
Predict the products of the following reaction
\[{{C}_{6}}{{H}_{5}}-CO-C{{H}_{3}}\xrightarrow{NaOH/{{I}_{2}}}\]
Answer
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Hint:The reaction in which the reactant is being reacted with the sodium hydroxide and iodine is known as the Haloform reaction. This is also generally known as the Iodoform test. This test is called the iodoform test because in this reaction or the test the iodoform is produced as one of the products. The only condition of this test is that this test is given positive only by those compounds which have methyl ketone in their compound.
Complete step-by-step answer: So let us first understand the basic haloform reaction. so in the haloform reaction the methyl ketone is reacted with the halogen and the aqueous solution of the sodium hydroxide. In this reaction the ketone is oxidised to the sodium salt of the acid but in this salt the one carbon atom is less than that of ketone. And in this same reaction the formation of haloform takes place. So this reaction is also known as the iodoform test. When the haloform reaction is carried out with the sodium hydroxide and the iodine the formation of yellow precipitate of the iodine takes place. So this yellow precipitate of iodine is used to determine whether the compound contains the methyl ketone or not. So now let us complete the reaction given. So the product will be iodoform and the sodium salt of the acid with one carbon atom less so that will be \[{{C}_{6}}{{H}_{5}}COONa\]. Now let us complete the reaction:
\[{{C}_{6}}{{H}_{5}}-CO-C{{H}_{3}}\xrightarrow{NaOH/{{I}_{2}}}{{C}_{6}}{{H}_{5}}COONa+CH{{I}_{3}}+{{H}_{2}}O\]
The products formed are iodoform and the \[{{C}_{6}}{{H}_{5}}COONa\].
Note: The iodoform is considered as the pale yellow coloured volatile substance which ahs the distinctive odour. Like the iodoform compound there are compounds such as bromoform and chloroform also. The formation of them undergoes the same mechanism as that of iodoform. While performing the iodoform reaction the corrosiveness of the iodine must be taken care.
Complete step-by-step answer: So let us first understand the basic haloform reaction. so in the haloform reaction the methyl ketone is reacted with the halogen and the aqueous solution of the sodium hydroxide. In this reaction the ketone is oxidised to the sodium salt of the acid but in this salt the one carbon atom is less than that of ketone. And in this same reaction the formation of haloform takes place. So this reaction is also known as the iodoform test. When the haloform reaction is carried out with the sodium hydroxide and the iodine the formation of yellow precipitate of the iodine takes place. So this yellow precipitate of iodine is used to determine whether the compound contains the methyl ketone or not. So now let us complete the reaction given. So the product will be iodoform and the sodium salt of the acid with one carbon atom less so that will be \[{{C}_{6}}{{H}_{5}}COONa\]. Now let us complete the reaction:
\[{{C}_{6}}{{H}_{5}}-CO-C{{H}_{3}}\xrightarrow{NaOH/{{I}_{2}}}{{C}_{6}}{{H}_{5}}COONa+CH{{I}_{3}}+{{H}_{2}}O\]
The products formed are iodoform and the \[{{C}_{6}}{{H}_{5}}COONa\].
Note: The iodoform is considered as the pale yellow coloured volatile substance which ahs the distinctive odour. Like the iodoform compound there are compounds such as bromoform and chloroform also. The formation of them undergoes the same mechanism as that of iodoform. While performing the iodoform reaction the corrosiveness of the iodine must be taken care.
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