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Distillation of phenol with zinc-dust gives:
(A) Benzene
(B) Diphenyl - zinc
(C) Diphenyl Ether
(D) none of the above

Answer
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408.9k+ views
The product was discovered by Kekule. It has the molecular formula \[{{C}_{6}}{{H}_{6}}\].
It is an aromatic compound as it obeys Huckel’s rule of \[(4n+2)\pi \]electrons. It is a very stable compound due to resonance.

Complete step by step answer:
Distillation of phenol with zinc-dust gives benzene and ZnO as a side-product.
The reaction proceeds as follows:
Detailed Mechanism:
- Zn is in the +2 oxidation state.
- The phenol gets converted into phenoxide ion and the released proton accepts an electron from Zn forming hydrogen radical. Due to the heating, there is homolytic fission of C of the phenyl ring and\[{{O}^{-}}\].
- Then\[{{O}^{-}}\]forms an electron from Zn and forms an oxide ion. In this way, zinc forms zinc oxide, and the phenyl radical produced forms a bond with hydrogen radical to form Benzene and bi-phenyl.




















So, the correct option is A.

Note: The yield of this reaction used for benzene formation is lower.
Homolytic fission is the equal splitting of a pair of electrons between two separated atoms.
It is a reduction reaction of phenol to benzene by zinc dust. So, zinc is a reducing agent.
Benzene formed is volatile in nature. It is separated by fractional distillation.