Exercises — Electronic effects — inductive (+I - −I), mesomeric - resonance (+M - −M), hyperconjugation, electromeric

• Inductive (±I): pull/push through sigma bonds, partial, permanent, dies out after ~3 atoms.
• Mesomeric / Resonance (±M): whole electron pair delocalised through a pi system or lone pair, permanent, travels the full conjugated path.
• Hyperconjugation: donation from an α C–H sigma bond into an adjacent empty p-orbital or pi bond — "no-bond resonance". Count the α-H.
• Electromeric (±E): temporary, complete pi-pair shift that happens only when a reagent shows up.

The word α-hydrogen means: a hydrogen sitting on the carbon that is directly next to the special site (the cation, the double bond, the C=O). See Carbocation Stability and Rearrangements and Resonance and Delocalisation if these feel shaky.

The rule: a group that pulls electrons off the chain is −I; a group that pushes electrons onto the chain is +I.

• (a) $-{\text{CH}}_3$ — alkyl groups push → +I.
• (b) $-\text{Cl}$ — chlorine is electronegative, it pulls → −I.
• (c) $-{\text{COO}}^{-}$ — carries a full negative charge it wants to shed, so it pushes → +I.
• (d) $-{\text{NO}}_2$ — nitrogen carries ${\delta }^{+}$ and oxygens pull hard, so it drains the chain → −I.

The sign trick from the parent note: a group that donates a lone pair into the ring is +M; a group with a pi bond to a greedy (electronegative) atom that sucks density out is −M.

• (a) $-{\text{NH}}_2$ — N has a lone pair it pushes in → +M.
• (b) $-\text{CHO}$ — the C=O pi bond pulls density out → −M.
• (c) $-\text{OR}$ — O has lone pairs to donate → +M.
• (d) $-\text{CN}$ — the C≡N pi bond drains density → −M.

The electromeric effect (±E). It is a temporary, complete shift of a pi pair triggered by the incoming reagent. Remove the reagent — the pi pair snaps back. This is the tug-of-war rope that only gets yanked when someone shows up to grab it.

Each Cl is −I: it pulls electron density away, helping spread the negative charge of the conjugate base $-{\text{COO}}^{-}$ formed after losing ${\text{H}}^{+}$. More spreading = more stable base = stronger acid. More Cl atoms = more pulling. ${\text{Cl}}_3\text{CCOOH}>{\text{Cl}}_2\text{CHCOOH}>{\text{ClCH}}_2\text{COOH}>{\text{CH}}_3\text{COOH}$ Confirmed by $p{K}_a$ (smaller = stronger): $0.66<1.29<2.86<4.76$. See Acidity and Basicity of Organic Compounds.

Hyperconjugation counts α-hydrogens — H on carbons directly attached to the positive carbon.

• Ethyl ${\text{CH}}_3{\text{-CH}}_2^{+}$: the positive carbon has one neighbouring carbon bearing 3 α-H → 3 α-H.
• Isopropyl $({\text{CH}}_3{)}_2{\text{CH}}^{+}$: the positive carbon has two neighbouring CH₃, each with 3 H → 6 α-H. More α-H ⇒ more no-bond resonance structures ⇒ positive charge more spread out ⇒ isopropyl is more stable. This is exactly why $3^{\circ }>2^{\circ }>1^{\circ }$. See Carbocation Stability and Rearrangements.

When ${\text{H}}^{+}$ adds first, the double pi pair shifts to grab it — that is the +E (electromeric) effect, triggered by the reagent. Two carbocations are possible:

• ${\text{H}}^{+}$ onto C1 → cation on C2 = a secondary cation (2 α-H-bearing neighbours).
• ${\text{H}}^{+}$ onto C2 → cation on C1 = a primary cation (fewer α-H). The more stable secondary cation forms preferentially (more hyperconjugation + more +I from two alkyl groups). ${\text{Br}}^{-}$ then attacks C2 → 2-bromopropane. This is Markovnikov: "H to the carbon with more H's." See Markovnikov Addition.

Chlorine wears two hats at once:

• −I (through sigma): it drains electron density from the whole ring → the ring reacts slower → deactivating.
• +M (lone pair into pi): its lone pair delocalises specifically onto the ortho and para carbons, making those positions the least electron-poor → ortho/para directing.

−I dominates the overall speed (so the ring is deactivated), but +M decides the position (so o/p wins over meta). Both are true simultaneously — no contradiction. See Directing Effects in Electrophilic Aromatic Substitution.

Basicity = how available the N lone pair is to grab a proton.

• Methylamine: $-{\text{CH}}_3$ is +I, pushing density onto N → lone pair richer, more available → strongest base.
• Ammonia: bare N, nothing helping or hurting → middle.
• Aniline: the ring pulls the lone pair into itself by +M resonance → lone pair tied up, less available → weakest base. ${\text{CH}}_3{\text{NH}}_2>{\text{NH}}_3>{\text{C}}_6{\text{H}}_5{\text{NH}}_2$ ($p{K}_b$ roughly $3.4<4.7<9.4$; smaller $p{K}_b$ = stronger base.) See Acidity and Basicity of Organic Compounds.

Both lose ${\text{H}}^{+}$ to give a phenoxide ($-{\text{O}}^{-}$) whose charge is delocalised into the ring.

• In p-nitrophenol, the $-{\text{NO}}_2$ at the para position is −M and −I: it pulls the negative charge further off the oxygen and onto itself (an extra resonance structure places the charge on the nitro oxygens).
• More charge spread ⇒ more stable phenoxide ⇒ stronger acid.

Remember: lower $p{K}_a$ = stronger acid. Since p-nitrophenol has the lower $p{K}_a$, it is the stronger acid: $p{K}_a(\text{p-nitrophenol})\approx 7.1<p{K}_a(\text{phenol})\approx 10.0$ So p-nitrophenol is the stronger acid, exactly as the resonance argument predicts.

Judge each conjugate base's stability:

• Ethanol → ethoxide ${\text{CH}}_3{\text{CH}}_2{\text{O}}^{-}$: the +I of the ethyl group pushes charge onto O, making it unstable → weakest acid.
• Water → hydroxide ${\text{OH}}^{-}$: no push, no delocalisation → next.
• Phenol → phenoxide: charge delocalised over the ring by resonance → more stable → stronger acid.
• Acetic acid → acetate ${\text{CH}}_3{\text{COO}}^{-}$: charge shared equally over two oxygens by resonance → very stable → strongest here.

Increasing acidity: $\text{ethanol}<\text{water}<\text{phenol}<\text{acetic acid}$ By $p{K}_a$ (higher = weaker): $16>15.7>10.0>4.76$. See Resonance and Delocalisation.

First, why some C–H bonds don't count. Hyperconjugation needs a C–H sigma bond that is adjacent to the pi bond, i.e. on a carbon next to the double-bond carbons — an α carbon. The C–H bonds on the sp² double-bond carbons themselves are vinylic (in the plane of the pi bond) and lie in the wrong geometry to overlap sideways with the pi cloud, so they give no hyperconjugative stabilisation. We therefore count only H on carbons one bond away from the C=C.

• Ethene: the only H present sit on the two sp² carbons themselves (vinylic) — there are no α carbons at all → 0 α-H available for hyperconjugation.
• Propene: one ${\text{CH}}_3$ attached to the C=C → 3 α-H.
• 2-methyl-2-butene: two ${\text{CH}}_3$ on one side (6 H) + one ${\text{CH}}_3$ on the other (3 H) → 9 α-H.

More α-H ⇒ more hyperconjugation ⇒ lower energy ⇒ more stable: $\text{ethene}<\text{propene}<\text{2-methyl-2-butene}$

First, what the symbols mean. Write ${\delta }_n$ (read "delta-sub-$n$") for the size of the partial charge that Cl's pull manages to induce on the $n$-th carbon down the chain — a small fraction of a full electron charge, measured in the same units as any charge. Then ${\delta }_1$ is that partial charge on the first carbon (the one bearing Cl), the largest one. The empirical decay rule from the parent note is ${\delta }_n\approx {\delta }_1\cdot k^{n-1},k\approx \frac{1}{3},$ meaning each extra sigma bond passes on only about one-third of the partial charge it received. So ${\delta }_1$ = starting partial charge, ${\delta }_n$ = the shrunken partial charge $n-1$ bonds later, and $k$ = the per-bond shrink factor.

Why the acidity order: Cl is −I, stabilising the conjugate base by pulling charge. The closer Cl sits to $-{\text{COO}}^{-}$, the larger the partial charge ${\delta }_n$ that still reaches it. Plain butanoic acid has no Cl at all, so it is weakest.

Counting bonds from Cl to the acid carbon: at $\alpha$ the Cl-bearing carbon is bond 1 ($n=1$), at $\beta$ it is bond 2 ($n=2$), at $\gamma$ bond 3 ($n=3$). Using ${\delta }_n={\delta }_1{k}^{n-1}$:

• $\alpha$ ($n=1$): factor $k^0=1$.
• $\beta$ ($n=2$): factor $k^1=\frac{1}{3}\approx 0.333$.
• $\gamma$ ($n=3$): factor $k^2=\frac{1}{9}\approx 0.111$.

So from $\alpha$ to $\gamma$ the reach shrinks by $\frac{k^2}{k^0}=\frac{1}{9}\approx 0.111$ — about a 9-fold drop. That is why $\gamma$-Cl is barely more acidic than plain butanoic acid.

Plausible — yes. In vinyl chloride the Cl lone pair can donate into the adjacent C=C pi system (+M), giving a resonance structure with partial double-bond character between C and Cl. Partial double bond = shorter, stronger C–Cl bond and reduced reactivity (that's why vinyl/aryl halides resist nucleophilic substitution). In chloroethane there is no adjacent pi system, so no +M, so the C–Cl is an ordinary single bond — longer and more reactive. The effect responsible is the mesomeric (+M) donation of the halogen lone pair. See Resonance and Delocalisation.

Count the help each cation gets:

• ${\text{CH}}_3^{+}$ (methyl): no α-H, no adjacent alkyl, no pi system → nothing stabilising → least stable.
• ${\text{CH}}_3{\text{CH}}_2^{+}$ (ethyl): 3 α-H (hyperconjugation) + one +I alkyl group pushing electrons in.
• $({\text{CH}}_3{)}_3{\text{C}}^{+}$ (tert-butyl): 9 α-H (strong hyperconjugation) + three +I alkyl groups → much more stabilisation than ethyl.
• ${\text{C}}_6{\text{H}}_5{\text{CH}}_2^{+}$ (benzyl): the positive charge is delocalised into the benzene ring by full resonance — whole electron pairs shift, spreading the charge over several ring carbons (ortho and para). Genuine resonance beats hyperconjugation, so benzyl is the most stable even though it has only 2 α-H.

Final order (least → most stable): ${\text{CH}}_3^{+}<{\text{CH}}_3{\text{CH}}_2^{+}<({\text{CH}}_3{)}_3{\text{C}}^{+}<{\text{C}}_6{\text{H}}_5{\text{CH}}_2^{+}$ Rule of thumb honoured: resonance (benzyl) > inductive + hyperconjugation stack (tert-butyl) > single alkyl (ethyl) > nothing (methyl). See Carbocation Stability and Rearrangements.