The major products C and D formed in the following reaction respectively are:${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2{\text{OC(CH}}_3{)}_3\stackrel{\text{Excess HI, }\Delta }{\to }\text{C}+\text{D}$

When an ether is treated with excess $\text{HI}$, both alkyl groups are converted into alkyl iodides.The ether is $n$ -propyl $t$ -butyl ether.Step 1: Protonation of oxygen.Step 2: Since one group is tertiary ($t$ -butyl), the reaction proceeds via $S_{N}1$ mechanism, forming a stable $t$ -butyl carbocation. This reacts with ${\text{I}}^{-}$ to form ${\text{I-C(CH}}_3{)}_3$. The other fragment is $n$ -propyl alcohol.Step 3: Because $\text{HI}$ is in excess, the $n$ -propyl alcohol further reacts with $\text{HI}$ to form $n$ -propyl iodide (${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{I}$).Products: ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{I}$ and $({\text{CH}}_3{)}_3\text{CI}$.