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I figured that the first reaction just adds a bromine to the meta position but I don't know what step ii) does. The only thing I can think of is the base gets rid of the alpha hydrogen next to the carbonyl. Then perhaps the negative carbon now attacks iodine? But that seems a bit off to me.


$\ce{I2}/\ce{OH-}$ will perform the iodoform reaction on the the product of previous reaction. The base deprotonates an alpha hydrogen, and the resulting enolate nucleophile attacks iodine. After three substitutions, the $\ce{CI3}$ group is a passable leaving group, and a nucleophilic acyl substitution occurs. The reaction is done in basic conditions, so the carboxylic acid is deprotonated. The leaving group is a strong base, so it is protonated. $\ce{CHI3}$ will be formed along with the conjugate base of m-bromo-benzoic acid. Here is a generic version:

$$\ce{RC(O)CH3 + 3I2 + 4OH- -> RCO2- + 3I- + 3H2O + CHI3}$$

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