The reaction sequence given below gives produ | vedclass

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(D) The reaction sequence is as follows:$1$. The starting material is a mono-methyl ester of a dicarboxylic acid, $HO_2C-(CH_2)_4-CO_2Me$.$2$. Treatme

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$MeO_2C-(CH_2)_4-Br$

Vedclass · Answer

(D) The reaction sequence is as follows:$1$. The starting material is a mono-methyl ester of a dicarboxylic acid, $HO_2C-(CH_2)_4-CO_2Me$.$2$. Treatment with $Ag_2O$ converts the carboxylic acid group $(-COOH)$ into its silver salt $(-COO^-Ag^+)$.$3$. Subsequent treatment with $Br_2$ in $CCl_4$ is the Borodine-Hunsdiecker reaction, which decarboxylates the silver salt and replaces the carboxyl group with a bromine atom.$4$. The reaction is: $HO_2C-(CH_2)_4-CO_2Me$ $\xrightarrow{Ag_2O} AgO_2C-(CH_2)_4-CO_2Me$ $\xrightarrow{Br_2, CCl_4} Br-(CH_2)_4-CO_2Me$.$5$. Thus, the product $R$ is $MeO_2C-(CH_2)_4-Br$.

The reaction sequence given below gives product $R$. The structure of the product $R$ is:

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