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In the **Borodine–Hunsdiecker reaction**, the silver salt of a carboxylic acid is treated with which reagent?
A$KMnO_4$
B$Br_2/CCl_4$
C$NaOH$
D$Cl_2/h
u$
Answer & Solution
Correct answer: B. $Br_2/CCl_4$
Hunsdiecker: $RCOOAg + Br_2 \xrightarrow{CCl_4} RBr + CO_2 + AgBr$. The reaction converts a $C_n$ acid silver salt into a $C_{n-1}$ alkyl bromide.
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