Question

Write short notes on the following. (i) Cross aldol condensation (ii) Stephen reaction (iii) Etard reaction. (2+1½+1½=5)
OR
How will you obtain (write chemical equations only): (i) Ethanoic anhydride from Ethanoic acid (ii) Benzoic acid from Ethyl benzoate (iii) Benzamide from benzoic acid. (2+1½+1½=5)

Hint:

Option 1: aldol between two different carbonyls (one lacking \(\alpha\)-H gives a single product); nitrile + SnCl\(_2\)/HCl then hydrolysis gives an aldehyde; toluene + chromyl chloride gives benzaldehyde. Option 2: dehydrate the acid with P\(_2\)O\(_5\); hydrolyse the ester then acidify; make the ammonium salt and heat.

Answer 1

Option 1

(i) Cross aldol condensation: Think of it as the aldol reaction with a twist: the two carbonyl partners are not identical. The base first pulls off an \(\alpha\)-hydrogen from whichever molecule has one, generating a nucleophilic carbanion. That carbanion attacks the electrophilic carbonyl carbon of the partner molecule, giving a \(\beta\)-hydroxy aldehyde/ketone that dehydrates on heating to an \(\alpha,\beta\)-unsaturated product. The reaction is preparatively clean only when one partner cannot form a carbanion (has no \(\alpha\)-H). Benzaldehyde is the classic such partner:
\[ C_6H_5CHO + CH_3CHO \xrightarrow{OH^-} C_6H_5CH=CHCHO + H_2O \]
If both partners carry \(\alpha\)-hydrogens, four different products (two self-aldols + two cross-aldols) form, which limits its use.

(ii) Stephen reaction: This is essentially a controlled partial reduction of the C\(\equiv\)N triple bond. The mild reducing system SnCl\(_2\) + HCl stops at the imine (aldimine) stage rather than going all the way to a primary amine. Acidic work-up hydrolyses the imine to a carbonyl:
\[ RCN \xrightarrow{SnCl_2,\,HCl} [RCH=NH] \xrightarrow{H_2O/H^+} RCHO + NH_3 \]
It is a convenient route to convert an acid-derived nitrile into the corresponding aldehyde with the carbon skeleton unchanged.

(iii) Etard reaction: Here the strong oxidant chromyl chloride selectively attacks the benzylic \(-CH_3\) of toluene but is trapped as an insoluble chromium complex before it can over-oxidise. Careful aqueous hydrolysis then releases the aldehyde:
\[ C_6H_5CH_3 \xrightarrow{CrO_2Cl_2} \text{brown complex} \xrightarrow{H_2O} C_6H_5CHO \]
The formation of the stable intermediate is what prevents further oxidation to benzoic acid.

Option 2

(i) Removing water from two acid molecules gives the anhydride:
\[ 2\,CH_3COOH \xrightarrow{P_2O_5,\,\Delta} (CH_3CO)_2O + H_2O \]
(ii) Saponify the ester, then acidify the carboxylate salt:
\[ C_6H_5COOC_2H_5 \xrightarrow{NaOH} C_6H_5COONa \xrightarrow{HCl} C_6H_5COOH \]
(iii) Convert the acid to its ammonium salt and dehydrate it thermally to the amide:
\[ C_6H_5COOH \xrightarrow{NH_3} C_6H_5COONH_4 \xrightarrow{\Delta,\,-H_2O} C_6H_5CONH_2 \]