The major products C and D formed in the following reaction respectively are: $CH_3CH_2CH_2-O-C(CH_3)_3 + \text{Excess } HI \xrightarrow{\Delta} C + D$

The reaction of an ether with hydrogen iodide (HI) involves the cleavage of the C-O bond.

1. Mechanism (First Cleavage): The ether contains a primary propyl group and a tertiary butyl group. When one of the alkyl groups is tertiary, the reaction proceeds via an $S_N1$ mechanism due to the stability of the tertiary carbocation . Protonation of the oxygen is followed by the cleavage of the bond between oxygen and the tertiary carbon, forming the stable tert-butyl carbocation and propyl alcohol. The iodide ion attacks the carbocation to form tert-butyl iodide ($I-C(CH_3)_3$).
2. Effect of Excess HI: The reaction mixture contains Excess HI. The alcohol formed in the first step (propyl alcohol, $H_3C-CH_2-CH_2-OH$) reacts further with HI. The hydroxyl group is replaced by iodine via nucleophilic substitution : $R-OH + HI \rightarrow R-I + H_2O$. This converts propyl alcohol to propyl iodide ($H_3C-CH_2-CH_2-I$).
3. Conclusion: Both alkyl groups are converted to their corresponding iodides.

Products: $H_3C-CH_2-CH_2-I$ + $I-C(CH_3)_3$