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Chemistry · Organic Chemistry · Chapter notes

Inductive and Mesomeric Effects · +I, −I, +M and −M

GOC notes on the inductive and mesomeric effects: what +I, −I, +M and −M actually mean, which groups pull and which push, why induction fades after three carbons while mesomerism does not, which effect wins when a group shows both, and how the pKa data proves it.

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In short

The sign is written from the molecule's point of view, never as a charge: minus means the rest of the molecule loses electron density, plus means it gains. The letter says which wire the effect travels on — I along sigma bonds, fading to nothing after about three carbons, and M through the pi system, which does not fade because the electrons are genuinely delocalised.

Contents
  1. 01Two wires, and they behave differently
  2. 02Reading the sign
  3. 03When one group does both: which wins?
  4. 04The proof: watch the pKa fall
The whole note in one line
The sign tells you what happens to the rest of the molecule.
Minus means the molecule loses electron density. Plus means it gains. The letter tells you which wire the effect travels on: I along sigma bonds, M through the pi cloud.
01

Two wires, and they behave differently

Induction fades with distance. Mesomerism does not, because the electrons are genuinely delocalised.TWO DIFFERENT WIRES. THEY DO NOT BEHAVE THE SAME WAY.INDUCTIVE (I) runs along the sigma skeletonClCCCCstrongweakerfaintgonethe pull dies away after about three carbonsMESOMERIC (M) runs through the pi cloudNH2the lone pair isshared across theWHOLE ring, atfull strengthno fading with distanceInduction needs only a sigma bond. Mesomerism needs conjugation: a lone pair or pi bond next to a pi system.
Induction fades with distance. Mesomerism does not, because the electrons are genuinely delocalised.
Inductive (I)Mesomeric (M)
Travels throughsigma bondsthe pi system, or a lone pair next to it
Needsa bond and an electronegativity differenceconjugation
With distancefades fast, gone after about 3 carbonsdoes not fade
Electronsonly shifted, never fully movedgenuinely delocalised
02

Reading the sign

A greedy group is negative. A generous group is positive. The sign is never a charge.THE SIGN IS WRITTEN FROM THE MOLECULE'S POINT OF VIEWMINUS ( -I and -M )NO2Cthe groupTAKESthe molecule LOSES electron densityelectron withdrawingPLUS ( +I and +M )CH3Cthe groupGIVESthe molecule GAINS electron densityelectron donatingMinus means the rest of the molecule loses. Plus means it gains. Never read the sign as a charge.
A greedy group is negative. A generous group is positive. The sign is never a charge.
Groups that PULL

−I (by electronegativity):
−NO2 > −CN > −COOH > −F > −Cl > −Br > −I > −OCH3 > −OH > −C6H5

−M (a multiple bond to an electronegative atom):
−NO2, −CN, −CHO, −COR, −COOH, −COOR, −CONH2, −SO3H

Groups that PUSH

+I (alkyl groups, and anions):
−C(CH3)3 > −CH(CH3)2 > −CH2CH3 > −CH3, and −O−, −COO−

+M (a lone pair to donate):
−NH2, −NHR, −NR2, −OH, −OR, −SH, −SR, and the halogens

How to spot which effect a group can show
  • Every group has an I effect. Every bond has some electronegativity difference.
  • A group shows M only if it sits on a pi system and has a lone pair or pi bond of its own.
  • Lone pair to give away, pointing into the ring: +M. Multiple bond to an electronegative atom, pulling out: −M.
03

When one group does both: which wins?

Most exam questions live in this table. The halogen row is the one that catches people.WHEN ONE GROUP DOES BOTH, WHICH ONE WINS?-NH2-I+M+M is far strongerACTIVATING, ortho and para directing-OH-I+M+M still winsACTIVATING, ortho and para directing-Cl-I+M-I wins on strengthDEACTIVATING, but still ortho and para-NO2-I+Mboth pull the same waySTRONGLY DEACTIVATING, meta directingHalogens are the famous exception: -I decides the RATE (slower), +M decides the PLACE (ortho and para).For NO2 there is no contest, because -I and -M both pull in the same direction.
Most exam questions live in this table. The halogen row is the one that catches people.
The halogen exception, asked almost every year

Halogens are deactivating yet ortho and para directing, which sounds contradictory. It is not. The strong −I makes the whole ring poorer, so the reaction is slower. But the weak +M still feeds electron density specifically to the ortho and para positions, so when it does react, it reacts there. Rate and position are decided by different effects.

04

The proof: watch the pKa fall

Top: more pulling groups, stronger acid. Bottom: the same group moved away, and the effect fades.THE EFFECT IS REAL AND MEASURABLE: WATCH THE pKa FALLacetic acidpKa 4.76no helpchloroacetic acidpKa 2.86one -I chlorinedichloroacetic acidpKa 1.29two chlorinestrichloroacetic acidpKa 0.65three chlorinesmore -I groups, more stable anion, stronger acidbutanoic acidpKa 4.82no chlorine4-chloro (far)pKa 4.52three bonds away3-chloropKa 4.05two bonds away2-chloro (next door)pKa 2.86one bond awaysame chlorine, moved further away: the effect fades to almost nothing
Top: more pulling groups, stronger acid. Bottom: the same group moved away, and the effect fades.
Turning the effect into an answer
  • Acid strength: −I and −M groups stabilise the negative conjugate base, so the acid gets stronger. +I groups destabilise it, so the acid gets weaker.
  • Base strength: +I makes an amine stronger. In aniline the lone pair is delocalised into the ring, so it is far weaker than ammonia.
  • Carbocation stability: +I and +M feed the positive centre, so 3° > 2° > 1°.
  • Ring substitution: +M activates, ortho and para. −M deactivates, meta.
Two mistakes worth avoiding
  • Reading the sign as a charge, or applying induction past three carbons.
  • Forgetting that −OH and −NH2 are −I as well as +M. With no ring to conjugate with, only the −I is left, which is why ethanol is more acidic than ethane.
The sixty second drill
  1. Lone pair or multiple bond? That decides whether M is possible at all. Then ask: greedy or generous?
  2. If both effects exist and disagree, say which wins and what each one controls.

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