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The haloform reaction using I2 and NaOH is referred to as the "iodoform" test for methyl ketones
The haloform reaction using I2 and NaOH is referred to as the "iodoform" test for methyl ketones. The test also gives positive results for compounds containing the -CH(OH)CH3 group. This results from the oxidation of the alcohol to the methyl ketone in the first stage. Write a balanced equation for the conversion of C6H5CHOHCH3 to the methyl ketone in the presence of I2 and NaOH. Identify which species is being oxidized and which is being reduced.
Expert Solution
The basic reaction is an oxidation. The alcohol is the "reduced" form of the compound while the methylketone is the "oxidized" form. How do we know this? Notice the loss of 2 H atoms from the alcohol. Typically, when compounds lose hydrogens, they undergo oxidation. Also, notice the oxidation state of the oxygen. It goes from an alcohol having a single bond to a carbonyl group having a double bond. This is a classic sign of an oxidation process.
Therefore, the secondary alcohol is being oxidized in the process.
O
||
C6H5CHOHCH3 ----------> C6H5CCH3
The balanced equation is:
O
||
C6H5CHOHCH3 + I2 ----------> C6H5CCH3 + 2 HI
I2 is the oxidizing agent. It gets reduced to HI in the process.
If you kept the process going, then the OH- would react as follows:
O O
|| ||
C6H5CCH3 + I2 + NaOH -------> C6H5CCH2I + H2O + NaI
O O
|| ||
C6H5CCH2I + I2 + NaOH -------> C6H5CCHI2 + H2O + NaI
O O
|| ||
C6H5CCHI2 + I2 + NaOH -------> C6H5CCI3 + H2O + NaI
O O
|| ||
C6H5CCI3 + OH- -------> C6H5CO- + H2O + CHI3
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