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(A) For reaction $(III)$: The starting material is $o$-chloromethyl methyl benzoate. Treatment with $KCN$ gives $o$-cyanomethyl methyl benzoate. Hydro

Vedclass · Accepted Answer

$1, 2$

Vedclass · Answer

(A) For reaction $(III)$: The starting material is $o$-chloromethyl methyl benzoate. Treatment with $KCN$ gives $o$-cyanomethyl methyl benzoate. Hydrolysis $(H_3O^+, \Delta)$ yields $o$-carboxymethyl benzoic acid $(T)$. Reduction with $LiAlH_4$ gives the diol,and cyclization with $conc. H_2SO_4$ yields the ether $(R)$. Thus,$(III)$ matches $(T)$ and $(R)$.For reaction $(IV)$: The starting material is dimethyl $o$-phenylenediacetate. Reduction with $LiAlH_4$ gives the diol $(Q)$. Cyclization with $conc. H_2SO_4$ gives the ether $(R)$. Thus,$(IV)$ matches $(Q)$ and $(R)$.For reaction $(I)$: The starting material is $2$-(cyanomethyl)benzaldehyde ethylene acetal. $DiBAL-H$ followed by $dil. HCl$ gives the dialdehyde. $NaBH_4$ reduces it to the diol $(Q)$. Cyclization with $conc. H_2SO_4$ gives the ether $(R)$. Thus,$(I)$ matches $(Q)$ and $(R)$.For reaction $(II)$: The starting material is $o$-allylbenzoic acid. Ozonolysis $(O_3/Zn, H_2O)$ gives the aldehyde-acid. $NaBH_4$ reduces the aldehyde to alcohol $(S)$. Cyclization with $conc. H_2SO_4$ gives the lactone $(U)$. Thus,$(II)$ matches $(S)$ and $(U)$.Question $(1)$: $(III)$ matches $(T, R)$ and $(IV)$ matches $(Q, R)$. Option $(1)$ is $(III, T, R)$ and Option $(2)$ is $(IV, Q, R)$. Both are correct combinations.Question $(2)$: $(I)$ matches $(Q, R)$ and $(II)$ matches $(S, U)$. Option $(1)$ is $(I, Q, R)$ and Option $(2)$ is $(II, S, U)$. Both are correct combinations.Therefore,the correct answer is $(1, 2)$.

Column-$I$ (Name)	Column-$II$ (Formula)	Column-$III$ (Example)
$(A)$ Ketone	$(i)$ $R-CO-OH$	$(p)$ $CH_3CHO$
$(B)$ Carboxylic acid	$(ii)$ $R-CO-H$	$(q)$ $CH_3COOH$
$(C)$ Amide	$(iii)$ $R-CO-R'$	$(r)$ $CH_3CONH_2$
$(D)$ Aldehyde	$(iv)$ $R-CO-NH_2$	$(s)$ $CH_3COC_2H_5$

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