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: Short notes

(i) Cannizzaro's reaction: Aldehydes that have no α-hydrogen undergo self oxidation and reduction (disproportionation) when warmed with concentrated alkali. One molecule is oxidised to a carboxylate salt while the other is reduced to a primary alcohol.
Example (benzaldehyde): 2 C₆H₅CHO + NaOH(conc.) → C₆H₅CH₂OH + C₆H₅COONa
Here benzaldehyde gives benzyl alcohol (reduction) and sodium benzoate (oxidation). Formaldehyde behaves similarly: 2 HCHO + NaOH → CH₃OH + HCOONa.

(ii) Aldol condensation: Aldehydes or ketones having at least one α-hydrogen react in the presence of dilute alkali (dil. NaOH) to give a β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol).
2 CH₃CHO →(dil. NaOH) CH₃CH(OH)CH₂CHO (3-hydroxybutanal, the 'aldol').
On heating, the aldol loses water to give an α,β-unsaturated carbonyl compound: CH₃CH(OH)CH₂CHO →(Δ) CH₃CH=CHCHO (but-2-enal) + H₂O.

(iii) Gattermann-Koch reaction: When benzene (or a benzene derivative) is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous AlCl₃ (with a little CuCl), it gives benzaldehyde. It is a formylation that puts a -CHO group on the ring.
C₆H₆ + CO + HCl →(anhyd. AlCl₃/CuCl) C₆H₅CHO

Option 2: Preparations (equations)

(i) Benzaldehyde from Toluene (Etard reaction): C₆H₅CH₃ + CrO₂Cl₂ →(CS₂) chromium complex →(H₃O⁺) C₆H₅CHO.
(ii) Benzamide from Benzoic acid: C₆H₅COOH →(SOCl₂) C₆H₅COCl →(NH₃) C₆H₅CONH₂.
(iii) Phthalimide from Phthalic acid: Phthalic acid →(Δ, -H₂O) phthalic anhydride →(NH₃, Δ) phthalimide (the cyclic imide).
(iv) m-Nitrobenzaldehyde from Benzaldehyde: C₆H₅CHO + conc. HNO₃ →(conc. H₂SO₄) m-O₂N-C₆H₄-CHO. The -CHO group is a deactivating, meta-directing group, so nitration occurs at the meta position.
(v) Benzaldehyde from Benzene (Gattermann-Koch): C₆H₆ + CO + HCl →(anhyd. AlCl₃/CuCl) C₆H₅CHO.