Question

7 multiple answer 1 point which of the following factors favor an s_n2 reaction mechanism? polar aprotic solvent strong nucleophile tertiary alkyl halide polar protic solvent previous

Explanation:

• Polar aprotic solvent: These solvents do not hydrogen - bond with nucleophiles. This allows the nucleophile to be more "free" to attack the electrophilic carbon in the \(S_N2\) reaction. For example, acetone (\(CH_3COCH_3\)) is a polar aprotic solvent. In \(S_N2\) reactions, the lack of hydrogen - bonding with the nucleophile means that the nucleophile's reactivity is not significantly diminished.
• Strong nucleophile: In an \(S_N2\) reaction, the nucleophile attacks the electrophilic carbon simultaneously as the leaving group departs. A strong nucleophile has a high affinity for the electrophilic carbon. For instance, \(OH^-\) is a stronger nucleophile than \(H_2O\) in many \(S_N2\) scenarios. The rate of an \(S_N2\) reaction is directly proportional to the concentration of the nucleophile (\(rate = k[Nu^-][R - L]\)), so a more reactive (stronger) nucleophile will increase the rate of the \(S_N2\) reaction.
• Tertiary alkyl halide: In \(S_N2\) reactions, steric hindrance is a major factor. A tertiary alkyl halide (\(R_3C - X\)) has three alkyl groups attached to the carbon bearing the leaving group (\(X\)). These alkyl groups create significant steric hindrance, making it difficult for the nucleophile to approach the electrophilic carbon. \(S_N2\) reactions are favored for primary (\(RCH_2 - X\)) and methyl (\(CH_3 - X\)) alkyl halides (less steric hindrance).
• Polar protic solvent: Polar protic solvents (e.g., water (\(H_2O\)), methanol (\(CH_3OH\))) can hydrogen - bond with nucleophiles. This solvates the nucleophile, reducing its reactivity. The nucleophile is "trapped" in a shell of solvent molecules, and it has to shed some of these solvent molecules to attack the electrophilic carbon, which is an energy - consuming process. So, polar protic solvents are more favorable for \(S_N1\) reactions (where the rate - determining step is the formation of a carbocation and the nucleophile attacks the carbocation after it is formed, and the solvated nucleophile can still attack the planar carbocation) rather than \(S_N2\) reactions.

Answer:

A. Polar aprotic solvent, B. Strong nucleophile