Question: In \[Ph-CH(OH)C{{H}_{3}}\xrightarrow{SOC{{l}_{2}}}Ph-CH(Cl)C{{H}_{3}}+S{{O}_{2}}+HCl\] Which of th...

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

Solution

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.