Question

multiple choice 1 point which of the following conditions would not favor an s_n1 reaction? a good leaving group a polar protic solvent a strong nucleophile a tertiary alkyl halide

Explanation:

Step1: Analyze \(S_N1\) reaction mechanism

\(S_N1\) is a two - step reaction. The first step is the formation of a carbocation intermediate (\(R - L
ightarrow R^{+}+L^{-}\), where \(L\) is the leaving group). The second step is the attack of the nucleophile on the carbocation (\(R^{+}+Nu^{-}
ightarrow R - Nu\)).

Step2: Evaluate each option

• A good leaving group: Facilitates the first step (\(R - L

ightarrow R^{+}+L^{-}\)) of the \(S_N1\) reaction. A better leaving group (more stable as an anion) makes it easier for the substrate to ionize into a carbocation. For example, \(I^{-}\) is a better leaving group than \(Cl^{-}\) in alkyl halides.

• A polar protic solvent: Stabilizes the carbocation intermediate (through solvation) and the leaving group (as an anion). Polar protic solvents (e.g., water, ethanol) have \(OH\) or \(NH\) groups that can hydrogen - bond with the leaving group and stabilize the carbocation.
• A strong nucleophile: In \(S_N1\) reaction, the rate - determining step is the formation of the carbocation (\(R - L

ightarrow R^{+}+L^{-}\)). The nucleophile attacks the carbocation in the second (fast) step. A strong nucleophile has no effect on the rate of the \(S_N1\) reaction (since it does not participate in the rate - determining step). In fact, strong nucleophiles favor \(S_N2\) reactions (where the nucleophile attacks the substrate in the rate - determining step).

• A tertiary alkyl halide: Forms a more stable carbocation (\(3^{\circ}\) carbocation is more stable than \(2^{\circ}\), \(1^{\circ}\) due to hyper - conjugation and inductive effects). For example, \((CH_3)_3C - Cl\) forms \((CH_3)_3C^{+}\) which is highly stabilized.

Answer:

C. A strong nucleophile