IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

madhusudan.chavan

HO¾CH₂¾CH₂¾CH₂¾OH + O=S(¾Cl)¾Cl ®

H¾O⁺¾CH₂¾CH₂¾CH₂¾OH

ï - HCl

O^-¾S(¾Cl)¾Cl ®

CH₂¾CH₂¾CH₂

ï ï

O¾S(=O) OH

ï

Cl ®

CH₂¾CH₂¾CH₂

ï ï

O¾S(=O)¾O⁺¾H

ï -HCl

Cl ®

CH₂¾CH₂¾CH₂

ï ï

O¾S(=O)¾O

As explained in the earlier mechanism for formation of corresponding chloride the intermediate is formed by the reaction of thionyl chloride instead of being attacked by Cl^- by S_N², the lone pair of electron on second OH attacks the sulphur atom to form the intermediate as shown above which loses the HCl to give six membered ring.This is the possible mechanism for the formation of the product shown by you.

Thank you for correcting me. Initially it appears to be a simple reaction of thionyl chloride and alcoholic OH. If the reaction is with only one mole of thionyl chloride there 3-chloro ?1-propanol should be formed along with the above mentioned product. The % of each product will depend on the reaction condition. I have my own doubt about the stability of the cyclic product as sulphur is attached to three more electronegative atoms.

The problems posted by you are really challenging , keep it up; do share such problems with me.

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