Major product formed in the following reaction is a mixture of:

We need to determine the major product formed when the given ether reacts with hydrogen iodide ($$HI$$).

Key Concept: The reaction of an asymmetric ether with a strong acid like $$HI$$ proceeds via protonation of the ether oxygen, followed by a nucleophilic substitution ($$S_N1$$ or $$S_N2$$) reaction by the iodide ion ($$I^-$$).

The given reactant is cyclohexylmethyl t-butyl ether, where an oxygen atom connects a cyclohexylmethyl group and a tertiary butyl ($$t\text{-butyl}$$) group.

Step 1: Protonation of the ether oxygen

The lone pair of electrons on the ether oxygen attacks the proton of $$HI$$, forming a protonated ether intermediate and releasing an iodide ion ($$I^-$$):

$$\text{R}_1\text{-O-R}_2 + HI \rightarrow \text{R}_1\text{-}\overset{\text{H}}{\overset{+}{\text{O}}}\text{-R}_2 + I^-$$

Step 2: Cleavage of the C-O bond via $$S_N1$$ pathway

Because one of the alkyl groups attached to the oxygen is a tertiary butyl group, the reaction proceeds via an $$S_N1$$ mechanism. The leaving group is cyclohexylmethanol, and a highly stable tertiary butyl carbocation ($$(CH_3)_3C^+$$) is formed stable due to hyperconjugation and inductive effects.

$$\text{Cyclohexylmethyl-}\overset{\text{H}}{\overset{+}{\text{O}}}\text{-C}(CH_3)_3 \rightarrow \text{Cyclohexylmethyl-OH} + ^{+}C(CH_3)_3$$

Step 3: Attack of the nucleophile ($$I^-$$)

The stable tertiary carbocation is then rapidly attacked by the remaining iodide nucleophile ($$I^-$$) to yield tert-butyl iodide:

$$^{-}C(CH_3)_3 + I^- \rightarrow I\text{-C}(CH_3)_3$$

Therefore, the major products formed in this reaction mixture are cyclohexylmethanol and tert-butyl iodide.

Answer: Option D — Cyclohexylmethanol and tert-butyl iodide