Question

Write chemical equations of three methods of preparation and two chemical properties of benzaldehyde.
OR
Write short notes on the following: (i) Aldol condensation (ii) Cross aldol condensation (iii) Cannizzaro reaction (iv) Hell-Volhard-Zelinsky reaction (v) Kolbe electrolysis.

Hint:

Benzaldehyde has no alpha-hydrogen, so think Etard/Gattermann-Koch for preparation and oxidation/Cannizzaro for reactions. For the notes, sort the reactions by whether an alpha-hydrogen is present.

Answer 1

Option 1 (alternative presentation) - making and reactions of benzaldehyde.

Preparation, Method 1 (Rosenmund reduction): Benzoyl chloride is reduced with hydrogen over palladium poisoned with BaSO₄, which stops the reduction at the aldehyde stage.
\(C_6H_5COCl + H_2 \xrightarrow{Pd/BaSO_4} C_6H_5CHO + HCl\)
Method 2 (side-chain chlorination then hydrolysis): Toluene is chlorinated in light to benzal chloride, which is then hydrolysed.
\(C_6H_5CH_3 \xrightarrow{Cl_2,\ h\nu} C_6H_5CHCl_2 \xrightarrow{H_2O} C_6H_5CHO\)
Method 3 (Gattermann-Koch formylation): \(C_6H_6 + CO + HCl \xrightarrow{AlCl_3/CuCl} C_6H_5CHO\)

Chemical property 1 - nucleophilic addition of HCN: The polar carbonyl adds HCN to form a cyanohydrin.
\(C_6H_5CHO + HCN \rightarrow C_6H_5CH(OH)CN\)
Chemical property 2 - benzoin condensation: Two molecules of benzaldehyde, with KCN as catalyst, couple to give benzoin.
\(2C_6H_5CHO \xrightarrow{KCN} C_6H_5CH(OH)COC_6H_5\)

Option 2 (alternative wording) - the five named reactions.
Aldol condensation: Base removes an alpha-hydrogen to form an enolate carbanion that adds to the carbonyl carbon of a second molecule; the resulting beta-hydroxy carbonyl dehydrates on warming. Acetaldehyde gives but-2-enal.
Cross aldol condensation: The same idea but between two different aldehydes/ketones; up to four products form, so it is practical only when one reactant lacks an alpha-hydrogen (benzaldehyde with acetaldehyde gives cinnamaldehyde).
Cannizzaro reaction: Limited to carbonyls with no alpha-hydrogen; concentrated alkali drives a hydride transfer from one molecule to another, so one is oxidised to a carboxylate and the other reduced to a primary alcohol. Example: \(2HCHO + NaOH \rightarrow CH_3OH + HCOONa\).
Hell-Volhard-Zelinsky reaction: Red phosphorus first converts the acid to its acid halide, whose alpha-position is halogenated, and the alpha-halo acid is then regenerated. Example: \(CH_3CH_2COOH \xrightarrow{Br_2/red\ P} CH_3CHBrCOOH\).
Kolbe electrolysis: At the anode the carboxylate loses an electron and CO₂, and the two alkyl radicals couple to a symmetrical alkane; hydrogen is set free at the cathode. Example: \(2CH_3COO^- \rightarrow C_2H_6 + 2CO_2 + 2e^-\).
\(\boxed{\text{Five reactions summarised}}\)