Question

In \[Ph-CH(OH)C{{H}_{3}}\xrightarrow{SOC{{l}_{2}}}Ph-CH(Cl)C{{H}_{3}}+S{{O}_{2}}+HCl\]
Which of the following acts as a leaving group?
A. \[O{{H}^{-}}\]
B. \[C{{l}^{-}}\]
C. \[S{{O}_{2}}\]
D.

Answer 1

Hint: Alcohols react with thionyl chloride and form their chloro derivatives (alkyl or aryl halides) as the products. This reaction is called chlorination. Chlorination is a best example for substitution reactions.

Complete step by step answer:
* In the given reaction aryl halide is reacting with thionyl chloride and forming its halide derivative and Sulphur dioxide and hydrochloric acid as by products.

\[Ph-CH(OH)C{{H}_{3}}\xrightarrow{SOC{{l}_{2}}}Ph-CH(Cl)C{{H}_{3}}+S{{O}_{2}}+HCl\]

* In the question it mentioned that we have found the leaving group in the reaction.
* To know about the leaving group we should write the mechanism of the given reaction.
The mechanism of the given reaction is as follows.

* In the first step, a lone pair of electrons on oxygen on alcohol attacks on the electron deficient Sulphur atom.
* Later to stabilize the hydrogen atom of alcohol comes out as a hydrogen ion, in continuation the negative charge on the oxygen attached to Sulphur is getting stabilized by donating electrons to Sulphur in the process chloride ion comes out.
* The liberated chloride ion attacks on the carbon which is attached to Sulphur through oxygen.
* At that time in the process of stabilization Sulphur dioxide was liberated as a by-product.
* In the given reaction the liberated product is Sulphur dioxide.
So, the correct option is C.

Note: All of us think that in the given reaction the hydroxyl group is converting into halogen in the product. So, the liberated product is the hydroxide group (\[O{{H}^{-}}\]). But through mechanism only we can say which group is going to substitute in the reaction.