# Why do we obtain racemic iodoalkane in nucleophilic displacement reactions?

In my textbook, the following statement is given :

Attempts to prepare optically active iodides by nucleophilic displacement on optically active bromides using the iodide ion normally produces a racemic mixture of iodoalkanes.

The reason given is that this is due to the following equilibrium:

My doubts:

1. $$\ce{I-}$$ is a stronger nucleophile than $$\ce{Br-}$$. Then, why does the backward reaction even occur?

2. Is the reaction SN2?

3. Does this happen for other types of nucleophilic displacements also?

Book: Arihant Organic Chemistry (DPP Problems)

• This suggests the reaction has some SN1 character. Formation of a carbocation leads to loss of stereochemical integrity. – Waylander Mar 11 '18 at 17:06