How will you convert:
(i) Propanone to Propene,
(ii) Benzoic acid to Benzaldehyde,
(iii) Benzene to m-Nitroacetophenone?
(i) Propanone to Propene,
(ii) Benzoic acid to Benzaldehyde,
(iii) Benzene to m-Nitroacetophenone?
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Solution and Explanation
• Conversion via reduction and dehydration.
• Aldehyde reactivity depends on steric hindrance and electronic effects.
• Aldol condensation requires $\alpha$-hydrogen atoms.
Step 1: Chemical Conversions.
(i) Propanone $\xrightarrow{NaBH_4}$ Propan-2-ol $\xrightarrow{conc. H_2SO_4, \Delta}$ Propene. (ii) Benzoic acid $\xrightarrow{SOCl_2}$ Benzoyl chloride $\xrightarrow{H_2, Pd/BaSO_4}$ Benzaldehyde (Rosenmund Reduction). (iii) Benzene $\xrightarrow{CH_3COCl/AlCl_3}$ Acetophenone $\xrightarrow{conc. HNO_3/H_2SO_4}$ m-Nitroacetophenone.
Step 2: Reactivity and Aldol identification.
(b)(i) Reactivity towards HCN: Reactivity decreases with increased steric hindrance and +I effect of alkyl groups. Order: Di-tert-butyl ketone < Propanone < Acetaldehyde. (b)(ii) Aldol Condensation: Only compounds with $\alpha$-hydrogens react. $HCHO$ and $C_6H_5CHO$ have no $\alpha$-H. $CH_3CHO$ and Cyclohexanone undergo Aldol condensation.
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\((v)\)
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