1°, 2°, 3° Amines with HNO₂

Primary (1°) Amine

(Just for comparison)

• Aliphatic:

$$RNH_2 + HNO_2 \rightarrow ROH + N_2 \uparrow + H_2O$$

Brisk effervescence (N₂ gas)

• Aromatic: forms diazonium salt

2° (Secondary) Amine

Reaction:

$$R_2NH + HNO_2 \rightarrow R_2N–N=O + H_2O$$ Product: N-nitrosamine

Observation:

• Yellow oily liquid
• No gas evolution

✔ Used as a test for 2° amines

3° (Tertiary) Amine

(A) Aliphatic 3° Amine:

$$R_3N + HNO_2 \rightarrow R_3NH^+ NO_2^-$$​

Forms ammonium nitrite salt

Observation:

• No visible reaction
• No gas

(B) Aromatic 3° Amine:

Undergoes electrophilic substitution (ring reaction)

Example: aniline derivatives like dimethylaniline$$\text{Ar–N(CH₃)₂} + HNO_2 \rightarrow p\text{-nitroso compound}$$

✔ Green-colored solution (often)

Summary Table

Quick Trick

• 1° → Gas (N₂)
• 2° → Nitroso (yellow oil)
• 3° → No gas

Correct Reagent

We use NaNO₂ + HCl, but the actual reacting species is HNO₂ (nitrous acid)

Why?

• HNO₂ is unstable, so it is not stored directly
• It is generated in situ:

$$NaNO_2 + HCl \rightarrow HNO_2 + NaCl$$

Conclusion

✔ In reaction/mechanism → write HNO₂
✔ In practical/lab conditions → write NaNO₂ + HCl

emperature is very important in reactions with nitrous acid (HNO₂ / NaNO₂ + HCl).

Required Temperature

✔ Primary (1°) Aromatic Amines

0–5°C (ice-cold conditions)

• Needed to form stable diazonium salt
• Above 5°C → diazonium salt decomposes

Primary (1°) Aliphatic Amines

Also done at 0–5°C

• But diazonium salt is unstable → immediately decomposes to alcohol + N₂

Secondary (2°) Amines

0–5°C preferred

• Forms N-nitrosamine
• Low temperature prevents side reactions

Tertiary (3°) Amines

• Aliphatic: 0–5°C (no major reaction, just salt formation)
• Aromatic: slightly higher temp possible, but usually cold conditions used

Summary Table

tephen Reaction (Stephen Aldehyde Synthesis) – Mechanism (JEE/NEET Concept)

Definition:
Conversion of nitriles (R–C≡N) into aldehydes (R–CHO) using SnCl₂/HCl, followed by hydrolysis.

Overall Reaction

$$R-C \equiv N \xrightarrow[\text{HCl}]{\text{SnCl}_2} R-CH=NH \cdot HCl \xrightarrow{\text{H}_2O} R-CHO$$

Stepwise Mechanism

Step 1: Formation of Iminium Salt (Reduction Step)

• SnCl₂ (mild reducing agent) donates electrons.
• Nitrile gets partially reduced.
• Formation of iminium chloride salt:

$$R-C \equiv N \xrightarrow{\text{SnCl}_2/HCl} R-CH=NH \cdot HCl$$

Key point:

• Triple bond (C≡N) → double bond (C=NH)
• Controlled reduction (not all the way to amine)

Step 2: Hydrolysis of Iminium Salt

• Iminium salt reacts with water.
• Produces aldehyde + ammonium salt

$$R-CH=NH \cdot HCl + H_2O \rightarrow R-CHO + NH_4Cl$$R−CH=NH⋅HCl+H2​O→R−CHO+NH4​Cl

Important Points (Exam-Oriented)

• Works best for alkyl nitriles (aryl nitriles give poor yield).
• Stops at aldehyde stage (unlike strong reduction → amine).
• SnCl₂/HCl = selective mild reducing system
• Intermediate = iminium salt (not imine directly)

Shortcut Trick (Memory Tip)

“Nitrile + Mild Reduction = Aldehyde”
Think: Stephen = Stops Early (Aldehyde, not amine)

Example

$$CH_3CN \xrightarrow{\text{SnCl}_2/HCl} CH_3CHO$$

Oxidation of a benzylic methyl group (–CH₃ attached to benzene) to an aldehyde (–CHO) using
Chromyl chloride in a non-aqueous solvent (like CCl₄).

Mechanism (Stepwise)

Formation of Etard Complex

• Chromyl chloride attacks benzylic hydrogen
• Forms a brown complex (Etard complex)

$$Ar-CH_3 \rightarrow Ar-CH(OCrOCl_2)_2$$

Hydrolysis of Complex

• On hydrolysis (H₂O), the complex breaks
• Gives aldehyde

$$Ar-CH(OCrOCl_2)_2 \xrightarrow{H_2O} Ar-CHO$$

Key Idea

Unsaturated ethers (enol ethers) on acidic hydrolysis give carbonyl compounds (aldehydes/ketones).

Why does this happen?

Unsaturated ethers are basically enol forms of carbonyl compounds.

General structure:$$R – CH = CH – OR’$$

Under acidic conditions:

1. Protonation of ether oxygen
2. Formation of unstable enol
3. Tautomerism (enol → carbonyl)

🔹 Important Reaction

Example:$$CH_2 = CH – OCH_3 \xrightarrow{H^+, H_2O} CH_3CHO + CH_3OH$$

Product:

• Aldehyde/Ketone (carbonyl compound)
• Alcohol

🔹 Mechanism Shortcut (Exam Trick ⚡)

• Break C–O bond
• Form enol
• Enol → carbonyl (keto form)

One-line Concept (for revision)

Unsaturated ether + acidic hydrolysis → enol → carbonyl compound

Important for Exams

• Only vinyl / allyl ethers show this behavior
• Due to keto-enol tautomerism
• Very common in mechanism-based questions

Cannizzaro Reaction (JEE/NEET Concepts)

Definition:
Cannizzaro reaction is a disproportionation reaction in which an aldehyde without α-hydrogen undergoes self oxidation–reduction in the presence of strong base (NaOH/KOH).

General Reaction

$$2RCHO + OH^- \rightarrow RCH_2OH + RCOO^-$$

(One molecule is reduced → alcohol, other oxidized → carboxylate salt)

• Key Conditions (VERY IMPORTANT for JEE)

• Aldehyde must have NO α-H (alpha hydrogen)
• Strong base: conc. NaOH or KOH
• Usually occurs with:
  • Formaldehyde (HCHO)
  • Benzaldehyde (C₆H₅CHO)

Example

$$2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa$$

(Benzaldehyde → benzyl alcohol + sodium benzoate)

Mechanism (Conceptual Steps)

1. Nucleophilic attack of OH⁻ on aldehyde → alkoxide intermediate
2. Hydride transfer (H⁻ shift) from one molecule to another
3. Formation of:
   • Alcohol (reduction)
   • Carboxylate ion (oxidation)

Types of Cannizzaro Reaction

1. Self Cannizzaro
   Same aldehyde reacts
   Example: benzaldehyde
2. Cross Cannizzaro (Important!)
   Two different aldehydes
   • One should be formaldehyde (HCHO) (best reducing agent)

$$HCHO + C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + HCOONa$$

(Formaldehyde gets oxidized, other gets reduced)

This is a high-weightage concept from alcohols + organometallics.

General Reaction:

$$\text{Epoxide} + RMgX \xrightarrow{\text{dry ether}} \text{alkoxide} \xrightarrow{H_3O^+} \text{alcohol}$$

Core Mechanism:

• Grignard reagent (RMgX) behaves like R⁻ (strong nucleophile)
• Attacks epoxide → ring opening (SN2 type)
• Final step: acidic hydrolysis → alcohol

Golden Rules (VERY IMPORTANT):

Attack Position:

✔ Always attacks LESS substituted carbon (SN2 mechanism)

Product Type:

✔ Alcohol is formed at the carbon where O⁻ was present
✔ Nature depends on epoxide structure:

• Ethylene oxide → Primary alcohol
• Substituted epoxide → Secondary / tertiary alcohol

Chain Extension Trick:

✔ Adds +2 carbons (only with ethylene oxide clearly)
✔ With other epoxides → chain increases but depends on structure

Important Cases:

1. Ethylene Oxide:

$$RMgX \rightarrow R-CH_2-CH_2-OH$$

Always primary alcohol
+2 carbon extension (fixed)

2. Unsymmetrical Epoxide (e.g., Propylene oxide):

$$RMgX \rightarrow R-CH_2-CH(R’)-OH$$

Attack at less hindered carbon
Usually secondary alcohol

• JEE/NEET Quick Summary:

✔ RMgX = nucleophile (R⁻)
✔ Epoxide opening = SN2 reaction
✔ Attack = less substituted carbon
✔ Product = alcohol after H₃O⁺

HO–CH₂–CH₂–OH + HIO₄ → 2 HCHO (formaldehyde) + HIO₃ + H₂O

Explanation (2–3 marks)

• Periodic acid cleaves –CHOH–CHOH (vicinal diol)
• C–C bond breaks
• Each carbon forms a carbonyl compound
• Since both carbons are –CH₂OH → gives 2 molecules of formaldehyde

A. CH₃(CH₂)₃NH₂ → (i) Alkaline hydrolysis

B. CH₃C≡CH → (ii) With KOH (alcohol) and CHCl₃ produces bad smell

C. CH₃CH₂COOCH₃ → (iii) Gives white ppt. with ammoniacal AgNO₃

D. CH₃CH(OH)CH₃ → (iv) With Lucas reagent cloudiness