Predict the products of the given reaction.
$C H_{3} - O - C H (C H_{3})_{2} + H I Heated \to$

I C H_{2} O C H (C H_{3})_{2}

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C H_{3} I + (C H_{3})_{2} C H O H

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C H_{3} O H + (C H_{3})_{2} C H I

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Solution

The correct option is C $C H_{3} I + (C H_{3})_{2} C H O H$
Ethers can be reactive under drastic reaction conditions (high temperature, high concentration) due to the cleavage of the C-O bond.
In first step of the reaction, protonation of ether $(O)$ group take place.
In second step, $S_{N} 2$ reaction take place, where $I^{-}$ nucleophile will attack the less hindered site to form methyl iodide and propan-2-ol. Hence option (c) is correct. The cleavage reaction is shown below:

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(B)

C H_{3} - C \equiv C H H_{2} S O_{4}, H_{2} O - ------- \to H g S O_{4} (A) \to (B)

{(C H_{3})}_{3} C - O - C H_{2} C H_{3} + H I (1 m o l e) h e a t - --------- \to

C H_{3} - O | | C H Z n - H g - ----- \to C o n c . H C l

C H_{3} - O - C H {(C H_{3})}_{2}

H I

\to

C H_{3} - C l + K C N \to ?

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