Question

Identify 'X' and 'Y' in the following reaction.
\(C_{3}H_{7}-O-C_{3}H_{7}+HI\ (\text{excess})\xrightarrow[\text{Hot}]{\Delta }X+Y\)

• A. \(C_{3}H_{7}OH\) and \(H_{2}O\)
• B. \(C_{3}H_{7}\) and \(H_{2}O\)
• C. \(C_{3}H_{7}I\) and \(C_{3}H_{7}OH\)
• D. \(C_{3}H_{7}I\) and \(H_{2}O\)

Hint:

Excess hot HI cleaves both C-O bonds of the ether to give an alkyl iodide and water.

Answer 1

The Correct Option is D

Step 1: Overall reaction
One mole of ether needs two moles of HI. $\text{R-O-R}+2\text{HI}\rightarrow2\text{R-I}+\text{H}_2\text{O}$

Step 2: Substitute R
Here $\text{R}=\text{C}_3\text{H}_7$, so two moles of propyl iodide and one mole of water form.

Step 3: Match
The pair of propyl iodide and water appears in the last option.

Final Answer:
The products are C3H7I and H2O. \[ \boxed{\text{(D) }\text{C}_{3}\text{H}_{7}\text{I and H}_{2}\text{O}} \]