Iodide ion reacts with chloromethane to displace chloride ion in a common organic substitution reaction:

${\mathbf{I}}^{\mathbf{-}}\mathbf{+}\mathbf{CH}\mathbf{3}\mathbf{CI}\mathbf{\to }\mathbf{CH}\mathbf{3}\mathbf{I}\mathbf{+}{\mathbf{CI}}^{\mathbf{-}}$

(a) Draw a wedge-bond structure of chloroform and indicate the most effective direction of ${\mathbf{I}}^{\mathbf{-}}$attack.

(b) The analogous reaction with 2-chlorobutane [Figure P16.107(b)] results in a major change in specific rotation as measured by polarimetry. Explain, showing a wedge-bond structure of the product.

(c) Under different conditions, 2-chlorobutane loses ${\mathbf{CI}}^{\mathbf{-}}$in a rate-determining step to form a planar intermediate [Figure P16.107(c)]. This cationic species reacts with HI and then loses H to form a product that exhibits no optical activity. Explain, showing a wedge-bond structure.

From the rear side, the iodide ion approaches the electrophilic carbon atom. Iodine forms a new sigma bond by donating a pair of electrons. The following is the mechanism for this reaction, as shown by the arrow:

From the rear side, the iodide ion approaches the electrophilic carbon atom. Iodine forms a new sigma bond by donating a pair of electrons. The stereochemistry of the result is inverted due to the backside assault of the entering nucleophile (here iodide), which inverts the geometry at the electrophilic carbon. The following is the curved arrow mechanism for the given reaction:

The nucleophile hits either the front or rear side of the reaction due to the planar intermediate. As a result, the final product is a racemic combination.