Question

In the reaction, (CH₃)₃C − O − CH₃ + HI → Products, CH₃OH and (CH₃)₃CCl are the products and not CH₃I and (CH₃)₃COH. It is because

• (A) In step 2 of the reaction, the departure of leaving group (HO–CH3) creates less stable carbocation
• (B) The reaction follows an SN1 mechanism
• (C) In step 2 of the reaction, the departure of leaving group (HO–CH3) creates more stable carbonation
• (D) The reaction follows SN2 mechanism

Answer 1

The Correct Option is B

The reaction of (CH₃)₃C − O − CH₃ with HI leads to the formation of CH₃OH and (CH₃)₃CCl instead of CH₃I and (CH₃)₃COH due to the type of mechanism it follows. This reaction proceeds via an SN₁ mechanism. In an SN₁ reaction, the rate-determining step involves the formation of a carbocation intermediate. The stability of the carbocation is crucial for the progression of the reaction. In this case, the (CH₃)₃C⁺ carbocation is more stable compared to any carbocation that would be formed from CH₃ group. Therefore, the continuing steps naturally lead to the products of CH₃OH and (CH₃)₃CCl. This stability plays a key role in favoring the pathway that produces the observed products rather than alternative products like CH₃I and (CH₃)₃COH

Hence, the correct answer is: (B) The reaction follows an SN₁ mechanism

Answer 2

The reaction of (CH₃)₃C - O - CH₃ with HI involves the cleavage of an ether bond. The products are CH₃OH and (CH₃)₃CI, not CH₃I and (CH₃)₃COH.

Here's why:

1. Protonation: The first step is the protonation of the ether oxygen by HI, forming an oxonium ion:

(CH₃)₃C - O⁺H - CH₃ I^-

1. Cleavage: The next step involves the cleavage of either the (CH₃)₃C-O bond or the CH₃-O bond by the iodide ion (I^-).

The key factor is the stability of the carbocation formed during the cleavage. If the (CH₃)₃C-O bond breaks, it forms a tertiary carbocation ((CH₃)₃C⁺), which is highly stable. If the CH₃-O bond breaks, it forms a primary carbocation (CH₃⁺), which is much less stable.

Therefore, the reaction favors the formation of the more stable tertiary carbocation, leading to the formation of (CH₃)₃CI and CH₃OH.

Also, since a tertiary carbocation is formed, the reaction proceeds via an SN1 mechanism.

Let's analyze each option:

• (A) In step 2 of the reaction, the departure of leaving group (HO–CH₃) creates less stable carbocation: This is incorrect. The leaving group creates a stable tertiary carbonation.
• (B) The reaction follows an SN1 mechanism: This is correct.
• (C) In step 2 of the reaction, the departure of leaving group (HO–CH₃) creates more stable carbocation: This is correct.
• (D) The reaction follows SN2 mechanism: This is incorrect. Tertiary carbocations favor SN1 reactions.

The best explanation is a combination of (B) and (C). However, since the question asks for a single reason, (C) is the most comprehensive answer.

The correct answer is:

Option 3: (B) The reaction follows an SN1 mechanism