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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 $(1, 2)$.

List-$I$	List-$II$
$(a)$ $R-COCl \to R-CHO$	$(i)$ $Br_2 / NaOH$
$(b)$ $R-CH_2-COOH \to R-CH(Cl)-COOH$	$(ii)$ $H_2 / Pd-BaSO_4$
$(c)$ $R-CONH_2 \to R-NH_2$	$(iii)$ $Zn(Hg) / \text{Conc. } HCl$
$(d)$ $R-CO-CH_3 \to R-CH_2-CH_3$	$(iv)$ $Cl_2 / \text{Red } P, H_2O$

Similar Questions

The product of the following reaction is:

Which compound on oxidation with acidic $KMnO_4$ gives Benzoic acid?

What is the product $C$ in the following reaction?
$CH_3Br$ $\xrightarrow{KCN} A$ $\xrightarrow{H_3O^+} B$ $\xrightarrow{LiAlH_4} C$

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