Part A Draw a reasonable mechanism for the following reaction: OH CH3 H2O -CH3 Br Draw all missing reactants and/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. Add charges needed. Electron-flow arrows should start on an atom or a bond and should end on an atom, bond, or location where a new bond should be created. Post-Lecture Assignment 6c mechanism Practice Problem: Nucleophilic Substitution Reactions + ? N CH, H2O) N s 1 :OH 11 CH,
The Correct Answer and Explanation is:
Assumed Reaction Setup
From the parts you provided:
- One of the starting compounds is an alkyl bromide (–CH₃–Br).
- Water (H₂O) is the solvent/nucleophile.
- The product has an –OH (hydroxyl group) substituted where Br was.
- You mention SN1, which is unimolecular nucleophilic substitution.
Complete Reaction Example
Let’s assume the substrate is tert-butyl bromide:
(CH₃)₃C–Br
With water as solvent/nucleophile:
(CH₃)₃C–Br + H₂O → (CH₃)₃C–OH + HBr
Mechanism: SN1 Pathway
Step 1: Formation of Carbocation (Rate-Determining Step)
- The C–Br bond breaks heterolytically.
- Br⁻ leaves, forming a tertiary carbocation, which is stabilized by hyperconjugation and inductive effects from the three methyl groups.
(CH3)3C–Br→(CH3)3C++Br−(CH₃)₃C–Br → (CH₃)₃C⁺ + Br⁻
Step 2: Nucleophilic Attack by Water
- Water (a weak nucleophile) attacks the electrophilic carbocation.
(CH3)3C++H2O→(CH3)3C–OH2+(CH₃)₃C⁺ + H₂O → (CH₃)₃C–OH₂⁺
Step 3: Deprotonation
- Another water molecule deprotonates the protonated alcohol to form the neutral alcohol.
(CH3)3C–OH2++H2O→(CH3)3C–OH+H3O+(CH₃)₃C–OH₂⁺ + H₂O → (CH₃)₃C–OH + H₃O⁺
Explanation
This reaction follows an SN1 mechanism, which stands for “substitution nucleophilic unimolecular.” The reaction proceeds in three distinct steps. First, the leaving group, which is bromide (Br⁻), dissociates from the carbon atom to form a carbocation intermediate. This is the rate-determining step and occurs slowly, especially if the carbon is not well stabilized. In this case, a tertiary carbocation forms, which is very stable due to hyperconjugation and the inductive effect of adjacent methyl groups. These effects help delocalize the positive charge, making the carbocation more stable and thus favoring the SN1 mechanism.
Once the carbocation is formed, the next step is nucleophilic attack. Water, a neutral molecule but with lone pairs on the oxygen, acts as a nucleophile and attacks the carbocation. Since water is a weak nucleophile, the reaction proceeds well only because the carbocation is highly reactive.
The product of this attack is a protonated alcohol (an oxonium ion), where the oxygen has three bonds and a positive charge. In the final step, a second molecule of water acts as a base and deprotonates this intermediate, yielding the neutral alcohol product and regenerating a hydronium ion (H₃O⁺).
This reaction is favored in polar protic solvents like water that stabilize ions and can facilitate carbocation formation. Since the rate-determining step involves only the substrate, the reaction rate depends only on the concentration of the alkyl halide, not the nucleophile.
