Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i

Exercises · 9.1

Q. Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(i)

(CH_3)_2CHNH_2

(ii)

CH_3(CH_2)_2NH_2

(iii)

CH_3NHCH(CH_3)_2

(iv)

(CH_3)_3CNH_2

(v)

C_6H_5NHCH_3

(vi)

(CH_3CH_2)_2NCH_3

(vii)

m-BrC_6H_4NH_2

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Concept understanding — Nucleophilic Substitution Reactions

Nucleophilic Addition – From Intuition to Precision

Imagine you have a molecule with a carbon–oxygen double bond — a carbonyl group ( $\ce{C=O}$ ). That oxygen is greedy for electrons; it pulls them away from carbon, leaving the carbon slightly positive ( $\delta^+$ ) and the oxygen slightly negative ( $\delta^-$ ). Now, if you bring a species that is rich in electrons (a nucleophile, meaning "nucleus-loving"), it will naturally be attracted to that electron-deficient carbon. The nucleophile attacks the carbon, the $\pi$ bond breaks, and the oxygen picks up a proton (or some other electrophile) to become stable. That, in a nutshell, is nucleophilic addition.

Note

The key idea: a nucleophile adds across a polar multiple bond (usually $\ce{C=O}$ or $\ce{C≡N}$ ), breaking the $\pi$ bond and forming two new sigma bonds.

The Precise Statement

Nucleophilic addition is a reaction in which a nucleophile (an electron-rich species) forms a sigma bond with an electrophilic carbon atom of a polar multiple bond (typically a carbonyl group, $\ce{C=O}$ , or a nitrile, $\ce{C≡N}$ ), while the $\pi$ bond breaks. The resulting intermediate then captures a proton (or another electrophile) to give a neutral product.

In general form:

\ce{R2C=O + Nu^- ->[H^+] R2C(OH)Nu}

The nucleophile ( $\ce{Nu^-}$ ) attacks the carbonyl carbon; the oxygen becomes negatively charged; then a proton ( $\ce{H^+}$ ) from the medium attaches to the oxygen, yielding an alcohol.

Why It Happens – The Driving Force

The carbonyl carbon is electrophilic because:

Oxygen is more electronegative than carbon, so the $\ce{C=O}$ bond is polarised: $\ce{C^{\delta+}=O^{\delta-}}$ .
The $\pi$ bond is weaker than a $\sigma$ bond, so it can break relatively easily.

A nucleophile (like $\ce{OH^-}$ , $\ce{CN^-}$ , or $\ce{NH3}$ ) has a lone pair or a negative charge. It seeks positive centres. The attack forms a new $\sigma$ bond, and the $\pi$ electrons move entirely to oxygen, creating an alkoxide ion ( $\ce{R2C-O^-}$ ). This intermediate is then quenched by a proton.

Watch out

A common mistake: thinking the nucleophile attacks the oxygen. No — oxygen is already electron-rich; the nucleophile goes to the carbon because it is electron-deficient.

A Concrete Example – Addition of HCN to a Ketone

Take acetone ( $\ce{CH3COCH3}$ ) and hydrogen cyanide ( $\ce{HCN}$ ). In the presence of a base, $\ce{CN^-}$ (the nucleophile) attacks the carbonyl carbon:

\ce{CH3COCH3 + CN^- -> CH3C(O^-)(CN)CH3}

The alkoxide intermediate then picks up a proton from $\ce{HCN}$ (or from water) to give a cyanohydrin:

\ce{CH3C(O^-)(CN)CH3 + H^+ -> CH3C(OH)(CN)CH3}

The product is acetone cyanohydrin. Notice: two new sigma bonds formed ( $\ce{C-CN}$ and $\ce{O-H}$ ), and the $\pi$ bond is gone.

What Makes a Good Nucleophile?

Strong nucleophiles are usually negatively charged or have lone pairs:

$\ce{OH^-}$ , $\ce{CN^-}$ , $\ce{NH2^-}$ , $\ce{CH3O^-}$ , $\ce{H^-}$ (from hydride reagents like $\ce{NaBH4}$ or $\ce{LiAlH4}$ )
Neutral but polarisable: $\ce{NH3}$ , $\ce{H2O}$ (weaker, but can add under acidic conditions)

Tip

In exams, remember: hydride ion ( $\ce{H^-}$ ) is a nucleophile in reductions (e.g., $\ce{NaBH4}$ reduces aldehydes/ketones to alcohols via nucleophilic addition).

The Big Picture – Why This Matters

Nucleophilic addition is the fundamental reaction of carbonyl compounds. It is how:

Aldehydes and ketones form alcohols (with $\ce{NaBH4}$ or $\ce{LiAlH4}$ )
Cyanohydrins are made (important in organic synthesis)
Grignard reagents ( $\ce{RMgX}$ ) add to carbonyls to form new carbon–carbon bonds
Hemiacetals and acetals form (key in carbohydrate chemistry)

Every time you see a $\ce{C=O}$ group, think: this carbon is a target for nucleophiles.

Final Answer

Nucleophilic addition is a reaction where an electron-rich nucleophile attacks the electrophilic carbon of a polar multiple bond (typically $\ce{C=O}$ or $\ce{C≡N}$ ), breaking the $\pi$ bond and forming two new sigma bonds — one to the nucleophile and one to a proton (or other electrophile). The driving force is the polarity of the $\ce{C=O}$ bond and the stability gained by forming stronger sigma bonds.

Nucleophilic addition reactions of carbonyl compounds form a central part of the NCERT Class 12 Chemistry chapter on aldehydes and ketones, and questions on cyanohydrin formation or the role of hydride nucleophiles like NaBH4 are common in CBSE boards and JEE Main. Students searching "nucleophilic addition mechanism class 12 chemistry important questions" will find this carbonyl-carbon-attack explanation is the standard NCERT approach.

Why this formula?

Nucleophilic Substitution Reactions: Why the Key Formulas Hold

Nucleophilic substitution reactions are a cornerstone of organic chemistry. Instead of just memorizing the rate laws, let's understand why they arise from the molecular events.

1. The Two Main Mechanisms: A Tale of Timing

The key formulas (rate laws) for nucleophilic substitution come directly from how many molecules are involved in the rate-determining step (RDS) — the slowest step that controls the overall reaction speed.

SN1: Unimolecular — The Leaving Group Goes First

The Rate Law:

\text{Rate} = k[\text{RX}]

Why?

The reaction happens in two steps:

Slow step (RDS): The C–X bond breaks spontaneously, forming a carbocation intermediate. Only the substrate ( $\text{RX}$ ) is involved.

\text{RX} \xrightarrow{\text{slow}} \text{R}^+ + \text{X}^-

Fast step: The nucleophile ( $\text{Nu}^-$ ) attacks the carbocation.

\text{R}^+ + \text{Nu}^- \xrightarrow{\text{fast}} \text{RNu}

Since the slow step depends only on the concentration of $\text{RX}$ , the rate law has no dependence on $[\text{Nu}^-]$ . The nucleophile arrives after the carbocation is formed — it cannot affect the speed of the first step.

Key insight: The rate is determined by how easily the leaving group leaves, not by how fast the nucleophile attacks.

SN2: Bimolecular — Simultaneous Attack and Departure

The Rate Law:

\text{Rate} = k[\text{RX}][\text{Nu}^-]

Why?

The reaction occurs in one concerted step:

The nucleophile attacks the carbon from one side at the same time as the leaving group departs from the opposite side.
Both $\text{RX}$ and $\text{Nu}^-$ must collide with the correct orientation and sufficient energy.

The rate depends on the frequency of productive collisions between the two molecules. This is directly proportional to the product of their concentrations:

\text{Rate} \propto [\text{RX}] \times [\text{Nu}^-]

Key insight: Both partners are involved in the transition state simultaneously — if either is missing, the reaction cannot proceed.

2. The Transition State: Why the Formulas Are Not Just "Given"

For SN2, the transition state has a pentavalent carbon (five bonds partially formed/broken). The energy barrier depends on steric hindrance — bulkier groups around carbon make it harder for the nucleophile to approach, which is why SN2 is favored at primary carbons.

For SN1, the transition state of the slow step involves only bond breaking (no bond formation yet). The carbocation stability (tertiary > secondary > primary) determines the activation energy — this is why SN1 is favored at tertiary carbons.

3. The Temperature Dependence: The Arrhenius Equation

Both rate constants $k$ follow the Arrhenius equation:

k = A e^{-E_a / RT}

$A$ = frequency factor (how often collisions occur with correct orientation)
$E_a$ = activation energy (the energy barrier for the RDS)
$R$ = gas constant
$T$ = temperature

This is not a separate formula — it explains why the rate constants change with temperature. Higher $T$ increases the fraction of molecules with energy $\geq E_a$ , speeding up the reaction.

4. Summary: The "Why" in One Table

Mechanism	Rate Law	Why? (Molecular Reason)
SN1	$\text{Rate} = k[\text{RX}]$	Slow step involves only the substrate breaking apart. Nucleophile waits.
SN2	$\text{Rate} = k[\text{RX}][\text{Nu}^-]$	Both molecules must collide in the single, concerted step.

Final takeaway: The formulas are not arbitrary — they are direct consequences of which molecules are present in the slowest step. Always ask: "What is happening in the rate-determining step?" The answer gives you the rate law.

Amines are classified by the number of carbon groups attached to the nitrogen atom. The IUPAC names and classifications for the seven compounds are: (i) Propan-2-amine, primary;

(ii) Propan-1-amine, primary;

(iii) N-Methylpropan-2-amine, secondary;

(iv) 2-Methylpropan-2-amine, primary;

(v) N-Methylaniline, secondary;

(vi) N-Ethyl-N-methylethanamine, tertiary;

(vii) 3-Bromoaniline, primary.

The Core Idea: What Makes an Amine Primary, Secondary, or Tertiary?

In organic chemistry, the classification of amines is based only on the number of carbon-containing groups directly bonded to the nitrogen atom. It has nothing to do with the carbon skeleton of the rest of the molecule.

Primary (1°): Nitrogen is bonded to one carbon group and two hydrogens ( $RNH_2$ ).
Secondary (2°): Nitrogen is bonded to two carbon groups and one hydrogen ( $R_2NH$ ).
Tertiary (3°): Nitrogen is bonded to three carbon groups and no hydrogens ( $R_3N$ ).

The "R" groups can be alkyl or aryl. The key is to look at the nitrogen, count its attachments, and ignore the rest.

Watch out

A common mistake is to classify based on the carbon atom that is attached to the nitrogen. For example, $(CH_3)_3CNH_2$ has a tertiary carbon, but the nitrogen is only attached to one carbon. It is a primary amine. Always look at the nitrogen, not the carbon.

Step-by-Step Solution

1. $(CH_3)_2CHNH_2$

Structure: The nitrogen is attached to a single isopropyl group ( $(CH_3)_2CH-$ ) and two hydrogens.
Classification: One carbon group on N → Primary amine.
IUPAC Name: The parent chain is propane. The $NH_2$ group is on carbon-2. The name is Propan-2-amine (common name: Isopropylamine).

2. $CH_3(CH_2)_2NH_2$

Structure: The nitrogen is attached to a straight propyl chain ( $CH_3CH_2CH_2-$ ) and two hydrogens.
Classification: One carbon group on N → Primary amine.
IUPAC Name: The parent chain is propane. The $NH_2$ group is on carbon-1. The name is Propan-1-amine (common name: n-Propylamine).

3. $CH_3NHCH(CH_3)_2$

Structure: The nitrogen is attached to a methyl group ( $CH_3-$ ), an isopropyl group ( $(CH_3)_2CH-$ ), and one hydrogen.
Classification: Two carbon groups on N → Secondary amine.
IUPAC Name: The larger alkyl group (isopropyl) forms the base name "propan-2-amine". The smaller methyl group is treated as a substituent on the nitrogen, prefixed with "N-". The name is N-Methylpropan-2-amine.

4. $(CH_3)_3CNH_2$

Structure: The nitrogen is attached to a single tert-butyl group ( $(CH_3)_3C-$ ) and two hydrogens.
Classification: One carbon group on N → Primary amine.
IUPAC Name: The parent chain is propane (three carbons). The $NH_2$ group and the two methyl branches are all on carbon-2. The name is 2-Methylpropan-2-amine (common name: tert-Butylamine).

Tip

For branched alkyl groups, the IUPAC name is derived from the longest continuous carbon chain. Here, the longest chain is propane, with a methyl substituent at the 2-position.

5. $C_6H_5NHCH_3$

Structure: The nitrogen is attached to a phenyl group ( $C_6H_5-$ ), a methyl group ( $CH_3-$ ), and one hydrogen.
Classification: Two carbon groups on N → Secondary amine.
IUPAC Name: The base compound is aniline ( $C_6H_5NH_2$ ). The methyl group on the nitrogen is indicated by "N-methyl". The name is N-Methylaniline.

6. $(CH_3CH_2)_2NCH_3$

Structure: The nitrogen is attached to two ethyl groups ( $CH_3CH_2-$ ) and one methyl group ( $CH_3-$ ). There are no hydrogens on the nitrogen.
Classification: Three carbon groups on N → Tertiary amine.
IUPAC Name: Pick one of the ethyl groups as the parent chain ("ethanamine"); the remaining two substituents on nitrogen — the other ethyl group and the methyl group — are cited as N-substituents, in alphabetical order. The name is N-Ethyl-N-methylethanamine.

Watch out

This compound has only two ethyl groups and one methyl group on nitrogen (molecular formula $C_5H_{13}N$ ). It is not the same compound as triethylamine, $(C_2H_5)_3N$ (formula $C_6H_{15}N$ , three ethyl groups) — do not call this compound "N,N-diethylethanamine" or "triethylamine"; those names describe a different molecule with one more carbon.

7. $m-BrC_6H_4NH_2$

Structure: The nitrogen is attached to a single bromophenyl group ( $BrC_6H_4-$ ) and two hydrogens.
Classification: One carbon group on N → Primary amine.
IUPAC Name: The base compound is aniline. The bromine atom is a substituent on the benzene ring at the meta (m) position. The name is 3-Bromoaniline (or m-Bromoaniline).

✓Final answer

The IUPAC names and classifications are: (i) Propan-2-amine, primary; (ii) Propan-1-amine, primary; (iii) N-Methylpropan-2-amine, secondary; (iv) 2-Methylpropan-2-amine, primary; (v) N-Methylaniline, secondary; (vi) N-Ethyl-N-methylethanamine, tertiary; (vii) 3-Bromoaniline, primary.

Method: IUPAC Nomenclature & Classification of Amines

Method Name: IUPAC Naming + Functional Group Classification

Steps:

Identify the longest carbon chain attached to the nitrogen atom.
- For primary amines ( $RNH_2$ ), the chain includes the $NH_2$ group.
- For secondary ( $R_2NH$ ) and tertiary ( $R_3N$ ) amines, treat the largest alkyl group as the parent chain.
Number the chain so that the carbon attached to $NH_2$ (or $N$ ) gets the lowest possible number.
Write the IUPAC name:
- Primary amines: alkanamine (e.g., propan-1-amine).
- Secondary/tertiary: treat as N-substituted derivatives of the parent amine.
Classify based on the number of carbon groups directly bonded to nitrogen:
- Primary (1°): $R-NH_2$
- Secondary (2°): $R_2NH$
- Tertiary (3°): $R_3N$

Applying the method:

(i) $(CH_3)_2CHNH_2$

Parent chain: 3 carbons (propane) with $NH_2$ on carbon 2.
IUPAC: Propan-2-amine
Classification: Primary (1°) — one alkyl group attached to N.

(ii) $CH_3(CH_2)_2NH_2$

Parent chain: 3 carbons, $NH_2$ on terminal carbon.
IUPAC: Propan-1-amine
Classification: Primary (1°)

(iii) $CH_3NHCH(CH_3)_2$

Parent amine: the larger group (isopropyl) gives the parent chain — propane, with N on carbon 2; the methyl group is an N-substituent.
IUPAC: N-Methylpropan-2-amine
Classification: Secondary (2°) — two alkyl groups on N.

(iv) $(CH_3)_3CNH_2$

Parent chain: 3 carbons (propane) — the longest chain through the carbon bearing $NH_2$ ; both the $NH_2$ and a methyl group sit on carbon 2.
IUPAC: 2-Methylpropan-2-amine (common: tert-butylamine)
Classification: Primary (1°) — only one alkyl group attached to N.

(v) $C_6H_5NHCH_3$

Parent: aniline ( $C_6H_5NH_2$ ) with a methyl group on N.
IUPAC: N-Methylaniline (or N-methylbenzenamine)
Classification: Secondary (2°)

(vi) $(CH_3CH_2)_2NCH_3$

Parent: ethanamine — one of the two ethyl groups (the largest groups on N) is the parent chain; the remaining ethyl and the methyl become N-substituents.
IUPAC: N-Ethyl-N-methylethanamine
Classification: Tertiary (3°) — three alkyl groups on N.
⚠️ Do not call this compound N,N-diethylethanamine or triethylamine — those names describe $(C_2H_5)_3N$ , a different molecule with six carbons.

(vii) $m-BrC_6H_4NH_2$

Parent: aniline with bromine at meta position.
IUPAC: 3-Bromoaniline (or m-bromoaniline)
Classification: Primary (1°) — $NH_2$ directly attached to benzene.

Final Summary Table:

Compound	IUPAC Name	Classification
$(CH_3)_2CHNH_2$	Propan-2-amine	Primary
$CH_3(CH_2)_2NH_2$	Propan-1-amine	Primary
$CH_3NHCH(CH_3)_2$	N-Methylpropan-2-amine	Secondary
$(CH_3)_3CNH_2$	2-Methylpropan-2-amine	Primary
$C_6H_5NHCH_3$	N-Methylaniline	Secondary
$(CH_3CH_2)_2NCH_3$	N-Ethyl-N-methylethanamine	Tertiary
$m-BrC_6H_4NH_2$	3-Bromoaniline	Primary

Key takeaway: Classification depends only on the number of carbon groups directly bonded to nitrogen, not on the total number of carbons in the molecule.

🧠 Understanding the Core Concept First

This question is pure nomenclature and classification. Two rules decide everything:

Classification: an amine is primary (1°), secondary (2°) or tertiary (3°) according to how many carbon atoms are bonded directly to the nitrogen — nothing else matters.
Naming: build the IUPAC name from the longest chain (for 1° amines) or the largest group on nitrogen (for 2°/3° amines); everything else on N becomes an N-substituent.

Get these two rules straight and every part of this question follows.

✗ Common Mistake #1: Confusing the alkyl group with the amine class

The Mistake

Students see $(CH_3)_2CHNH_2$ and think:

"Two methyl groups on the carbon → secondary amine."

✓ How to Avoid

Classification is based on the number of carbon atoms directly attached to the nitrogen atom, not on the carbon skeleton.

Primary (1°) amine: N is attached to one carbon (alkyl or aryl).
Secondary (2°) amine: N is attached to two carbons.
Tertiary (3°) amine: N is attached to three carbons.

Example

$(CH_3)_2CHNH_2$

N is attached to one C (the isopropyl group).
✓ Primary amine (1°).

✗ Common Mistake #2: Writing the wrong IUPAC name for branched amines

The Mistake

For $(CH_3)_2CHNH_2$ , students write:

"2-methylpropan-1-amine" ✗ (wrong because the NH₂ is on the middle carbon).

✓ How to Avoid

Find the longest carbon chain that includes the carbon attached to the –NH₂ group.
Number the chain so that the carbon with –NH₂ gets the lowest locant.
Use the suffix -amine (or -anamine for systematic IUPAC).

Correct Name

$(CH_3)_2CHNH_2$ → Propan-2-amine (or isopropylamine in common name).

Chain = 3 carbons (propane).
NH₂ on C2 → propan-2-amine.

✗ Common Mistake #3: Misidentifying aryl amines as primary/secondary

The Mistake

For $C_6H_5NHCH_3$ , students say:

"It's a primary amine because the benzene ring is attached." ✗

✓ How to Avoid

Count only the atoms directly bonded to N:

N is bonded to:
- one C from benzene ring
- one C from methyl group
That’s two carbons → secondary (2°) amine.

Correct Classification

$C_6H_5NHCH_3$ → Secondary amine (N-methylaniline).

✗ Common Mistake #4: Forgetting to name substituents on nitrogen in secondary/tertiary amines

The Mistake

For $(CH_3CH_2)_2NCH_3$ , students write:

"N,N-diethylmethylamine" ✗ (wrong order or missing N-).

✓ How to Avoid

Treat the largest alkyl group as the parent chain.
Name smaller groups attached to N as N-alkyl substituents.
List substituents alphabetically (ignoring di-, tri-).

Correct Name

$(CH_3CH_2)_2NCH_3$ → N-Ethyl-N-methylethanamine.

Parent = ethanamine (one of the two ethyl groups — the largest groups on N — is the parent chain).
The remaining ethyl and methyl groups become N-substituents, listed alphabetically: N-ethyl-N-methyl.
⚠️ Never write “N,N-diethylethanamine” or “triethylamine” here — those names describe $(C_2H_5)_3N$ , a different (six-carbon) compound.

✗ Common Mistake #5: Misplacing the locant for substituents on the ring in aromatic amines

The Mistake

For $m-BrC_6H_4NH_2$ , students write:

"3-bromoaniline" ✓ (correct) but then classify it as secondary ✗.

✓ How to Avoid

The –NH₂ group is directly attached to the benzene ring.
N is bonded to only one carbon (from the ring).
✓ Primary amine (1°).

Correct Name & Class

$m-BrC_6H_4NH_2$ → 3-Bromoaniline (or m-bromoaniline) → Primary amine.

📋 Quick Reference Table

Compound	IUPAC Name	Class
$(CH_3)_2CHNH_2$	Propan-2-amine	1°
$CH_3(CH_2)_2NH_2$	Propan-1-amine	1°
$CH_3NHCH(CH_3)_2$	N-Methylpropan-2-amine	2°
$(CH_3)_3CNH_2$	2-Methylpropan-2-amine	1°
$C_6H_5NHCH_3$	N-Methylaniline	2°
$(CH_3CH_2)_2NCH_3$	N-Ethyl-N-methylethanamine	3°
$m-BrC_6H_4NH_2$	3-Bromoaniline	1°

🎯 Final Exam Tip

Always draw the structure around the nitrogen atom — count the bonds from N to carbon atoms. That single step eliminates 90% of classification errors.

←Previous section · summarySummary →Next question9.2 — Give one chemical test to distinguish between the following pairs of c…

Q.Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

The Core Idea: What Makes an Amine Primary, Secondary, or Tertiary?

Step-by-Step Solution

1. (CH3)2CHNH2(CH_3)_2CHNH_2(CH3​)2​CHNH2​

2. CH3(CH2)2NH2CH_3(CH_2)_2NH_2CH3​(CH2​)2​NH2​

3. CH3NHCH(CH3)2CH_3NHCH(CH_3)_2CH3​NHCH(CH3​)2​

4. (CH3)3CNH2(CH_3)_3CNH_2(CH3​)3​CNH2​

5. C6H5NHCH3C_6H_5NHCH_3C6​H5​NHCH3​

6. (CH3CH2)2NCH3(CH_3CH_2)_2NCH_3(CH3​CH2​)2​NCH3​

7. m−BrC6H4NH2m-BrC_6H_4NH_2m−BrC6​H4​NH2​

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1. $(CH_3)_2CHNH_2$

2. $CH_3(CH_2)_2NH_2$

3. $CH_3NHCH(CH_3)_2$

4. $(CH_3)_3CNH_2$

5. $C_6H_5NHCH_3$

6. $(CH_3CH_2)_2NCH_3$

7. $m-BrC_6H_4NH_2$

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