Arrange the following compounds in order from fastest to slowest $S_{\mathrm{N}}2$ reaction rate when reacting with ${\mathrm{Br}}_{-}$ as the nucleophile: 1-bromopropane, 2-bromopropane, 2-bromo-2-methylpropane.

• A. 2-bromo-2-methylpropane > 1-bromopropane > 2-bromopropane
• B. 2-bromopropane > 1-bromopropane > 2-bromo-2-methylpropane
• C. 1-bromopropane > 2-bromopropane > 2-bromo-2-methylpropane
• D. 2-bromo-2-methylpropane > 2-bromopropane > 1-bromopropane

Answer: (C)

Verified step by step guidance

1. Identify the type of alkyl halide for each compound: 1-bromopropane is a primary alkyl halide, 2-bromopropane is a secondary alkyl halide, and 2-bromo-2-methylpropane is a tertiary alkyl halide.
2. Recall that SN2 reactions proceed via a backside attack mechanism and are sterically hindered by bulky groups near the electrophilic carbon; thus, primary alkyl halides react faster than secondary, which react faster than tertiary.
3. Consider the nucleophile, Br⁻, which is a strong nucleophile favoring SN2 reactions, so the reaction rate will mainly depend on the steric hindrance around the carbon bonded to bromine.
4. Rank the compounds based on steric hindrance: primary (1-bromopropane) has the least hindrance, secondary (2-bromopropane) has moderate hindrance, and tertiary (2-bromo-2-methylpropane) has the most hindrance, slowing the SN2 reaction.
5. Conclude that the order of SN2 reaction rates from fastest to slowest is: 1-bromopropane > 2-bromopropane > 2-bromo-2-methylpropane.