Question

What is the product of
 $ {(C{H_3})_3}C - O{C_2}{H_5}\xrightarrow{{HI}} $

Answer 1

Hint: When ether is treated with hydrogen iodide then the products formed are as alkyl halide and alcohol. The tertiary alkyl group will form alkyl halide and the primary or secondary alkyl group will form alkyl halide.

Complete step by step solution
First of all we will talk about the alkanes, alkenes and alkynes.
Alkanes: The compounds which are formed by carbon and hydrogen and have only a single bond between the carbon-carbon atoms, are known as alkanes. For example: The first member of the alkane family is ethane $ ({H_3}C - C{H_3}) $ . The general formula of the alkane group is $ {C_n}{H_{(2n + 2)}} $ .
Alkenes: The compounds which are formed by carbon and hydrogen and have at least one double bond along with a single bond between the carbon-carbon atoms, are known as alkenes. For example: The first member of the alkene family is ethene $ ({H_2}C = C{H_2}) $ . The general formula of the alkene group is $ {C_n}{H_{2n}} $ .
Alkynes: The compounds which are formed by carbon and hydrogen and have at least one triple bond along with a single bond between the carbon-carbon atoms, are known as alkynes. For example: The first member of the alkyne family is ethyne $ (HC \equiv CH) $ . The general formula of the alkyne group is $ {C_n}{H_{(2n - 2)}} $ .
Ethers: When a function group is as $ R - O - R $ this is known as ethers.
For the reaction of ether with hydroiodic acid the product formed is as: when one alkyl group in ether is methyl and another is primary or secondary alkyl then one product will be methyl iodide and other will be alcohol based on the alkyl group.
And if in the ether one alkyl group is methyl and other is tertiary alkyl then product will be methanol and other will be tertiary iodide.
Here we are given with $ {(C{H_3})_3}C - O{C_2}{H_5} $ as reactant and we have to write the reaction of this reactant with that of hydrogen iodide. The reaction is as follows:
 $ {(C{H_3})_3}C - O{C_2}{H_5}\xrightarrow{{HI}}{(C{H_3})_3}CI + {C_2}{H_5}OH $ .
Here the products formed are tertiary halide (tert-butyl iodide) and primary alcohol (ethyl alcohol). This is because the stability of iodide ions is high and stability of tertiary carbocation is high. So they react with each other to form tertiary iodide.

Note
Carbocation: The carbon containing positive charge is known as carbocation. The stability of carbocation is as follows: tertiary carbocation is more stable than secondary than primary carbocation i.e. $ {3^ \circ } > {2^ \circ } > {1^ \circ } $ .