##(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.
