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Concept 1: Iupac Nomenclature

Question. Match the following:

Answer:
The correct matched pairs are:

In simple words: This matching couples the structure of different amines with their corresponding IUPAC names.

Exam Tip: Be sure to distinguish between primary, secondary, and tertiary amine IUPAC names where substituents are prefixed with 'N-'.

Question. Write the IUPAC name of following:
a. C6H5-NH-CH3 b. CH3-NH-C2H5 c.(C6H5)2-N-CH3

Answer:

In simple words: For secondary and tertiary amines, the larger alkyl group is treated as the parent, while the smaller attached group is prefixed with 'N-'.

Exam Tip: When naming tertiary aromatic amines, write the N-substituents in alphabetical order before the parent aniline name.

Question. The IUPAC name of (CH3)2-N-C2H5:
(a) N,N-Diethylethanamine
(b) N,N-Dimethylethanamine
(c) N,N-Ethylmethylmethanamine
(d) Dimethylmethanamine
Answer: (b) N,N-Dimethylethanamine
In simple words: Since the longest continuous carbon chain attached to nitrogen contains two carbons (ethane), and two methyl groups are also attached, it is named N,N-dimethylethanamine.

Exam Tip: Always identify and name the longest chain first to find the correct parent amine name.

Question. Choose the appropriate answer of the following:
a) IUPAC name of CH3-NH-CH3
(a) Ethylmethylamine
(b) Methylethylamine
(c) N-Methylethanamine
(d) N-Ethylmethanamine
Answer: (c) N-Methylethanamine
In simple words: The options correspond to N-methylethanamine (derived from the structure \( \text{CH}_3\text{-NH-C}_2\text{H}_5 \) which contains a typographical error in the formula printed in the question).

Exam Tip: Pay attention to formula-option mismatches; \( \text{CH}_3\text{-NH-CH}_3 \) would strictly be N-methylmethanamine, but the options suggest a ethyl group substituent.

Question. Common name of CH3CH2NH2 ?
(a) Ethylamine
(b) Ehanamine
(c) Dimethylamine
(d) None of the options
Answer: (a) Ethylamine
In simple words: The common name for this primary amine containing two carbons is ethylamine.

Exam Tip: The common name of aliphatic amines is written as alkyl group + amine, while the IUPAC name ends in "-anamine".

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Question. IUPAC name of (C6H5)2NH ?
(a) Diphenylamine
(b) N-Phenylbenzenamine
(c) 1,2-Diphenylamine
(d) All of the options
Answer: (b) N-Phenylbenzenamine
In simple words: Diphenylamine is the common name of the compound, whereas N-Phenylbenzenamine is its systematic IUPAC name.

Exam Tip: For aromatic amines, the benzene ring with the amine group is named as benzenamine.

Question. Common name of (CH3)2CH-NH2
(a) Isopropylamine
(b) Ethaemethanamine
(c) Methaneehaneamine
(d) 2-Methylethanamine
Answer: (a) Isopropylamine
In simple words: The common name is isopropylamine because the amino group is attached to an isopropyl group.

Exam Tip: The IUPAC name for this compound is propan-2-amine, while the common name is isopropylamine.

Question. Arrange the following alkyl groups in decreasing order
Methyl,Ethyl,Isopropyl,n-Butyl

Answer: Isopropyl > n-Butyl > Ethyl > Methyl (based on the decreasing order of +I inductive effect) or n-Butyl > Isopropyl > Ethyl > Methyl (based on decreasing carbon chain length). Primarily, in terms of +I inductive effect, the order is: Isopropyl > n-Butyl > Ethyl > Methyl.
In simple words: This list ranks alkyl groups by their capacity to donate electrons, where secondary groups like isopropyl have a stronger effect than primary chains.

Exam Tip: Electron-donating (+I) inductive strength increases with the degree of carbon branching and size.

Concept 2: Basic Character of Amines

Question. Wtite the relation between Basicity of Amine & pKb

Answer: The basicity of an amine is inversely proportional to its \( pK_b \) value. A higher \( K_b \) value or a lower \( pK_b \) value signifies a stronger basic character. \[ pK_b = -\log_{10} K_b \]
In simple words: An amine with a lower pKb value is a stronger base, while a higher pKb indicates a weaker base.

Exam Tip: Use the direct inverse proportion to explain similar theoretical questions.

Question. Arrange the following in decreasing order of pKb values
C2H5NH2 , C6H5-NH-CH3 , (C2H5)2NH, C6H5NH2

Answer: The decreasing order of \( pK_b \) values is: \[ \text{C}_6\text{H}_5\text{NH}_2 > \text{C}_6\text{H}_5\text{-NH-CH}_3 > \text{C}_2\text{H}_5\text{NH}_2 > (\text{C}_2\text{H}_5)_2\text{NH} \] A higher \( pK_b \) value corresponds to a weaker basic character, so aniline (having delocalized electrons) has the highest value, while diethylamine is the strongest base.
In simple words: Aniline is the weakest base and has the highest pKb value, while diethylamine is the strongest base and has the lowest pKb value.

Exam Tip: Decreasing order of pKb values implies arranging compounds from the weakest base to the strongest base.

Question. Arrange the following Amines in increasing order of Basic character.
a. CH3NH2 , (CH3)2NH, .(CH3)3N in Aq.Solution & in Gaseous Phase
b. C2H5NH2 , (C2H5)2NH , (C2H5)3N in Aq.Solution

Answer:
a. For methyl-substituted amines:
- In Aqueous Solution: \( (\text{CH}_3)_3\text{N} < \text{CH}_3\text{NH}_2 < (\text{CH}_3)_2\text{NH} \)
- In Gaseous Phase: \( \text{CH}_3\text{NH}_2 < (\text{CH}_3)_2\text{NH} < (\text{CH}_3)_3\text{N} \)

b. For ethyl-substituted amines in Aqueous Solution:
- In Aqueous Solution: \( \text{C}_2\text{H}_5\text{NH}_2 < (\text{C}_2\text{H}_5)_3\text{N} < (\text{C}_2\text{H}_5)_2\text{NH} \)
In simple words: Basicity in the gaseous phase depends only on the inductive effect. In water, solvation and steric effects make secondary amines the most basic.

Exam Tip: Memorize the aqueous basicity orders: 2° > 1° > 3° for methyl groups and 2° > 3° > 1° for ethyl groups.

Question. Which one is more Basic & Why?
C6H5NH2 or C2H5NH2

Answer: Ethanamine (\( \text{C}_2\text{H}_5\text{NH}_2 \)) is more basic than aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)). In aniline, the unshared lone pair of electrons on nitrogen is involved in resonance with the benzene ring, making it less available for protonation. In ethylamine, the +I effect of the ethyl group increases electron density on nitrogen, making its lone pair highly available.
In simple words: Ethylamine is a stronger base because its nitrogen has an easily sharable pair of electrons, whereas aniline's electrons are pulled into the benzene ring.

Exam Tip: Support your answer by mentioning the resonance structures of aniline showing lone pair delocalization.

Question. Name the factors affecting the Basicity of Amines in Aq.Solution & in Gaseous Phase .

Answer:
- In Gaseous Phase: Only the Inductive effect (+I effect of alkyl groups) affects basicity.
- In Aqueous Solution: Basicity is governed by:
1. Inductive effect (+I effect)
2. Solvation effect (hydration through hydrogen bonding)
3. Steric hindrance of alkyl groups
In simple words: In gas, only the push of electrons matters. In water, the size of the molecule and hydrogen bonding with water molecules also play a role.

Exam Tip: Clearly distinguish between gas phase and aqueous phase factors when answering this question.

Question. Match the following .

Answer:
The correct matched pairs are:

In simple words: Secondary aliphatic amines are the most basic and have the lowest pKb values, while aromatic amines (Benzamine) are the weakest bases with the highest pKb values.

Exam Tip: Stronger bases have smaller pKb values and larger Kb values.

Concept-3: Boiling Point of Amines

Question. Write the Factors Affecting the Boiling Point of Amines

Answer: The factors affecting boiling point are:
1. Intermolecular Hydrogen Bonding: Primary and secondary amines form polar intermolecular hydrogen bonds which increase the boiling point. Tertiary amines cannot form these bonds.
2. Molecular Mass: Boiling points increase with larger molecular mass due to stronger van der Waals dispersion forces.
3. Branching: Increased branching in isomeric amines decreases the surface area and weakens van der Waals forces, lowering the boiling point.
In simple words: Boiling points increase with stronger intermolecular forces, such as hydrogen bonding and van der Waals interactions.

Exam Tip: Be sure to discuss both polar hydrogen bonds and non-polar dispersion forces to provide a complete answer.

Question. Why is Primary Amine have higher Boiling Point than that Sec & Tert-Amines ?

Answer: Primary amines have two hydrogen atoms directly bonded to the nitrogen atom, allowing them to form a highly extensive network of intermolecular hydrogen bonds. Secondary amines have only one N-H bond, which leads to weaker hydrogen bonding, while tertiary amines have no N-H bonds and cannot form intermolecular hydrogen bonds. Therefore, primary amines require more energy to separate and have the highest boiling points.
In simple words: Primary amines have more hydrogen atoms to form strong hydrogen bonds, holding molecules tightly together.

Exam Tip: Focus on the number of active N-H bonds present in each class of amine in your explanation.

Question. Arrange the following Amines in decreasing order of B.P. 1O , 2O , 3O

Answer: The decreasing order of boiling points is:
\( 1^\circ \text{ Amine} > 2^\circ \text{ Amine} > 3^\circ \text{ Amine} \)
In simple words: Primary amines boil at the highest temperatures, followed by secondary, and tertiary amines boil at the lowest.

Exam Tip: Polar N-H hydrogen bonds decrease in number from primary to tertiary amines, dictating this exact physical trend.

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Question. Match the following

Answer:
The correct matched pairs are:

In simple words: Alcohols have the highest boiling points due to stronger O-H bonds, followed by primary, secondary, and tertiary amines, while non-polar alkanes have the lowest.

Exam Tip: Alcohols have higher boiling points than amines of comparable molecular mass because oxygen is more electronegative than nitrogen, making O-H bonds more polar than N-H bonds.

Question. Why are alcohols have higher B.P than that of amines of comparable molecular mass ?

Answer: Oxygen is more electronegative than nitrogen, which makes the O-H bonds in alcohols more polar than the N-H bonds in amines. This results in stronger and more extensive intermolecular hydrogen bonding in alcohols compared to amines of comparable molecular mass.
In simple words: The O-H bond in alcohols is more polar than the N-H bond in amines, making hydrogen bonds in alcohols stronger.

Exam Tip: Cite the difference in electronegativity between Oxygen (3.5) and Nitrogen (3.0).

Concept-2: To Distinguish Between Pri, Sec & Tert -Amines

Question. Distinguish between the (a) CH3NH2 and (CH3)2NH (b) Aniline & N-Methyl Aniline (c) Sec- Amine & Tert-Amine

Answer:
(a) \( \text{CH}_3\text{NH}_2 \) (Primary) and \( (\text{CH}_3)_2\text{NH} \) (Secondary):
- **Carbylamine test:** \( \text{CH}_3\text{NH}_2 \) gives a foul-smelling methyl isocyanide on heating with chloroform and alc. KOH. \( (\text{CH}_3)_2\text{NH} \) does not.
- **Hinsberg's test:** \( \text{CH}_3\text{NH}_2 \) forms a precipitate soluble in KOH, whereas \( (\text{CH}_3)_2\text{NH} \) forms a precipitate insoluble in KOH.

(b) Aniline (Primary aromatic) and N-Methyl Aniline (Secondary aromatic):
- **Azo-dye test:** Aniline forms an orange-red dye with nitrous acid followed by phenol coupling. N-Methyl aniline does not.
- **Carbylamine test:** Aniline gives a positive test with a foul smell, whereas N-methyl aniline does not.

(c) Secondary Amine & Tertiary Amine:
- **Hinsberg's test:** Secondary amines react to form a product insoluble in KOH, whereas tertiary amines do not react at all.
- **Nitrous acid test:** Secondary amines form a yellow oily compound, whereas tertiary amines dissolve to form a soluble salt.
In simple words: Primary amines can be distinguished by their smell in the carbylamine test, while secondary and tertiary amines are best separated using Hinsberg's test.

Exam Tip: Hinsberg's reagent (benzene sulphonyl chloride) is the primary test used to differentiate 1°, 2°, and 3° amines.

Concept-3 : Name Reactions

Question. (i)_ Carbyl amine reaction (ii) Sandmayer reaction (iii) Gatterman reaction

Answer:
(i) Carbylamine reaction: Aliphatic or aromatic primary amines, when heated with chloroform and alcoholic KOH, produce extremely foul-smelling alkyl/aryl isocyanides. \[ \text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{R-NC} + 3\text{KCl} + 3\text{H}_2\text{O} \] (ii) Sandmeyer reaction: Replacement of the diazonium group in benzenediazonium salt by chlorine, bromine, or cyano group using cuprous salts (\( \text{Cu}_2\text{Cl}_2 \box \), \( \text{Cu}_2\text{Br}_2 \), \( \text{Cu}_2(\text{CN})_2 \)). \[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{Cu}_2\text{Cl}_2 / \text{HCl}} \text{C}_6\text{H}_5\text{Cl} + \text{N}_2 \] (iii) Gattermann reaction: Replacement of the diazonium group by treating the salt with copper powder in the presence of halogen acid. \[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{Cu / HCl}} \text{C}_6\text{H}_5\text{Cl} + \text{N}_2 + \text{CuCl} \]
In simple words: Carbylamine reaction tests for primary amines with a bad odor. Sandmeyer and Gattermann reactions both introduce halogens onto benzene rings using copper.

Exam Tip: Sandmeyer reaction generally yields better results than the Gattermann reaction.

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