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The product of the Hunsdiecker reaction on $CH_3CH_2COOAg$ with $Br_2/CCl_4$ is

A$CH_3CH_2COBr$
B$CH_3Br$
C$CH_3CH_2Br$
D$CH_3CH_2CH_2Br$
Answer & Solution
Correct answer: C. $CH_3CH_2Br$
Hunsdiecker loses one carbon (as $CO_2$): $CH_3CH_2COOAg + Br_2 \to CH_3CH_2Br + CO_2 + AgBr$.
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