(ii) Identify the compounds, which would undergo Aldol condensation:
\( \mathrm{HCHO} \), \( \mathrm{C_6H_5CHO} \), \( \mathrm{CH_3CHO} \), and the cyclic ketone (a six-membered carbon ring with one C=O) shown in the figure below
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Use a two-step route for each conversion. For HCN reactivity, less steric crowding means more reactive. Only compounds with an alpha hydrogen give aldol condensation.
Step 1: Convert propanone to propene (a-i): First reduce the ketone to the alcohol. Sodium borohydride (\( \mathrm{NaBH_4} \)) or \( \mathrm{H_2/Ni} \) gives propan-2-ol. Then dehydrate the alcohol with conc. \( \mathrm{H_2SO_4} \) on heating to about 443 K. \[ \mathrm{CH_3COCH_3 \xrightarrow{NaBH_4} CH_3CH(OH)CH_3 \xrightarrow[443\ K]{conc.\ H_2SO_4} CH_3CH=CH_2} \]
Step 2: Convert benzoic acid to benzaldehyde (a-ii): Change the acid to the acid chloride with thionyl chloride. Then do a Rosenmund reduction with hydrogen over Pd-BaSO4. \[ \mathrm{C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl \xrightarrow{H_2,\ Pd-BaSO_4} C_6H_5CHO} \] The poisoned catalyst stops the reduction at the aldehyde stage.
Step 3: Convert benzene to m-nitroacetophenone (a-iii): Do the Friedel-Crafts acylation first, with \( \mathrm{CH_3COCl} \) and anhydrous \( \mathrm{AlCl_3} \), to get acetophenone. The acetyl group is meta directing, so nitration with conc. \( \mathrm{HNO_3} \) and conc. \( \mathrm{H_2SO_4} \) puts \( \mathrm{NO_2} \) at the meta position. \[ \mathrm{C_6H_6 \xrightarrow[AlCl_3]{CH_3COCl} C_6H_5COCH_3 \xrightarrow{conc.\ HNO_3/H_2SO_4} m\text{-}O_2NC_6H_4COCH_3} \] Nitration cannot be done first, because nitrobenzene does not undergo Friedel-Crafts acylation.
Step 4: Reactivity towards HCN (b-i): HCN adds by nucleophilic attack on the carbonyl carbon. Bulky groups block the attack, and alkyl groups push electrons to the carbon and make it less positive. Acetaldehyde has one methyl group and one hydrogen, so it is the least crowded and the most reactive. Acetone has two methyl groups, so it is less reactive. Di-tert-butyl ketone has two very bulky groups, so it is the least reactive. \[ \mathrm{(CH_3)_3C{-}CO{-}C(CH_3)_3 < (CH_3)_2C=O < CH_3CHO} \]
Step 5: Aldol condensation (b-ii): Aldol condensation needs at least one alpha hydrogen, that is, a hydrogen on the carbon next to the C=O. \( \mathrm{HCHO} \) has no alpha carbon and \( \mathrm{C_6H_5CHO} \) has no alpha hydrogen. So these two do not react. \( \mathrm{CH_3CHO} \) has three alpha hydrogens, and cyclohexanone has four alpha hydrogens. So these two undergo aldol condensation.
Final Answer: (a) (i) reduce, then dehydrate. (ii) \( \mathrm{SOCl_2} \), then Rosenmund reduction. (iii) acylation, then nitration. (b) (i) di-tert-butyl ketone < acetone < acetaldehyde. (ii) \( \mathrm{CH_3CHO} \) and cyclohexanone. \[ \boxed{\text{(b-ii) } \mathrm{CH_3CHO} \text{ and cyclohexanone}} \]
