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Ben Norris
Ben Norris
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I think, that I2/OH- will form IO3- and that$\ce{I2}/\ce{OH-}$ will perform Iodoform testthe iodoform reaction on the the product of previous reaction. CHCl3The 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:

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

I think, that I2/OH- will form IO3- and that will perform Iodoform test on the the product of previous reaction. CHCl3 will be formed along with m-bromo-benzoic acid.

$\ce{I2}/\ce{OH-}$ 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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CodeBlooded
CodeBlooded
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I think, that I2/OH- will form IO3- and that will perform Iodoform test on the the product of previous reaction. CHCl3 will be formed along with m-bromo-benzoic acid.