Amines are important class of organic compounds which are basic in nature. They are generally derived by replacing one or more hydrogen atoms of ammonia molecule by alkyl / aryl group(s). In nature, they are generally found in proteins, vitamins, alkaloids and hormones. Two biologically active compounds namely adrenaline and ephedrine contain secondary amino group and are used to increase blood pressure.
Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore, sp³ hybridized and the geometry of amine is pyramidal. Due to the presence of unshared pair of electrons, the angle C–N–E (where E is C or H) is less than 109.5°.
Inductive effects, solvation effects, steric effects etc. affect the basic strength of amines. Amines are classified as primary (1°), secondary (2°), and tertiary (3°) depending upon the number of hydrogen atoms replaced by alkyl or aryl groups in ammonia molecule. Amines can be prepared from various compounds like reduction of nitro, nitriles, amides etc. compounds. Amines are engaged in intermolecular hydrogen bonding. Aliphatic and aromatic primary amines show positive carbylamine reaction. A colourless crystalline solid benzene diazonium chloride can be obtained from aniline by reacting with sodium nitrite and hydrochloric acid at 273–278 K.
Question: 1
Select the correct order of boiling points of isomeric amines:
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For isomeric amines:
\[
1^\circ > 2^\circ > 3^\circ
\]
in boiling point because intermolecular hydrogen bonding decreases in the same order.
Concept:
Boiling point of amines depends mainly on intermolecular hydrogen bonding and molecular shape.
Primary amines form the strongest intermolecular hydrogen bonding, secondary amines form less extensive hydrogen bonding, while tertiary amines cannot form intermolecular hydrogen bonding through N-H bonds.
Step 1: Consider primary amine.
\[
n\text{-}C_4H_9NH_2
\]
contains two N-H bonds and forms strong intermolecular hydrogen bonding.
Hence it has the highest boiling point.
Step 2: Consider secondary amine.
\[
(C_2H_5)_2NH
\]
contains only one N-H bond.
Hydrogen bonding is weaker than in primary amines.
Step 3: Consider tertiary amine.
\[
C_2H_5N(CH_3)_2
\]
contains no N-H bond.
Intermolecular hydrogen bonding is absent.
Hence it has the lowest boiling point.
Step 4: Arrange in decreasing order.
\[
n\text{-}C_4H_9NH_2>(C_2H_5)_2NH>C_2H_5N(CH_3)_2
\]
\[
\boxed{\text{Option (A)}}
\]
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Question: 2
Identify the tertiary amine (\(3^\circ\)) from the following compounds.
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A tertiary amine contains three carbon groups attached to nitrogen and no N-H bond.
Concept:
A tertiary amine is formed when all three hydrogen atoms of ammonia are replaced by alkyl or aryl groups.
Step 1: Examine Ethylenediamine.
Contains two \(-NH_2\) groups.
Hence it is not a tertiary amine.
Step 2: Examine \(N,N\)-Dimethylaniline.
Structure:
\[
C_6H_5N(CH_3)_2
\]
Nitrogen is attached to one phenyl group and two methyl groups.
No hydrogen is attached to nitrogen.
Therefore it is a tertiary amine.
\[
\boxed{N,N\text{-Dimethylaniline}}
\]
Step 3: Examine remaining compounds.
Both \(p\)-toluidine and 2,4,6-tribromoaniline contain \(-NH_2\) group.
Therefore they are primary amines.
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Question: 3
Identify the correct decreasing order of basic strength of amines in aqueous solution.
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In aqueous medium:
\[
2^\circ > 3^\circ > 1^\circ > NH_3
\]
for aliphatic amines because both inductive effect and solvation must be considered.
Concept:
In aqueous solution, basic strength depends on both electron donating inductive effect and solvation of the conjugate acid.
Step 1: Consider inductive effect.
Alkyl groups increase electron density on nitrogen and increase basicity.
Step 2: Consider solvation effect.
The protonated secondary amine is better solvated than protonated tertiary amine.
Therefore secondary amines become strongest bases in water.
Step 3: Write the order.
\[
(C_2H_5)_2NH>(C_2H_5)_3N>C_2H_5NH_2>NH_3
\]
Hence
\[
\boxed{\text{Option (D)}}
\]
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Question: 4
The final structure of amine produced by Hoffmann degradation of \(m\)-bromobenzamide is:
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Hoffmann bromamide degradation:
\[
RCONH_2 \rightarrow RNH_2
\]
One carbon atom is lost during the reaction.
Concept:
Hoffmann bromamide degradation converts an amide into an amine containing one carbon atom less.
\[
RCONH_2
\xrightarrow{Br_2/KOH}
RNH_2
\]
Step 1: Identify the starting amide.
The substrate is \(m\)-bromobenzamide.
\[
m\text{-}BrC_6H_4CONH_2
\]
Step 2: Apply Hoffmann degradation.
The carbonyl carbon is removed during the reaction.
The aromatic ring remains unchanged.
\[
m\text{-}BrC_6H_4CONH_2
\rightarrow
m\text{-}BrC_6H_4NH_2
\]
Step 3: Identify the product.
The product is \(m\)-bromoaniline.
\[
\boxed{m\text{-Bromoaniline}}
\]
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Question: 5
What would be the structure of final product X in the following chemical reaction?
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Primary aromatic amines form diazonium salts, secondary amines form N-nitrosoamines, while tertiary aromatic amines undergo para-nitrosation.
Concept:
Tertiary aromatic amines do not form diazonium salts with nitrous acid. Instead, electrophilic substitution occurs on the aromatic ring producing para-nitroso derivatives.
Step 1: Identify the nature of the amine.
\[
N,N\text{-Dimethylaniline}
\]
is a tertiary aromatic amine.
There is no N-H bond present.
Step 2: Action of nitrous acid.
Nitrous acid generates the electrophile \(NO^+\).
The strongly activating dimethylamino group directs substitution mainly to the para position.
Step 3: Formation of nitroso derivative.
The nitroso group enters predominantly at the para position.
\[
N,N\text{-Dimethylaniline}
\rightarrow
p\text{-Nitroso-}N,N\text{-dimethylaniline}
\]
Step 4: Final answer.
\[
\boxed{p\text{-Nitroso-}N,N\text{-dimethylaniline}
}
\]
