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

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Question

(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 $C_6H_5COOC_2H_5$.

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

Similar Questions

The products $C$ and $D$ are given by the following reaction sequence:
$CH_3CH_2COCH_3$ $\xrightarrow[(ii) H_3O^+]{(i) CH_3MgBr} A$ $\xrightarrow[358 \ K]{20\% H_3PO_4} B \text{ (Major)}$ $\xrightarrow[(ii) Zn/H_2O]{(i) O_3} C + D$

Which one of the following compounds does not give a primary alcohol on reduction?

Which among the following compounds does not exhibit the iodoform test?

Assertion : Hydroxyketones are not directly used in Grignard reaction.
Reason : Grignard reagents react with hydroxyl group.

An organic compound with the molecular formula $C_9H_{10}O$ forms a $2,4-DNP$ derivative,reduces Tollen's reagent and undergoes Cannizzaro reaction; on vigorous oxidation it gives $1,2-$benzenedicarboxylic acid. The structure of the organic compound is:

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