Question

In iodination, for preparation of iodomethane compound used is
$\left( 1 \right)HI{O_3}$
$\left( 2 \right)HgO$
$\left( 3 \right)Both\left( 1 \right)and\left( 2 \right)$
$\left( 4 \right)HI$

Answer 1

Hint:The formation of an alkyl halide involves the interaction between the components having the interactive character between the alkyl group with that of the halogen. Formation of free halogen in the reaction is important for the process of interaction.

Complete step by step answer:
In the process of iodination in organic elements, the addition of iodine takes place at the different presence of aqueous hydrogen iodide. The conversion starts with the reagent methanol when it is subjected to treatment with the hydrogen iodide. The resultant major product which is formed here is iodomethane, which means an association of one iodine residue with a methyl group.
Therefore, the given chemical reaction due to the action of hydrogen iodide is:
\[C{H_3}OH + HI \to C{H_3}I + {H_2}O\]
The aqueous solution of hydrogen iodide shows the dissociation of iodine residue and the formation of ${H^ + }$ ions. The Iodide residue is associated with the methyl group and hence the hydrogen ion is associated with the $O{H^ - }$ ions. The association of these two ions result in the formation of ${H_2}O$ as one of the products of the reaction. The other product is formed when the association of one iodine residue with a methyl group takes place resulting in the formation of iodomethane.
Therefore, among the given compounds the preparation of iodomethane or iodomethane is possible in the presence of $\left( 4 \right)HI$ among the reactants.

Note:
The formation of alkyl iodide can take place because of interaction between the partially positive residue with that of the partially negative one. The attack of ions takes place due to the interaction between positive with negative charges. This results in the formation of a specific compound.