Which of the following reagents is the best choice to convert an alkyl bromide to an alcohol via an S$N_2$ mechanism?

• A. Aqueous ${\mathrm{NaOH}}_{\mathrm{}}$
• B. ${\mathrm{NaOEt}}_{\mathrm{}}$ in ethanol
• C. Water ($H_{2}O$) only
• D. Concentrated $H_2\mathrm{SO}{4}_{\mathrm{}}$

Answer: (A)

Verified step by step guidance

1. Identify the goal: converting an alkyl bromide (a good leaving group) into an alcohol via an S_N2 mechanism, which involves a nucleophilic substitution where the nucleophile attacks the electrophilic carbon and displaces the leaving group in one step.
2. Recall that for an S_N2 reaction, the nucleophile should be strong and negatively charged to effectively attack the carbon bearing the leaving group. Also, the solvent should ideally be polar aprotic to favor the S_N2 pathway, but aqueous conditions can still work if the nucleophile is strong enough.
3. Evaluate each reagent option: - Aqueous NaOH provides hydroxide ions (OH⁻), a strong nucleophile that can attack the alkyl bromide to form the alcohol. - NaOEt in ethanol provides ethoxide ions (EtO⁻), which are strong nucleophiles but will produce an ether rather than an alcohol. - Water (H₂O) is a weak nucleophile and would lead to a slower reaction. - Concentrated H₂SO₄ is acidic and would likely lead to elimination or other side reactions rather than substitution.
4. Conclude that aqueous NaOH is the best choice because it supplies a strong nucleophile (OH⁻) that can perform a clean S_N2 substitution on the alkyl bromide to yield the alcohol.
5. Remember that the S_N2 mechanism proceeds with inversion of configuration at the carbon center, so if the alkyl bromide is chiral, the product alcohol will have inverted stereochemistry.