An ester (A) with molecular formula C_9H_{10} | vedclass

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(A) $1$. Ester $(A)$ is $C_6H_5COOC_2H_5$ (Ethyl benzoate). $2$. Reaction with excess $CH_3MgBr$ followed by acid workup yields a tertiary alcohol: $C

Vedclass · Accepted Answer

$C_6H_5COOC_2H_5$

Vedclass · Answer

(A) $1$. Ester $(A)$ is $C_6H_5COOC_2H_5$ (Ethyl benzoate). $2$. Reaction with excess $CH_3MgBr$ followed by acid workup yields a tertiary alcohol: $C_6H_5C(OH)(CH_3)_2$. $3$. Dehydration of this alcohol with $H_2SO_4$ gives the olefin $(B)$,which is $C_6H_5-C(CH_3)=CH_2$ ($2$-phenylpropene). $4$. Ozonolysis of $(B)$ $(C_6H_5-C(CH_3)=CH_2)$ yields acetophenone $(C_6H_5COCH_3)$,which has the formula $C_8H_8O$. $5$. Acetophenone contains a methyl ketone group $(-COCH_3)$,which gives a positive iodoform test. $6$. Therefore,the ester $(A)$ is $C_6H_5COOC_2H_5$.

$I. HNO_3$	$II. NH_2NH_2 / OH^-$
$III. 2[Ag(NH_3)_2]^+$	$IV. NaOH$

An ester $(A)$ with molecular formula $C_9H_{10}O_2$ was treated with excess of $CH_3MgBr$ and the complex so formed was treated with $H_2SO_4$ to give an olefin $(B)$. Ozonolysis of $(B)$ gave a ketone with molecular formula $C_8H_8O$ which shows positive iodoform test. The structure of $(A)$ is

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