Question

Assertion (A) : \( \mathrm{(CH_3)_3C - O - CH_3} \) gives \( \mathrm{(CH_3)_3C - I} \) and \( \mathrm{CH_3OH} \) on reaction with HI.

Reason (R) : The reaction occurs by \( \mathrm{S_N1} \) mechanism.

• A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Hint:

A tertiary alkyl ether is cleaved by HI through a stable carbocation, so iodine joins the tertiary carbon. The SN1 mechanism explains this product.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Ethers are cleaved by HI. The way they break depends on the groups attached to oxygen.
If one group is tertiary, the reaction follows the \( \mathrm{S_N1} \) path. If both groups are primary or methyl, it follows \( \mathrm{S_N2} \).

Step 2: Mechanism for tert-butyl methyl ether:
First, oxygen picks up a proton from HI and becomes an oxonium ion. \[ \mathrm{(CH_3)_3C{-}O{-}CH_3 + HI \rightarrow (CH_3)_3C{-}\overset{+}{O}H{-}CH_3 + I^-} \]
Next, the C-O bond to the tertiary carbon breaks. This gives the stable tertiary carbocation \( \mathrm{(CH_3)_3C^+} \) and methanol.
Finally, the iodide ion attacks the carbocation. \[ \mathrm{(CH_3)_3C^+ + I^- \rightarrow (CH_3)_3C{-}I} \]

Step 3: Check the Assertion.
The products are \( \mathrm{(CH_3)_3C{-}I} \) and \( \mathrm{CH_3OH} \). So the Assertion is TRUE.

Step 4: Check the Reason.
The reaction goes through a carbocation, which is the \( \mathrm{S_N1} \) path. So the Reason is TRUE.

Step 5: Link between A and R.
The \( \mathrm{S_N1} \) path explains why iodine ends up on the tertiary carbon and methanol is released. The iodide does not attack the methyl carbon. So R correctly explains A.

Final Answer:
Both A and R are true and R explains A, so the answer is option (A).\[ \boxed{\text{(A)}} \]