Activating vs deactivating groups; ortho - para vs meta directors; reactivity order: Exercises

Figure — Activating vs deactivating groups; ortho - para vs meta directors; reactivity order

Recall The three rules we lean on the whole page

Rule A — rate: a group that pushes electron density INTO the ring speeds up EAS; a group that pulls it OUT slows it. Rule B — direction: donors direct ortho/para; withdrawers direct meta; halogens are the one exception (deactivating BUT o/p). Rule C — why: attack happens where the positive charge of the arenium ion lands on the carbon next to a group that likes positive charge (donor) — or far from a group that hates it (withdrawer).

Prerequisite tools if any word above is new: Inductive effect (+I / −I), Resonance / Mesomeric effect (+M / −M), Carbocation stability, Benzene structure and aromaticity.

Level 1 — Recognition

Recall Solution L1.1

(a) $- N H_{2}$ : nitrogen lone pair donated into the ring → strong $+ M$ → activator. (b) $- N O_{2}$ : pulls electron density out through both $σ$ ( $- I$ ) and $π$ ( $- M$ ) → deactivator. (c) $- C H_{3}$ : alkyl group pushes weakly through $σ$ bonds → $+ I$ → weak activator. (d) $- C l$ : strong $- I$ (electronegative) beats its weak $+ M$ for rate → deactivator (but see L2, it still directs o/p).

Recall Solution L1.2

(a) $- O H$ : donor ( $+ M$ ) → o/p. (b) $- C N$ : withdrawer ( $- M$ ) → meta. (c) $- B r$ : halogen exception → o/p. (d) $- S O_{3} H$ : withdrawer ( $- M$ ) → meta.

Level 2 — Application

Recall Solution L2.1

$- C H_{3}$ is a weak activator, o/p director. So $- N O_{2}$ enters ortho and para:

o-nitrotoluene (2-position)
p-nitrotoluene (4-position) — usually the major product, because para has no steric clash with the methyl next door. Rate is faster than benzene (ring enriched by $+ I$ ).

Recall Solution L2.2

$- C l$ : deactivating ( $- I$ dominates rate) so slower than benzene, but o/p directing (weak $+ M$ fixes position). Products: o-dichlorobenzene and p-dichlorobenzene.

Recall Solution L2.3

In the arenium ion, positive charge sits on the carbons ortho and para to the point of attack. For para attack, one of those charged carbons is the one bearing $- O H$ . Oxygen's lone pair then donates onto that carbon to make a $C = O +$ (oxocarbenium) form — an extra, very stable resonance structure (see figure below). For meta attack, the charged carbons never coincide with the $- O H$ carbon, so oxygen can't help → no bonus resonance form → meta disfavoured.

Level 3 — Analysis

Recall Solution L3.1

Slot each on the activation scale ( $- N H_{2} > - C H_{3} > - H > - X > - N O_{2}$ ): $aniline > toluene > benzene > bromobenzene > nitrobenzene$ Reasons: $- N H_{2}$ strong $+ M$ activator; $- C H_{3}$ weak $+ I$ activator; benzene the reference; $- B r$ mild deactivator ( $- I$ ); $- N O_{2}$ strong deactivator ( $- I$ and $- M$ ).

Recall Solution L3.2

Activation strength = how eagerly the atom hands its lone pair into the ring. Nitrogen is less electronegative than oxygen, so it holds its lone pair more loosely and donates more generously → stronger $+ M$ → stronger activation. Oxygen grips its lone pair tighter (more electronegative), donating less. Hence $- N H_{2} > - O H$ . (Compare with Acidity of phenol vs alcohols: the same lone-pair generosity of oxygen into the ring explains phenol's acidity too.)

Recall Solution L3.3

For ortho or para attack, one resonance form of the arenium ion places the $+$ charge directly on the carbon carrying $- N O_{2}$ . But that carbon is already $δ^{+}$ (nitrogen pulls its density away, $- I / - M$ ). Two positive charges side by side = strongly destabilized — that resonance form is forbidden, so o/p intermediates are poor. For meta attack, the charged carbons never touch the $- N O_{2}$ carbon, so no clash → meta is the least-bad path. Result: m-nitro product, and the whole reaction is slow because the ring is electron-poor everywhere.

Level 4 — Synthesis

Recall Solution L4.1

Ask each group where it wants the electrophile:

$- C H_{3}$ (o/p director, C1) → wants positions 2 and 6 (its ortho) and C4 (blocked).
$- N O_{2}$ (meta director, C4) → wants positions 2 and 6 (its meta). Both agree on C2/C6. The activator ( $- C H_{3}$ ) also wins the rate argument and reinforces those spots. New group enters ortho to methyl (position 2 or 6), i.e. between the two substituents' preferred zone → 2-substituted-4-nitrotoluene.

Recall Solution L4.2

Both are o/p directors, but the stronger activator dominates because it controls the faster pathway. $- O C H_{3}$ (strong $+ M$ ) beats $- C l$ (weak $+ M$ , mostly $- I$ ). So the electrophile goes ortho/para to $- O C H_{3}$ . Among those, positions that are also ortho/para to $- C l$ and not sterically blocked are preferred. Rule: when two o/p directors disagree, the stronger activator calls the shot.

Recall Solution L4.3

The final relationship is meta, and only $- N O_{2}$ is a meta director. So install $- N O_{2}$ first:

Nitrate benzene → nitrobenzene.
Brominate → $- B r$ enters meta to $- N O_{2}$ → m-bromonitrobenzene. ✔ Reverse order would fail: bromobenzene is o/p directing, so nitration would give o/p-bromonitrobenzene, not meta. Install the meta-director first when you need a meta product.

Level 5 — Mastery

Recall Solution L5.1

Rate is set in the rate-determining step by how electron-rich the whole ring is on average. Chlorine's strong $- I$ (high electronegativity) drains density → ring poorer → slower than benzene.
Direction is set by which transition-state/intermediate is most stabilized. Only when $E^{+}$ attacks o/p does the arenium-ion positive charge land on the $- C l$ carbon, letting chlorine's lone pair ( $+ M$ ) donate and stabilize that intermediate. Meta gets no such help. So even the deactivated ring reacts preferentially o/p. One-word answer: slower. Two different effects, two different jobs: $- I$ owns the rate, $+ M$ owns the position.

Recall Solution L5.2

Both substituents are o/p directors, but $- O H$ (strong $+ M$ activator) outranks $- C H_{3}$ (weak $+ I$ ). So direction is set by $- O H$ : electrophile enters ortho to $- O H$ (C2 or C6; para to $- O H$ is C4, already occupied by $- C H_{3}$ ). Product: 2-hydroxy-5-methylbenzenesulphonic acid (sulphonate ortho to $- O H$ ). Rate: both groups activate → ring is electron-rich → faster than benzene.

Recall Solution L5.3

$- N H_{2}$ : strong $+ M$ activator → fastest.
benzene: reference.
$- N H_{3}^{+}$ : full positive charge, no lone pair to donate, strong $- I$ → strong deactivator.
$- N O_{2}$ : strong $- I / - M$ deactivator. Comparing the two deactivators: $- N H_{3}^{+}$ and $- N O_{2}$ are both meta-directing, strongly deactivating; $- N H_{3}^{+}$ (full $+$ , pure $- I$ ) and $- N O_{2}$ (partial $δ^{+}$ with $- M$ ) are close, with $- N H_{3}^{+}$ typically the most deactivated of common groups. So: $aniline > benzene > nitrobenzene > anilinium$

Recall One-line self-test (say each before revealing)

Rate is controlled by which effect for halogens? ::: The inductive $- I$ effect (dominates, so slower). Direction is controlled by which effect for halogens? ::: The weak $+ M$ resonance effect (still o/p). To get a meta disubstituted product, install which director first? ::: The meta director (e.g. $- N O_{2}$ ) first. When two o/p directors conflict, who wins? ::: The stronger activator (controls the faster pathway). Why does aniline "lose" its activating power in nitration acid? ::: It is protonated to $- N H_{3}^{+}$ , a deactivating meta-director.