NCERT Exercise 12.15 Solved

Detailed "2-step" conversion strategies for Aldehydes, Ketones, and Carboxylic Acids. Master reagents like Grignard, Rosenmund Catalyst, and Aldol Condensation.

Board Exam Important Class 12

(i) Propanone to Propene

$$ \text{CH}_3\text{COCH}_3 \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{CH(OH)CH}_3 $$ $$ \xrightarrow[\Delta]{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{CH}=\text{CH}_2 $$

Strategy: Reduce ketone to secondary alcohol, then dehydrate to alkene.

(ii) Benzoic acid to Benzaldehyde

$$ \text{Ph-COOH} \xrightarrow{\text{SOCl}_2} \text{Ph-COCl} $$ $$ \xrightarrow[\text{Pd-BaSO}_4]{\text{H}_2} \text{Ph-CHO} $$

Strategy: Convert acid to acid chloride, then use Rosenmund Reduction to stop reduction at aldehyde stage.

(iii) Ethanol to 3-Hydroxybutanal

$$ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\text{573 K}]{\text{Cu}} \text{CH}_3\text{CHO} $$ $$ \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO} $$

Strategy: Mild oxidation to ethanal (acetaldehyde), followed by Aldol Condensation.

(iv) Benzene to m-Nitroacetophenone

$$ \text{C}_6\text{H}_6 \xrightarrow[\text{AlCl}_3]{\text{CH}_3\text{COCl}} \text{Ph-COCH}_3 $$ $$ \xrightarrow{\text{conc. HNO}_3/\text{H}_2\text{SO}_4} \text{m-NO}_2\text{-C}_6\text{H}_4\text{-COCH}_3 $$

Strategy: Friedel-Crafts Acylation first. The -COCH₃ group is meta-directing, guiding the nitro group to the correct position.

(v) Benzaldehyde to Benzophenone

$$ \text{Ph-CHO} \xrightarrow[\text{2. H}_3\text{O}^+]{\text{1. PhMgBr}} \text{Ph-CH(OH)-Ph} $$ $$ \xrightarrow{\text{CrO}_3 \text{ or PCC}} \text{Ph-CO-Ph} $$

Strategy: Grignard reaction (Phenyl Magnesium Bromide) creates a secondary alcohol. Oxidation converts it to a ketone.

(vi) Bromobenzene to 1-Phenylethanol

$$ \text{Ph-Br} \xrightarrow[\text{Ether}]{\text{Mg}} \text{PhMgBr} $$ $$ \xrightarrow[\text{2. H}_3\text{O}^+]{\text{1. CH}_3\text{CHO}} \text{Ph-CH(OH)-CH}_3 $$

Strategy: Convert to Grignard reagent. Reacting Grignard with Acetaldehyde yields a secondary alcohol.

(vii) Benzaldehyde to 3-Phenylpropan-1-ol

$$ \text{PhCHO} + \text{CH}_3\text{CHO} \xrightarrow[\Delta]{\text{dil. NaOH}} \text{PhCH=CHCHO} $$ $$ \xrightarrow{\text{H}_2/\text{Ni}} \text{Ph-CH}_2\text{CH}_2\text{CH}_2\text{OH} $$

Strategy: Cross-Aldol condensation with acetaldehyde gives Cinnamaldehyde. Catalytic hydrogenation reduces both the double bond and the aldehyde to alcohol.

(viii) Benzaldehyde to $\alpha$-Hydroxyphenylacetic acid

$$ \text{Ph-CHO} \xrightarrow{\text{HCN}} \text{Ph-CH(OH)-CN} $$ $$ \xrightarrow{\text{H}_3\text{O}^+} \text{Ph-CH(OH)-COOH} $$

Strategy: Nucleophilic addition of HCN forms a Cyanohydrin. Hydrolysis of the -CN group converts it to -COOH.

(ix) Benzoic acid to m-Nitrobenzyl alcohol

$$ \text{Ph-COOH} \xrightarrow{\text{HNO}_3/\text{H}_2\text{SO}_4} \text{m-NO}_2\text{-C}_6\text{H}_4\text{-COOH} $$ $$ \xrightarrow{\text{B}_2\text{H}_6 \text{ or LiAlH}_4} \text{m-NO}_2\text{-C}_6\text{H}_4\text{-CH}_2\text{OH} $$

Strategy: Nitration first (-COOH is meta directing). Then reduce the carboxylic acid to alcohol (Diborane is preferred as it doesn't reduce the nitro group easily).

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NCERT Ex 12.15 Solved: Aldehydes & Ketones Conversions | ChemCa