Question

(a) Write short notes on the following: (i) Cannizzaro's reaction (ii) Aldol condensation (iii) Gattermann-Koch reaction. (2+1½+1½=5)
OR
How will you obtain (write chemical equations only): (i) Benzaldehyde from Toluene (ii) Benzamide from Benzoic acid (iii) Phthalimide from Phthalic acid (iv) m-nitrobenzaldehyde from Benzaldehyde (v) Benzaldehyde from Benzene. (1+1+1+1+1=5)

Hint:

For Option 1 recall three named aldehyde reactions: disproportionation of aldehydes with no α-H (Cannizzaro), α-H self-addition then dehydration (aldol), and CO+HCl with AlCl₃ ring formylation (Gattermann-Koch). For Option 2 think Etard, acid to acid chloride to amide, cyclic imide, meta-directing -CHO, and Gattermann-Koch.

Answer 1

Option 1 (alternative explanation):

(i) Cannizzaro's reaction: A base-induced disproportionation shown only by aldehydes lacking an α-hydrogen (HCHO, C₆H₅CHO, (CH₃)₃CCHO). Brief mechanism: OH^− adds to the carbonyl carbon of one molecule to form a tetrahedral alkoxide, which then transfers a hydride ion (H^−) to the carbonyl carbon of a second molecule; the first is oxidised to the acid salt and the second is reduced to the alcohol.
2 HCHO + NaOH → HCOONa + CH₃OH.

(ii) Aldol condensation: The α-carbon of an aldehyde or ketone is acidic. Dilute base pulls off an α-H to give an enolate (carbanion), which attacks the carbonyl carbon of another molecule, forming a β-hydroxy carbonyl compound (the aldol). Gentle heating dehydrates it to an α,β-unsaturated carbonyl.
Acetaldehyde: 2 CH₃CHO → CH₃CH(OH)CH₂CHO →(−H₂O) CH₃CH=CHCHO.
Acetone (ketol): 2 (CH₃)₂CO → (CH₃)₂C(OH)CH₂COCH₃ (4-hydroxy-4-methylpentan-2-one).

(iii) Gattermann-Koch reaction: A ring formylation. A mixture of CO and HCl behaves like the unstable acid chloride of formic acid (HCOCl); with anhydrous AlCl₃/CuCl it delivers a -CHO group to benzene to make benzaldehyde: C₆H₆ + CO + HCl → C₆H₅CHO.

Option 2 (alternative routes):
(i) Toluene to benzaldehyde by the Etard reaction with chromyl chloride: C₆H₅CH₃ + CrO₂Cl₂, then hydrolysis → C₆H₅CHO. Side-chain oxidation stops at the aldehyde.
(ii) Benzoic acid to benzamide: warm the acid with ammonia to the ammonium salt, which loses water on strong heating: C₆H₅COOH + NH₃ → C₆H₅COONH₄ →(Δ, −H₂O) C₆H₅CONH₂.
(iii) Phthalic acid to phthalimide: heating with ammonia first gives the anhydride and then the cyclic imide (phthalimide), with loss of water.
(iv) Benzaldehyde to m-nitrobenzaldehyde: electrophilic nitration (HNO₃/H₂SO₄); as -CHO withdraws electrons and is meta-directing, the -NO₂ enters position 3.
(v) Benzene to benzaldehyde: direct formylation by the Gattermann-Koch method (CO, HCl, anhyd. AlCl₃/CuCl).