Why does adding sodium iodide or potassium iodide to 1-Methyl-1,4-cyclohexanediol make diiodoalkane. I thought only halogen plus alcohol made a dihaloalkane. And I thought potassium or sodium halide plus alcohol made a haloalkane (and sodium surface and water)
Why does adding sodium iodide or potassium iodide to 1-Methyl-1,4-cyclohexanediol make diiodoalkane. I thought only halogen plus alcohol made a dihaloalkane. And I thought potassium or sodium halide plus alcohol made a haloalkane (and sodium surface and water)
It wouldn’t form the diiodoalkane if you used just the alcohol and NaI/KI. You would need to also add 50% H2SO4, which reacts with the salt to form HI in situ, which is the iodinating agent in the reaction.
If you just add a halogen to the alcohol, you don’t form a haloalkane. However, you can make haloalkane by refluxing an alcohol with the corresponding halogen and red phosphorus, but this is because the reaction produces the corresponding phosphorus trihalide, which is the halogenating agent.
It wouldn’t form the diiodoalkane if you used just the alcohol and NaI/KI. You would need to also add 50% H2SO4, which reacts with the salt to form HI in situ, which is the iodinating agent in the reaction.
If you just add a halogen to the alcohol, you don’t form a haloalkane. However, you can make haloalkane by refluxing an alcohol with the corresponding halogen and red phosphorus, but this is because the reaction produces the corresponding phosphorus trihalide, which is the halogenating agent.
I understand it makes a hydrogen iodide in situ, but again doesn’t that form a haloalkane instead of a dihaloalkane because u need to halide ions to replace two hydrogens.
I understand it makes a hydrogen iodide in situ, but again doesn’t that form a haloalkane instead of a dihaloalkane because u need to halide ions to replace two hydrogens.
I understand it makes a hydrogen iodide in situ, but again doesn’t that form a haloalkane instead of a dihaloalkane because u need to halide ions to replace two hydrogens.
The alcohol in the question is a diol, so it has two -OH groups that can be substituted.