NCERT Solutions for Class 12 Chemistry Chapter 9 – Amines (English Medium)
🧠 Get NCERT Solutions for Class 12 Chemistry Chapter 9 – Amines with accurate answers and step-by-step explanations in an exam-friendly format. This chapter covers classification (1°, 2°, 3° amines), IUPAC nomenclature, preparation methods (reduction, ammonolysis, Gabriel synthesis), basic nature (inductive/resonance effects), diazotization & diazonium salts, Sandmeyer/Gattermann reactions, and important tests like Hinsberg test and carbylamine test. Perfect for CBSE boards and NEET/JEE revision. ✅
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Q.1:
Write IUPAC names of the following compounds and classify them into
primary,
secondary
and
tertiary
amines.
(i) (CH3)2CHNH2
(ii) CH3(CH2)2NH2
(iii) CH3NHCH(CH3)2
(iv) (CH3)3CNH2
(v) C6H5NHCH3
(vi) (CH3CH2)2NCH3
(vii) m–BrC6H4NH2
(i) (CH3)2CHNH2
(ii) CH3(CH2)2NH2
(iii) CH3NHCH(CH3)2
(iv) (CH3)3CNH2
(v) C6H5NHCH3
(vi) (CH3CH2)2NCH3
(vii) m–BrC6H4NH2
Answer ✅
(i)
(CH3)2CHNH2
👉 IUPAC: propan-2-amine (also written as 2-aminopropane)
👉 Class: Primary (1°) amine
👉 IUPAC: propan-2-amine (also written as 2-aminopropane)
👉 Class: Primary (1°) amine
(ii)
CH3(CH2)2NH2
👉 IUPAC: propan-1-amine (1-aminopropane)
👉 Class: Primary (1°) amine
👉 IUPAC: propan-1-amine (1-aminopropane)
👉 Class: Primary (1°) amine
(iii)
CH3NHCH(CH3)2
👉 IUPAC: N-methylpropan-2-amine
👉 Class: Secondary (2°) amine
👉 IUPAC: N-methylpropan-2-amine
👉 Class: Secondary (2°) amine
(iv)
(CH3)3CNH2
👉 IUPAC: 2-methylpropan-2-amine
👉 Class: Primary (1°) amine
👉 IUPAC: 2-methylpropan-2-amine
👉 Class: Primary (1°) amine
(v)
C6H5NHCH3
👉 IUPAC: N-methylbenzenamine (common name: N-methylaniline)
👉 Class: Secondary (2°) amine
👉 IUPAC: N-methylbenzenamine (common name: N-methylaniline)
👉 Class: Secondary (2°) amine
(vi)
(CH3CH2)2NCH3
👉 IUPAC: N-ethyl-N-methylethanamine
👉 Class: Tertiary (3°) amine
👉 IUPAC: N-ethyl-N-methylethanamine
👉 Class: Tertiary (3°) amine
(vii)
m–BrC6H4NH2
👉 IUPAC: 3-bromobenzenamine (common name: m-bromoaniline)
👉 Class: Primary (1°) amine
👉 IUPAC: 3-bromobenzenamine (common name: m-bromoaniline)
👉 Class: Primary (1°) amine
Explanation (Step by Step) 👉
✅ How to classify amines (fast trick):• Primary (1°): R–NH2 (N attached to one carbon group)
• Secondary (2°): R–NH–R' (N attached to two carbon groups)
• Tertiary (3°): R–N(R')–R'' (N attached to three carbon groups)
✅ IUPAC naming tips:
• Choose the longest carbon chain attached to N as the parent (…amine).
• If extra groups are attached to nitrogen, write them with N- (like N-methyl, N-ethyl).
• For aromatic: benzenamine is IUPAC for aniline.
Did You Know? 🤔📌
✅ Aniline
= benzenamine
in IUPAC, and substituents on the ring are numbered like benzene
(meta = 3-).✅ Tertiary amines have no N–H bond, so they cannot form intermolecular hydrogen bonding like primary/secondary amines do (this often affects boiling points).
Q.12:
Why cannot aromatic primary amines
be prepared by Gabriel phthalimide synthesis?
Answer ✅
✅ Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because this method works through an
SN2 substitution,
and
aryl halides (Ar–X)
do not undergo SN2 reaction easily.
Explanation (Step by Step) 👉
1) What Gabriel synthesis needs
In Gabriel synthesis, potassium phthalimide reacts with an alkyl halide
(R–X):👉 Phthalimide anion + R–X → N-alkyl phthalimide (via SN2) → hydrolysis → R–NH2
✅ So the key step is SN2 attack on the carbon attached to halogen.
2) Why aryl halides fail
In aryl halides
(Ar–X),
the carbon–halogen bond has
partial double bond character
due to resonance with the benzene ring.✅ This makes the C–X bond shorter and stronger, so it is hard to break.
Also, the halogen-bearing carbon is sp2 hybridized.
✅ SN2 needs a backside attack, but backside attack is not feasible on an sp2 carbon inside a planar ring system.
👉 Therefore, phthalimide anion cannot replace X from Ar–X, so aromatic primary amines are not formed.
Did You Know? 🤔📌
✅ Gabriel synthesis is best for making pure
primary aliphatic amines
(R–NH2).✅ It also generally fails with secondary/tertiary alkyl halides because SN2 is blocked by steric hindrance.
Q.14:
Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Answer ✅
(i)
Amines are less acidic because after losing
H+
they form an
amide ion (R–NH−),
which is less stable than the
alkoxide ion (R–O−)
formed by alcohols.
(ii) Primary amines have higher boiling points because they can form strong intermolecular hydrogen bonding (they have N–H bonds), while tertiary amines cannot.
(iii) Aliphatic amines are stronger bases because alkyl groups donate electrons (+I effect), but in aromatic amines the lone pair on nitrogen is delocalized into the benzene ring (−R effect), so it is less available to accept H+.
(ii) Primary amines have higher boiling points because they can form strong intermolecular hydrogen bonding (they have N–H bonds), while tertiary amines cannot.
(iii) Aliphatic amines are stronger bases because alkyl groups donate electrons (+I effect), but in aromatic amines the lone pair on nitrogen is delocalized into the benzene ring (−R effect), so it is less available to accept H+.
Explanation (Step by Step) 👉
(i) Amines are less acidic than alcohols
When they lose a proton:
✅ Amine:
👉 R–NH2 ⇌ R–NH− + H+ (amide ion)
✅ Alcohol:
👉 R–OH ⇌ R–O− + H+ (alkoxide ion)
Now compare stability:
In amide ion, negative charge is on N.
In alkoxide ion, negative charge is on O.
✅ Oxygen is more electronegative than nitrogen, so it holds negative charge more easily.
👉 Therefore, alkoxide ion is more stable, so alcohol can lose H+ more easily (more acidic).
✅ Amide ion is less stable → amines are less acidic.
When they lose a proton:
✅ Amine:
👉 R–NH2 ⇌ R–NH− + H+ (amide ion)
✅ Alcohol:
👉 R–OH ⇌ R–O− + H+ (alkoxide ion)
Now compare stability:
In amide ion, negative charge is on N.
In alkoxide ion, negative charge is on O.
✅ Oxygen is more electronegative than nitrogen, so it holds negative charge more easily.
👉 Therefore, alkoxide ion is more stable, so alcohol can lose H+ more easily (more acidic).
✅ Amide ion is less stable → amines are less acidic.
(ii) Primary amines have higher boiling point than tertiary amines
✅ Primary amines (R–NH2) have N–H bonds, so they form intermolecular H-bonds:
👉 R–NH···N–R (hydrogen bonding)
✅ This strong attraction needs extra energy to break → higher boiling point.
✅ Tertiary amines (R3N) have no N–H bond, so they cannot hydrogen bond among themselves strongly.
👉 Weaker attractions → lower boiling point.
✅ Primary amines (R–NH2) have N–H bonds, so they form intermolecular H-bonds:
👉 R–NH···N–R (hydrogen bonding)
✅ This strong attraction needs extra energy to break → higher boiling point.
✅ Tertiary amines (R3N) have no N–H bond, so they cannot hydrogen bond among themselves strongly.
👉 Weaker attractions → lower boiling point.
(iii) Aliphatic amines are stronger bases than aromatic amines
✅ Basic strength depends on how easily N can donate its lone pair to H+.
Aliphatic amines:
Alkyl groups show +I effect (electron donating).
✅ This increases electron density on N, making lone pair more available → stronger base.
Aromatic amines (like aniline):
Nitrogen lone pair gets delocalized into benzene ring by resonance (−R effect).
✅ So lone pair is less available to accept H+ → weaker base.
✅ Basic strength depends on how easily N can donate its lone pair to H+.
Aliphatic amines:
Alkyl groups show +I effect (electron donating).
✅ This increases electron density on N, making lone pair more available → stronger base.
Aromatic amines (like aniline):
Nitrogen lone pair gets delocalized into benzene ring by resonance (−R effect).
✅ So lone pair is less available to accept H+ → weaker base.
Did You Know? 🤔📌
✅ Acidity/base concepts often boil down to
stability of ions:
more stable ion = easier formation.✅ Hydrogen bonding is a major reason why many molecules (like alcohols and 1°/2° amines) have higher boiling points than similar-sized molecules without H-bonding.
✅ A quick memory trick:
👉 Alkyl amine = stronger base (lone pair free)
👉 Aryl amine = weaker base (lone pair shared with ring)