Q.Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
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 (C=O). That oxygen is greedy for electrons; it pulls them away from carbon, leaving the carbon slightly positive (δ+) and the oxygen slightly negative (δ−). 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 π bond breaks, and the oxygen picks up a proton (or some other electrophile) to become stable. That, in a nutshell, is nucleophilic addition.
The key idea: a nucleophile adds across a polar multiple bond (usually C=O or C≡N), breaking the π 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, C=O, or a nitrile, C≡N), while the π bond breaks. The resulting intermediate then captures a proton (or another electrophile) to give a neutral product.
In general form:
RX2C=O+NuX−HX+RX2C(OH)Nu
The nucleophile (NuX−) attacks the carbonyl carbon; the oxygen becomes negatively charged; then a proton (HX+) 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 C=O bond is polarised: CXδ+=OXδ−.
- The π bond is weaker than a σ bond, so it can break relatively easily.
A nucleophile (like OHX−, CNX−, or NHX3) has a lone pair or a negative charge. It seeks positive centres. The attack forms a new σ bond, and the π electrons move entirely to oxygen, creating an alkoxide ion (RX2C−OX−). This intermediate is then quenched by a proton.
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 (CHX3COCHX3) and hydrogen cyanide (HCN). In the presence of a base, CNX− (the nucleophile) attacks the carbonyl carbon:
CHX3COCHX3+CNX−CHX3C(OX−)(CN)CHX3
The alkoxide intermediate then picks up a proton from HCN (or from water) to give a cyanohydrin:
CHX3C(OX−)(CN)CHX3+HX+CHX3C(OH)(CN)CHX3
The product is acetone cyanohydrin. Notice: two new sigma bonds formed (C−CN and O−H), and the π bond is gone.
What Makes a Good Nucleophile?
Strong nucleophiles are usually negatively charged or have lone pairs:
- OHX−, CNX−, NHX2X−, CHX3OX−, HX− (from hydride reagents like NaBHX4 or LiAlHX4)
- Neutral but polarisable: NHX3, HX2O (weaker, but can add under acidic conditions)
In exams, remember: hydride ion (HX−) is a nucleophile in reductions (e.g., NaBHX4 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 NaBHX4 or LiAlHX4)
- Cyanohydrins are made (important in organic synthesis)
- Grignard reagents (RMgX) add to carbonyls to form new carbon–carbon bonds
- Hemiacetals and acetals form (key in carbohydrate chemistry)
Every time you see a 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 C=O or C≡N), breaking the π 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 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:
Rate=k[RX]
Why?
The reaction happens in two steps:
- Slow step (RDS): The C–X bond breaks spontaneously, forming a carbocation intermediate. Only the substrate (RX) is involved.
RXslowR++X−
- Fast step: The nucleophile (Nu−) attacks the carbocation.
R++Nu−fastRNu
Since the slow step depends only on the concentration of RX, the rate law has no dependence on [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:
Rate=k[RX][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 RX and 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:
Rate∝[RX]×[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=Ae−Ea/RT
- A = frequency factor (how often collisions occur with correct orientation)
- Ea = 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 ≥Ea, speeding up the reaction.
4. Summary: The "Why" in One Table
| Mechanism | Rate Law | Why? (Molecular Reason) |
|---|---|---|
| SN1 | Rate=k[RX] | Slow step involves only the substrate breaking apart. Nucleophile waits. |
| SN2 | Rate=k[RX][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.
Concept: IUPAC nomenclature and classification of amines — an amine is primary, secondary or tertiary according to the number of carbon groups attached directly to the nitrogen atom.
Reasoning:
- Primary amine: Nitrogen is attached to one carbon (alkyl or aryl) and two hydrogens. General: RNH2 or ArNH2.
- Secondary amine: Nitrogen is attached to two carbons and one hydrogen. General: R2NH or R−NH−R′.
- Tertiary amine: Nitrogen is attached to three carbons and no hydrogen. General: R3N.
Step-by-step classification:
- (i) (CH3)2CHNH2: Isopropyl group attached to NH2 → Primary. IUPAC: Propan-2-amine.
- (ii) CH3(CH2)2NH2: Propyl group attached to NH2 → Primary. IUPAC: Propan-1-amine.
- (iii) CH3NHCH(CH3)2: Methyl and isopropyl groups on N → Secondary. IUPAC: N-methylpropan-2-amine.
- (iv) (CH3)3CNH2: tert-Butyl group attached to NH2 → Primary. IUPAC: 2-Methylpropan-2-amine.
- (v) C6H5NHCH3: Phenyl and methyl groups on N → Secondary. IUPAC: N-methylaniline.
- (vi) (CH3CH2)2NCH3: Two ethyl and one methyl group on N → Tertiary. IUPAC: N-ethyl-N-methylethanamine.
- (vii) m−BrC6H4NH2: Bromine at meta position on benzene ring with NH2 → Primary. IUPAC: 3-Bromoaniline.
The classifications are: (i) Primary,
(ii) Primary,
(iii) Secondary,
(iv) Primary,
(v) Secondary,
(vi) Tertiary,
(vii) Primary.
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 (RNH2).
- Secondary (2°): Nitrogen is bonded to two carbon groups and one hydrogen (R2NH).
- Tertiary (3°): Nitrogen is bonded to three carbon groups and no hydrogens (R3N).
The "R" groups can be alkyl or aryl. The key is to look at the nitrogen, count its attachments, and ignore the rest.
A common mistake is to classify based on the carbon atom that is attached to the nitrogen. For example, (CH3)3CNH2 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. (CH3)2CHNH2
- Structure: The nitrogen is attached to a single isopropyl group ((CH3)2CH−) and two hydrogens.
- Classification: One carbon group on N → Primary amine.
- IUPAC Name: The parent chain is propane. The NH2 group is on carbon-2. The name is Propan-2-amine (common name: Isopropylamine).
2. CH3(CH2)2NH2
- Structure: The nitrogen is attached to a straight propyl chain (CH3CH2CH2−) and two hydrogens.
- Classification: One carbon group on N → Primary amine.
- IUPAC Name: The parent chain is propane. The NH2 group is on carbon-1. The name is Propan-1-amine (common name: n-Propylamine).
3. CH3NHCH(CH3)2
- Structure: The nitrogen is attached to a methyl group (CH3−), an isopropyl group ((CH3)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. (CH3)3CNH2
- Structure: The nitrogen is attached to a single tert-butyl group ((CH3)3C−) and two hydrogens.
- Classification: One carbon group on N → Primary amine.
- IUPAC Name: The parent chain is propane (three carbons). The NH2 group and the two methyl branches are all on carbon-2. The name is 2-Methylpropan-2-amine (common name: tert-Butylamine).
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. C6H5NHCH3
- Structure: The nitrogen is attached to a phenyl group (C6H5−), a methyl group (CH3−), and one hydrogen.
- Classification: Two carbon groups on N → Secondary amine.
- IUPAC Name: The base compound is aniline (C6H5NH2). The methyl group on the nitrogen is indicated by "N-methyl". The name is N-Methylaniline.
6. (CH3CH2)2NCH3
- Structure: The nitrogen is attached to two ethyl groups (CH3CH2−) and one methyl group (CH3−). 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.
This compound has only two ethyl groups and one methyl group on nitrogen (molecular formula C5H13N). It is not the same compound as triethylamine, (C2H5)3N (formula C6H15N, 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−BrC6H4NH2
- Structure: The nitrogen is attached to a single bromophenyl group (BrC6H4−) 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).
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 (RNH2), the chain includes the NH2 group.
- For secondary (R2NH) and tertiary (R3N) amines, treat the largest alkyl group as the parent chain.
-
Number the chain so that the carbon attached to NH2 (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−NH2
- Secondary (2°): R2NH
- Tertiary (3°): R3N
Applying the method:
(i) (CH3)2CHNH2
- Parent chain: 3 carbons (propane) with NH2 on carbon 2.
- IUPAC: Propan-2-amine
- Classification: Primary (1°) — one alkyl group attached to N.
(ii) CH3(CH2)2NH2
- Parent chain: 3 carbons, NH2 on terminal carbon.
- IUPAC: Propan-1-amine
- Classification: Primary (1°)
(iii) CH3NHCH(CH3)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) (CH3)3CNH2
- Parent chain: 3 carbons (propane) — the longest chain through the carbon bearing NH2; both the NH2 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) C6H5NHCH3
- Parent: aniline (C6H5NH2) with a methyl group on N.
- IUPAC: N-Methylaniline (or N-methylbenzenamine)
- Classification: Secondary (2°)
(vi) (CH3CH2)2NCH3
- 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 (C2H5)3N, a different molecule with six carbons.
(vii) m−BrC6H4NH2
- Parent: aniline with bromine at meta position.
- IUPAC: 3-Bromoaniline (or m-bromoaniline)
- Classification: Primary (1°) — NH2 directly attached to benzene.
Final Summary Table:
| Compound | IUPAC Name | Classification |
|---|---|---|
| (CH3)2CHNH2 | Propan-2-amine | Primary |
| CH3(CH2)2NH2 | Propan-1-amine | Primary |
| CH3NHCH(CH3)2 | N-Methylpropan-2-amine | Secondary |
| (CH3)3CNH2 | 2-Methylpropan-2-amine | Primary |
| C6H5NHCH3 | N-Methylaniline | Secondary |
| (CH3CH2)2NCH3 | N-Ethyl-N-methylethanamine | Tertiary |
| m−BrC6H4NH2 | 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 (CH3)2CHNH2 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
(CH3)2CHNH2
- 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 (CH3)2CHNH2, 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
(CH3)2CHNH2 → 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 C6H5NHCH3, 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
C6H5NHCH3 → Secondary amine (N-methylaniline).
✗ Common Mistake #4: Forgetting to name substituents on nitrogen in secondary/tertiary amines
The Mistake
For (CH3CH2)2NCH3, 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
(CH3CH2)2NCH3 → 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 (C2H5)3N, a different (six-carbon) compound.
✗ Common Mistake #5: Misplacing the locant for substituents on the ring in aromatic amines
The Mistake
For m−BrC6H4NH2, 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−BrC6H4NH2 → 3-Bromoaniline (or m-bromoaniline) → Primary amine.
📋 Quick Reference Table
| Compound | IUPAC Name | Class |
|---|---|---|
| (CH3)2CHNH2 | Propan-2-amine | 1° |
| CH3(CH2)2NH2 | Propan-1-amine | 1° |
| CH3NHCH(CH3)2 | N-Methylpropan-2-amine | 2° |
| (CH3)3CNH2 | 2-Methylpropan-2-amine | 1° |
| C6H5NHCH3 | N-Methylaniline | 2° |
| (CH3CH2)2NCH3 | N-Ethyl-N-methylethanamine | 3° |
| m−BrC6H4NH2 | 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.
- TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQQ.The reagent which is used to distinguish C6H5N(CH3)2 and (C2H5)2NH is (anhy = anhydrous, conc = concentrated, alc = alcoholic) (A) C6H5SO2Cl (B) Anhy. ZnCl2 ∣ conc. HCl (C) CrO3 ∣ H2SO4 (D) CHCl3 ∣ alc. KOH
›Reveal solutionSolution
The pair is a tertiary amine and a secondary amine. Only Hinsberg's reagent, C6H5SO2Cl (option A), separates them: the secondary amine gives an alkali-insoluble sulphonamide, the tertiary amine does not react.
The concept first
Before choosing a reagent, classify the two compounds:
- C6H5N(CH3)2 (N,N-dimethylaniline): the nitrogen carries three carbon groups → tertiary amine, no N–H.
- (C2H5)2NH (diethylamine): the nitrogen carries two carbon groups and one H → secondary amine, one N–H.
So you need a test that distinguishes 2∘ from 3∘ — i.e. a test that keys off the presence of an N–H bond, not off basicity.
Hinsberg's test does exactly that. Benzenesulphonyl chloride acylates the nitrogen, but only if there is an N–H available:
- 1∘ amine → C6H5SO2NHR. The N–H left over is made acidic by the two electron-withdrawing S=O groups → the sulphonamide dissolves in KOH/NaOH.
- 2∘ amine → C6H5SO2NR2. No N–H remains, so the sulphonamide is a solid that is insoluble in alkali.
- 3∘ amine → no reaction (no N–H to substitute); the amine remains as an oily layer, dissolving only on adding acid.
Step-by-step
- Test option (D), CHCl3 + alc. KOH (carbylamine). It gives the foul-smelling isocyanide only with primary amines. Both compounds here are non-primary → both give a negative test → cannot distinguish. ✗
- Test option (B), anhy. ZnCl2 + conc. HCl (Lucas reagent). This is a test for alcohols (distinguishing 1∘/2∘/3∘ ROH by turbidity). Amines are not alcohols. ✗
- Test option (C), CrO3/H2SO4. An oxidising mixture; it does not give two clearly different, characteristic observations for these two amines. ✗
- Test option (A), C6H5SO2Cl (Hinsberg).
- With (C2H5)2NH (2∘):
C6H5SO2Cl+(C2H5)2NH→C6H5SO2N(C2H5)2+HCl
a **solid sulphonamide, insoluble in aqueous KOH** — a clear positive observation.- With C6H5N(CH3)2 (3∘): no reaction — the amine is recovered unchanged. Two visibly different outcomes → the reagent distinguishes them. ✓
✓Final answerBenzenesulphonyl chloride (Hinsberg's reagent) reacts with the secondary amine but not with the tertiary amine, so the correct option is (A).
ANSWER: A
- TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQQ.An organic compound C7H9N on reduction with reagent X gave Y. Reaction of Y with p-toluene sulphonyl chloride gave Z which is insoluble in alkali. X and Y respectively are (A) LiAlH4 , C6H5−CH2−NH2 (benzylamine) (B) NaBH4 , C6H5−NHCH3 (N-methylaniline) (C) H2∣Ni , C6H5−CH2−NH2 (benzylamine) (D) H2∣Ni , C6H5−NHCH3 (N-methylaniline)
›Reveal solutionSolution
The alkali-insoluble benzenesulphonamide tells us Y must be a secondary amine, which fixes Y=C6H5NHCH3; a secondary amine of this shape arises from catalytic reduction of an isocyanide, so X=H2/Ni. Option (D).
The concept: why alkali-solubility identifies the class of amine
When an amine reacts with a sulphonyl chloride (Hinsberg's reagent, here p-toluenesulphonyl chloride):
- 1° amine RNH2 gives RNH−SO2Ar. The N–H left on nitrogen sits between an electron-withdrawing SO2 group and the ring, so it is acidic — the sulphonamide dissolves in NaOH forming a salt.
- 2° amine R2NH gives R2N−SO2Ar. Nitrogen now has no hydrogen at all, nothing to ionise, so the product is insoluble in alkali.
- 3° amine does not react at all.
So "Z is insoluble in alkali" is a direct statement that Y is a secondary amine.
Step-by-step
- Use the Hinsberg clue. Z insoluble in alkali ⇒ Y has no N–H after sulphonylation ⇒ Y is 2°.
- Screen the options. C6H5CH2NH2 (benzylamine) is a primary amine — its tosylamide C6H5CH2NH−SO2C6H4CH3 still has an acidic N–H and would dissolve in KOH. So options (A) and (C) are out. Y must be C6H5NHCH3.
- Now identify the reduction. A secondary amine bearing an N−CH3 group is the hallmark product of reducing an isocyanide (carbylamine):
C6H5−NC+4[H] H2/Ni C6H5−NH−CH3
Contrast with a nitrile, R−C≡N+4[H]→R−CH2−NH2, which always gives a primary amine — that is the benzylamine route, already ruled out.
4. Pick the reagent. NaBH4 is too mild for this reduction (it does not reduce nitriles/isocyanides), so option (B) fails. Catalytic hydrogenation, H2∣Ni, is the standard reagent.
5. Confirm the sequence:
substrate H2/Ni C6H5NHCH3 (Y) TsCl C6H5N(CH3)SO2C6H4CH3 (Z), no N–H⇒alkali-insoluble ✓
✓Final answerX is H2∣Ni and Y is N-methylaniline (C6H5NHCH3), so the correct option is (D).
ANSWER: D
- TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQQ.Identify the set, in which X and Y are correctly matched (A) NH2OH, Hydrazone (B) NH2NH2, Semicarbazone (C) C6H5NH2, Schiff base (D) RNH2, Oxime
›Reveal solutionSolution
The question asks which pair of reagent (X) and product (Y) is correctly matched. The correct match is aniline (C6H5NH2) with Schiff base, so option (C) is correct.
Concept & Intuition
This is a classic organic chemistry matching problem about carbonyl derivatives. Each reagent (X) reacts with a carbonyl compound (aldehyde or ketone) to give a specific nitrogen-containing derivative. The key is to recall the functional group of the product formed:
- Oximes come from hydroxylamine (NH2OH).
- Hydrazones come from hydrazine (NH2NH2).
- Semicarbazones come from semicarbazide (NH2NHCONH2).
- Schiff bases (imines) come from primary amines (RNH2), especially aromatic ones like aniline.
Let’s check each option step by step.
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Option (A): NH2OH → Hydrazone
Hydroxylamine (NH2OH) reacts with a carbonyl to form an oxime (with a C=NOH group), not a hydrazone. Hydrazones come from hydrazine. So this is incorrect.
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Option (B): NH2NH2 → Semicarbazone
Hydrazine (NH2NH2) gives a hydrazone (C=NNH2). Semicarbazones require semicarbazide (NH2NHCONH2). So this is incorrect.
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Option (C): C6H5NH2 → Schiff base
Aniline (C6H5NH2) is a primary aromatic amine. It reacts with aldehydes/ketones to form an imine (also called a Schiff base), with the general structure C6H5N=CR2. This is correct.
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Option (D): RNH2 → Oxime
A generic primary amine (RNH2) forms an imine (Schiff base), not an oxime. Oximes require hydroxylamine. So this is incorrect.
Watch outA common mistake is confusing the product names: remember that “oxime” always comes from hydroxylamine, “hydrazone” from hydrazine, “semicarbazone” from semicarbazide, and “Schiff base” from a primary amine.
TipA quick mnemonic: Hydroxylamine → Oxime (both have “O”); Hydrazine → Hydrazone; Semicarbazide → Semicarbazone; Amine → Azomethine (Schiff base).
✓Final answerThe correct option is (C).
ANSWER: C
- TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQQ.Identify what is Y in the following reaction sequence?
[!FORMULA] CH3−CO−NH2Br2NaOH (aq)X(i) NaNO2+HCl(ii) H2OY
(A) CH3NH2 (B) CH3CONHBr (C) CH3OH (D) BrCH2CONH2›Reveal solutionSolution
Acetamide undergoes Hofmann bromamide degradation to methylamine (X); nitrous acid converts this primary aliphatic amine into an unstable diazonium ion that immediately loses NX2 and picks up water, giving methanol — option (C).
The concept first: why aliphatic diazonium salts die instantly
Diazotisation (NaNOX2+HCl) makes the same −NX2X+ group from any primary amine. The difference is stability. In an aryl diazonium salt the −NX2X+ is conjugated with the ring, so at 273–278 K it survives long enough to be used in coupling/Sandmeyer reactions. In an alkyl diazonium ion there is no such delocalisation, and NX2 is an outstanding leaving group — so the ion falls apart the moment it forms, giving a carbocation that the solvent (water) captures. That is why primary aliphatic amines simply effervesce and give alcohols with nitrous acid.
Step-by-step
Step 1 — Identify X. Acetamide with bromine in aqueous alkali is the textbook Hofmann bromamide degradation:
CHX3CONHX2+BrX2+4NaOHCHX3NHX2+NaX2COX3+2NaBr+2HX2O
So X=CHX3NHX2 (methanamine). Notice the carbon count drops from 2 to 1 — the carbonyl carbon leaves as carbonate.
Step 2 — Diazotise X.
CHX3NHX2NaNOX2+HClCHX3−N+≡N ClX−
Step 3 — Let it decompose in water.
CHX3−NX2X+CHX3X++NX2↑
CHX3X++HX2OCHX3OH+HX+
So Y=CHX3OH.
Step 4 — Rule out the others. (A) CHX3NHX2 is X, not Y. (B) CHX3CONHBr is only the intermediate N-bromoamide inside step 1. (D) would require ring/α-bromination, which BrX2/NaOH on an amide does not do.
✓Final answerY is methanol, CHX3OH, so the correct option is (C).
ANSWER: C
- TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQQ.An amine (X) reacts with p-toluene sulphonyl chloride to give the product Y, which is insoluble in alkali. The product of X with benzoyl chloride is (A) CH3CH2CH(NH2)−COC6H5 (B) CH3CH2CH2N(CH3)COCH2C6H5 (C) CH3CH2NHCOC6H5 (D) CH3CH2N(CH3)COC6H5
›Reveal solutionSolution
A sulphonamide that is insoluble in alkali can only come from a secondary amine, so X is CHX3CHX2NHCHX3; benzoyl chloride then acylates its nitrogen to give CHX3CHX2N(CHX3)COCX6HX5 — option (D).
The concept first: why alkali-solubility fingerprints the amine class
p-Toluenesulphonyl chloride (a Hinsberg-type reagent) reacts with the amine's lone pair, replacing an N−H hydrogen with the bulky −SOX2Ar group:
- Primary amine R−NHX2 → R−NH−SOX2Ar. One N−H remains. That hydrogen is acidic, because the resulting anion is stabilised by the strongly electron-withdrawing sulphonyl group. Hence the product dissolves in KOH/NaOH.
- Secondary amine RX2NH → RX2N−SOX2Ar. No N−H is left. There is no acidic proton to remove, so the product is insoluble in alkali. ✓
- Tertiary amine RX3N → no reaction (there is no N−H to substitute in the first place), so no product Y at all.
The stem says a product Y forms and is insoluble in alkali. Both facts together force X to be secondary.
Step-by-step
Step 1 — Classify X. Product formed ⇒ not tertiary. Product alkali-insoluble ⇒ not primary. Therefore X is a 2∘ amine.
Step 2 — Benzoylation (Schotten–Baumann). Benzoyl chloride CX6HX5COCl acylates the amine nitrogen:
RX2NH+CX6HX5COClRX2N−CO−CX6HX5+HCl
The amide nitrogen ends up carrying both original alkyl groups plus the benzoyl group — i.e. it is a tertiary (N,N-disubstituted) amide with no N−H.
Step 3 — Test each option against "benzoyl on a 2∘ nitrogen".
- (A) CHX3CHX2CH(NHX2)COCX6HX5 — the benzoyl is on carbon, and a free −NHX2 survives. Wrong on both counts.
- (B) CHX3CHX2CHX2N(CHX3)COCHX2CX6HX5 — the acyl group here is −COCHX2CX6HX5 (phenylacetyl), not benzoyl −COCX6HX5.
- (C) CHX3CHX2NHCOCX6HX5 — this amide still has an N−H, so it must have come from a primary amine (CHX3CHX2NHX2). But a primary amine would have given an alkali-soluble Y. Contradiction.
- (D) CHX3CHX2N(CHX3)COCX6HX5 — nitrogen carries ethyl + methyl + benzoyl: exactly what you get by benzoylating the secondary amine CHX3CHX2−NH−CHX3. ✓
Step 4 — Back-substitute and check. With X=CHX3CHX2NHCHX3, the sulphonamide would be CHX3CHX2N(CHX3)SOX2CX6HX4CHX3 — no N−H, insoluble in alkali. ✓ Everything is consistent.
✓Final answerX is the secondary amine N-methylethanamine, whose benzoylation gives CHX3CHX2N(CHX3)COCX6HX5, so the correct option is (D).
ANSWER: D
- TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQQ.Lithium nitrate on heating gives (A) Li2O+NO2 (B) Li2O+NO2+O2 (C) LiNO2+O2 (D) Li2O2+NO2+O2
›Reveal solutionSolution
Lithium nitrate decomposes differently from other alkali metal nitrates because of the small size and high polarising power of Li⁺; it gives Li2O, NO2, and O2, so the correct option is (B).
Concept & Intuition
Most alkali metal nitrates (like NaNO₃, KNO₃) decompose on strong heating to give the nitrite and oxygen:
2MNO3→2MNO2+O2.
But lithium is an exception. Because Li⁺ is very small, it has a high charge density and strongly polarises the nitrate ion. This destabilises the nitrate, causing it to break down further — lithium nitrite (LiNO₂) itself is unstable at high temperature and decomposes to lithium oxide, nitrogen dioxide, and oxygen. So the final products are not simply nitrite + O₂, but oxide + NO₂ + O₂.
Step-by-step reasoning
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General trend for alkali metal nitrates
For Na, K, Rb, Cs:
2MNO3Δ2MNO2+O2
The nitrite is stable at the decomposition temperature.
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Lithium’s anomaly
Li⁺ is much smaller than other alkali ions. Its high polarising power weakens the N–O bonds in the nitrate ion, so decomposition occurs at a lower temperature and proceeds further.
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First step – formation of nitrite
Initially, LiNO₃ does form LiNO₂ and O₂:
2LiNO3→2LiNO2+O2
But LiNO₂ is not stable at the temperature needed for decomposition.
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Second step – further decomposition of LiNO₂
Lithium nitrite decomposes to lithium oxide, nitrogen dioxide, and oxygen:
2LiNO2→Li2O+NO2+NO
However, NO reacts immediately with O₂ to give NO₂:
2NO+O2→2NO2
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Overall net reaction
Combining the steps:
4LiNO3→2Li2O+4NO2+O2
Dividing by 2 gives the simplest form:
2LiNO3→Li2O+2NO2+21O2
So the products are Li2O, NO2, and O2.
Watch outA common mistake is to assume lithium behaves like sodium or potassium and pick option (C) LiNO2+O2. But LiNO₂ is thermally unstable — it does not survive the heating.
TipRemember the “lithium exception” for nitrates, carbonates, and hydroxides: small cation → greater polarisation → more extensive decomposition.
✓Final answerThe correct option is (B).
ANSWER: B
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