(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
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:
(CH3)3C - O+H - CH3 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:
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
\(\text{Reaction of aniline with conc. HNO}_3 \text{ and conc. H}_2\text{SO}_4 \text{ at 298 K will produce 47\% of:}\)