The product of the Hunsdiecker reaction on $CH_3CH_2COOAg$ with $Br_2/CCl_4$ is
- $CH_3CH_2COBr$
- $CH_3Br$
- $CH_3CH_2Br$
- $CH_3CH_2CH_2Br$
Show answer and solution
Correct answer: C. $CH_3CH_2Br$
Hunsdiecker loses one carbon (as $CO_2$): $CH_3CH_2COOAg + Br_2 \to CH_3CH_2Br + CO_2 + AgBr$.