Question

Identify the product obtained in following reaction.
\(\mathrm{C}_{3}\mathrm{H}_{7}-\mathrm{O}-\mathrm{C}_{3}\mathrm{H}_{7}+\mathrm{HI}_{\text{ excess }}\xrightarrow[\mathrm{Hot}]{\Delta }X+Y\)

• A. \(\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{OH}\) and \(\mathrm{H}_{2}\mathrm{O}\)
• B. \(\mathrm{C}_{3}\mathrm{H}_{7}\) and \(\mathrm{H}_{2}\mathrm{O}\)
• C. \(\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{I}\) and \(\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{OH}\)
• D. \(\mathrm{C}_{3}\mathrm{H}_{7}\mathrm{I}\) and \(\mathrm{H}_{2}\mathrm{O}\)

Hint:

With excess HI, both carbon-oxygen bonds of the ether are cleaved, giving two molecules of alkyl iodide and water.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Ethers are cleaved by hydrogen halides. With excess HI and heating, the cleavage goes on until both alkyl groups become alkyl iodides.

Step 2: First step
\(\text{C}_3\text{H}_7\text{-O-C}_3\text{H}_7+\text{HI}\rightarrow\text{C}_3\text{H}_7\text{I}+\text{C}_3\text{H}_7\text{OH}\)

Step 3: Second step
The propanol formed reacts with the excess HI. \(\text{C}_3\text{H}_7\text{OH}+\text{HI}\rightarrow\text{C}_3\text{H}_7\text{I}+\text{H}_2\text{O}\)

Step 4: Compare the options
The overall products are propyl iodide and water.

Final Answer:
The products are propyl iodide and water. \[ \boxed{\text{(D) }\text{C}_{3}\text{H}_{7}\text{I and H}_{2}\text{O}} \]