101. Examine the effects listed below.
| Row | Structural or chemical change | Given effect on the nitrogen lone pair |
|---|
| P | Attachment of electron-releasing alkyl groups | Electron density is always withdrawn from nitrogen |
| Q | Conjugation with a carbonyl group | The lone pair becomes completely localised on nitrogen |
| R | Severe crowding around nitrogen | Approach of an electrophile may be hindered |
| S | Protonation of nitrogen | An additional free lone pair is produced |
The valid row is:
ⓐ. P only
ⓑ. Q only
ⓒ. R only
ⓓ. R and S only
Correct Answer: R only
Explanation: Steric crowding can obstruct the approach of an electrophile to the nitrogen lone pair, so row R is valid. Electron-releasing alkyl groups generally push electron density toward nitrogen rather than withdrawing it. Conjugation with a carbonyl group delocalises the lone pair instead of localising it completely. Protonation uses the lone pair to form a new nitrogen–hydrogen bond. Lone-pair availability depends on both electronic effects and physical accessibility.
102. Acylation of an amine generally makes the nitrogen less nucleophilic because:
ⓐ. its lone pair becomes delocalised toward the carbonyl group
ⓑ. nitrogen becomes less electronegative than carbon
ⓒ. the carbonyl oxygen transfers a second lone pair to nitrogen
ⓓ. all carbon–nitrogen bonds become non-polar
Correct Answer: its lone pair becomes delocalised toward the carbonyl group
Explanation: Acylation places nitrogen directly next to a carbonyl group. The nitrogen lone pair can overlap with the carbonyl system and become delocalised through resonance. A delocalised pair is less available for direct donation to a new electrophile. The resulting amide nitrogen is therefore much less nucleophilic than the nitrogen of a comparable amine. The reduced reactivity arises from electron delocalisation rather than from loss of the carbon–nitrogen bond.
103. A neutral amine is converted into its protonated ammonium ion. Its immediate nucleophilicity at nitrogen decreases mainly because:
ⓐ. the nitrogen atom is removed from the molecule
ⓑ. protonation creates a second unshared pair
ⓒ. the attached carbon groups become strongly electron deficient after protonation
ⓓ. protonation consumes the lone pair in the new nitrogen–hydrogen bond
Correct Answer: protonation consumes the lone pair in the new nitrogen–hydrogen bond
Explanation: A neutral amine attacks many electrophiles by donating its nitrogen lone pair. During protonation, that pair is donated to \(\mathrm{H^+}\) and becomes part of a new nitrogen–hydrogen bond. The protonated nitrogen no longer possesses the same freely available lone pair. It is also positively charged, making further electron-pair donation unfavourable. Protonation can therefore switch off or greatly reduce the nucleophilic behaviour of the nitrogen centre.
104. In a substitution reaction where access to the electrophilic carbon is strongly restricted by steric effects, methylamine may react faster than tert-butylamine because:
ⓐ. methylamine is less hindered during electrophile approach
ⓑ. tert-butylamine has no nitrogen lone pair
ⓒ. methylamine is a quaternary ammonium ion
ⓓ. tert-butylamine carries a permanent positive charge at nitrogen
Correct Answer: methylamine is less hindered during electrophile approach
Explanation: Both methylamine and tert-butylamine are neutral primary amines with a lone pair on nitrogen. The bulky tert-butyl group occupies substantial space around the nitrogen centre. This crowding can make close approach to an electrophilic carbon more difficult. Methylamine has a much smaller methyl group and offers less steric resistance. The comparison shows why nucleophilicity cannot always be predicted from electron-releasing effects alone.
105. A molecule of water approaches trimethylamine. The most favourable hydrogen-bond arrangement is:
ⓐ. \(\mathrm{O-H\cdots N(CH_3)_3}\), with water as donor
ⓑ. \(\mathrm{(CH_3)_3N-H\cdots O}\), with trimethylamine donating
ⓒ. \(\mathrm{O\cdots C-H}\), with nitrogen not participating
ⓓ. \(\mathrm{N-H\cdots N}\), with both molecules acting as amines
Correct Answer: \(\mathrm{O-H\cdots N(CH_3)_3}\), with water as donor
Explanation: Trimethylamine has an available lone pair on nitrogen. It does not contain an \(\mathrm{N-H}\) bond, so it cannot donate an ordinary nitrogen-centred hydrogen bond. Water contains polar \(\mathrm{O-H}\) bonds and can donate a hydrogen bond. The hydrogen of water is attracted toward the electron-rich nitrogen lone pair. Thus, a tertiary amine can interact strongly with water even though it cannot hydrogen-bond to another tertiary amine in the same donor–acceptor manner.
106. Hydrogen bonding in amines is generally weaker than in comparable alcohols because:
ⓐ. nitrogen has no lone pair in an amine
ⓑ. nitrogen is less electronegative than oxygen
ⓒ. alcohols contain ionic \(\mathrm{O-H}\) bonds
ⓓ. amines cannot form intermolecular attractions
Correct Answer: nitrogen is less electronegative than oxygen
Explanation: Oxygen is more electronegative than nitrogen. An \(\mathrm{O-H}\) bond is therefore generally more strongly polarised than an \(\mathrm{N-H}\) bond. The hydrogen in an alcohol carries a stronger partial positive character and participates in stronger hydrogen bonding. Amines still form hydrogen bonds when suitable donor and acceptor sites are present. The lower electronegativity of nitrogen makes these interactions weaker than the corresponding oxygen-centred interactions in comparable alcohols.
107. Examine the hydrogen-bonding descriptions below.
| Row | Amine class | \(\mathrm{N-H}\) bond present | Ordinary self hydrogen-bond donation | Hydrogen-bond acceptance through nitrogen |
|---|
| P | Primary | Yes | Yes | Yes |
| Q | Secondary | No | No | Yes |
| R | Tertiary | Yes | Yes | No |
| S | Quaternary ammonium ion | No | Yes | Yes |
The valid row is:
ⓐ. Q only
ⓑ. R only
ⓒ. P only
ⓓ. P and S only
Correct Answer: P only
Explanation: A neutral primary amine contains at least one \(\mathrm{N-H}\) bond and one nitrogen lone pair. It can therefore donate and accept hydrogen bonds, making row P valid. A secondary amine normally contains one \(\mathrm{N-H}\) bond, so row Q incorrectly states that none is present. A tertiary amine has no \(\mathrm{N-H}\) bond but retains a lone pair, reversing the assignments in row R. A quaternary ammonium ion lacks a nitrogen lone pair and cannot serve as an ordinary lone-pair acceptor.
108. The most suitable boiling-point order for ethanol, ethylamine, and ethane is:
ⓐ. ethane \(>\) ethylamine \(>\) ethanol
ⓑ. ethanol \(>\) ethylamine \(>\) ethane
ⓒ. ethylamine \(>\) ethanol \(>\) ethane
ⓓ. ethanol \(>\) ethane \(>\) ethylamine
Correct Answer: ethanol \(>\) ethylamine \(>\) ethane
Explanation: Ethanol forms strong oxygen-centred intermolecular hydrogen bonds. Ethylamine also forms hydrogen bonds, but nitrogen is less electronegative than oxygen, so its hydrogen bonding is generally weaker. Ethane lacks an appropriate strongly polar bond and does not form ordinary hydrogen-bonded networks. It is held together mainly by weaker dispersion forces. The increasing strength of intermolecular association gives the order ethanol, then ethylamine, then ethane.
109. Among primary amines having the same molecular formula, increased branching usually lowers the boiling point because branching:
ⓐ. converts the primary amine into a tertiary amine
ⓑ. removes both nitrogen–hydrogen bonds
ⓒ. makes nitrogen more electronegative than oxygen
ⓓ. reduces surface contact and dispersion forces
Correct Answer: reduces surface contact and dispersion forces
Explanation: Branching produces a more compact molecular shape. Compact molecules generally have less effective surface contact with neighbouring molecules than their less-branched isomers. Reduced contact weakens London dispersion attractions. The amine may still possess the same number of \(\mathrm{N-H}\) bonds and remain capable of hydrogen bonding. The boiling-point decrease arises mainly from molecular shape and packing rather than from a change in amine degree.
110. A graph plots boiling point on the vertical axis and carbon-atom count on the horizontal axis for a homologous series of straight-chain primary amines. The most reasonable overall shape is:
ⓐ. a horizontal line because every member contains \(\mathrm{NH_2}\)
ⓑ. a downward trend because longer chains weaken all attractions
ⓒ. an upward trend as dispersion forces strengthen with chain size
ⓓ. an alternating line determined only by odd or even carbon count
Correct Answer: an upward trend as dispersion forces strengthen with chain size
Explanation: \( \textbf{Graph variables:} \)
The horizontal axis represents the number of carbon atoms.
The vertical axis represents boiling point.
\( \textbf{Feature held broadly constant:} \)
Every member is a straight-chain primary amine with an \(\mathrm{-NH_2}\) group.
\( \textbf{Change across the series:} \)
Carbon count and molar mass increase from left to right.
\( \textbf{Intermolecular consequence:} \)
Larger electron clouds produce stronger London dispersion forces.
\( \textbf{Energy consequence:} \)
More thermal energy is required to separate the larger molecules.
\( \textbf{Expected graph direction:} \)
Boiling point therefore increases overall as carbon count increases.
\( \textbf{Final interpretation:} \)
The graph should show an upward trend, although individual values need not lie on a perfectly straight line.
111. Three isomeric amines P, Q, and R have the formula \(\mathrm{C_3H_9N}\). P has two \(\mathrm{N-H}\) bonds, Q has one \(\mathrm{N-H}\) bond, and R has none. Their most likely decreasing boiling-point order is:
ⓐ. P \(>\) Q \(>\) R
ⓑ. R \(>\) Q \(>\) P
ⓒ. Q \(>\) P \(>\) R
ⓓ. R \(>\) P \(>\) Q
Correct Answer: P \(>\) Q \(>\) R
Explanation: P is a primary amine because it has two nitrogen-bonded hydrogen atoms. Q is secondary and retains one \(\mathrm{N-H}\) bond. R is tertiary and has no \(\mathrm{N-H}\) bond. Since the three compounds are isomeric, their molar masses are equal, making hydrogen-bonding capacity the major distinguishing factor. The extent of self-association decreases from P to Q to R, giving the boiling-point order P \(>\) Q \(>\) R.
112. Consider the following statements about water solubility of neutral amines.
Statement I: Primary, secondary, and tertiary amines can accept hydrogen bonds from water.
Statement II: Solubility generally decreases as the hydrophobic carbon portion becomes larger.
Statement III: A tertiary amine must contain an \(\mathrm{N-H}\) bond to dissolve in water.
Select the applicable combination.
ⓐ. I only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I and II only
Correct Answer: I and II only
Explanation: Neutral primary, secondary, and tertiary amines ordinarily possess a nitrogen lone pair. Each class can therefore accept hydrogen bonds from water. As the hydrocarbon portion grows, the non-polar contribution becomes increasingly important and opposes hydration. Statement III is not valid because a tertiary amine can accept hydrogen bonds without containing an \(\mathrm{N-H}\) bond. Water solubility depends on the balance between the polar amino group and the hydrophobic carbon framework.
113. A graph shows water solubility on the vertical axis and carbon-chain length on the horizontal axis for a homologous series of straight-chain primary amines. The most reasonable overall trend is:
ⓐ. solubility remains constant because every member contains \(\mathrm{NH_2}\)
ⓑ. solubility decreases as the carbon chain becomes longer
ⓒ. solubility rises because each additional carbon forms another hydrogen bond
ⓓ. solubility first becomes zero and then rises sharply
Correct Answer: solubility decreases as the carbon chain becomes longer
Explanation: \( \textbf{Axes:} \)
The horizontal axis represents increasing carbon-chain length.
The vertical axis represents water solubility.
\( \textbf{Common polar feature:} \)
Every member contains one primary amino group capable of interacting with water.
\( \textbf{Changing structural feature:} \)
The non-polar hydrocarbon portion grows as carbon atoms are added.
\( \textbf{Hydration effect:} \)
The amino group continues to form hydrogen bonds with water.
\( \textbf{Hydrophobic effect:} \)
The larger carbon chain becomes progressively less compatible with the polar solvent.
\( \textbf{Overall balance:} \)
The hydrophobic contribution increasingly outweighs the benefit of the single amino group.
\( \textbf{Graph interpretation:} \)
The plotted solubility therefore shows an overall downward trend with increasing chain length.
114. Methylamine is much more soluble in water than aniline mainly because:
ⓐ. aniline has no nitrogen lone pair
ⓑ. methylamine is an ionic compound in its pure state
ⓒ. aniline has a much larger hydrophobic phenyl group
ⓓ. aniline contains no polar carbon–nitrogen bond
Correct Answer: aniline has a much larger hydrophobic phenyl group
Explanation: Both methylamine and aniline contain a nitrogen lone pair that can interact with water. Methylamine has only a small methyl group attached to nitrogen. Aniline contains a large aromatic phenyl ring that is comparatively non-polar and hydrophobic. The phenyl ring greatly reduces the overall compatibility of the molecule with water. Aniline is therefore less water soluble than lower aliphatic amines despite retaining a polar amino group.
115. Conversion of a tertiary amine \(\mathrm{R_3N}\) into the more water-soluble ionic form is represented by:
ⓐ. \(\mathrm{R_3N+H^+\rightarrow R_3NH^+}\)
ⓑ. \(\mathrm{R_3N+OH^-\rightarrow R_3NOH^-}\)
ⓒ. \(\mathrm{R_3N\rightarrow R_2NH+R}\)
ⓓ. \(\mathrm{R_3N+H_2O\rightarrow R_3N=O+H_2}\)
Correct Answer: \(\mathrm{R_3N+H^+\rightarrow R_3NH^+}\)
Explanation: A tertiary amine contains an available nitrogen lone pair even though it has no \(\mathrm{N-H}\) bond. The lone pair can accept a proton from an acid. Protonation produces the substituted ammonium ion \(\mathrm{R_3NH^+}\). The charged ion interacts strongly with polar water molecules through ion–dipole attractions. Increased water solubility therefore does not require the original neutral amine to possess a nitrogen-bonded hydrogen.
116. An organic solution contains a neutral amine mixed with a neutral hydrocarbon. Shaking the solution with dilute hydrochloric acid transfers the amine mainly into the aqueous layer because:
ⓐ. the hydrocarbon reacts to form a soluble chloride salt
ⓑ. hydrochloric acid removes the carbon chain from the amine
ⓒ. the amine loses nitrogen and becomes an alcohol
ⓓ. protonation converts the amine into a water-soluble salt
Correct Answer: protonation converts the amine into a water-soluble salt
Explanation: The neutral amine initially has sufficient non-polar character to remain in the organic phase. Dilute hydrochloric acid donates a proton to the nitrogen lone pair. The product is an ammonium chloride salt carrying a positive charge on nitrogen. Ionic species are strongly hydrated and preferentially enter the aqueous phase. The neutral hydrocarbon does not form a comparable salt and therefore remains mainly in the organic layer.
117. After an amine has been transferred into an aqueous layer as \(\mathrm{RNH_3^+Cl^-}\), treatment with excess sodium hydroxide followed by extraction with an organic solvent will mainly:
ⓐ. convert the amine salt into an amide
ⓑ. keep the ammonium ion permanently in the aqueous layer
ⓒ. oxidise the carbon chain to a carboxylic acid
ⓓ. liberate the neutral amine into the organic layer
Correct Answer: liberate the neutral amine into the organic layer
Explanation: Hydroxide ion removes a proton from the substituted ammonium ion. This acid–base step regenerates the neutral amine.
\[
\mathrm{RNH_3^+ + OH^- \rightarrow RNH_2 + H_2O}
\]
The neutral amine is less strongly hydrated than its ionic salt. It can therefore be extracted into a suitable organic solvent. Acidification and basification provide a reversible method for moving an amine between aqueous and organic phases.
118. The broad physical-state trend across an increasing homologous series of alkylamines is:
ⓐ. solids at low mass, liquids at intermediate mass, and gases at high mass
ⓑ. gases at low mass, liquids at intermediate mass, and solids at high mass
ⓒ. liquids at low mass, gases at intermediate mass, and solids at high mass
ⓓ. gases at low mass, solids at intermediate mass, and liquids at high mass
Correct Answer: gases at low mass, liquids at intermediate mass, and solids at high mass
Explanation: Small alkylamines have relatively low molar masses and comparatively weak overall dispersion forces. Several lower members therefore have low boiling points and may occur as gases under ordinary conditions. Intermediate members have stronger intermolecular attractions and are commonly liquids. Still larger members possess stronger dispersion forces and may occur as solids. The trend reflects increasing molecular size rather than a change in the basic amine functional group.
119. A freshly purified aromatic amine is colourless but gradually becomes darker on standing in air. The most reasonable explanation is:
ⓐ. the amine is converted completely into elemental carbon
ⓑ. nitrogen evaporates while the aromatic ring remains
ⓒ. exposure to air may produce coloured oxidation products
ⓓ. all aromatic amines are naturally black solids
Correct Answer: exposure to air may produce coloured oxidation products
Explanation: Pure amines are commonly colourless when freshly prepared. Some aromatic amines are sensitive to oxygen in the air. Slow oxidation can generate small amounts of coloured products or complex mixtures. These impurities cause the sample to darken even though the original amine may be colourless. The colour change is therefore an ageing or oxidation observation rather than an essential colour of every aromatic amine.
120. A colourless, volatile compound has a strong fish-like odour and dissolves in dilute hydrochloric acid to give a clear aqueous solution. The combined observations are most consistent with:
ⓐ. a non-polar hydrocarbon that does not react with acids
ⓑ. a lower amine that forms a water-soluble ammonium salt
ⓒ. a high-molar-mass amide that is permanently ionic
ⓓ. an aromatic hydrocarbon undergoing complete oxidation
Correct Answer: a lower amine that forms a water-soluble ammonium salt
Explanation: Volatility and a characteristic fish-like odour are commonly associated with lower amines. The decisive additional observation is dissolution in dilute hydrochloric acid. An amine accepts a proton and forms an ionic ammonium chloride salt. The salt is strongly hydrated and dissolves readily in the aqueous medium. The observations together support an amine assignment more strongly than odour or appearance considered alone.
