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Amines and Amides: Structure, Basicity, and Reactions

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Question 1Question →

Which of the following organic compounds acts as a weak base in an aqueous solution due to the presence of an unshared pair of electrons on its nitrogen atom?

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Answer: Methylamine (CH3NH2CH_3NH_2)

Answer

Methylamine (CH3NH2CH_3NH_2)
Methylamine (CH3NH2CH_3NH_2) is an aliphatic amine. The nitrogen atom retains a localized lone pair of electrons that readily accepts a proton from water to form a methylammonium ion and a hydroxide ion, demonstrating basic behavior in aqueous solution.

Step-by-Step Solution

1
Identify the functional groups present in each compound.
Methylamine (CH3NH2CH_3NH_2) is an amine; ethanamide (CH3CONH2CH_3CONH_2) is an amide; ethanoic acid (CH3COOHCH_3COOH) is a carboxylic acid; ethanol (CH3CH2OHCH_3CH_2OH) is an alcohol.
Basic properties in organic nitrogen compounds depend on the availability of the nitrogen lone pair.
2
Evaluate the availability of the unshared pair of electrons on the nitrogen atom.
In primary aliphatic amines like methylamine, the lone pair on nitrogen is readily available to accept a proton (H+H^+). In amides like ethanamide, resonance delocalizes the nitrogen lone pair toward the carbonyl oxygen, removing its basic character.
Proton acceptance (Lewis/Brønsted-Lowry basicity) requires an available lone pair.

Key Concept

Basicity of Amines versus Amides
Question 2Question →

Match each nitrogen-containing organic compound on the left with its correct structural classification on the right.

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Items

Methylamine (CH3NH2CH_3NH_2)
Ethanamide (CH3CONH2CH_3CONH_2)
Phenylamine (C6H5NH2C_6H_5NH_2)
Dimethylamine ((CH3)2NH(CH_3)_2NH)

Matches

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Answer

Methylamine pairs with Primary aliphatic amine; Ethanamide pairs with Neutral organic amide; Phenylamine pairs with Primary aromatic amine; Dimethylamine pairs with Secondary aliphatic amine.
Each compound matches its unique chemical definition: Methylamine is a 1∘1^\circ aliphatic amine, Ethanamide is a neutral amide, Phenylamine is a 1∘1^\circ aromatic amine, and Dimethylamine is a 2∘2^\circ aliphatic amine.

Step-by-Step Solution

1
Examine the functional groups and substituents attached to nitrogen in each compound.
Methylamine (CH3NH2CH_3NH_2) and Phenylamine (C6H5NH2C_6H_5NH_2) each have one organic group attached (1∘1^\circ). Dimethylamine ((CH3)2NH(CH_3)_2NH) has two organic groups attached (2∘2^\circ). Ethanamide (CH3CONH2CH_3CONH_2) has a carbonyl group (C=OC=O) linked directly to nitrogen.
The number of alkyl/aryl groups determines amine degree (1∘,2∘,3∘1^\circ, 2^\circ, 3^\circ), while a carbonyl-nitrogen bond defines an amide.
2
Classify by aliphatic, aromatic, or neutral amide characteristics.
Methylamine contains an alkyl group (1∘1^\circ aliphatic amine). Phenylamine contains a benzene ring (1∘1^\circ aromatic amine). Dimethylamine has two alkyl groups (2∘2^\circ aliphatic amine). Ethanamide is an amide and exhibits neutral aqueous behavior due to lone pair resonance delocalization.
Structure and electronic delocalization determine both classification and relative basicity.

Key Concept

Classification of amines (primary, secondary, aromatic, aliphatic) and amides.
Question 3Question →

Match each nitrogen-containing organic compound on the left with the statement on the right that accurately accounts for its aqueous basicity and lone-pair electronic behavior.

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Items

Dimethylamine, (CH3)2NH(CH_3)_2NH
Phenylamine, C6H5NH2C_6H_5NH_2
Ethanamide, CH3CONH2CH_3CONH_2
Triethylamine, (C2H5)3N(C_2H_5)_3N

Matches

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Answer

Dimethylamine matches the statement describing higher aqueous basicity than ammonia due to inductive donation and solvation; Phenylamine matches the statement describing reduced basicity from aromatic resonance delocalization; Ethanamide matches the statement describing neutrality caused by carbonyl resonance; Triethylamine matches the statement describing steric hindrance affecting conjugate acid solvation.
The correct matches reflect fundamental physical-organic chemistry principles governing nitrogen basicity: Dimethylamine combines inductive donation with high conjugate acid solvation stability; Phenylamine suffers basicity loss from aromatic resonance delocalization; Ethanamide lone-pair delocalization into the carbonyl group yields a neutral compound; Triethylamine basicity in water is moderated by steric crowding that interferes with hydration of the ammonium cation.

Step-by-Step Solution

1
Analyze the electronic structure of Dimethylamine ((CH3)2NH(CH_3)_2NH).
Two methyl groups supply electron density via +I+I inductive effects, enhancing nitrogen lone-pair availability, while the secondary cation remains readily solvated by water.
Secondary aliphatic amines are generally the strongest bases in aqueous media.
2
Analyze the resonance interactions in Phenylamine (C6H5NH2C_6H_5NH_2).
The unshared electron pair on nitrogen participates in resonance with the benzene ring, lowering lone-pair availability.
Aromatic amines are significantly weaker bases than ammonia and aliphatic amines.
3
Examine the functional group characteristics of Ethanamide (CH3CONH2CH_3CONH_2).
Resonance delocalization between nitrogen's lone pair and the adjacent C=OC=O double bond (O=C−N↔−O−C=N+O=C-N \leftrightarrow ^-O-C=N^+) deprives nitrogen of basic character.
Amides behave as neutral organic compounds in aqueous solution.
4
Evaluate steric effects in Triethylamine ((C2H5)3N(C_2H_5)_3N).
Three ethyl groups create steric crowding around the nitrogen cation, hindering stabilization through hydration in water.
In aqueous solution, tertiary aliphatic amines are often weaker bases than secondary aliphatic amines due to solvation factors.

Key Concept

Relative basicity of aliphatic amines, aromatic amines, and amides governed by inductive, resonance, and solvation steric effects.
Estimated Time:2m 0s
Question 4Question →

Match each organic nitrogen compound on the left with its corresponding relative basicity or acid-base structural property on the right.

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Items

Phenylamine (C6H5NH2C_6H_5NH_2)
Methylamine (CH3NH2CH_3NH_2)
Ethanamide (CH3CONH2CH_3CONH_2)
Dimethylamine ((CH3)2NH(CH_3)_2NH)

Matches

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Answer

Phenylamine matches with being weaker than ammonia due to aromatic delocalization; Methylamine matches with being stronger than ammonia due to +I inductive effect; Ethanamide matches with being neutral due to carbonyl resonance; Dimethylamine matches with being stronger than methylamine in aqueous solution.
Phenylamine is less basic than ammonia because its lone pair is delocalized across the aromatic ring. Methylamine is more basic than ammonia due to inductive electron donation by the methyl group. Ethanamide is neutral because its lone pair participates in resonance with the carbonyl group. Dimethylamine is more basic than methylamine in water due to two electron-donating methyl groups.

Step-by-Step Solution

1
Analyze how structural features influence nitrogen lone pair availability.
Electron-donating alkyl groups (+I effect) enhance lone pair availability (increasing basicity), while electron-withdrawing groups or resonance delocalization decrease lone pair availability (decreasing basicity).
Lewis/Brønsted-Lowry basicity of nitrogen compounds depends directly on lone pair availability to accept a proton.
2
Evaluate phenylamine and ethanamide.
Phenylamine delocalizes its lone pair into the benzene ring, making it weaker than NH3NH_3. Ethanamide delocalizes its lone pair into the C=OC=O double bond, making it neutral in aqueous solution.
Resonance delocalization significantly stabilizes the unprotonated state and lowers basicity.
3
Compare methylamine and dimethylamine.
Methylamine has one alkyl group increasing basicity over NH3NH_3. Dimethylamine has two alkyl groups supplying greater electron density, making it more basic than methylamine in aqueous solution.
Inductive electron donation by methyl groups stabilizes the positive conjugate ammonium ion.

Key Concept

Relative basicity of amines and amides based on inductive and resonance effects
Question 5Question →

Unlike aliphatic amines, amides such as ethanamide (CH3CONH2CH_3CONH_2) do not exhibit basic properties in aqueous solution. Which of the following best explains this neutral behavior?

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Answer: The lone pair of electrons on the nitrogen atom is delocalized by resonance with the adjacent carbonyl group.

Answer

The lone pair of electrons on the nitrogen atom is delocalized by resonance with the adjacent carbonyl group.
The neutral character of amides is due to resonance stabilization. The non-bonding lone pair of electrons on the nitrogen atom overlaps with the π\pi-system of the adjacent carbonyl group (C=OC=O). This delocalization significantly decreases the electron density on nitrogen, preventing it from accepting a proton (H+H^+) and acting as a base.

Step-by-Step Solution

1
Identify the functional group and structural elements of ethanamide (CH3CONH2CH_3CONH_2).
Ethanamide contains an amino group (−NH2-NH_2) directly attached to a carbonyl group (−C=O-C=O).
Understanding the adjacent arrangement of the carbonyl carbon and nitrogen is essential for evaluating electronic effects.
2
Analyze the availability of the nitrogen lone pair for protonation.
The nitrogen atom has a lone pair of electrons, but it interacts with the π\pi-orbital of the carbonyl group, forming a resonance structure: CH3−C(O−)=NH2+CH_3-C(O^-)=NH_2^+.
Basicity depends on the availability of a lone pair to accept a proton (H+H^+). Delocalization drastically reduces lone pair availability.
3
Conclude the acid-base nature of the molecule.
Because the lone pair is delocalized, ethanamide cannot readily act as a proton acceptor, rendering it neutral in aqueous solution.
This explains the contrast between basic amines (localized lone pair) and neutral amides (delocalized lone pair).

Key Concept

Neutrality of amides due to resonance delocalization of the nitrogen lone pair into the adjacent carbonyl group
Estimated Time:45s
Question 6Question →

Primary aliphatic amines react with nitrous acid (HNO2HNO_2) at room temperature to yield an alcohol along with rapid effervescence. Which gas is evolved during this reaction?

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Answer: Nitrogen gas (N2N_2)

Answer

Nitrogen gas (N2N_2)
Primary aliphatic amines react with cold nitrous acid (HNO2HNO_2) to form an unstable diazonium compound that decomposes rapidly at room temperature. This decomposition produces an alcohol, water, and liberates nitrogen gas (N2N_2), which is observed as effervescence.

Step-by-Step Solution

1
Identify the functional group and reagent
The reaction involves a primary aliphatic amine (R−NH2R-NH_2) and nitrous acid (HNO2HNO_2).
Nitrous acid is generated in situ from sodium trioxonitrate(III) (NaNO2NaNO_2) and dilute hydrochloric acid (HClHCl).
2
Write the chemical reaction equation
R−NH2+HNO2→R−OH+N2↑+H2OR-NH_2 + HNO_2 \rightarrow R-OH + N_2 \uparrow + H_2O
The aliphatic diazonium salt formed as an intermediate is highly unstable and decomposes immediately at room temperature.
3
Determine the gas released
The effervescence observed is due to the release of nitrogen gas (N2N_2).
Deamination of aliphatic primary amines yields nitrogen gas as the characteristic gaseous product.

Key Concept

Reaction of primary aliphatic amines with nitrous acid to yield alcohols and nitrogen gas
Estimated Time:1m 0s
Question 7Question →

An organic compound XX with the molecular formula C3H9NC_3H_9N reacts with cold nitrous acid (HNO2HNO_2) at 0−5∘C0-5^\circ\text{C} to produce a yellow, oily liquid without the evolution of nitrogen gas. Which of the following is the IUPAC name of compound XX?

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Answer: NN-methylethanamine

Answer

The correct compound is NN-methylethanamine because secondary aliphatic amines react with nitrous acid (HNO2HNO_2) at low temperatures to produce insoluble, yellow oily NN-nitrosamines without liberating nitrogen gas.
Secondary aliphatic amines such as NN-methylethanamine react with cold nitrous acid (HNO2HNO_2) to form NN-nitrosamines. These compounds are insoluble in water and appear as yellow oily liquids. Because no aliphatic diazonium intermediate breaks down to release gas, no nitrogen gas effervescence is observed.

Step-by-Step Solution

1
Classify the given structural isomers of C3H9NC_3H_9N by amine degree.
Propan-1-amine and propan-2-amine are primary (1∘1^\circ) amines; NN-methylethanamine is a secondary (2∘2^\circ) amine; N,NN,N-dimethylmethanamine is a tertiary (3∘3^\circ) amine.
Amine classification determines the distinct reaction pathway and observable products with nitrous acid.
2
Analyze the reaction behavior of each amine class with nitrous acid (HNO2HNO_2) at 0−5∘C0-5^\circ\text{C}.
Primary amines evolve N2N_2 gas and form alcohols; secondary amines form yellow oily NN-nitrosamines without gas evolution; tertiary amines form soluble nitrite salts.
Nitrous acid is used as a qualitative reagent to distinguish between 1∘1^\circ, 2∘2^\circ, and 3∘3^\circ amines.
3
Match the observation (yellow oily liquid, no nitrogen gas) to the correct compound.
The observation corresponds uniquely to a secondary amine, which is NN-methylethanamine.
Only secondary amines undergo nitrosation at the nitrogen atom to form neutral, oily NN-nitrosamine layers.

Key Concept

Distinction tests for primary, secondary, and tertiary amines using nitrous acid (HNO2HNO_2)
Question 8Question →

Match each organic reaction involving an amine or amide on the left with its corresponding principal product on the right.

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Items

Reduction of propanamide (CH3CH2CONH2CH_3CH_2CONH_2) using LiAlH4LiAlH_4 in dry ether
Reaction of propanamide (CH3CH2CONH2CH_3CH_2CONH_2) with bromine (Br2Br_2) in aqueous KOHKOH
Alkaline hydrolysis of propanamide (CH3CH2CONH2CH_3CH_2CONH_2) by boiling with aqueous NaOHNaOH
Acylation of methylamine (CH3NH2CH_3NH_2) using ethanoyl chloride (CH3COClCH_3COCl)

Matches

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Answer

Reduction of propanamide with LiAlH4LiAlH_4 pairs with propylamine; Reaction of propanamide with Br2/KOHBr_2/KOH pairs with ethylamine; Alkaline hydrolysis of propanamide pairs with sodium propanoate and ammonia; Acylation of methylamine with ethanoyl chloride pairs with NN-methylethanamide.
Each reaction pair is determined by its specific mechanistic pathway: LiAlH4LiAlH_4 reduces the carbonyl group to methylene (retaining carbon count to form propylamine); Br2/KOHBr_2/KOH undergoes Hofmann degradation to lose the carbonyl carbon (forming ethylamine); basic hydrolysis cleaves the C−NC-N bond (yielding sodium propanoate and ammonia); and acylation of methylamine with ethanoyl chloride produces the substituted amide (NN-methylethanamide).

Step-by-Step Solution

1
Analyze the reduction reaction of primary amides.
Reducing CH3CH2CONH2CH_3CH_2CONH_2 with LiAlH4LiAlH_4 reduces the C=OC=O bond to a −CH2−-CH_2- group without altering the total carbon count, yielding CH3CH2CH2NH2CH_3CH_2CH_2NH_2 (propylamine).
Amide reduction retains the full carbon skeleton.
2
Identify the reaction of primary amides with Br2Br_2 and KOHKOH.
This is Hofmann degradation, which removes the carbonyl carbon (C=OC=O) as carbonate, reducing the carbon length by 1. Propanamide (3 carbons) yields ethylamine (2 carbons).
Hofmann degradation shortens the carbon chain by one atom.
3
Examine the basic hydrolysis of amides.
Nucleophilic attack of OH−OH^- on the carbonyl carbon of propanamide cleaves the amide bond to generate propanoate anion (forming sodium propanoate with Na+Na^+) and ammonia gas.
Base hydrolysis of amides yields a carboxylate salt and ammonia.
4
Examine the nucleophilic substitution between methylamine and ethanoyl chloride.
The nitrogen lone pair of methylamine attacks ethanoyl chloride, releasing HClHCl to form a secondary amide, NN-methylethanamide (CH3CONHCH3CH_3CONHCH_3).
Primary amines undergo acylation to form secondary amides.

Key Concept

Chemical Reactions and Interconversions of Amines and Amides
Estimated Time:1m 30s
Amines and Amides: Structure, Basicity, and Reactions Practice Questions — JAMB UTME | Examkin