How to order the reactivity of t-butyl iodide and neopentyl bromide for SN2

Question

The relative leaving group ability of iodide is better than that of bromide but steric crowding is more important as the neucleophile will attack the back lobe of carbon atom.
So how should I order the steric hinderance produced by t-butyl and neopentyl groups?

Answer 1

The order will be $\text{neopentyl bromide} > \text{t-butyl bromide}$. This is just an addition to @user377386's answer. From Solomons and Fryhle's Organic Chemistry, the relative rates of reaction (S_N2) of different kinds of halide are (for the same halogen): $$ \begin{array}{c|c} \textbf{Type of halide} & \textbf{Relative Rate} \\ \hline \text{Primary Halide} & 1 \\ \text{Secondary Halide} & 0.03 \\ \text{Neopentyl Halide} & 0.00001 \\ \text{Tertiary Halide} & \approx 0 \\ \end{array} $$

You can easily see that neopentyl halides and tertiary halides are almost unreactive towards S_N2 reaction, so it makes little sense to compare them. Neither of them is going to react by S_N2 ever, they will instead react through S_N1 mechanism. The slightly increased leaving group ability of iodine does not matter much. If you insist on comparing though, neopentyl halide will be better at S_N2.

Answer 2

The order would be neo-pentyl bromide > tert-butyl iodide . This is because SN2 has concerted mechanism. Although, better is the leaving group, higher is the reaction rate for SN2, but, if the substrate is bulky, it has more tendency to follow SN1 mechanism because of bulkiness of the substrate which hinders the backside attack of the nucleophile. In case of tert-butyl iodide, the carbocation formed is very stable, so SN1 tendency is much more than SN2.