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Question

In which of the following reactions the inversion of configuration takes place?

(I) pentan2ol + SOCl2
(II) pentan2ol + COCl2
(III) pentan2ol + (A)+SOCl2 Pyridine−−−−
(IV) pentan2ol + COCl2 Pyridine−−−−

A
(I),(III)
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B
(I),(II)
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C
(III),(IV)
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D
(I),(IV)
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Solution

The correct option is C (III),(IV)
Retention of stereochemistry with SOCl2 or COCl2 alone, inversion with SOCl2 and pyridine, both reactions form the “chlorosulfite” intermediate. But when pyridine (a nucleophile) is present, it can attack the chlorosulfite, displacing chloride ion and forming a charged intermediate.
Now, if the leaving group departs, forming a carbocation, there’s no lone pair nearby on the same face that can attack. Chloride attacks the carbon from the backside, leading to inversion of configuration and formation of a CCl bond through SN2 mechanism. Reaction III and IV both proceed by SN2 mechanism and lead to inversion.

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