Question

In the following reaction, X is:
\[{C_2}{H_5} - O - {C_2}{H_5} + 4P\text{(red phosphorus)} \to 2X + {H_2}O\]
A.Ethane
B.Ethylene
C.Butane
D.Propane

Answer 1

Hint: As we know that the ether is very less reactive amongst other functional groups. This reaction will require protonation of the oxygen atom followed by the attack of the nucleophile iodide ion to break the strong \[C - O\] single bond.

Complete step by step answer:
This reaction involves the cleavage of \[C - O\] bond in ethers but ethers are least reactive functional groups. We are required to react to hydrogen halides under drastic conditions to make a reaction to occur.
We can complete the given reaction as follows,
\[{C_2}{H_5} - O - {C_2}{H_5} + 4P\xrightarrow{{HI}}2{C_2}{H_6} + {H_2}O\]
X is ethane can split as \[C{H_3} - C{H_3}\], when diethyl ether is heated with a mixture of red P and HI, which can reduce one molecule of ether to two molecules of ethane.

So, the correct answer is Option A.

Additional Information:
1.If alkyl aryl ethers are allowed to react with hydrogen halides at drastic conditions, then \[C - O\] bond between alkyl carbon and oxygen will break because bond between aryl carbon and oxygen will be more stable in comparison.
2.In these types of reaction, reactivity of hydrogen halides is as follows; \[HI \geqslant HBr \geqslant HCl\]
3.The mechanism of this reaction involves protonation of the ether oxygen then the nucleophilic attack of halides breaks the bond by \[{S_N}2\] mechanism.
4.If asymmetric ether is reacted with hydrogen halides, then the alkyl group that has least steric hindrance will form halide. Means tendency to form halide of alkyl groups will be primary, secondary, tertiary.

Note:
Do not think that as protonation of the ether can be done with HI, it will react to give product at lower temperatures. The nucleophilic attack to break the \[C - O\] bond of ether always requires drastic conditions