Draw mechanism for this reaction: Interactive 3D display mode Edit the reaction by drawing all steps in the appropriate boxes and connecting them with reaction arrows. Add charges where needed. Electron flow arrows should start on an atom Or bond and should end on an atom_ bond or Iocation where new bond should be created: H: Cont # ci ‘0: {CI_ ci
The Correct Answer and Explanation is:
To provide a detailed mechanism for your reaction and explain it clearly, I’ll need to interpret what seems to be an incomplete or jumbled reaction description. You mentioned H: Cont # ci ‘0: {CI_ ci, which appears to contain fragments of a chemical reaction involving chlorine (Cl), perhaps in a nucleophilic substitution or elimination context.
Let’s assume you are referring to a nucleophilic substitution reaction where a nucleophile attacks an alkyl halide (R–Cl), leading to the displacement of the chlorine atom. This is a common SN2 (bimolecular nucleophilic substitution) mechanism.
Mechanism for SN2 Reaction:
Reactants:
- A primary alkyl halide (e.g., CH₃CH₂Cl)
- A strong nucleophile (e.g., OH⁻ or :OH)
Products:
- A substituted product (e.g., CH₃CH₂OH)
- A leaving group (Cl⁻)
Step-by-step Mechanism (with curved arrows):
- The nucleophile (OH⁻) uses its lone pair to attack the electrophilic carbon in the CH₃CH₂Cl molecule. This carbon is partially positive because it is bonded to the electronegative chlorine atom.
- As the nucleophile attacks from the backside, the C–Cl bond starts to break.
- The electrons from the C–Cl bond move onto the Cl atom, releasing it as a chloride ion (Cl⁻).
- The result is a new C–O bond and the formation of ethanol (CH₃CH₂OH), along with Cl⁻.
Electron Flow Arrows:
- Arrow from lone pair on O in OH⁻ to the carbon attached to Cl.
- Arrow from C–Cl bond to Cl atom, indicating departure of Cl⁻.
Explanation:
This reaction follows a concerted one-step SN2 mechanism, where the nucleophile attacks the carbon center at the same time as the leaving group (Cl⁻) leaves. Because of this simultaneous action, there is no intermediate. The backside attack causes an inversion of configuration at the carbon center, which is crucial if the carbon is chiral. This reaction is favored with primary alkyl halides and strong nucleophiles, as steric hindrance is minimal. The transition state is unstable and high in energy, so solvents and the nature of the nucleophile and leaving group significantly affect the rate. Chloride is a good leaving group due to its ability to stabilize the negative charge after departure.
