Synthesis of Phenols
You can prepare phenols in large quantities by the pyrolysis of the sodium salt of benzene sulfonic acid, by the Dow process, and by the air oxidation of cumene. Each of these processes is described below. You can also prepare small amounts of phenol by the peroxide oxidation of phenylboronic acid and the hydrolysis of diazonium salts.
Pyrolysis of sodium benzene sulfonate
In this process, benzene sulfonic acid is reacted with aqueous sodium hydroxide. The resulting salt is mixed with solid sodium hydroxide and fused at a high temperature. The product of this reaction is sodium phenoxide, which is acidified with aqueous acid to yield phenol.



Dow process
In the Dow process, chlorobenzene is reacted with dilute sodium hydroxide at 300°C and 3000 psi pressure. The following figure illustrates the Dow process.



Air oxidation of cumene
The air oxidation of cumene (isopropyl benzene) leads to the production of both phenol and acetone, as shown in the following figure. The mechanisms for the formation and degradation of cumene hydroperoxide require closer looks, which are provided following the figure.




Cumene hydroperoxide formation.The formation of the hydroperoxide proceeds by a free radical chain reaction. A radical initiator abstracts a hydrogen-free radical from the molecule, creating a tertiary free radical. The creation of the tertiary free radical is the initial step in the reaction.




In the next step, the free radical is attracted to an oxygen molecule. This attraction produces the hydroperoxide free radical.




Finally, the hydroperoxide free radical abstracts a hydrogen free radical from a second molecule of cumene to form cumene hydroperoxide and a new tertiary free radical.




Cumene hydroperoxide degradation.The degradation of the cumene hydroperoxide proceeds via a carbocation mechanism. In the first step, a pair of electrons on the oxygen of the hydroperoxide's “hydroxyl group” is attracted to a proton of the H3O+molecule, forming an oxonium ion.




Next, the oxonium ion becomes stabilized when the positively charged oxygen leaves in a water molecule. This loss of a water molecule produces a new oxonium ion.




A phenide ion shift to the oxygen atom (which creates a tertiary carbocation) stabilizes the positively charged oxygen. (A phenide ion is a phenyl group with an electron bonding pair available to form a new bond to the ring.)




The carbocation is stabilized by an acid-base reaction with a water molecule, leading to the formation of an oxonium ion.




The loss of a proton stabilizes the oxonium ion.




Next, a proton is picked up by the ether oxygen in an acid-base reaction, yielding a new oxonium ion.




The positively charged ether oxygen pulls the electrons in the oxygen-carbon bond toward itself, thus delocalizing the charge over both of the atoms. The partial positive charge on the carbon attracts the nonbonding electron pair from the oxygen of the OH group, allowing the electrons in the original oxygen-carbon bond to be released back to the more electronegative oxygen atom.




Finally, a proton is lost from the protonated acetone molecule, leading to the formation of acetone.


Thanks & Regards
Rinkoo Gupta
AskIITians Faculty