Q.Identify the following compounds 'A' to 'J' (write only structural formula): [½×10]

A run of standard carbonyl/carboxylic-acid transformations — HVZ bromination, cyanide substitution, SeO2 oxidation, ester hydrolysis, side-chain oxidation, nitration, acid-chloride formation, ketonic decarboxylation, oxime dehydration, and gem-dihalide hydrolysis.

(i) $CH_3CH_2COOH$ (propanoic acid) $\xrightarrow{\text{red P}/Br_2}$ A: this is the Hell-Volhard-Zelinsky (HVZ) reaction, which brominates the $\alpha$-carbon of a carboxylic acid: A = $CH_3$-$CHBr$-$COOH$ (2-bromopropanoic acid). A $\xrightarrow{KCN}$ B: the $\alpha$-bromo acid undergoes nucleophilic substitution ($S_N2$) of Br by $CN^-$: B = $CH_3$-$CH(CN)$-$COOH$ (2-cyanopropanoic acid).

(ii) $CH_3COCH_3$ (acetone) $\xrightarrow{SeO_2/CH_3COOH}$ C: Selenium dioxide oxidation (Riley oxidation) converts an active methyl/methylene group adjacent to a carbonyl into a second carbonyl, giving the 1,2-diketo compound: C = $CH_3$-$CO$-$CHO$ (methylglyoxal/pyruvaldehyde).

(iii) $CH_3COOC_2H_5$ (ethyl acetate) $\xrightarrow{(i)\,aq.\,NaOH/\Delta\ (ii)\,HCl}$ D: base hydrolysis (saponification) followed by acidification hydrolyses the ester back to the free acid: D = $CH_3COOH$ (acetic acid), along with ethanol.

(iv) Toluene $\xrightarrow{(i)\,alk.\,KMnO_4\ (ii)\,H_3O^+}$ E: vigorous oxidation of the benzylic side-chain (regardless of chain length) converts it fully to a $-COOH$ group directly on the ring: E = $C_6H_5COOH$ (benzoic acid).

(v) Benzaldehyde $\xrightarrow{conc.\,HNO_3/conc.\,H_2SO_4}$ F: nitration of benzaldehyde; the $-CHO$ group is a deactivating, meta-directing substituent, so the incoming $-NO_2$ enters the meta position: F = m-nitrobenzaldehyde, m-$O_2N$-$C_6H_4$-$CHO$.

(vi) Benzoic acid $\xrightarrow{SOCl_2}$ G: thionyl chloride converts a carboxylic acid to the corresponding acid chloride, releasing $SO_2$ and HCl: G = $C_6H_5COCl$ (benzoyl chloride).

(vii) $(CH_3COO)_2Ca + (HCOO)_2Ca \xrightarrow{\Delta}$ H: dry (pyrolytic) distillation of a mixture of calcium acetate and calcium formate performs a crossed ketonic decarboxylation, giving an aldehyde (rather than the symmetrical ketone acetone that pure calcium acetate alone would give): H = $CH_3CHO$ (acetaldehyde).

(viii) Benzaldehyde $\xrightarrow{(i)\,NH_2OH/H^+\ (ii)\,P_2O_5}$ I: the aldehyde first condenses with hydroxylamine to form an oxime ($C_6H_5CH=NOH$), which is then dehydrated by $P_2O_5$ to give the nitrile: I = $C_6H_5CN$ (benzonitrile).

(ix) $C_6H_5CHCl_2 \xrightarrow{H_3O^+}$ J: hydrolysis of a benzylic gem-dihalide (both halogens on the same carbon) regenerates the parent carbonyl compound: J = $C_6H_5CHO$ (benzaldehyde).

OR (x) Benzaldehyde to cinnamic acid: Perkin reaction — benzaldehyde is heated with acetic anhydride in the presence of sodium acetate (base): $C_6H_5CHO + (CH_3CO)_2O \xrightarrow{CH_3COONa,\ \Delta} C_6H_5$-$CH=CH$-$COOH$ (cinnamic acid) $+ CH_3COOH$.

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