Question

When mixed ethers are treated with HI, the iodine atom goes to -----alkyl group.

Answer 1

Hint: When the ethers especially the mixed ones which consists of two different alkyl groups attached to it are treated with the HI, then the iodine atom will go to that alkyl group which is simpler and has least molecular weight. Now you can easily answer the given statement accordingly.

Complete answer:
First of all let’s discuss what an ether is. Ether are the organic compounds which consist of the functional group as R-O-R in its compounds.
When the same group i.e. methyl group or ethyl group or any other group is attached to the functional group -O-, then that ethers are known as the identical ethers.
But when one attached an alkyl group to the functional group -O- is methyl and the other attached alkyl group to the functional group -O- is ethyl, then that ethers are known as the mixed ethers.
Now considering the statement as;
 When the mixed ethers are made to react with the hydrogen iodide i.e. HI, then the iodine part from the hydrogen iodide will go to that alkyl group of the ether which is simpler one and less steric hindered.
Let us consider the given example as;-
When ethyl methyl ether reacts with HI, then the iodine atom goes to the methyl alkyl group because it is simpler than the ethyl alkyl group and results in the formation of methyl iodide and ethanol. The reaction is supposed to take place as;-
$C{{H}_{3}}O{{C}_{2}}{{H}_{5}}+HI\to C{{H}_{3}}I+{{C}_{2}}{{H}_{5}}OH$
So, thus When mixed ethers are treated with HI, the iodine atom goes to a simpler alkyl group.

Note:
Ethers are the compounds which have an ethereal sweet-smell and are colorless. They are highly volatile in nature and are insoluble in water and has the lowest boiling point of all i.e. from alcohols, ketones and carboxylic acids but their boiling points are somewhat similar to the alkanes.