is added. The amine is:
ⓐ. primary
ⓑ. secondary
ⓒ. tertiary
ⓓ. a quaternary ammonium ion
Correct Answer: tertiary
Explanation: A tertiary amine contains no \(\mathrm{N-H}\) bond and therefore does not form the ordinary Hinsberg sulphonamide. It remains as the unreacted neutral amine in the alkaline reaction mixture. Addition of dilute acid protonates its nitrogen lone pair and produces a water-soluble ammonium salt. A primary amine would give an alkali-soluble sulphonamide, while a secondary amine would give an alkali-insoluble sulphonamide. A quaternary ammonium ion is already permanently charged and does not show this neutral-amine acid extraction sequence.
302. An unknown amine gives a positive carbylamine test. Its Hinsberg product dissolves in excess alkali and reappears on acidification. With nitrous acid, the amine produces brisk nitrogen evolution and an alcohol. The amine is:
ⓐ. a secondary aliphatic amine
ⓑ. a primary aromatic amine
ⓒ. a primary aliphatic amine
ⓓ. a tertiary aliphatic amine
Correct Answer: a primary aliphatic amine
Explanation: A positive carbylamine test shows that the compound is a primary amine. The alkali-soluble Hinsberg product confirms the presence of a primary sulphonamide containing an acidic \(\mathrm{N-H}\) bond. These two tests alone do not distinguish primary aliphatic and primary aromatic amines. Nitrous acid provides the decisive evidence. A primary aliphatic amine forms an unstable aliphatic diazonium intermediate that rapidly loses \(\mathrm{N_2}\) and gives an alcohol. A primary aromatic amine instead forms a comparatively stable diazonium salt at low temperature.
303. Review the qualitative-test records below.
| Row | Amine class | Carbylamine test | Hinsberg behaviour | Nitrous-acid behaviour |
|---|
| P | Primary aliphatic | Positive | Product soluble in alkali | \(\mathrm{N_2}\) evolution and alcohol formation |
| Q | Secondary | Negative | Insoluble sulphonamide | \(\mathrm{N}\)-nitrosoamine formation |
| R | Tertiary aliphatic | Negative | No ordinary sulphonamide; amine dissolves in acid | No primary-amine nitrogen evolution |
| S | Primary aromatic | Negative | Product soluble in alkali | Diazonium salt at low temperature |
The completely valid rows are:
ⓐ. P and S only
ⓑ. Q and R only
ⓒ. P, Q and S only
ⓓ. P, Q and R only
Correct Answer: P, Q and R only
Explanation: Row P correctly describes a primary aliphatic amine. Row Q correctly combines the negative carbylamine test, insoluble secondary sulphonamide, and nitrosoamine formation. Row R also gives the characteristic behaviour of a tertiary aliphatic amine. Row S is incorrect because primary aromatic amines such as aniline give a positive carbylamine test. The other observations in S are compatible with a primary aromatic amine, but one incorrect entry invalidates the row.
304. Use the following passage.
An unknown amine gives no carbylamine reaction. With benzenesulphonyl chloride, it forms a neutral solid that remains insoluble in alkali. On treatment with nitrous acid, it gives a yellow oily product without brisk evolution of nitrogen gas.
The unknown is most reasonably:
ⓐ. a primary aliphatic amine
ⓑ. a secondary amine
ⓒ. a tertiary aliphatic amine
ⓓ. a primary aromatic amine
Correct Answer: a secondary amine
Explanation: The negative carbylamine test rules out primary amines. Formation of a sulphonamide shows that the amine contains an \(\mathrm{N-H}\) bond and can react with Hinsberg reagent. Insolubility of that sulphonamide in alkali indicates that no acidic nitrogen hydrogen remains in the product. This is characteristic of a secondary amine. The yellow oily nitrosoamine formed with nitrous acid provides independent confirmation. A tertiary amine would not form the ordinary Hinsberg sulphonamide.
305. Acylation can distinguish a secondary amine from a tertiary amine because:
ⓐ. both classes lack an \(\mathrm{N-H}\) bond and therefore show no acylation difference
ⓑ. a tertiary amine forms an ordinary stable amide, whereas a secondary amine does not
ⓒ. a secondary amine has an \(\mathrm{N-H}\) bond and forms an amide, whereas a tertiary amine lacks that bond
ⓓ. both classes form identical amides because only the nitrogen lone pair is required
Correct Answer: a secondary amine has an \(\mathrm{N-H}\) bond and forms an amide, whereas a tertiary amine lacks that bond
Explanation: Both secondary and tertiary amines fail the carbylamine test, so that test cannot separate them. A secondary amine still possesses one hydrogen attached to nitrogen. It can undergo acyl substitution and form an \(N,N\)-disubstituted amide. A tertiary amine has no nitrogen-bonded hydrogen and does not give the corresponding ordinary stable amide by this route. Acylation therefore provides a useful distinction between the two classes.
306. A mixture contains methylamine, dimethylamine, and trimethylamine. Complete neutralisation shows that the mixture contains \(0.200\ \mathrm{mol}\) of total amines. The carbylamine test produces \(0.0500\ \mathrm{mol}\) of isocyanide, while complete Hinsberg reaction consumes \(0.140\ \mathrm{mol}\) of benzenesulphonyl chloride. Using molar masses \(31.0\), \(45.0\), and \(59.0\ \mathrm{g\,mol^{-1}}\), respectively, the amounts of the three amines and the total mixture mass are:
ⓐ. \(0.0500\), \(0.0600\), \(0.0900\ \mathrm{mol}\); \(9.56\ \mathrm{g}\)
ⓑ. \(0.0500\), \(0.0900\), \(0.0600\ \mathrm{mol}\); \(9.14\ \mathrm{g}\)
ⓒ. \(0.0900\), \(0.0500\), \(0.0600\ \mathrm{mol}\); \(8.86\ \mathrm{g}\)
ⓓ. \(0.0600\), \(0.0900\), \(0.0500\ \mathrm{mol}\); \(9.02\ \mathrm{g}\)
Correct Answer: \(0.0500\), \(0.0900\), \(0.0600\ \mathrm{mol}\); \(9.14\ \mathrm{g}\)
Explanation: Let the amounts of methylamine, dimethylamine, and trimethylamine be \(x\), \(y\), and \(z\).
Only methylamine gives the carbylamine test.
\[
x=0.0500\ \mathrm{mol}
\]
Primary and secondary amines react with Hinsberg reagent.
\[
x+y=0.140
\]
\[
y=0.140-0.0500=0.0900\ \mathrm{mol}
\]
The total amine amount is
\[
x+y+z=0.200
\]
\[
z=0.200-0.0500-0.0900
\]
\[
z=0.0600\ \mathrm{mol}
\]
\( \textbf{Mass of methylamine:} \)
\[
m_1=0.0500\times31.0=1.55\ \mathrm{g}
\]
\( \textbf{Mass of dimethylamine:} \)
\[
m_2=0.0900\times45.0=4.05\ \mathrm{g}
\]
\( \textbf{Mass of trimethylamine:} \)
\[
m_3=0.0600\times59.0=3.54\ \mathrm{g}
\]
\[
m_{\text{total}}=1.55+4.05+3.54=9.14\ \mathrm{g}
\]
307. Two unknown amines P and Q both give negative carbylamine tests. A graph plots moles of stable acyl derivative formed against moles of ethanoyl chloride added to equal initial amounts of P and Q. Curve P rises with slope \(1\) and reaches a plateau, whereas curve Q remains at zero. Which assignment is most reasonable?
ⓐ. P is secondary and Q is tertiary
ⓑ. P is tertiary and Q is secondary
ⓒ. P is primary and Q is tertiary
ⓓ. P is tertiary and Q is primary
Correct Answer: P is secondary and Q is tertiary
Explanation: Both samples are negative in the carbylamine test, so neither is a primary amine. A secondary amine contains one \(\mathrm{N-H}\) bond and forms a stable amide with ethanoyl chloride. Its product amount initially increases in a \(1:1\) relation with added acylating reagent. A tertiary amine lacks an \(\mathrm{N-H}\) bond and does not give the ordinary stable amide product. Curve P therefore represents the secondary amine, while the zero-product curve Q represents the tertiary amine.
308. Why should odour alone not be used to identify an unknown amine?
ⓐ. amines of the same class always have indistinguishable odours
ⓑ. only volatile primary amines produce a detectable odour
ⓒ. odour changes predictably enough to determine exact molecular structure
ⓓ. odour varies with concentration, impurities, and observer perception
Correct Answer: odour varies with concentration, impurities, and observer perception
Explanation: Many low-molar-mass amines have strong or fish-like odours, but such descriptions are not structurally specific. Perceived odour depends on concentration, volatility, impurities, and individual sensitivity. Different amines can smell similar, and the same amine may smell different in mixtures. Direct smelling can also be unsafe. Reliable identification should instead use controlled reactions such as carbylamine, Hinsberg, nitrous-acid, acylation, and salt-solubility tests.
309. Aliphatic diazonium ions decompose much more readily than arenediazonium ions mainly because:
ⓐ. they lack the resonance stabilisation available to arenediazonium ions
ⓑ. aliphatic diazonium ions contain no nitrogen atoms
ⓒ. arenediazonium ions cannot lose molecular nitrogen
ⓓ. alkyl groups stabilise aliphatic diazonium ions more strongly than aryl groups
Correct Answer: they lack the resonance stabilisation available to arenediazonium ions
Explanation: An aliphatic diazonium group is attached to an \(sp^3\)-carbon framework and is not stabilised by conjugation with an aromatic ring. Loss of \(\mathrm{N_2}\) is strongly favourable because molecular nitrogen is exceptionally stable. The remaining carbon centre can then react rapidly with water or undergo rearrangement. Arenediazonium ions receive stabilisation from the aromatic system and can persist at low temperature. This difference explains the immediate gas evolution observed with primary aliphatic amines.
310. Assertion: Under ideal stoichiometric conditions, one mole of a primary aliphatic monoamine can produce one mole of nitrogen gas with nitrous acid.
Reason: The two nitrogen atoms present in the amine-derived diazonium group leave together as one molecule of \(\mathrm{N_2}\).
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Nitrous acid supplies the second nitrogen atom needed to form the diazonium intermediate. The original amine nitrogen and the nitrosating nitrogen become part of the diazonium group. Decomposition releases these two atoms together as one molecule of \(\mathrm{N_2}\). Consequently, one mole of a monofunctional primary aliphatic amine ideally gives one mole of nitrogen gas. The Reason directly accounts for the stated mole ratio.
311. The table lists standard nitrous-acid products.
| Row | Starting amine | Standard major organic product | Gas evolved |
|---|
| P | \(\mathrm{CH_3NH_2}\) | \(\mathrm{CH_3OH}\) | \(\mathrm{N_2}\) |
| Q | \(\mathrm{C_2H_5NH_2}\) | \(\mathrm{CH_2{=}CH_2}\) only | No gas |
| R | \(\mathrm{(CH_3)_2CHNH_2}\) | \(\mathrm{(CH_3)_2CHOH}\) | \(\mathrm{N_2}\) |
| S | \(\mathrm{C_6H_5NH_2}\) | \(\mathrm{C_6H_5OH}\) immediately at \(0\text{–}5^\circ\mathrm{C}\) | \(\mathrm{N_2}\) immediately |
The valid rows are:
ⓐ. Q and S only
ⓑ. P and R only
ⓒ. P, Q and R only
ⓓ. R and S only
Correct Answer: P and R only
Explanation: Methylamine gives methanol with nitrogen evolution, so P is valid. Ethylamine is represented as giving ethanol, not ethene only, and nitrogen is evolved, making Q invalid. Propan-\(2\)-amine gives propan-\(2\)-ol in the standard representation and releases nitrogen, so R is valid. Aniline forms a benzenediazonium salt at low temperature rather than immediately giving phenol and nitrogen under the stated conditions. Only P and R are completely correct.
312. Use the following passage.
A primary aliphatic amine is diazotised in aqueous acid. The diazonium intermediate loses nitrogen rapidly. In a simple substrate, water capture gives the corresponding alcohol. In a more highly substituted or unsymmetrical substrate, the carbon intermediate may rearrange or follow competing pathways.
Which conclusion is most appropriate?
ⓐ. Alcohol formation is standard, but rearranged or mixed products may occur
ⓑ. Every primary aliphatic amine gives one pure alcohol without any competing process
ⓒ. Nitrogen evolution proves that no carbon intermediate is formed
ⓓ. Rearrangement converts the reaction into an ordinary secondary-amine nitrosation
Correct Answer: Alcohol formation is standard, but rearranged or mixed products may occur
Explanation: The general classroom equation represents replacement of \(\mathrm{-NH_2}\) by \(\mathrm{-OH}\) with release of nitrogen. This description works well for simple product prediction. Mechanistically, however, loss of nitrogen can create a highly reactive carbon centre. Such an intermediate may rearrange if a more stable structure can be formed. Elimination or other competing reactions may also occur in some cases. The standard alcohol product should therefore be understood as the principal representation rather than an absolute guarantee of a single product for every substrate.
313. Nitrous acid is supplied in an amount of \(0.0800\ \mathrm{mol}\), and increasing amounts of a primary aliphatic monoamine are added. A graph plots the volume of dry nitrogen produced at STP against moles of amine added. Assuming quantitative reaction and \(22.4\ \mathrm{L\,mol^{-1}}\) at STP, the graph is described by:
ⓐ. The volume rises with slope \(44.8\ \mathrm{L\,mol^{-1}}\) and reaches \(3.584\ \mathrm{L}\)
ⓑ. The volume rises with slope \(22.4\ \mathrm{L\,mol^{-1}}\) to \(1.792\ \mathrm{L}\), then plateaus
ⓒ. The volume remains zero until \(0.0800\ \mathrm{mol}\) of amine is added and then rises vertically
ⓓ. The volume decreases linearly because nitrous acid is consumed
Correct Answer: The volume rises with slope \(22.4\ \mathrm{L\,mol^{-1}}\) to \(1.792\ \mathrm{L}\), then plateaus
Explanation: One mole of primary aliphatic monoamine gives one mole of nitrogen gas.
Before \(0.0800\ \mathrm{mol}\) of amine has been added, the amine is limiting.
Therefore,
\[
n(\mathrm{N_2})=n(\mathrm{amine})
\]
At STP,
\[
V(\mathrm{N_2})=22.4\,n(\mathrm{amine})
\]
The initial slope is therefore
\[
22.4\ \mathrm{L\,mol^{-1}}
\]
Nitrous acid can react with at most
\[
0.0800\ \mathrm{mol}
\]
of amine.
The maximum gas volume is
\[
V_{\max}=0.0800\times22.4
\]
\[
V_{\max}=1.792\ \mathrm{L}
\]
Beyond this point, nitrous acid is limiting and the graph reaches a plateau.
314. The \(6.05\ \mathrm{g}\) mixture contains methylamine and dimethylamine. Complete neutralisation requires \(150\ \mathrm{mL}\) of \(1.00\ \mathrm{mol\,L^{-1}}\) hydrochloric acid. When a separate identical mixture is treated with excess nitrous acid, \(1.12\ \mathrm{L}\) of nitrogen is obtained at STP. Assuming \(22.4\ \mathrm{L\,mol^{-1}}\) at STP, the amounts of methylamine and dimethylamine and the mass percentage of methylamine are:
ⓐ. \(0.1000\ \mathrm{mol}\), \(0.0500\ \mathrm{mol}\), and \(51.2\%\)
ⓑ. \(0.0500\ \mathrm{mol}\), \(0.1000\ \mathrm{mol}\), and \(74.4\%\)
ⓒ. \(0.0500\ \mathrm{mol}\), \(0.1000\ \mathrm{mol}\), and \(25.6\%\)
ⓓ. \(0.1000\ \mathrm{mol}\), \(0.0500\ \mathrm{mol}\), and \(25.6\%\)
Correct Answer: \(0.0500\ \mathrm{mol}\), \(0.1000\ \mathrm{mol}\), and \(25.6\%\)
Explanation: Let the amounts of methylamine and dimethylamine be \(x\) and \(y\), respectively.
Both amines consume one mole of hydrochloric acid per mole.
\[
n(\mathrm{HCl})=1.00\times0.150
\]
\[
n(\mathrm{HCl})=0.150\ \mathrm{mol}
\]
Therefore,
\[
x+y=0.150
\]
Only methylamine is a primary aliphatic amine and produces nitrogen with nitrous acid.
\[
n(\mathrm{N_2})=\frac{1.12}{22.4}
\]
\[
n(\mathrm{N_2})=0.0500\ \mathrm{mol}
\]
The mole ratio between methylamine and nitrogen is \(1:1\).
\[
x=0.0500\ \mathrm{mol}
\]
Hence,
\[
y=0.1500-0.0500=0.1000\ \mathrm{mol}
\]
\( \textbf{Mass of methylamine:} \)
\[
m=0.0500\times31.0=1.55\ \mathrm{g}
\]
\[
\%\mathrm{CH_3NH_2}=\frac{1.55}{6.05}\times100
\]
\[
\%\mathrm{CH_3NH_2}=25.6\%
\]
The calculated dimethylamine mass is \(0.1000\times45.0=4.50\ \mathrm{g}\), giving the stated total mass.
315. A primary aliphatic amine labelled as \(\mathrm{R^{15}NH_2}\) is treated with nitrous acid prepared using \(\mathrm{^{14}NO_2^-}\). The reaction is carried out in water enriched with \(\mathrm{H_2^{18}O}\). Which labelled products are most reasonably expected in the standard alcohol-forming pathway?
ⓐ. \(\mathrm{^{15}N_2}\) and \(\mathrm{RO^{16}H}\)
ⓑ. \(\mathrm{^{14}N_2}\) and \(\mathrm{R^{18}OH}\)
ⓒ. mixed \(\mathrm{^{15}N^{14}N}\) and \(\mathrm{RO^{16}H}\)
ⓓ. mixed \(\mathrm{^{15}N^{14}N}\) and \(\mathrm{R^{18}OH}\)
Correct Answer: mixed \(\mathrm{^{15}N^{14}N}\) and \(\mathrm{R^{18}OH}\)
Explanation: The original amine nitrogen becomes one nitrogen atom of the transient diazonium group. Nitrous acid supplies the second nitrogen atom. Loss of the diazonium group therefore produces a mixed nitrogen molecule containing one \(\mathrm{^{15}N}\) atom and one \(\mathrm{^{14}N}\) atom. After nitrogen leaves, water captures the reactive carbon centre. The oxygen atom in the alcohol consequently comes from the labelled water in the stated pathway. The expected products are mixed \(\mathrm{^{15}N^{14}N}\) and \(\mathrm{R^{18}OH}\).
316. Four sealed flasks P, Q, R, and S contain equal amounts of methylamine, ethylamine, propan-\(1\)-amine, and butan-\(1\)-amine, respectively. Each amine is treated quantitatively with excess nitrous acid under identical conditions. The resulting nitrogen volumes are described by:
ⓐ. All four flasks produce equal amounts of nitrogen gas
ⓑ. Flask S produces four times as much nitrogen as flask P
ⓒ. Nitrogen production increases directly with the molar mass of the amine
ⓓ. Only flask P produces nitrogen because it contains the smallest alkyl group
Correct Answer: All four flasks produce equal amounts of nitrogen gas
Explanation: All four compounds are monofunctional primary aliphatic amines. One mole of each amine ideally produces one mole of molecular nitrogen. The identity and molar mass of the alkyl group do not alter this basic stoichiometric ratio. Equal amounts of the four amines therefore give equal amounts of \(\mathrm{N_2}\). Their alcohol products have different molar masses, but their nitrogen yields are identical under quantitative conditions.
317. A \(4.50\ \mathrm{g}\) sample of ethylamine is treated with nitrous acid. Only \(80.0\%\) of the amine reacts, and \(90.0\%\) of the nitrogen formed is collected over water at \(300\ \mathrm{K}\). The total pressure is \(1.00\ \mathrm{bar}\), and the vapour pressure of water is \(0.040\ \mathrm{bar}\). Calculate the collected wet-gas volume. Use \(M(\mathrm{C_2H_5NH_2})=45.0\ \mathrm{g\,mol^{-1}}\) and \(R=0.08314\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}\).
ⓐ. \(1.44\ \mathrm{L}\)
ⓑ. \(1.87\ \mathrm{L}\)
ⓒ. \(2.08\ \mathrm{L}\)
ⓓ. \(2.34\ \mathrm{L}\)
Correct Answer: \(1.87\ \mathrm{L}\)
Explanation: \( \textbf{Initial ethylamine amount:} \)
\[
n(\mathrm{amine})=\frac{4.50}{45.0}
\]
\[
n(\mathrm{amine})=0.100\ \mathrm{mol}
\]
\( \textbf{Amount that reacts:} \)
\[
n_{\text{reacted}}=0.100\times0.800
\]
\[
n_{\text{reacted}}=0.0800\ \mathrm{mol}
\]
One mole of reacted primary aliphatic amine gives one mole of nitrogen.
\[
n(\mathrm{N_2,\ formed})=0.0800\ \mathrm{mol}
\]
\( \textbf{Amount collected:} \)
\[
n(\mathrm{N_2,\ collected})=0.0800\times0.900
\]
\[
n(\mathrm{N_2,\ collected})=0.0720\ \mathrm{mol}
\]
The partial pressure of dry nitrogen is
\[
P_{\mathrm{N_2}}=1.00-0.040
\]
\[
P_{\mathrm{N_2}}=0.960\ \mathrm{bar}
\]
Use
\[
P_{\mathrm{N_2}}V=nRT
\]
\[
V=\frac{(0.0720)(0.08314)(300)}{0.960}
\]
\[
V=1.87\ \mathrm{L}
\]
The water-vapour correction is essential because the measured total pressure is not the nitrogen pressure.
318. The standard reaction of a secondary amine with nitrous acid is:
ⓐ. \(\mathrm{R_2NH+HNO_2\rightarrow R_2NOH+N_2}\)
ⓑ. \(\mathrm{R_2NH+HNO_2\rightarrow RNH_2+RNO_2}\)
ⓒ. \(\mathrm{R_2NH+HNO_2\rightarrow R_2N^+H_2NO_2^-}\)
ⓓ. \(\mathrm{R_2NH+HNO_2\rightarrow R_2N-N{=}O+H_2O}\)
Correct Answer: \(\mathrm{R_2NH+HNO_2\rightarrow R_2N-N{=}O+H_2O}\)
Explanation: A secondary amine contains one hydrogen attached to nitrogen. Nitrosation replaces this hydrogen by a nitroso group. The resulting product is an \(N\)-nitrosoamine with the general structure \(\mathrm{R_2N-N{=}O}\). Water is formed as the other product in the overall equation. Molecular nitrogen is not released because both nitrogen atoms remain in the nitrosoamine.
319. Assertion: A secondary amine gives a negative carbylamine test but can form an \(N\)-nitrosoamine with nitrous acid.
Reason: A secondary amine lacks the primary \(\mathrm{-NH_2}\) group required for the carbylamine reaction but retains one \(\mathrm{N-H}\) bond that can be replaced during nitrosation.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The carbylamine reaction is characteristic of primary amines and requires the \(\mathrm{-NH_2}\) framework. A secondary amine has only one hydrogen attached to nitrogen and therefore fails that test. The remaining \(\mathrm{N-H}\) bond can participate in nitrosation. It is replaced by the \(\mathrm{-N{=}O}\) unit to form an \(N\)-nitrosoamine. The structural distinction stated in the Reason explains both parts of the Assertion.
320. Compare the predicted reactions below.
| Row | Secondary amine | Predicted nitrous-acid product | Rapid \(\mathrm{N_2}\) evolution |
|---|
| P | \(\mathrm{(CH_3)_2NH}\) | \(\mathrm{(CH_3)_2N-N{=}O}\) | No |
| Q | \(\mathrm{(C_2H_5)_2NH}\) | \(\mathrm{(C_2H_5)_2N-N{=}O}\) | No |
| R | \(\mathrm{C_6H_5NHCH_3}\) | \(\mathrm{C_6H_5N(CH_3)-N{=}O}\) | No |
| S | \(\mathrm{(CH_3)_2NH}\) | \(\mathrm{CH_3OH}\) | Yes |
The valid rows are:
ⓐ. P and S only
ⓑ. Q and S only
ⓒ. P, Q and R only
ⓓ. P, Q and S only
Correct Answer: P, Q and R only
Explanation: Dimethylamine forms \(N\)-nitrosodimethylamine and does not release nitrogen rapidly. Diethylamine behaves similarly and gives \(N\)-nitrosodiethylamine. \(N\)-Methylaniline is also a secondary amine and forms the corresponding \(N\)-nitroso derivative. Row S incorrectly assigns primary-aliphatic behaviour to dimethylamine. The valid rows are therefore P, Q, and R.
