Answer the following by appropriately matchin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $(1, 2)$.

$A$. Formic acid	$I$. Rancid butter
$B$. Acetic acid	$II$. Vinegar
$C$. Butyric acid	$III$. Red ants

Similar Questions

The final product $(III)$ obtained in the reaction sequence is:
$CH_3-CH_2-COOH$ $\xrightarrow{PCl_3} I$ $\xrightarrow{C_6H_6/AlCl_3} II$ $\xrightarrow{NH_2-NH_2/\text{base/heat}} III$

What will be $X$ and $Y$ in the following reactions?

Match the name of the acid from List-$I$ with the source from which it is derived.
$A$. Formic acid $I$. Rancid butter
$B$. Acetic acid $II$. Vinegar
$C$. Butyric acid $III$. Red ants

In the reaction $CH_3COOH$ $\xrightarrow{LiAlH_4} (A)$ $\xrightarrow{I_2 + NaOH} (B)$ $\xrightarrow{Ag \text{ (Dust)}} (C)$,the final product $(C)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module