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

1 Answer

#(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:

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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.

Chemical structures of compound molecules.

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