Which of the following compounds will undergo an Sn2 reaction most readily: # (CH_3)_3C CH_2I# or #(CH_3)_2CHI#?

#(CH_3)_2CHI# will undergo an #S_N2# reaction more readily than#(CH_3)_3C CH_2I# .

To make this question less complicated, it is helpful to draw the structures of both compounds as shown in the image below:

Take a look at the carbon atom bound directly to the iodine.

For #(CH_3)_2CHI#(isopropyl iodide), the carbon is bound to the iodine, one hydrogen, and two other carbons. This is called a secondary halide (secondary meaning bound to two carbons).

An incoming nucleophile will often react with whatever electrophile it can reach most easily. In technical terms, a secondary halide is more sterically hindered than a primary halide, so #S_N2# will occur more readily at the primary halide.

We should therefore expect the isopropyl iodide to have the slower reaction rate.

In #(CH_3)_3C CH_2I# (neopentyl iodide), in addition to iodine, the carbon atom is bound to two hydrogen atoms and only one other carbon. This is called a primary halide (primary meaning bound to only one carbon).

You would therefore expect this compound to have the fastest #"S"_"N"2# reaction rate. But there is a complication.

The bulky t-butyl group prevents backside attack by the nucleophile.

The steric hindrance is so effective that isopropyl iodide reacts almost 3000 times as fast as neopentyl iodide.