Question

For the given pair of compounds, compare which compound is better $S_{N}2$ substrate.

2-methyl-1-iodopropane, t-butyliodide

• A. 2-methyl-1-iodopropane
• B. t-butyliodide
• C. Both show $S_{N}1$
• D. Both show similar reactivity

Answer 1

Answer: (A)

2-methyl-1-iodopropane

2-methyl-1-iodopropane is a primary $\left(1^{\circ }\right)$ alkyl halide and t-butyl iodide is a tertiary $\left(3^o\right)$ alkyl halide. 2-methyl-1-iodopropane undergoes $S_{N}2$ reaction faster because incoming nucleophile favourably undergoes backside attack to the less-hindered substrate because it leads to leads to the formation of the less-crowded transition state. The transition state during the attack of t-butyl iodide is more crowded because of 3 methyl groups which is unfavourable.