Saponification (basic hydrolysis) of C_6H_5-C | vedclass

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Question

(A) Saponification of an ester $R-CO-OR'$ with aqueous base follows $Acyl$ $Oxygen$ $Cleavage$ ($AAC_2$ mechanism).The hydroxide ion $(OH^-)$ attacks

Vedclass · Accepted Answer

$C_6H_5-COO^- + CH_3-^{18}OH$

Vedclass · Answer

(A) Saponification of an ester $R-CO-OR'$ with aqueous base follows $Acyl$ $Oxygen$ $Cleavage$ ($AAC_2$ mechanism).The hydroxide ion $(OH^-)$ attacks the carbonyl carbon to form a tetrahedral intermediate.This intermediate collapses to release the alkoxide ion $(R'O^-)$.The alkoxide ion quickly abstracts a proton from the carboxylic acid to form the alcohol $(R'OH)$.In the given ester $C_6H_5-CO-^{18}OCH_3$,the $^{18}O$ isotope is present in the alkoxy group $(-^{18}OCH_3)$.Therefore,the $^{18}O$ isotope will end up in the resulting alcohol,$CH_3-^{18}OH$.

$a$. $\text{Toluene} \xrightarrow[(ii) H^+/H_2O]{(i) CrO_2Cl_2/CS_2}$	$p$. $\text{Kolbe's electrolysis}$
$b$. $CH_3COOK \xrightarrow{\text{electrolysis}}$	$q$. $\text{Etard reaction}$
$c$. $CH_3Br \xrightarrow{Na, \text{dry ether}}$	$r$. $\text{Wurtz reaction}$
$d$. $\text{Benzene} \xrightarrow{CH_3Cl, AlCl_3}$	$s$. $\text{Friedel-Crafts alkylation}$

Saponification (basic hydrolysis) of $C_6H_5-CO-^{18}OCH_3$ will yield. [$^{18}O$ = mass $-18$ isotope of oxygen]

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