Which of the following compounds will undergo an $S_{\mathrm{N}}2$ reaction most readily?

• A. ${\mathrm{(CH}}_3{)}_2\mathrm{CHBr}$
• B. ${\mathrm{CH}}_3\mathrm{Br}$
• C. ${\mathrm{(CH}}_3{)}_3\mathrm{CBr}$
• D. ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}$

Answer: (B)

Verified step by step guidance

1. Identify the type of alkyl halide in each compound: primary, secondary, or tertiary. This classification is crucial because it influences the mechanism and rate of the SN2 reaction.
2. Recall that SN2 reactions proceed via a backside attack and are sterically hindered by bulky groups. Therefore, primary alkyl halides react fastest, secondary are slower, and tertiary alkyl halides rarely undergo SN2 due to steric hindrance.
3. Analyze each compound: (CH\3)\2CHBr is a secondary alkyl halide, CH\3Br is a primary alkyl halide, (CH\3)\3CBr is a tertiary alkyl halide, and CH\3CH\2Br is a primary alkyl halide.
4. Compare the primary alkyl halides CH\3Br and CH\3CH\2Br. Both are primary, but CH\3Br is a methyl halide, which has even less steric hindrance than a primary alkyl halide, making it more reactive in SN2.
5. Conclude that the compound with the least steric hindrance (methyl halide CH\3Br) will undergo SN2 most readily due to easier backside attack by the nucleophile.