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

Glycerol + excess HI → 2-iodopropane (isopropyl iodide) + I₂ + H₂O

---

Explanation

• First, –OH groups are replaced by –I
• Then elimination and rearrangement occur
• Finally, the most stable secondary alkyl iodide is formed

Ethylene glycol (ethane-1,2-diol)

---

• Ethylene glycol has two –OH groups, so it forms strong hydrogen bonding.
• It lowers the freezing point of water (freezing point depression).
• It also raises the boiling point, preventing overheating.

---

✔ Used as antifreeze because it prevents water from freezing in cold conditions.

---

Other than ethylene glycol, there are a few compounds used as antifreeze:

---

Other Antifreeze Compounds

1. Propylene glycol (propane-1,2-diol)

• Less toxic than ethylene glycol
• Used in safer antifreeze formulations (food-grade systems)

---

2. Glycerol (glycerine, propane-1,2,3-triol)

• Has three –OH groups → strong H-bonding
• Good freezing point depression
• Used in older or special applications

---

3. Methanol / Ethanol

• Also lower freezing point
• But ❌ not preferred because:
  • Highly volatile
  • Flammable
  • Evaporate easily

---

Degree of Unsaturation (DU) / Index of Hydrogen Deficiency (IHD) Formula

The Degree of Unsaturation (DU) tells you the total number of rings and π (pi) bonds present in an organic compound.

General Formula

Where:

• C = Number of carbon atoms
• H = Number of hydrogen atoms
• N = Number of nitrogen atoms
• X = Number of halogen atoms (F, Cl, Br, I)
• O, S, Se = Ignored (they do not affect DU)

---

Shortcut Formula

---

Effect of Different Atoms

Atom	Effect on DU Calculation
C	Count normally
H	Subtract
N	Add 1 H equivalent (+1)
F, Cl, Br, I	Treat as H
O, S	Ignore

---

Meaning of DU

DU	Possible Structure
0	Saturated compound (no ring, no π bond)
1	One ring or one double bond
2	Two double bonds, or one triple bond, or one ring + one double bond
3	Three π bonds/rings in any combination
4	Benzene ring (1 ring + 3 double bonds = 4 DU)

1. Picric acid has three –NO₂ groups which are strong electron-withdrawing groups.
2. These groups stabilize the conjugate base (picrate ion) by resonance and –I effect.
3. Formic acid has only one –COOH group, so its conjugate base is less stabilized.
4. Greater stabilization of conjugate base ⇒ higher acidity, so picric acid is stronger.

HCOOH (Picric Acid)

Picric acid is soluble in NaHCO₃ solution?

Picric acid (2,4,6-trinitrophenol) is a strong acid due to the presence of three –NO₂ groups, which strongly withdraw electrons and stabilize the phenoxide ion. Because of this high acidity, it can react even with a weak base like NaHCO₃.

Reaction:
Picric acid + NaHCO₃ → Sodium picrate (soluble) + CO₂ + H₂O

So, it dissolves in NaHCO₃ with effervescence of CO₂ gas.