3.4.2—Nucleophilic substitution

The nucleophile donates a pair to carbon while the leaving group departs with its bonding pair. Deduce the product by replacing the leaving group with the nucleophile.

For CH₃CH₂Br + OH⁻, the C–O bond forms as the C–Br bond breaks, giving CH₃CH₂OH + Br⁻. Account for charge and every atom in the product; the leaving group takes the bonding pair rather than departing as a neutral bromine atom.

Representative question

[Maximum number: 4]

Explain the mechanism of the reaction, using curly arrows to represent the movement of electron pairs.

curly arrow from lone pair/negative charge on O in OH to C attached to Br curly arrow from C-Br bond to Br

transition state showing negative charge AND partial bonds products ( $\mathrm{Br}^{-}$ AND $\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}$ )

Award [3 max] if $S_{N} 1$ mechanism is given.

Accept curly arrows in the transition state.

Do not penalize if HO and Br are not at $180^{\circ}$ .

Accept NaBr as part of the products only if $\mathrm{Na}^{+}$ is shown at the start.

Retrieve the route: classify nucleophiles and electrophiles, show heterolysis, write substitution and addition mechanisms, map Lewis coordination, compare SN1/SN2, and restore aromaticity in benzene substitution.

Check electron-pair arrow origin and destination, leaving-group departure, intermediate identity, carbocation stability and the assessed mechanism boundary.

3.4.2—Nucleophilic substitution

Nucleophilic Substitution

Deducing Substitution Products

Assessment in practice

Electron-Pair Sharing Summary

Question 1