The reaction of an ether with hot concentrated HI involves the cleavage of the C-O bond. The mechanism depends on the nature of the alkyl groups attached to the oxygen.

1. Mechanism Check: The ether is 1-ethoxy-2-methylpropane. The alkyl groups attached to the oxygen are ethyl ($-CH_2CH_3$) and isobutyl ($-CH_2CH(CH_3)_2$). Both alkyl groups are primary.
2. SN2 Pathway: When both alkyl groups are primary or secondary, the reaction proceeds via an SN2 mechanism involving the attack of the nucleophile ($I^-$) on the less sterically hindered carbon atom.
3. Steric Hindrance: The ethyl group ($CH_3CH_2-$) is less sterically hindered than the isobutyl group ($(CH_3)_2CHCH_2-$), which has branching at the $\beta$-carbon.
4. Product Formation: The iodide ion attacks the ethyl group to form ethyl iodide ($CH_3CH_2I$). The proton ($H^+$) remains with the oxygen on the isobutyl group to form isobutyl alcohol ($(CH_3)_2CHCH_2OH$).

Therefore, the products are Isobutyl alcohol and Ethyl iodide.