The major products C and D formed in the following reaction respectively are:
$$CH_3CH_2CH_2-O-C(CH_3)_3 + e

Question

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

Accepted Answer

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

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