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The major product Z formed in the following sequence of reactions is :
C₂H₆ →(Cl₂/UV light) X (monochlorinated)
X →(NH₃) Y
Y →(i) NaNO₂/HCl, (ii) H₂O → Z
Options
1
C₂H₅−N=N−OH
2
C₂H₅OH (Ethanol)
3
C₂H₅NO₂
4
C₂H₅NH₂
Correct Answer
Option 2 : C₂H₅OH (Ethanol)
Step-by-Step Solution
1

Step 1 — C₂H₆ + Cl₂ →(UV) X:

Free radical halogenation. Ethane reacts with Cl₂ in UV light to give ethyl chloride (monochlorinated product).

C₂H₆ + Cl₂ →(UV) C₂H₅Cl (X) + HCl

2

Step 2 — C₂H₅Cl + NH₃ → Y:

Nucleophilic substitution (SN2). Ammonia acts as nucleophile, displaces Cl⁻.

C₂H₅Cl + NH₃ → C₂H₅NH₂ (Y) — Ethylamine (primary amine)

3

Step 3 — C₂H₅NH₂ + NaNO₂/HCl → diazonium salt → H₂O → Z:

Primary aliphatic amine + HNO₂ (formed from NaNO₂ + HCl) → unstable aliphatic diazonium salt.

C₂H₅NH₂ + HNO₂ → [C₂H₅N₂]⁺Cl⁻ → (H₂O) → C₂H₅OH (Z) + N₂ + HCl

Aliphatic diazonium salts are extremely unstable — they instantly decompose with loss of N₂ to give carbocation, which reacts with water to give alcohol.

X = C₂H₅Cl (Ethyl chloride)
Y = C₂H₅NH₂ (Ethylamine)
Z = C₂H₅OH (Ethanol) ← correct answer
Theory: Amines & Diazonium Chemistry
1. Free Radical Halogenation of Alkanes

Alkanes react with halogens (Cl₂, Br₂) in presence of UV light or heat via free radical mechanism. The reaction involves three steps: Initiation (Cl₂ → 2Cl• by UV), Propagation (Cl• abstracts H from alkane → alkyl radical; alkyl radical + Cl₂ → alkyl chloride + Cl•), and Termination (two radicals combine). Selectivity: Br₂ is more selective (tertiary > secondary > primary) while Cl₂ is less selective. UV light or high temperature is necessary — reaction does NOT occur in dark at room temperature.

2. Preparation of Amines

📌 From alkyl halide + NH₃: RX + NH₃ → RNH₂ (primary) + R₂NH (secondary) + R₃N (tertiary) + R₄N⁺X⁻ (quaternary) — mixture formed

📌 Gabriel phthalimide synthesis: Gives pure primary amine only

📌 Reduction of nitro compounds: ArNO₂ + 6H → ArNH₂ (Sn/HCl or Fe/HCl)

📌 Reduction of nitriles: RCN + 4H → RCH₂NH₂ (LiAlH₄)

📌 Hofmann bromamide: RCONH₂ + Br₂ + NaOH → RNH₂ (fewer carbons!)

📌 Reduction of amides: RCONH₂ + LiAlH₄ → RCH₂NH₂

3. Reaction of Primary Amines with HNO₂ (NaNO₂ + HCl)

HNO₂ = nitrous acid, formed in situ: NaNO₂ + HCl → HNO₂ + NaCl. Primary aliphatic amines (R-NH₂): Highly unstable diazonium salt [RN₂]⁺ formed instantly decomposes at room temperature itself → gives alcohol (R-OH) with water, or alkene, or other products with loss of N₂. Cannot be isolated. Primary aromatic amines (Ar-NH₂): Stable diazonium salt [ArN₂]⁺Cl⁻ forms at 0–5°C — can be isolated and used in coupling reactions (azo dyes). Aryl diazonium salts are stable below 5°C because resonance stabilises the positive charge.

4. Aliphatic vs Aromatic Diazonium Salts — Key Difference

📌 Aliphatic [RN₂]⁺: Unstable even at 0°C → instantly loses N₂ → carbocation → alcohol/alkene

📌 Aromatic [ArN₂]⁺: Stable at 0–5°C → can be stored, used in Sandmeyer, Balz-Schiemann, coupling reactions

📌 Stability difference: aryl cation (sp² hybridised) is stabilised by resonance with ring; alkyl cation is not

📌 This is why aliphatic primary amines give alcohol with NaNO₂/HCl at room temperature

5. Basicity of Amines

Amines are basic due to the lone pair on nitrogen. Basicity order in gas phase: tertiary > secondary > primary > NH₃. In aqueous solution: secondary > primary > tertiary > NH₃ (due to solvation effects — tertiary amine's bulky groups hinder solvation of the conjugate acid). Aromatic amines are much less basic than aliphatic amines because the lone pair on N is delocalised into the benzene ring, making it less available for protonation. Aniline (pKb ≈ 9·4) is far less basic than methylamine (pKb ≈ 3·4).

6. Hofmann's Rule vs Saytzeff's Rule in Elimination

When dehydrohalogenation can give multiple alkenes: Saytzeff's rule: more substituted (stable) alkene is the major product — applies with small base (like KOH in alcohol). Hofmann's rule: less substituted alkene is major product — applies with bulky base (like t-BuOK). In this reaction (C₂H₅Cl + alc. KOH): only one possible alkene (ethene), so no ambiguity. The alkyl chloride has only primary hydrogens — only one product possible.

7. Sandmeyer Reaction (for Aromatic Diazonium Salts)

Aromatic diazonium salts react with Cu₂X₂ (X = Cl, Br, CN) to replace -N₂⁺ with -X group — called Sandmeyer reaction. ArN₂⁺Cl⁻ + Cu₂Cl₂ → Ar-Cl + N₂. ArN₂⁺Cl⁻ + Cu₂Br₂ → Ar-Br + N₂. ArN₂⁺Cl⁻ + CuCN → Ar-CN + N₂. Balz-Schiemann: ArN₂⁺BF₄⁻ → Ar-F + N₂ + BF₃. Gattermann: ArN₂⁺ + HX + Cu → Ar-X (uses Cu powder, not Cu₂X₂). These are important methods to introduce halogens onto benzene ring — direct halogenation gives different orientation.

8. Coupling Reaction of Aromatic Diazonium Salts

Aromatic diazonium salts couple with phenols or tertiary aromatic amines to form azo compounds (R-N=N-R') — these are intensely coloured and used as dyes (azo dyes). Coupling occurs at the para position (or ortho if para is blocked). The reaction is electrophilic aromatic substitution where the diazonium ion is the electrophile. Example: benzenediazonium chloride + aniline → para-aminoazobenzene (orange-red dye). Congo red, Methyl orange are important azo dyes tested in NEET.

Frequently Asked Questions
1. Why does C₂H₅NH₂ give alcohol (not diazonium salt) with NaNO₂/HCl?
C₂H₅NH₂ is a primary aliphatic amine. It forms an aliphatic diazonium salt [C₂H₅N₂]⁺ which is extremely unstable even at 0°C. It immediately loses N₂ gas, forming a carbocation (C₂H₅⁺), which reacts with the water solvent to give ethanol (C₂H₅OH). Unlike aryl diazonium salts (stable below 5°C due to resonance), alkyl diazonium salts cannot be stored or used further.
2. What is the mechanism of free radical chlorination of ethane?
Initiation: Cl₂ + UV → 2Cl• (homolytic cleavage). Propagation step 1: Cl• + C₂H₆ → HCl + C₂H₅• (ethyl radical). Propagation step 2: C₂H₅• + Cl₂ → C₂H₅Cl + Cl• (cycle repeats thousands of times). Termination: Cl• + Cl• → Cl₂, or Cl• + C₂H₅• → C₂H₅Cl, or C₂H₅• + C₂H₅• → C₄H₁₀. The chain propagation steps regenerate the radical — this is why one photon can produce thousands of product molecules.
3. Why is aromatic diazonium salt stable but aliphatic is not?
Stability requires delocalisation of the positive charge. In [ArN₂]⁺: the positive charge is on N which is directly attached to the benzene ring. Resonance between N=N⁺ and the ring delocalises the charge, providing stability at 0–5°C. In [RN₂]⁺: there is no adjacent π system to delocalise the charge. The N₂ group is an excellent leaving group (N₂ is very stable gas) — so the C–N bond breaks instantly, releasing N₂ and forming a carbocation.
4. What are all products when a primary aliphatic amine reacts with HNO₂?
Products depend on the solvent and conditions: With water: R-OH (alcohol) + N₂ ↑ + HCl. Some alkene may also form (elimination of the carbocation). In some cases, rearranged products form (carbocation rearrangement). The reaction is not very useful synthetically because of this mixture — Gabriel synthesis is preferred for pure primary amines. The main products to remember for NEET: alcohol + N₂.
5. What is Gabriel phthalimide synthesis?
Purpose: prepare pure primary amines (no secondary or tertiary amine contamination). Steps: (1) Phthalimide + KOH → potassium phthalimide (K⁺ salt). (2) Potassium phthalimide + RX → N-alkylphthalimide (SN2 substitution). (3) N-alkylphthalimide + H₂NNH₂ (hydrazine) → RNH₂ + phthalhydrazide. Or: acid hydrolysis gives RNH₂ + phthalic acid. The phthalimide nitrogen can only be monoalkylated — so only primary amine is formed. Limitation: cannot make aromatic primary amines (aryl halides don't undergo SN2).
6. Why does Hofmann bromamide degradation give amine with one less carbon?
Hofmann bromamide: RCONH₂ + Br₂ + NaOH → RNH₂. The CONH₂ group (1C) is lost as CO₂ during the reaction. Mechanism: N-bromo amide → rearrangement (N→C migration of R group) → isocyanate (RNCO) → hydrolysis → RNH₂ + CO₂. Net result: amide has n carbons (including carbonyl), amine has n-1 carbons. Used when you need to degrade an amide to get a shorter-chain amine. Example: acetamide (CH₃CONH₂) → methylamine (CH₃NH₂).
7. What is the basicity order of methylamine, dimethylamine, trimethylamine, and aniline?
In aqueous solution (pKb values): dimethylamine (3·27) > methylamine (3·36) > trimethylamine (4·19) > ammonia (4·74) > aniline (9·40). Trend: secondary > primary > tertiary for alkyl amines in solution (solvation effect on trimethylamine). Aniline is far less basic because lone pair on N is delocalised into benzene ring via resonance, reducing its availability for protonation. This is a highly NEET-tested concept.
8. What happens when secondary aliphatic amine reacts with HNO₂?
Secondary amine (R₂NH) + HNO₂ → N-nitrosamine (R₂N-N=O) — a yellow oily liquid. These are carcinogenic compounds. N-nitrosamines are important in NEET as they demonstrate that secondary amines do NOT give diazonium salts. Tertiary aliphatic amines (R₃N) + HNO₂ → no stable product; they just form unstable salts that decompose. Tertiary aromatic amines undergo C-nitroso reaction on the benzene ring (ring nitrosation at para position).
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