What is the nucleophile in the reaction shown below?

Question

BlackTom AI · Accepted Answer

A nucleophile donates an electron pair to form a new bond. Here, the reaction proceeds by an SN1 pathway: the C–I bond heterolytically breaks first to give I– (leaving group) and a stabilized benzylic tertiary carbocation; then a lone pair donor attacks that carbocation.

1) Relevant concepts
- Nucleophile: electron-pair donor (Lewis base).
- Leaving group: species that departs with an electron pair (I– here).
- SN1 mechanism: stepwise; formation of carbocation followed by nucleophilic attack. Polar protic solvents like water favor SN1 and water can act as nucleophile and as base.

2) Step-by-step reasoning
- Step 1 (ionization): The benzylic tertiary iodide undergoes heterolysis: R–I → R+ (benzylic carbocation) + I–. I– is stabilized and leaves.
- Step 2 (nucleophilic attack): A water molecule uses one of its lone pairs to attack the carbocation, forming an oxonium intermediate R–OH2+.
- Step 3 (deprotonation): A second water molecule removes a proton from R–OH2+, yielding the neutral alcohol R–OH and H3O+ as product. I– remains in solution as counterion.

3) Intermediates and charge flow (no numeric calculation)
- R–I → R+ + I–
- R+ + H2O → R–OH2+ 
- R–OH2+ + H2O → R–OH + H3O+

4) Why H2O is the nucleophile
Water donates a lone pair to the carbocation in step 2, forming the new C–O bond. That is the defining nucleophilic action in this mechanism.

5) Why the other options are incorrect
- a. I–: This is the leaving group and a product, not the nucleophile attacking the carbocation.
- b. (starting alkyl iodide): The substrate is the electrophile (provides carbocation), not the nucleophile.
- c. H3O+: Hydronium is an acid (proton donor), not a nucleophile.
- e. OH–: Hydroxide is a strong nucleophile, but it is not present in this reaction; the solvent is neutral water, not OH–.

6) Conclusion
Because a water molecule attacks the carbocation and is deprotonated by another water, the nucleophile in the shown reaction is H2O, confirming answer d.

What is the nucleophile in the reaction shown below?Single choice

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