Question

When silver propionate is treated with $B{r_2}$ in $CC{l_4}$ as solvent, the product is
(1) $1 - $Bromopropane
(2) $2 - $Bromopropane
(3)$1,2 - $Dibromoethane
(4) Bromoethane

Answer 1

Hint: Hunsdiecker reaction is also known as Borodin reaction or Hunsdiecker – Borodin reaction in which carboxylation and halogenation both can be seen. In this reaction, the solvent used is carbon tetrachloride.

Complete answer:
When silver propionate is treated with bromine and carbon tetrachloride, the products obtained are silver bromide and bromoethane.
The reaction is as follows:
$C{H_3} - COOAg + B{r_2}\xrightarrow{{CC{l_4}}}C{H_3} - C{H_2}Br + AgBr + C{O_2}$
This reaction is known as the Hunsdiecker reaction.
The Hunsdiecker Reaction is a chemical reaction in which carboxylic acid silver salts combine with halogens to form an unstable intermediate, which then undergoes heat decarboxylation to produce alkyl halides as a final product.
Because the product has one fewer carbon atom than the starting material and a halogen atom is introduced in its place, it is an example of both a decarboxylation and a halogenation process.
Carbon tetrachloride serves as a solvent that is actually used for the smooth behaviour of the reaction. It is an inert solvent and the purpose of these inert solvent is to dissolve the reactants present in the reaction unlike the solvents like water, alcohol etc which plays a key role in the reaction as well.
Hence, the correct option is (4) Bromoethane.

Note:
The Hunsdiecker Reaction is a chemical reaction in which carboxylic acid silver salts react with halogens to form an unstable intermediate, which is then thermally decarboxylated to form an alkyl halide-like end product.