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