By which of these mechanism(s) would the following reaction proceed most favourably?[Fill in the blank]单项选择题
A
a. SN1
B
b. SN2
C
c. E1
D
d. E2

Question

By which of these mechanism(s) would the following reaction proceed most favourably?[Fill in the blank]

BlackTom AI · Accepted Answer

NaI in DMSO provides a strong nucleophile in a polar aprotic medium, which is the classic condition that favors an SN2 mechanism.

1. Relevant knowledge points and concepts
- SN2 is a bimolecular, concerted substitution: nucleophile attack and leaving-group departure occur simultaneously; rate = k[substrate][nucleophile]. SN2 is favored by strong nucleophiles and polar aprotic solvents.  
- SN1 is unimolecular, proceeds via a carbocation intermediate; favored by tertiary substrates and polar protic solvents that stabilize ions.  
- E2 is a bimolecular elimination requiring a strong base; E1 is unimolecular elimination via carbocation (favored by tertiary substrates and heat/weak bases).  
- I– is a good nucleophile but a relatively weak base; DMSO is polar aprotic (enhances nucleophilicity).

2. Step‑by‑step reasoning
- Identify substrate: the bromide is on a secondary cycloalkyl carbon (secondary alkyl bromide). Secondary substrates can undergo either SN2 or E2 depending on nucleophile/base and solvent.  
- Reagent/solvent: Na+I– in DMSO gives a high concentration of free I– (good nucleophile). DMSO (polar aprotic) does not solvate anions strongly, so it increases nucleophilicity of I– and disfavors carbocation stabilization (which SN1/E1 need).  
- Mechanistic outcome: I– will attack the electrophilic carbon from the backside while Br– leaves in one concerted step → SN2.

3. Intermediate/calculation steps
- No numeric calculation needed; mechanistic preference follows: SN2 favorable because both substrate (secondary, accessible) and conditions (strong nucleophile, polar aprotic solvent) meet SN2 requirements. Rate dependence: rate ∝ [substrate][I–], consistent with SN2.

4. Why the correct answer is correct
- The combination of a good nucleophile (I–) and polar aprotic solvent (DMSO) directly promotes bimolecular backside attack and disfavors formation of carbocations (SN1/E1). I– is not a strong base, so elimination (E2) is not favored.

5. Why other options are incorrect
- a. SN1: unlikely because substrate is secondary (not strongly carbocation-stabilizing) and DMSO does not stabilize ions well.  
- c. E1: requires carbocation formation and usually polar protic solvent and tertiary substrate or heat — not present.  
- d. E2: requires a strong base (e.g., t-BuO–); I– is a poor base, so substitution dominates.

6. Conclusion
These points confirm the reaction proceeds most favorably by b. SN2, matching the provided answer.

By which of these mechanism(s) would the following reaction proceed most favourably?[Fill in the blank]单项选择题

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