Exercises — Reagent classification — electrophiles, nucleophiles (hard - soft)

Ask one question each time: does it want to give an electron pair, or take one?

• $H^{+}$ → E. It is a bare proton, an empty 1s orbital, desperate to be filled.
• $N{H}_3$ → Nu. Nitrogen carries a lone pair to donate (neutral, but donation is what matters — not charge).
• $B{F}_3$ → E. Boron has only 6 electrons, an empty p-orbital; neutral yet electron-hungry.
• $C{N}^{-}$ → Nu. Negative charge + lone pairs = donor.
• The carbon of $\stackrel{{\delta }^{+}}{C}=O$ → E. Oxygen pulls electron density away, leaving carbon ${\delta }^{+}$ = electron-poor.
• $O{H}^{-}$ → Nu. Negative charge with lone pairs = donor.

Rule of thumb: small & tight = hard; big & loose = soft.

• $F^{-}$ → hard (tiny, cloud won't distort).
• $I^{-}$ → soft (huge, cloud smears easily). Going down group 17, $F^{-}<C{l}^{-}<B{r}^{-}<I^{-}$ = hard → soft.
• $L{i}^{+}$ → hard (small cation, high charge density).
• $A{g}^{+}$ → soft (large, diffuse d-electrons, very polarisable).
• $R{S}^{-}$ → soft (sulfur is a big, third-row donor; S/P/Se donors are soft, O/N/F donors are hard).

Step 1 — classify the electrophilic centre. In an $S_N2$ reaction the carbon under attack is a soft electrophilic site (large, polarisable transition state). Step 2 — classify the two ends of the ambident donor. $C{N}^{-}$ has a soft end (carbon, diffuse carbanion lobe) and a harder end (nitrogen). See Ambident Nucleophiles. Step 3 — pair like with like (HSAB). Soft carbon electrophile prefers the soft donor atom = C. Product: $R\text{–}CN$, an alkyl nitrile (C-bonded).

• $A{g}^{+}$ is soft. $F^{-}$ is hard, $I^{-}$ is soft.
• HSAB: soft prefers soft → $A{g}^{+}\cdots I^{-}$ is the stable match.
• Real-world confirmation: $AgI$ is famously insoluble (strong soft–soft bond), while $AgF$ is water-soluble (mismatched, weaker lattice preference toward hydration). The soft–soft pair wins.

The two ends of the enolate:

• O end: charge is localised, small, tight → hard donor.
• C end: charge sits in a diffuse $\pi$ system → soft donor. (See resonance in Inductive and Resonance Effects.)

Now match the electrophile:

• $RX$: the attacked carbon is a soft electrophile → soft–soft → bonds at C → C-alkylation. ✔
• $R_{3}SiCl$: silicon is a hard, oxophilic electrophile → hard–hard → bonds at O → silyl enol ether. ✔

Same nucleophile, opposite site — the partner's hardness decides.

Basicity ≠ nucleophilicity — they are different axes.

• Basicity = how well you grab a proton ($H^{+}$). $F^{-}$ is small and holds charge tightly → binds $H^{+}$ well → strong base.
• Nucleophilicity (in protic water) = how well you attack carbon kinetically. $F^{-}$ is trapped in a tight solvent cage (hard ion, strongly hydrogen-bonded to water) → sluggish. $I^{-}$ is soft, weakly solvated, and its polarisable cloud reaches the transition state early → faster. So $I^{-}$ wins on nucleophilicity while $F^{-}$ wins on basicity. No contradiction.

Step 1 — classify. $L{i}^{+}$ hard, $C{s}^{+}$ soft; $F^{-}$ hard, $I^{-}$ soft. Step 2 — build the HSAB-optimal pairs. Hard·hard $=LiF$; soft·soft $=CsI$. Step 3 — read the direction. The reactant side already contains both ideal pairs, so the equilibrium lies to the LEFT (toward $LiF+CsI$). See the energy picture below.

Score $+1$ for each end that is matched (hard–hard or soft–soft):

• $H^{+}(\text{hard})\cdots F^{-}(\text{hard})$ → matched → stable.
• $A{g}^{+}(\text{soft})\cdots I^{-}(\text{soft})$ → matched → stable.
• $H^{+}(\text{hard})\cdots I^{-}(\text{soft})$ → mismatch → less stable.
• $A{g}^{+}(\text{soft})\cdots F^{-}(\text{hard})$ → mismatch → less stable.

Order (most → least stable): the two matched pairs beat the two mismatched pairs: $H^{+}\cdots F^{-}\approx A{g}^{+}\cdots I^{-}>H^{+}\cdots I^{-}\approx A{g}^{+}\cdots F^{-}$ The two "$\approx$" halves may be split by the ionic term ($H^{+}\cdots F^{-}$ has the smallest $r$, so it takes the top slot on charge-density grounds).

Goal: force $C{N}^{-}$ to bond through its harder N end instead of its soft C end. Change: run the reaction under $A{g}^{+}$ (silver-assisted, $S_{N}1$-like) conditions — $A{g}^{+}$ pulls off $B{r}^{-}$, generating a free carbocation $R^{+}$. Why it flips the site:

• A naked carbocation is a harder, more ionic electrophilic centre than the soft $S_{N}2$ transition state.
• By HSAB, this harder electrophile now prefers the harder donor atom = N.
• Product: $R\text{–}NC$ (isocyanide).

Lesson: you don't change the nucleophile — you tune the electrophile's hardness (via mechanism $S_{N}1$ vs $S_{N}2$) to choose the site. This is the payoff of the whole hard/soft framework. (See Carbocation Stability for when $R^{+}$ is accessible.)

(a) Why electrophile: aluminium in $AlC{l}_3$ has only 6 electrons — an empty orbital. It accepts the lone pair on chlorine of $R\text{–}Cl$. Empty orbital ⇒ Lewis acid ⇒ electrophile (see Lewis Acids and Bases). (b) Hard or soft: $A{l}^{3+}$ is small with high charge (+3) → high charge density → hard acid. (c) The nucleophile in this step is the chlorine lone pair of $R\text{–}Cl$ (it donates into aluminium's empty orbital). The curved arrow runs Cl-lone-pair → Al; see Curved Arrow Notation.

Classify the electrophile: the acyl carbon of $C{H}_{3}COCl$ is a hard electrophilic centre — the carbon is ${\delta }^{+}$ from two electron-withdrawing atoms (the carbonyl O and the Cl), charge is localised, small, ionic-flavoured. Argument 1 (site match): hard electrophile → prefers the enolate's hard O end → O-acylation (an enol ester). Argument 2 (charge density): the localised ${\delta }^{+}$ acyl carbon interacts best with the localised negative charge on oxygen (large ionic term $\propto q_{+}{q}_{-}/r$, small $r$). Conclusion: major attack occurs at oxygen (O-acylation). Both the site-matching and the charge-density arguments agree — a robust prediction.

Curved arrow direction? ::: From nucleophile (source) to electrophile (sink). Hard prefers …? ::: Hard. Soft prefers soft. Softness trend $F^{-}\to I^{-}$? ::: Increases down the group (bigger, more polarisable). Why does $I^{-}$ beat $F^{-}$ as a nucleophile in water? ::: Soft + weakly solvated → reaches the $S_{N}2$ transition state easily. How to switch $C{N}^{-}$ from C-attack to N-attack? ::: Make the electrophile harder — use $A{g}^{+}/S_{N}1$ to give a bare $R^{+}$.