Question

\((CH_3)_3C-OC_2H_5\) on reaction with HI gives

• A. \((CH_3)_3C-I\) and \(C_2H_5-I\)
• B. \((CH_3)_3C-OH\) and \(C_2H_5-I\)
• C. \((CH_3)_3C-I\) and \(C_2H_5-OH\)
• D. \((CH_3)_3C-OH\) and \(C_2H_5-OH\)

Hint:

When an ether is cleaved by HI, the bond breaks so that the iodide attaches to the alkyl group that can form the more stable carbocation.

Answer 1

The Correct Option is C

Step 1: Identify the ether.
The compound \((CH_3)_3C-OC_2H_5\) is an ether with one tertiary group (tert-butyl) and one ethyl group.

Step 2: Rule for ether cleavage by HI.
When HI cleaves an ether, the iodide goes to the carbon that gives the more stable carbocation.

Step 3: Which carbocation is more stable?
The tert-butyl group forms a stable tertiary carbocation, so the C-O bond on that side breaks (SN1 path). The iodide attaches to the tert-butyl group.

Step 4: Decide the products.
So we get \((CH_3)_3C-I\), and the ethyl side leaves as \(C_2H_5-OH\) (ethanol).

Step 5: Conclusion.
The products are tert-butyl iodide and ethanol.

\[ \boxed{(CH_3)_3C\!-\!I \ \text{and}\ C_2H_5\!-\!OH} \]